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t.me/Ebooks_Encyclopedia27., , The secret of, , t.me/Magazines4all, , KOTA, , now at your Doorstep, , PHYSICS, 60 + 28, , DPP BOOKLET, , Topic-wise, , Chapter-wise, Tests for Concept, Checking &, Speed Building, , Ü Collection of 3100 + MCQs of all variety of questions, Ü Unique & innovative way of learning, Ü Detailed solutions to Topic-wise & Chapter-wise practice sheets, Ü Covers all important concepts of each topic, Ü As per latest pattern & syllabus, , Improves, your learning, by at least, , 20%
Page 3 : •, , t.me/Magazines4all, , Corporate Office : 45, 2nd Floor, Maharishi Dayanand Marg, Corner Market,, Malviya Nagar, New Delhi-110017, Tel. : 011-49842349 / 49842350, , Typeset by Disha DTP Team, , DISHA PUBLICATION, ALL RIGHTS RESERVED, © Copyright Publisher, No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the, publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have tried, and made our best efforts to provide accurate up-to-date information in this book., , For further information about books from DISHA,, Log on to www.dishapublication.com or email to
[email protected], , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , Daily Practice Problem (DPP) Sheets PHYSICS for, NEET/AIIMS/JIPMER (Kota's formula to Success), PREPARE, , ASSESS, , IMPROVE, , Assessment is the most integral part of a student’s preparation but still most of them avoid it. Only assessment can tell, where you stand and how you can improve from that point. So it is very important that you take the right assessment, which, is on the correct pattern, has the same level of difficulty as the actual exam and covers all the important concepts of the, subject., Disha Publication launches a first of its kind product which changed the way coaching was conducted in KOTA - the hub, of Engineering and Medical Entrance education in India., The book “Daily Practice Problem (DPP) Sheets for NEET/AIIMS” is precise, apt and tuned to all the requirements of a, NEET/AIIMS aspirant., KEY DIFFERENTIATING FEATURES OF THE DPP SHEETS, •, , Part A provides 60 DPP's with division of the complete NEET syllabus of Physics into 60 most important Topics., Each of the chapter has been broken into 2 or more topics., , •, , Part B consist of — Chapter-wise tests based on NCERT and NEET syllabus., , •, , Time Limit and Maximum Marks have been provided for each DPP Sheet/ topic. You must attempt each Sheet in, test like conditions following the time limits. Further to achieve perfect preparation in a topic or chapter one has, to score atleast 135 marks., , •, , Ultimate tool for Concept Checking & Speed Building., , •, , Collection of 3100 Standardised MCQ’s of all variety of NEW pattern questions – MCQ only one correct option, and Assertion-Reason., , •, , Unique & innovative way of learning. Whenever you have prepared a topic(Part A) or a chapter (Part B) just, attempt that worksheet., , •, , Do not refer the Solution Booklet until and unless you have made all the efforts to solve the DPP Sheets., , •, , Covers all important Concepts of each Topic in the form of different Questions in the DPP Sheets., , •, , As per latest pattern & syllabus of NEET/AIIMS JIPMER exam., , •, , Compliant to all boards of education., , No matter where you PREPARE from – a coaching or NCERT books or any other textbook/ Guide - Daily Practice Problem, Sheets provides you the right ASSESSMENT on each topic. Your performance provides you the right clues to IMPROVE, your concepts so as to perform better in the final examination., It is to be noted here that these are not tests but act as a checklist of student’s learning and ability to apply concepts to, different problems. Do proper analysis after you attempt each DPP sheet and try to locate your weak areas., It is our strong belief that if an aspirant works hard on the clues provided through each of the DPP sheets he/ she can, improve his/ her learning and finally the SCORE by at least 20%., , Χisha ∆xperts
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t.me/Magazines4all, , The book comprises of following two parts, Part A : Topic-wise DPP Sheets, Page No., Detailed Index, (i) to (iv), Topic-wise Sheets 1-60 (Each sheet 4 pages) 1 - 4, Solutions of Topic-wise Sheets, 1 - 158, Part B : Chapter-wise DPP Sheets, Detailed Index, Chapter-wise Sheets 1-28, Solutions of Chapter-wise Sheets, , (a) to (c), p-1 – p-112, S-1– S-115, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , INDEX/SYLLABUS, DPP-1Page No., , DPP-1, , PHYSICAL WORLD, UNITS & DIMENSIONS, , P1 - P 4, , DPP-2, , MEASUREMENTS (ERRORS), , P1 - P 4, , DPP-3, , MOTION IN A STRAIGHT LINE 1 (Distance, Displacement, Uniform & Non-uniform motion), , P1 - P 4, , DPP-4, , MOTION IN A STRAIGHT LINE 2 (Relative Motion & Motion Under Gravity), , P1 - P 4, , DPP-5, , VECTORS, , P1 - P 4, , DPP-6, , MOTION IN A PLANE-1 (Projectile Motion), , P1 - P 4, , DPP-7, , MOTION IN A PLANE-2 (Horizontal Circular Motion), , P1 - P 4, , DPP-8, , MOTION IN A PLANE-3 (Vertical Circular Motion, Relative Motion), , P1 - P 4, , DPP-9, , LAWS OF MOTION-1 (Newton's laws, momentum, pseudo force concept), , P1 - P 4, , DPP-10, , LAWS OF MOTION-2 (Blocks in contact, connected by string, pulley arrangement), , P1 - P 4, , DPP-11, , LAWS OF MOTION-3 (Friction), , P1 - P 4, , DPP-12, , WORK, ENERGY AND POWER-1 (Work by constant and variable forces, kinetic and potential energy,, work energy theorem), P1 - P 4, , DPP-13, , WORK, ENERGY AND POWER-2 (Conservation of momentum and energy, collision, rocket case), , P1 - P 4, , DPP-14, , CENTRE OF MASS AND ITS MOTION, , P1 - P 4, , DPP-15, , ROTATIONAL MOTION – 1: Basic concepts of rotational motion, moment of a force, torque, angular, momentum and its conservation with application, , P1 - P 4, , DPP-16, , ROTATIONAL MOTION-2 : Moment of inertia, radius of gyration, (values of moments of inertia simple, geometrical objects), P1 - P 4, , DPP-17, , ROTATIONAL MOTION - 3 : Rolling Motion, Parallel and perpendicular theorems and their, applications, Rigid body rotation, equations of rotational motion, , P1 - P 4, , DPP-18, , GRAVITATION - 1 (The Universal law of gravitation, Acceleration due to gravity and its variation with, altitude and depth, Kepler's law of planetary motion), P1 - P 4, , DPP-19, , GRAVITATION - 2 (Gravitational potential energy, Gravitational potential, Escape velocity & Orbital, velocity of a satellite, Geo-stationary satellites), P1 - P 4, , DPP-20, , MECHANICAL PROPERTIES OF SOLIDS, , P1 - P 4, , DPP-21, , FLUID MECHANICS, , P1 - P 4, , [ii]
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP-22, , THERMAL EXPANSION, CALORIMETRY AND CHANGE OF STATE, , P1 - P 4, , DPP-23, , HEAT TRANSFER & NEWTON’S LAW OF COOLING, , P1 - P 4, , DPP-24, , THERMODYNAMICS-1 (Thermal equilibrium, zeroth law of thermodynamics, concept of temperature,, Heat, work and internal energy, Different thermodynamic processes), P1 - P 4, , DPP-25, , THERMODYNAMICS-2 (1st and 2nd laws of thermodynamics, Reversible & irreversible processes,, Carnot engine and its efficiency), P1 - P 4, , DPP-26, , KINETIC THEORY, , DPP-27, , OSCILLATIONS-1 (Periodic motion - period, Frequency, Displacement as a function of time. Periodic, functions, Simple harmonic motion and its equation, Energy in S.H.M. - kinetic and potential energies), P1 - P 4, , DPP-28, , OSCILLATIONS-2 (Oscillations of a spring, simple pendulum, free, forced and damped oscillations,, Resonance), P1 - P 4, , DPP-29, , WAVES-1 (Wave motion, longitudinal and transverse waves, speed of a wave, displacement relation for a, progressive wave, principle of superposition of waves, reflection of waves), P1 - P 4, , DPP-30, , WAVES-2 (Standing waves in strings and organ pipes, Fundamental mode and harmonics, Beats, Doppler, effect in sound), P1 - P 4, , DPP-31, , PRACTICAL PHYSICS - 1, , P1 - P 4, , DPP-32, , ELECTROSTATICS-1 (Coulomb's law, electric field, field lines, Gauss's law), , P1 - P 4, , DPP-33, , ELECTROSTATICS-2 (Electric potential and potential difference, equipotential surfaces, electric dipole), P1 - P 4, , DPP-34, , ELECTROSTATICS -3 (Electrostatic Potential energy, conductors), , P1 - P 4, , DPP-35, , ELECTROSTATICS-4 (Capacitors, dielectrics), , P1 - P 4, , DPP-36, , CURRENT ELECTRICITY – 1 (Electric Current, drift velocity, Ohm's law, Electrical resistance, Resistances, of different materials, V-I characteristics of Ohm and non-ohmic conductors, electrical energy and power,, Electrical resistivity, Colour code of resistors, Temperature dependance of resistance), P1 - P 4, , DPP-37, , CURRENT ELECTRICITY – 2 Electrical cell and its internal resistance, Potential difference and E.M.F of a, cell, Combination of cells in series and in parallel, Kirchoff's laws and their applications, RC transient circuit,, Galvanometer, Ammeter, Voltmeter], P1 - P 4, , DPP-38, , CURRENT ELECTRICITY-3 : Wheatstone bridge, Meter bridge, Potentiometer-principle and its applications., P1 - P 4, , DPP-39, , MAGNETIC EFFECTS OF CURRENT-1 (Magnetic field due to current carrying wires, Biot savart law), P1 - P 4, , DPP-40, , MAGNETIC EFFECTS OF CURRENT-2 : (Motion of charge particle in a magnetic field, force between, current carrying wires.), P1 - P 4, , P1 - P 4, , [iii]
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DPP-41, , t.me/Magazines4all, , MAGNETIC EFFECTS OF CURRENT-3 (Magnetic dipole, Current carrying loop in magnetic, field,Galvanometer ), , EBD_7156, , t.me/Ebooks_Encyclopedia27., , P1 - P 4, , DPP-42, , MAGNETISM AND MATTER - 1 (Bar magnet as an equivalent solenoid, Magnetic field lines, Earth's, magnetic field and magnetic elements), P1 - P 4, , DPP-43, , MAGNETISM & MATTER-2 (Para, dia and ferro-magnetic substances, magnetic susceptibility and, permeability, Hysteresis, Electromagnets and permanent magnets.), P1 - P 4, , DPP-44, , ELECTROMAGNETIC INDUCTION-1 (Magnetic flux, Faraday's law of electromagnetic induction,, Lenz's law, motional e.m.f.), P1 - P 4, , DPP-45, , ELECTROMAGNETIC INDUCTION - 2 : Self inductance, mutual inductance, Growth and decay of, current in L.R. circuit, Transformer, Electric motor, Generator, P1 - P 4, , DPP-46, , ALTERNATING CURRENT - 1 (Alternating currents, peak and rms value of alternating current/voltage;, reactance and impedance, Pure circuits, LR, CR ac circuits.), P1 - P 4, , DPP-47, , ALTERNATING CURRENT - 2 (LCR series circuit, resonance, quality factor, power in AC circuits, wattless, and power current), P1 - P 4, , DPP-48, , EM WAVES, , P1 - P 4, , DPP-49, , RAY OPTICS-1 (Reflection on plane mirrors and curved mirrors), , P1 - P 4, , DPP-50, , RAY OPTICS - II (Refraction on plane surface, total internal reflection, prism), , P1 - P 4, , DPP-51, , RAY OPTICS - 3 (Refraction on curved surface lens, Optical instrument), , P1 - P 4, , DPP-52, , WAVE OPTICS - I (Interference of Light), , P1 - P 4, , DPP-53, , WAVE OPTICS - II (Diffraction and polarisation of light), , P1 - P 4, , DPP-54, , DUAL NATURE OF MATTER & RADIATION (Matter Waves, Photon, Photoelectric effect, X-ray), , P1 - P 4, , DPP-55, , ATOMS, , P1 - P 4, , DPP-56, , NUCLEI, , P1 - P 4, , DPP-57, , SEMICONDUCTOR ELECTRONICS - 1 (Semiconductors, LED, Photodiode, Zener diode), , P1 - P 4, , DPP-58, , SEMICONDUCTOR ELECTRONICS-2 (Junction transistor, transistor action, characteristics of a transistor,, transistor as an amplifier, logic gates), P1 - P 4, , DPP-59, , COMMUNICATION SYSTEMS, LASER, , P1 - P 4, , DPP-60, , PRACTICAL PHYSICS - 2, , P1 - P 4, , Solutions to Topic-wise DPP Sheets (1-60), , [iv], , 1-158
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 01, SYLLABUS : Physical World, Units & Dimensions, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 If L, C and R represent inductance, capacitance and, resistance respectively, then which of the following does, not represent dimensions of frequency?, (a), (c), , 1, RC, , 1, LC, , (b), , R, L, , (d), , C, L, , aZ, , Q.2 Number of particles crossing unit area perpendicular to, X-axis in unit time is given by n = - D, , RESPONSE GRID, , 1., , and n2 are number of particles per unit volume in the position, x1 and x2. Find dimensions of D called as diffusion constant., (a) [M 0 L T 2], (b) [M 0 L2 T –4], 0, –3, (c) [M L T ], (d) [M 0 L2 T –1], 2, Q.3 X = 3YZ find dimensions of Y in (MKSA) system, if X and, Z are the dimensions of capacity and magnetic field, respectively, (a) [M –3L –2T – 4A –1], (b) [ML– 2], –3, –2, 4, 4, (c) [M L T A ], (d) [M –3L– 2T 8A 4], a - kq, , P is pressure, Z is the distance,, e, b, k is Boltzmann constant and q is the temperature. The, dimensional formula of b will be, (a) [M 0L 2T 0], (b) [M 1L 2T 1], 1, 0, –1, (c) [M L T ], (d) [M 0L 2T –1], , Q.4 In the relation P =, , n2 - n1, , where n1, x2 - x1, , 2., , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 01, , 2, 1/ 2, , P éF ù, Q.5 The frequency of vibration of string is given by n = ê ú, 2l ë m û, , ., , Here P is number of segments in the string and l is the, length. The dimensional formula for m will be, (a) [M 0LT –1], (b) [ML 0T –1], –1, 0, (c) [ML T ], (d) [M 0L 0T 0], Q.6 What is the relationship between dyne and newton of force?, (a) 1 dyne = 10–5 newton (b) 1 dyne = 10–7 newton, (c) 1 dyne = 105 newton (d) 1 dyne = 107 newton, Q.7 The speed of light (c), gravitational constant (G) and, Planck's constant (h) are taken as the fundamental units in a, system. The dimensions of time in this new system should, be, (a) G1/2 h1/2 c –5/2, (b) G–1/2 h 1/2 c 1/2, (c) G1/2 h1/2 c –3/2, (d) G1/2 h 1/2 c 1/2, Q.8 If the constant of gravitation (G), Planck's constant (h) and, the velocity of light (c) be chosen as fundamental units., The dimensions of the radius of gyration is, (a) h1/2 c –3/2G 1/2, (b) h1/2 c 3/2 G 1/2, 1/2, –3/2, –1/2, (c) h c, G, (d) h–1/2 c –3/2 G 1/2, Q.9 The magnitude of any physical quantity, (a) depends on the method of measurement, (b) does not depend on the method of measurement, (c) is more in SI system than in CGS system, (d) directly proportional to the fundamental units of mass,, length and time, Q.10 The unit of Stefan's constant s is, (a) Wm–2 K–1, (b) Wm2K–4, –2, –4, (c) Wm K, (d) Wm–2K4, Q.11 In S = a + bt + ct2 , S is measured in metres and t in seconds., The unit of c is, (a) ms–2, (b) m, –1, (c) ms, (d) None, , RESPONSE, GRID, , Q.12 Wavelength of ray of light is 0.00006 m. It is equal to, (a) 6 microns, (b) 60 microns, (c) 600 microns, (d) 0.6 microns, Q.13 SI unit of permittivity is, (a) C2 m2 N–2, (b) C–1 m 2N –2, (c) C2 m2 N2, (d) C2 m–2 N –1, 1, e E2 (e0 = permittivity of free space, 2 0, and E = electric field) are, (a) MLT–1, (b) ML2 T–2, –1, –2, (c) ML T, (d) ML2 T–1, Q.15 Which of the following pairs is wrong?, (a) Pressure-Baromter, (b) Relative density-Pyrometer, (c) Temperature-Thermometer, (d) Earthquake-Seismograph, Q.16 A physical quantity x depends on quantities y and z as, follows: x = Ay + B tan Cz, where A, B and C are constants., Which of the following do not have the same dimensions?, (a) x and B, (b) C and z–1, (c) y and B/A, (d) x and A, Q.17 If the time period (T) of vibration of a liquid drop depends, on surface tension (S), radius (r) of the drop and density (r), of the liquid, then the expression of T is, , Q.14 The dimensions of, , (a) T = k rr 3 / S, (c), , T = k rr 3 / S1/ 2, , (b) T = k r1/ 2 r 3 / S, (d) None of these, , Q.18 The dimensional formula for Planck’s constant (h) is, (a) [ML–2T–3 ], (b) [M0L2T–2 ], (c) [M0L2T–1 ], (d) [ML–2T–2 ], Q.19 What are the dimensions of permeability (m0) of vaccum?, (a) MLT–2I2, (b) MLT–2I–2, (c) ML–1T–2I2, (d) ML–1T –2I–2, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 01, , 3, , Q.20 A small steel ball of radius r is allowed to fall under gravity, through a column of a viscous liquid of coefficient of, viscosity h. After some time the velocity of the ball attains a, constant value known as terminal velocity vT. The terminal, velocity depends on (i) the mass of the ball m, (ii) h, (iii) r, and (iv) acceleration due to gravity g. Which of the following, relations is dimensionally correct?, mg, hr, , (a), , vT µ, , (c), , vT µ hrmg, , (b), , vT µ, , hr, mg, , (d), , vT µ, , mgr, h, , Q.21 The equation of state of some gases can be expressed as, æ, a ö, ç P + 2 ÷ (V - b) = RT . Here P is the pressure, V is the, V ø, è, , volume, T is the absolute temperature and a, b and R are, constants. The dimensions of 'a' are, (a) ML5 T–2, (b) ML–1 T–2, (c) M0L3T0, , (d) M0L6T0, , Q.23 P represents radiation pressure, c represents speed of light and, S represents radiation energy striking unit area per sec. The non, zero integers x, y, z such that Px Sy cz is dimensionless are, (1) x = 1, (2) y = – 1, (3) z = 1, (4) x = – 1, Q.24 Which of the following pairs have same dimensions?, (1) Angular momentum and work, (2) Torque and work, (3) Energy and Young’s modulus, (4) Light year and wavelength, DIRECTION (Q.25-Q.27) : Read the passage given below and, answer the questions that follows :, Three of the fundamental constants of physics are the universal, gravitational constant, G = 6.7 × 10–11m3kg–1s–2, the speed, of light, c = 3.0 × 10 8 m/s, and Planck’s constant,, h = 6.6 × 10–34 kg m2 s–1., Q.25 Find a combination of these three constants that has the, dimensions of time. This time is called the Planck time and, represents the age of the universe before which the laws of, physics as presently understood cannot be applied., , DIRECTIONS (Q.22-Q.24) : In the following questions, more, than one of the answers given are correct. Select the correct, answers and mark it according to the following codes:, , (a), , Codes :, (a), , 1, 2 and 3 are correct, , (b) 1 and 2 are correct, , (c), , 2 and 4 are correct, , (d) 1 and 3 are correct, , Q.22 The frequency of vibration f of a mass m suspended from, a spring of spring constant k is given by a relation of the, type f = c mx ky, where c is a dimensionless constant. The, values of x and y are, 1, 2, , (1), , x=, , (3), , y=-, , RESPONSE, GRID, , 1, 2, , (2), , x=-, , (4), , y=, , 1, 2, , (c), , hG, c, , (b), , 4, , hG, c, , (d), , hG, c3, , hG, c5, , Q.26 Find the value of Planck time in seconds, (a) 1.3 × 10– 33 s, (b) 1.3 × 10– 43 s, (c) 2.3 × 10– 13 s, (d) 0.3 × 10– 23 s, Q.27 The energy of a photon is given by E = hc ., l, If l = 4 ´ 10 -7 m , the energy of photon is, , 1, 2, , 20., , 21., , 22., , 25., , 26., , 27., , (a) 3.0 eV, , (b) 4.5 eV, , (c) 2.10 eV, , (d) 3.95 eV, , Space for Rough Work, , 23., , 24.
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t.me/Magazines4all, DPP/ P 01, , 4, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason)., Each of these questions has four alternative choices, only one of, which is the correct answer. You have to select the correct choice., (a), (b), (c), (d), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is NOT, a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , Q.28 Statement -1 : Unit of Rydberg constant R is m–1, Statement -2 : It follows from Bohr’s formula, , RESPONSE GRID, , 28., , 29., , æ 1, 1 ö, v = Rç, - ÷ , where the symbols have their usual, 2, 2÷, çn, è 1 n2 ø, , meaning., Q.29 Statement -1: The time period of a pendulum is given by, the formula, T = 2p g / l ., Statement -2: According to the principle of homogeneity of, dimensions, only that formula is correct in which the, dimensions of L.H.S. is equal to dimensions of R.H.S., Q.30 Statement -1: L/R and CR both have same dimensions., Statement -2: L/R and CR both have dimension of time., , 30., , DAILY PRACTICE PROBLEM SHEET 1 - PHYSICS, Total Questions, 30, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), , Space for Rough Work, , 120, , 50, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 02, SYLLABUS : Measurements (Errors), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , Q.1 A wire has a mass 0.3 ± 0.003g, radius 0.5 ± 0.005 mm and, length 6 ± 0.06 cm. The maximum percentage error in the, measurement of its density is, (a) 1, (b) 2, (c) 3, (d) 4, Q.2 If 97.52 is divided by 2.54, the correct result in terms of, significant figures is, , measurement in a, b, c and d are 1%, 3%, 2% and 2%, respectiely. What is the percentage error in the quantity A, (a) 12%, (b) 7%, (c) 5%, (d) 14%, Q.4 A physical quantity is given by X = M a LbT c . The, percentage error in measurement of M, L and T are a, b and, g respectively. Then maximum percentage error in the, quantity X is, (a) aa+ bb + cg, (b) aa+ bb – cg, , (a) 38.4, (b) 38.3937, (c) 38.394, (d) 38.39, Q.3 A physical quantity A is related to four observable a, b, c, and d as follows, A =, , RESPONSE GRID, , 1., , a 2b3, c d, , the percentage errors of, , 2., , (c), , 3., , a b c, + +, a b g, , (d) None of these, , 4.
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t.me/Magazines4all, DPP/ P 02, , 2, Q.5 If the length of rod A is 3.25 ± 0.01 cm and that of B is, 4.19 ± 0.01 cm then the rod B is longer than rod A by, (a) 0.94 ± 0.00 cm, (b) 0.94 ± 0.01 cm, (c) 0.94 ± 0.02 cm, (d) 0.94 ± 0.005 cm, Q.6 If L = 2.331 cm, B = 2.1 cm, then L + B =, (a) 4.431 cm, (b) 4.43 cm, (c) 4.4 cm, (d) 4 cm, Q.7 The number of significant figures in all the given numbers, 25.12, 2009, 4.156 and 1.217 × 10–4 is, (a) 1, (b) 2, (c) 3, (d) 4, Q.8 In an experiment, the following observation's were, recorded: L = 2.820 m, M = 3.00 kg, l = 0.087 cm,, Diameter D = 0.041 cm. Taking g = 9.81 m/s2 using the, formula, Y = 4 MgL , the maximum percentage error in Y, pD 2l, is, (a) 7.96%, (b) 4.56% (c) 6.50% (d) 8.42%, Q.9 A physical parameter a can be determined by measuring, the parameters b, c, d and e using the relation a =, , b a cb, d g ed, , Q.10 The period of oscillation of a simple pendulum is given by, , l, where l is about 100 cm and is known to have, g, 1mm accuracy. The period is about 2s. The time of 100, oscillations is measured by a stopwatch of least count 0.1s., The percentage error in g is, (a) 0.1%, , (b) 1%, , (c) 0.2%, , (d) 0.8%, , RESPONSE, GRID, , Q.14 The resistance R =, , ., , If the maximum errors in the measurement of b, c, d and e, are b1%, c1%, d1%, and e1%, then the maximum error in, the value of a determined by the experiment is, (a) (b1+ c1+ d1+ e1)%, (b) (b1+ c1– d1– e1)%, (c) (ab1+ bc1– gd1– de1)%(d) (ab1+ bc1+ gd1+ de1)%, , T = 2p, , Q.11 The mean time period of second's pendulum is 2.00s and, mean absolute error in the time period is 0.05s. To express, maximum estimate of error, the time period should be, written as, (a) (2.00 ± 0.01)s, (b) (2.00 + 0.025) s, (c) (2.00 ± 0.05) s, (d) (2.00 ± 0.10) s, Q.12 Error in the measurement of radius of a sphere is 1%. The, error in the calculated value of its volume is, (a) 1%, (b) 3%, (c) 5%, (d) 7%, Q.13 The relative density of material of a body is found by, weighing it first in air and then in water. If the weight in air, is (5.00 ± 0.05) newton and weight in water is (4.00 ± 0.05), newton. Then the relative density along with the maximum, permissible percentage error is, (a) 5.0 ± 11%, (b) 5.0 ± 1%, (c) 5.0 ± 6%, (d) 1.25 ± 5%, V, where V = 100 ± 5 volts and, i, , i = 10 ± 0.2 amperes. What is the total error in R ?, (a) 5%, , (b) 7%, , (c) 5.2%, , (d), , 5, %, 2, , Q.15 The length of a cylinder is measured with a meter rod having, least count 0.1 cm. Its diameter is measured with vernier, calipers having least count 0.01 cm. Given that length is, 5.0 cm. and radius is 2.0 cm. The percentage error in the, calculated value of the volume will be, (a) 1%, (b) 2%, (c) 3%, (d) 4%, Q.16 According to Joule’s law of heating, heat produced H =, I2Rt, where I is current, R is resistance and t is time . If the, errors in the measurements of I,R. and t are 3%, 4% and, 6% respectively then error in the measurement of H is, (a) ± 17%, (b) ± 16%, (c) ± 19%, (d) ± 25%, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 02, , 3, , Q.17 A physical quantity P is given by P =, , 3, , A, , 1, B2, , -4, , 3, D2, , . The quantity, , C, which brings in the maximum percentage error in P is, (a) A, (b) B, (c) C, (d) D, Q.18 If there is a positive error of 50% in the measurement of, velocity of a body, then the error in the measurement of, kinetic energy is, (a) 25%, (b) 50%, (c) 100%, (d) 125%, Q.19 The random error in the arithmetic mean of 100, observations is x; then random error in the arithmetic mean, of 400 observations would be, 1, 1, x, x, (a) 4x, (b), (c) 2x, (d), 4, 2, Q.20 The percentage errors in the measurement of mass and, speed are 2% and 3% respectively. How much will be the, maximum error in the estimation of the kinetic energy, obtained by measuring mass and speed?, (a) 11%, (b) 8%, (c) 5%, (d) 1%, Q.21 The unit of percentage error is, (a) Same as that of physical quantity, (b) Different from that of physical quantity, (c) Percentage error is unitless, (d) Errors have got their own units which are different, from that of physical quantity measured, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, , RESPONSE, GRID, , Q.22 In the context of accuracy of measurement and significant, figures in expressing results of experiment, which of the, following is/are correct?, 1. Out of the two measurements 50.14 cm and 0.00025, ampere, the first one has greater accuracy, 2. Out of the two measurements 50.14 cm and 0.00025, ampere, the second has greater accuracy., 3. If one travels 478 km by rail and 397 m by road, the, total distance travelled is 875 km., 4. If one travels 697 m by rail and 478 km by road, the, total distance is 478 km., Q.23 A thin copper wire of length l metre increases in length by, 2% when heated through 10°C. Which is not the, percentage increase in area when a square copper sheet of, length l metre is heated through 10°C, (1) 12%, (2) 8%, (3) 16%, (4) 4%, Q.24 A body travels uniformly a distance of (13.8 ± 0.2) m in a, time (4.0 ± 0.3) s., 1., , Its velocity with error limit is (3.5 ± 0.31) ms–1, , 2., , Its velocity with error limit is (3.5 ± 0.11) ms–1, , 3., , Percentage error in velocity is ± 4%, , 4., , Percentage error in velocity is ± 9%, , DIRECTION (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, The internal radius of a 1m long resonance tube is measured as, 3 cm. A tuning fork of frequency 2000 Hz is used. The first, resonating length is measured as 4.6 cm and the second, resonating length is measured as 14.0 cm., Q.25 Calculate the maximum percentage error in measurement, of e., (a) 3.33%, (b) 2.23% (c) 4.33% (d) 5.33%, Q.26 Calculate the speed of sound at the room temperature., (a) 275 m/s, (b) 376 m/s, (c) 356 m/s, (d) 330 m/s, , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , Space for Rough Work
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t.me/Magazines4all, DPP/ P 02, , 4, Q.27 Calculate the end correction., (a) 0.2 cm, (b) 0.3 cm, (c) 0.1 cm, (d) 0.4 cm, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , RESPONSE GRID, , 27., , 28., , (c) Statement-1 is False, Statement-2 is True., (d) Statement-1 is True, Statement-2 is False., Q.28 Statement-1: Number of significant figures in 0.005 is, one and that in 0.500 is three., Statement-2 : This is because zero is not significant., Q.29 Statement-1: Out of three measurements l = 0.7 m;, l = 0.70 m and l = 0.700 m, the last one is most accurate., Statement-2: In every measurement, only the last, significant digit is not accurately known., Q.30 Statement-1: Parallex method cannot be used for, measuring distances of stars more than 100 light years, away., Statement-2: Because parallex angle reduces so much that, it cannot be measured accurately., , 29., , 30., , DAILY PRACTICE PROBLEM SHEET 2 - PHYSICS, Total Questions, 30, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), , Space for Rough Work, , 120, , 46, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 03, SYLLABUS : MOTION IN A STRAIGHT LINE 1 (Distance, Displacement, Uniform & Non-uniform motion), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A particle moving in a straight line covers half the distance, with speed of 3 m/s. The other half of the distance is, covered in two equal time intervals with speed of 4.5 m/s, and 7.5 m/s respectively. The average speed of the particle, during this motion is, (a) 4.0 m/s, (b) 5.0 m/s, (c) 5.5 m/s, (d) 4.8 m/s, Q.2 The acceleration of a particle is increasing linearly with, time t as bt. The particle starts from the origin with an initial, velocity v0. The distance travelled by the particle in time t, will be, , RESPONSE GRID, , 1., , 2., , 1, 1 3, v0 t + bt 2, (b) v0 t + bt, 3, 3, 1 3, 1 2, (d) v0t + bt, (c) v0 t + bt, 6, 2, Q.3 The motion of a body is given by the equation, , (a), , dv(t ), = 6.0 - 3v(t ), where v(t) is speed in m/s and t in sec., dt, , If body was at rest at t = 0, (a) The terminal speed is 4 m/s, (b) The speed varies with the time as v(t) = 2(1 – e–5t)m/s, (c) The speed is 0.1m/s when the acceleration is half the, initial value, (d) The magnitude of the initial acceleration is 6.0 m/s2, , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 03, , 2, Q.4 A particle of mass m moves on the x-axis as follows: it, starts from rest at t = 0 from the point x = 0 and comes to, rest at t = 1 at the point x = 1. No other information is, available about its motion at intermediate time (0 < t < 1)., If a denotes the instantaneous acceleration of the particle,, then, (a) a cannot remain positive for all t in the interval, 0 £ t £1, (b) | a | cannot exceed 2 at any point in its path, (c) | a | must be > 4 at some point or points in its path, (d) | a | = 2 at any point in its path., Q.5 A particle starts from rest. Its acceleration (a) versus time, (t) graph is as shown in the figure. The maximum speed of, the particle will be, a, , 10 m/s2, , t (s), , 11, , (a) 110 m/s, (b) 55 m/s, (c) 550 m/s, (d) 660 m/s, Q.6 A car accelerates from rest at a constant rate a for some, time, after which it decelerates at a constant rate b and, comes to rest. If the total time elapsed is t, then the maximum, velocity acquired by the car is, æ a 2 + b2 ö, ÷t, (a) çç, ÷, è ab ø, , æ a 2 - b2 ö, ÷t, (b) çç, ÷, è ab ø, abt, (a + b)t, (c), (d), a +b, ab, Q.7 A small block slides without friction down an inclined plane, starting from rest. Let Sn be the distance travelled from, time t = n – 1 to t = n. Then, (a), , 2n - 1, (b), 2n, , RESPONSE, GRID, , 2n + 1, 2n - 1, , Sn, is, S n +1, , (c), , 2n - 1, 2n + 1, , (d), , 2n, 2n + 1, , Q.8 A particle starts moving from the position of rest under a, constant acc. If it covers a distance x in t second, what, distance will it travel in next t second?, (a) x, (b) 2 x, (c) 3 x, (d) 4 x, Q.9 What will be the ratio of the distances moved by a freely, falling body from rest in 4th and 5th seconds of journey?, (a) 4 : 5 (b) 7 : 9, (c) 16 : 25 (d) 1 : 1, Q.10 If a ball is thrown vertically upwards with speed u, the, distance covered during the last t seconds of its ascent is, 1 2, 1, gt, (d) ut – gt2, 2, 2, Q.11 If the displacement of a particle is (2t2 + t + 5) meter, then, what will be acc. at t = 5 second?, (a) 21 m/s2, (b) 20 m/s2, 2, (c) 4 m/s, (d) 10 m/s2, Q.12 A particle moves along x-axis with acceleration a = a0 (1 – t/, T) where a0 and T are constants if velocity at t = 0 is zero, then find the average velocity from t = 0 to the time when a, = 0., a 0T, a 0T, a 0T, a 0T, (a), (b), (c), (d), 3, 2, 4, 5, Q.13 A point moves with uniform acceleration and v1, v2 and v3, denote the average velocities in the three successive, intervals of time t1, t2 and t3 . Which of the following, relations is correct ?, (a) (v1– v2) : (v2– v3) = (t1– t2) : (t2+ t3), (b) (v1– v2) : (v2– v3) = (t1+ t2) : (t2+ t3), (c) (v1– v2) : (v2– v3) = (t1– t2) : (t2 – t3), (d) (v1– v2) : (v2– v3) = (t1– t2) : (t2 – t3), Q.14 The position of a particle moving in the xy-plane at any, time t is given by x = (3t2 – 6t) metres, y = (t2 – 2t) metres., Select the correct statement about the moving particle from, the following, (a) The acceleration of the particle is zero at t = 0 second, (b) The velocity of the particle is zero at t = 0 second, (c) The velocity of the particle is zero at t = 1 second, (d) The velocity and acceleration of the particle are never, zero, , (a) (u+gt)t (b) ut, , (c), , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 03, , 3, , Q.15 Two cars A and B are travelling in the same direction with, velocities v1 and v2 (v1 > v2). When the car A is at a distance, d ahead of the car B, the driver of the car A applied the, brake producing a uniform retardation a. There will be no, collision when, (a), , d<, , ( v1 - v2 ) 2, 2a, , (b) d <, , v12 - v22, 2a, , (v1 - v2 ) 2, v 2 - v22, (d) d > 1, 2a, 2a, Q.16 A body travels for 15 second starting from rest with, constant acceleration. If it travels distances S1, S2 and S3, in the first five seconds, second five seconds and next five, seconds respectively the relation between S1, S2 and S3 is, (a) S1 = S2 = S3, (b) 5S1 = 3S2 = S3, 1, 1, 1, 1, (c) S1 = S2 = S3, (d) S1 = S2 = S3, 5, 3, 3, 5, Q.17 The position of a particle moving along the x-axis at certain, times is given below, , (c) d >, , t (s), , 0, , x ( m), , -2, , 1, 0, , 2, 6, , 3, 16, , Which of the following describes the motion correctly?, (a) Uniform, accelerated, (b) Uniform, decelerated, (c) Non-uniform, accelerated, (d) There is not enough data for generalization, Q.18 A body A moves with a uniform acceleration a and zero, initial velocity. Another body B, starts from the same point, moves in the same direction with a constant velocity v. The, two bodies meet after a time t. The value of t is, (a), , 2v, a, , (b), , (c), , v, 2a, , (d), , RESPONSE, GRID, , 15., 20., , v, a, , v, 2a, , 16., 21., , Q.19 A particle moves along x-axis as x = 4 (t – 2) + a (t – 2)2, Which of the following is true?, (a) The initial velocity of particle is 4, (b) The acceleration of particle is 2a, (c) The particle is at origin at t = 0, (d) None of these, Q.20 The displacement x of a particle varies with time t,, x = ae –at + bebt, where a, b, a and b are positive constants., The velocity of the particle will, (a) Go on decreasing with time, (b) Be independent of a and b, (c) Drop to zero when a = b, (d) Go on increasing with time, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21A particle moves as such acceleration is given by, a = 3 sin 4t, then :, (1) the acceleration of the particle becomes zero after, p, each interval of, second, 4, (2) the initial velocity of the particle must be zero, (3) the particle comes at its initial position after sometime, (4) the particle must move on a circular path, Q.22 A reference frame attached to the earth :, (1) is an inertial frame by definition, (2) cannot be an inertial frame because the earth is, revolving round the sun, (3) is an inertial frame because Newton’s laws are, applicable in this frame, (4) cannot be an inertial frame because the earth is, rotating about its own axis, , 17., 22., , Space for Rough Work, , 18., , 19.
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t.me/Magazines4all, DPP/ P 03, , 4, Q.23 If a particle travels a linear distance at speed v1 and comes, back along the same track at speed v2., (1) Its average speed is arithmetic mean (v1 + v2)/2, (2) Its average speed is harmonic mean 2 v1v2/(v1 + v2)/2, (3) Its average speed is geometric mean, (4) Its average velocity is zero, , v1v2, , DIRECTION (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, A particle moves along x-axis and its acceleration at any time t, is a = 2 sin (pt), where t is in seconds and a is in m/s2. The initial, velocity of particle (at time t = 0) is u = 0., Q.24., The distance travelled (in meters) by the particle from, time to t = 0 to t = 1s will be –, 1, 2, (a), (b), p, p, 4, (c), (d) None of these, p, Q.25 The distance travelled (in meters) by the particle from time, t = 0 to t = t will be –, 2, 2t, 2, 2t, sin pt (a), (b) - 2 sin pt +, 2, p, p, p, p, 2t, (c), (d) None of these, p, Q.26 The magnitude of displacement (in meters) by the particle, from time t = 0 to t = t will be –, , RESPONSE, GRID, , 23., , 24., , 28., , 29., , (a), , 2, 2, , sin pt -, , 2t, p, , 2, 2t, (b) - 2 sin pt +, p, p, , p, 2t, (c), (d) None of these, p, DIRECTIONS (Qs. 27-Q.29) : Each of these questions, contains two statements: Statement-1 (Assertion) and, Statement-2 (Reason). Each of these questions has four, alternative choices, only one of which is the correct answer., You have to select the correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : The position-time graph of a uniform motion, in one dimension of a body can have negative slope., Statement-2 : When the speed of body decreases with time,, the position-time graph of the moving body has negative slope., Q.28 Statement-1 : A body having non-zero acceleration can, have a constant velocity., Statement-2 : Acceleration is the rate of change of velocity., Q.29 Statement-1 : Displacement of a body may be zero when, distance travelled by it is not zero., Statement-2 : The displacement is the longest distance, between initial and final position., , 25., , 26., , DAILY PRACTICE PROBLEM SHEET 3 - PHYSICS, Total Questions, 29, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 27., , 116, , 48, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 04, SYLLABUS : MOTION IN A STRAIGHT LINE 2 (Relative Motion & Motion Under Gravity), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.19) : There are 19 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A stone is dropped from a minar of height h and it reaches, after t seconds on earth. From the same minar if two stones, are thrown (one upwards and other downwards) with the same, velocity u and they reach the earth surface after t1 and t2, seconds respectively, then, t1 + t2, 2, t = t12 t22, , (a) t = t1 - t2, , (b) t =, , (c) t = t1t2, , (d), , Q.2 A ball is projected upwards from a height h above the surface, of the earth with velocity v. The time at which the ball strikes, the ground is, , RESPONSE GRID, , 1., , 2., , (a), , v 2hg, +, g, 2, , (b), , vé, 2h ù, 1- 1+ ú, g êë, g û, , (c), , vé, 2 gh ù, ê1 + 1 + 2 ú, gë, v û, , (d), , vé, 2 2g ù, ê1 + v +, ú, gë, h û, , Q.3 A man throws balls with the same speed vertically upwards,, one after the other at an interval of 2 seconds. What should, be the speed of the throw so that more than two balls are in, the sky at any time? (Given g = 9.8 m/s2), (a) At least 0.8 m/s, (b) Any speed less than 19.6 m/s, (c) Only with speed 19.6 m/s, (d) More than 19.6 m/s, , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 04, , 2, Q.4 If a ball is thrown vertically upwards with speed u, the, distance covered during the last t second of its ascent is, 1 2, gt, 2, , (a), , 1, ut - gt 2 (c) (u – gt)t, 2, , (b), , (d) ut d, , Q.5 A ball is thrown vertically upwards. Which of the following, graphs represent velocity-time graph of the ball during its, flight? (air resistance is neglected), v, , v, , t, , (a), , t, , (b), , (a) h1 : h2 (b) h1 : h2 (c) h1 : 2h2 (d) 2h1 : h2, Q.10 Three different objects of masses m1, m2 and m3 are allowed, to fall from rest and from the same point 'O' along three, different frictionless paths. The speeds of the three objects,, on reaching the ground, will be in the ratio of, (a) m1 : m2 : m3, (b) m1 : 2m2 : 3m3, , v, , v, t, , (c), , t, , (d), , Q.8 A body is projected up with a speed 'u' and the time taken, by it is T to reach the maximum height H. Pick out the, correct statement, (a) It reaches H/2 in T/2 sec, (b) It acquires velocity u/2 in T/2 sec, (c) Its velocity is u/2 at H/2, (d) Same velocity at 2T, Q.9 Time taken by an object falling from rest to cover the height, of h1 and h2 is respectively t1 and t2 then the ratio of t1 to t2, is, , 1, , Q.6 A ball is dropped vertically from a height d above the, ground. It hits the ground and bounces up vertically to a, height d/2. Neglecting subsequent motion and air, resistance, its velocity v varies with the height h above the, ground is, v, , v, , d, , h, , (a), , v, , v, d, , (c), , h, , d, , (b), , h, , d, , (d), , h, , (d), , 1 1 1, :, :, m1 m2 m3, , Q.11 From the top of a tower, a particle is thrown vertically, downwards with a velocity of 10 m/s. The ratio of the, distances, covered by it in the 3rd and 2nd seconds of the, motion is (Take g = 10m/s2), (a) 5 : 7 (b) 7 : 5, (c) 3 : 6, (d) 6 : 3, Q.12 A body falls from a height h = 200 m. The ratio of distance, travelled in each 2 second during t = 0 to t = 6 second of, the journey is, (a) 1 : 4 : 9, (b) 1 : 2 : 4, (c) 1 : 3 : 5, (d) 1 : 2 : 3, Q.13 The effective acceleration of a body, when thrown upwards, with acceleration a will be :, (a), , Q.7 P, Q and R are three balloons ascending with velocities U,, 4U and 8U respectively. If stones of the same mass be, dropped from each, when they are at the same height, then, (a) They reach the ground at the same time, (b) Stone from P reaches the ground first, (c) Stone from R reaches the ground first, (d) Stone from Q reaches the ground first, , RESPONSE, GRID, , (c) 1 : 1 : 1, , a - g2, , (b), , a2 + g 2, , (c) (a – g), (d) (a + g), Q.14 An aeroplane is moving with a velocity u. It drops a packet, from a height h. The time t taken by the packet in reaching, the ground will be, (a), , æ 2g ö, ç ÷ (b), è h ø, , æ 2u ö, ç ÷, è g ø, , (c), , æ h ö, ç ÷, è 2g ø, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., Space for Rough Work, , (d), , æ 2h ö, ç ÷, è g ø, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 04, , 3, , Q.15 Two trains, each 50 m long are travelling in opposite direction, with velocity 10 m/s and 15 m/s. The time of crossing is, (a) 2s, (b) 4s, (c), , (d), , 2 3s, , 4 3s, , Q.16 A train of 150 metre length is going towards north direction, at a speed of 10 m/s. A parrot flies at the speed of 5 m/s, towards south direction parallel to the railway track. The, time taken by the parrot to cross the train is, (a) 12 sec, (b) 8 sec, (c) 15 sec, (d) 10 sec, Q.17 The distance between two particles is decreasing at the rate, of 6 m/sec. If these particles travel with same speeds and in, the same direction, then the separation increase at the rate, of 4 m/s. The particles have speeds as, (a) 5 m/sec; 1 m/sec, (b) 4 m/sec; 1 m/sec, (c) 4 m/sec; 2 m/sec, (d) 5 m/sec; 2 m/sec, Q.18 A train is moving towards east and a car is along north, both, with same speed. The observed direction of car to the, passenger in the train is, (a) East-north direction, (b) West-north direction, (c) South-east direction, (d) None of these, Q.19 An express train is moving with a velocity v1. Its driver, finds another train is moving on the same track in the same, direction with velocity v2. To escape collision, driver applies, a retardation a on the train, the minimum time of escaping, collision will be, (a), , v -v, t= 1 2, a, , (b), , (c) Both (a) and (b), , v 2 - v22, t1 = 1, 2, , (d) None of these, , DIRECTIONS (Q.20-Q.22) : In the following questions, more, than one of the answers given are correct. Select the correct, answers and mark it according to the following codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, , RESPONSE, GRID, , 15., 20., , (b) 1 and 2 are correct, (d) 1 and 3 are correct, , 16., 21., , Q.20 Two particles move simultaneously from two points A and B,, 300m apart. The particle at A, starts towards B with a velocity, of 25 m/s and that at B, moves normal to the former with a, velocity of 20 m/s., (1) The relative velocity of the particle at A, w.r.t. that at B is, 32.02 m/s, (2) The relative velocity of the particle at A, w.r.t. that at B is, 12.04 m/s, (3) They are closest to each other after 7.32 sec., (4) They are closest to each other after 4.25 sec., Q.21 A plane is to fly due north. The speed of the plane relative, to the air is 200 km/h, and the wind is blowing from west, to east at 90 km/h., (1) The plane should head in a direction of sin –1 (0.45), (2) The plane should head in a direction of sin –1 (0.60), (3) The relative velocity of plane w.r.t. ground is 179 km/h, (4) The relative velocity of plane w.r.t. ground is 149 km/h, Q.22 From the top of a multi-storeyed building 40m tall, a boy, projects a stone vertically upwards with an initial velocity, of 10 ms–1 such that it eventually falls to the ground., (1) After 4 s the stone will strike the ground, (2) After 2 s the stone will pass through the point from, where it was projected, (3) Its velocity when it strikes the ground is 30 m/s, (4) Its velocity when it strikes the ground is 40 m/s, DIRECTIONS (Q.23-Q.25) : Read the passage given below, and answer the questions that follows :, When an airplane flies, its total velocity with respect to the, ground is vtotal = vplane + vwind,, where vplane denotes the plane’s velocity through motionless air,, and vwind denotes the wind’s velocity. Crucially, all the quantities, in this equation are vectors. The magnitude of a velocity vector, is often called the “speed.”, Consider an airplane whose speed through motionless air is 100, meters per second (m/s). To reach its destination, the plane must, fly east., The “heading” of a plane is in the direction in which the nose of, the plane points. So, it is the direction in which the engines propel, the plane., , 17., 22., , Space for Rough Work, , 18., , 19.
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t.me/Magazines4all, DPP/ P 04, , 4, Q.23 If the plane has an eastward heading, and a 20 m/s wind, blows towards the southwest, then the plane’s speed is –, (a) 80 m/s, (b) more than 80 m/s but less than 100 m/s, (c) 100 m/s, (d) more than 100 m/s, Q.24 The pilot maintains an eastward heading while a 20 m/s, wind blows northward. The plane’s velocity is deflected, from due east by what angle?, 20, 100, -1 20, tan, 100, , -1, (a) sin, , -1, (b) cos, , (c), , (d) none, , 20, 100, , Q.25 Let f denote the answer of above question. The plane has, what speed with respect to the ground ?, (a) (100 m/s) sin f, (b) (100 m/s) cos f, (c), , 100 m/ s, sin f, , (d), , 100 m/ s, cosf, , DIRECTIONS (Qs. 26-Q.28) : Each of these questions, contains two statements: Statement-1 (Assertion) and, Statement-2 (Reason). Each of these questions has four, alternative choices, only one of which is the correct answer., You have to select the correct choice., , RESPONSE, GRID, , 23., , 24., , (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT, a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.26 Statement-1 : The magnitude of velocity of two boats, relative to river is same. Both boats start simultaneously, from same point on the bank may reach opposite bank, simultaneously moving along different paths., Statement-2 : For boats to cross the river in same time., The component of their velocity relative to river in, direction normal to flow should be same., Q.27 Statement-1 : The acceleration of a body of mass 2 kg, thrown vertically upwards is always constant., Statement-2 : A body of all mass group travels under, constant acceleration when only gravity acts on it., Q.28 Statement-1 : The velocity of a body A relative to the body, B is the sum of the velocities of bodies A and B if both, travel in opposite direction on a straight line., Statement-2 : The velocity of a body A relative to the body, B is the difference of the velocities of bodies A and B if, both travel in opposite direction on a straight line., , 25., , 26., , 27., , 28., , DAILY PRACTICE PROBLEM SHEET 4 - PHYSICS, Total Questions, 28, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 112, , 44, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 05, SYLLABUS : Vectors, , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The length of second’s hand in watch is 1 cm. The change, in velocity of its tip in 15 seconds is, (a) zero, , (b), , p, cm/sec, 30, , p, 30 2, , (a) zero, (c), , RESPONSE GRID, , cm/sec, , 2., , m/s2 N-E, , (d), , 2, 1, 2, , m/s 2 N-W, m/s 2S-W, , acts on a particle moving in the x-y plane. Starting from, the origin, the particle is taken along the positive x-axis to, the point (a, 0) and then parallel to the y-axis to the point, ur, (a, a). The total work done by the forces F on the particle, is, (a) – 2 Ka2, (b) 2 Ka2, 2, (c) – Ka, (d) Ka2, , (d), , 1., , 2, , 1, , ur, Q.3 A force F = - K ( yiˆ + xjˆ ) (where K is a positive constant), , p 2, cm/sec, 30, Q.2 A particle moves towards east with velocity 5 m/s. After, 10 seconds its direction changes towards north with same, velocity. The average acceleration of the particle is, , (c), , 1, , (b), , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 05, , 2, Q.4 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?, , A, 30°, 30 N, O, W, , q, , (a), , q, , (c) 60 3,30, , P, W, , r r r, (a) P = W tan q, (b) T + P + W = 0, (c) T 2 = P 2 + W 2, (d) T = P + W, Q.5 The speed of a boat is 5km/h in still water. It crosses a, river of width 1 km along the shortest possible path in 15, minutes. The velocity of the river water is, (a) 1 km/h, (b) 3 km/h, (c) 4 km/h, (d) 5 km/h, Q.6 A man crosses a 320 m wide river perpendicular to the, current in 4 minutes. If in still water he can swim with a, speed 5/3 times that of the current, then the speed of the, current in m/min is, (a) 30, (b) 40, (c) 50, (d) 60, Q.7 P, Q and R are three coplanar forces acting at a point and, are in equilibrium. Given P = 1.9318 kg wt, sin q1 = 0.9659,, the value of R is (in kg wt), 150°, P, , q2, , q1, , Q, , 30 3, 60, , (d) None of these, , Q.9 A boat is moving with a velocity 3iˆ + 4 ˆj with respect to, ground. The water in the river is moving with a velocity, – 3iˆ - 4 ˆj with respect to ground. The relative velocity of, the boat with respect to water is, (b) -6iˆ - 8 ˆj (c) 6iˆ + 8 ˆj (d) 5 2 iˆ, (a) 8 ˆj, Q.10 A person aiming to reach the exactly opposite point on the, bank of a stream is swimming with a speed of 0.5 m/s at an, angle of 120° with the direction of flow of water. The speed, of water in the stream is, (a) 1 m/s, (b) 0.5 m/s (c) 0.25 m/s(d) 0.433 m/s, Q.11 A man can swim with velocity v relative to water. He has to, cross a river of width d flowing with a velocity u (u > v)., The distance through which he is carried down stream by, the river is x. Which of the following statements is correct?, (a) If he crosses the river in minimum time x = du, v, du, (b) x cannot be less than, v, (c) For x to be minimum he has to swim in a direction making, ævö, ο, an angle of , sin,1 çç ÷÷÷ with the direction of the, çè u ø, 2, flow of water., (d) x will be maximum if he swims in a direction making, p, ævö, an angle of + sin -1 ç ÷ with direction of the flow of, 2, èuø, water., , R, , (a) 0.9659, , (b) 2, 1, (c) 1, (d), 2, Q.8 As shown in figure the tension in the horizontal cord is 30, N. The weight W and tension in the string OA in newton are, , RESPONSE, GRID, , (b), , 30 3,30, , 4., , 5., , 6., , 9., , 10., , 11., , Space for Rough Work, , 7., , 8., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., DPP/ P 05, , 3, , Q.12 A 120 m long train is moving towards west with a speed of, 10 m/s. A bird flying towards east with a speed of 5 m/s, crosses the train. The time taken by the bird to cross the, train will be, (a) 16 sec, (b) 12 sec (c) 10 sec (d) 8 sec, ur, Q.13 What is the value of linear velocity, if w = 3$i - 4 $j + k$ and, r, r = 5i$ - 6 $j + 6k$, (a) 6$i - 2 $j + 3k$, (b) 6$i - 2 $j + 8k$, , r, r, Q.20 Two forces F1 = 5iˆ + 10 ˆj - 20kˆ and F2 = 10iˆ - 5 ˆj - 15kˆ, r, r, act on a single point. The angle between F1 and F2 is nearly, (a) 30°, (b) 45°, (c) 60°, (d) 90°, Q.21 With respect to a rectangular cartesian coordinate system,, three vectors are expressed as, r, r, a = 4iˆ - ˆj, b = -3iˆ + 2 ˆj , and cr = - kˆ, , (c) 4$i - 13 $j + 6k$, (d) -18$i - 13 $j + 2k$, ur ur, ur ur, ur ur, Q.14 If | A ´ B | = 3 A . B, then the value of | A + B | is, , where iˆ, ˆj , kˆ are unit vectors, along the X, Y and Z-axis, respectively. The unit vectors r̂ along the direction of sum, of these vector is, 1 ˆ ˆ ˆ, 1 ˆ ˆ ˆ, (a) rˆ =, (b) rˆ =, (i + j - k ), (i + j - k ), 3, 2, 1, 1 ˆ ˆ ˆ, (c) rˆ = (iˆ - ˆj + kˆ), (d) rˆ =, (i + j + k ), 3, 2, , 1/ 2, , (a), , æ 2, 2 AB ö, çè A + B +, ÷, 3ø, , (b) A + B, , (c) ( A2 + B 2 + 3 AB )1/ 2 (d) ( A2 + B 2 + AB )1/ 2, ur, Q.15 Find the torque of a force F = -3$i + $j + 5k$ acting at a point, r, r = 7$i + 3 $j + k$, (a) 14$i - 38 $j + 16k$, (b) 4$i + 4 $j + 6k$, (c) 21$i + 4 $j + 4k$, (d) -14$i + 34 $j - 16k$, ur ur ur ur, ur, ur, Q.16 If | A ´ B | = | A . B |, then angle between A and B will be, (a) 30°, (b) 45°, (c) 60°, (d) 90°, ur, ur, $, $, $, Q.17 The vector P = ai + a j + 3k and Q = ai$ - 2 $j - k$ are, perpendicular to each other. The positive value of a is, (a) 3, (b) 4, (c) 9, (d) 13, Q.18 A particle moves from position 3$i + 2 $j - 6k$ to, 14$i + 13 $j + 9k$ due to a uniform force of (4i$ + $j + 3k$ ) N ., If the displacement in metres then work done will be, (a) 100 J, (b) 200 J (c) 300 J d) 250 J, ur, ur, Q.19 The three vectors A = 3iˆ - 2 ˆj + kˆ, B = iˆ - 3 ˆj + 5kˆ and, ur, C = 2iˆ + ˆj - 4kˆ form, (a) an equilateral triangle (b) isosceles triangle, (c) a right angled triangle (d) no triangle, , RESPONSE, GRID, , t.me/Magazines4all, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 A boy walks uniformally along the sides of a rectangular, park of size 400 m × 300 m, starting from one corner to, the other corner diagonally opposite. Which of the, following statements is correct?, (1) He has travelled a distance of 700 m, (2) His displacement is 500 m, (3) His velocity is not uniform throughout the walk, (4) His displacement is 700 m, r, r, ˆ B = iˆ - 3jˆ + 5kˆ and, Q.23 The three vectors A = 3iˆ - 2jˆ - k,, r, C = 2iˆ – ˆj - 4kˆ does not form, (1) an equilateral triangle (2) isosceles triangle, (3) a right angled triangle (4) no triangle, , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., Space for Rough Work
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t.me/Magazines4all, DPP/ P 05, , 4, r, r r r, Q.24 If for two vectors A and B, A ´ B < 0, which of the following, is not correct?, (1) They are perpendicular to each other, (2) They act at an angle of 60°, (3) They act at an angle of 30°, (4) They are parallel to each other, , DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, r, r, A = 2iˆ + ˆj + kˆ and B = iˆ + ˆj + kˆ are two vectors., r, Q.25 The unit vector perpendicular to A is, - ˆj - kˆ, - ˆj + kˆ, iˆ + kˆ, iˆ - kˆ, (c), (d), (b), (a), 2, 2, 2, 2, r, Q.26 The unit vector parallel to A is, 2iˆ - ˆj + 3kˆ, 2iˆ + ˆj + kˆ, (a), (b), 2, 6, ˆ, ˆ, ˆ, ˆ, 2i - j - k, 2i + ˆj - 2kˆ, (c), (d), 5, 6, r, Q.27 The unit vector perpendicular to B is, - ˆj - kˆ, - ˆj + kˆ, iˆ - kˆ, iˆ + kˆ, (a), (b), (c), (d), 3, 2, 3, 2, , RESPONSE, GRID, , 24., , 25., , DIRECTIONS (Q.28-Q.29) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason)., Each of these questions has four alternative choices, only one of, which is the correct answer. You have to select the correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., r r, r r, r, Q.28 Statement-1:If A + B = A - B , then angle between A, r, and B is 90°, r r r r, Statement-2 : A + B = B + A, Q.29 Statement-1 : The sum of two vectors can be zero., Statement-2 : Two vectors cancel each other, when they, are equal and opposite., , 26., , 27., , 28., , 29., , DAILY PRACTICE PROBLEM SHEET 5 - PHYSICS, Total Questions, 29, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 116, , 44, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 06, , SYLLABUS : MOTION IN A PLANE-1 (Projectile Motion), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The path followed by a body projected along y axis is given, by y = 3 x – (1/2) x2. If g = 10 m/s2 , then the initial, velocity of projectile will be – (x and y are in m), (a) 3, , 10 m/s, , (b) 2 10 m/s, , (c) 10 3 m/s, (d) 10 2 m/s, Q.2 When the angle of elevation of a gun are 60º and 30º, respectively, the height it shoots are h1 and h2 respectively,, h1/h2 equal to –, (a) 3/1, (b) 1/3, (c) 1/2, (d) 2/1, , RESPONSE GRID, , 1., , 2., , Q.3 If t1 be the time taken by a body to clear the top of a building, and t2 be the time spent in air, then t2 : t1 will be –, (a) 1 : 2 (b) 2 : 1, (c) 1 : 1, (d) 1 : 4, Q.4 The co-ordinates of a moving particle at any time t are given, by x = ct2 and y = bt2. The speed of the particle is, (a) 2t (c + b), , (b), , 2t, , c2 - b2, , (c) t c 2 + b 2, (d) 2t c 2 + b 2, Q.5 The height y and the distance x along the horizontal at plane, of the projectile on a certain planet (with no surrounding, atmosphere) are given by y = (8t – 5t2) metre and x = 6t, metre where t is in second. The velocity with which the, projectile is projected is, (a) 8 m/s, (b) 6 m/s, (c) 10 m/s, (d) Data is insufficient, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 06, , 2, Q.6 A body is thrown at an angle 30º to the horizontal with the, velocity of 30 m/s. After 1 sec, its velocity will be, (in m/s) (g = 10 m/s2), (a) 10 7 (b) 700 10 (c) 100 7 (d), 10, Q.7 A particle is moving in a plane with a velocity given by,, r, u = u0 î + (wa cos wt) ĵ , where î and ĵ are unit vectors, along x and y-axes respectively. If the particle is at the origin, at t = 0, then its distance from the origin at time t = 3p/, 2w will be, (a), , éæ 3pu ö 2, ù, 0, êç, ÷ + a2ú, êëè 2w ø, úû, , (b), , éæ 3pu 0 ö 2 ù, ÷+a ú, êç, ëè 2w ø, û, , éæ 3pu ö 2, ù, éæ 4pu ö 2, ù, 0, 0, ê, ú, êç, a, +, ç, ÷, ÷ + a2 ú, (c), (d), êëè 2w ø, úû, êëè 2w ø, úû, Q.8 A ball thrown by one player reaches the other in 2 sec. The, maximum height attained by the ball above the point of, projection will be about(a) 2.5 m (b) 5 m, (c) 7.5 m, (d) 10 m, Q.9 Rishabh and Bappy are playing with two different balls of, masses m and 2m respectively. If Rishabh throws his ball, vertically up and Bappy at an angle q, both of them stay in, our view for the same period. The height attained by the, two balls are in the ratio of, (a) 2 : 1 (b) 1 : 1, (c) 1 : cos q (d) 1 : sec q, Q.10 A projectile is thrown at an angle q and (900 – q) from the, same point with same velocity 98 m/s. The heights attained, by them, if the difference of heights is 50 m will be (in m), (a) 270, 220, (b) 300, 250, (c) 250, 200, (d) 200, 150, Q.11 A particle is projected with a velocity u so that its horizontal, range is twice the maximum height attained. The horizontal, range is, (a) u2/g, (b) 2u2 /3g (c) 4u2/5g, (d) u2/2g, Q.12 Mr C.P. Nawani kicked off a football with an initial speed, 19.6 m/s at a projection angle 45º. A receiver on the goal, line 67.4 m away in the direction of the kick starts running, , RESPONSE, GRID, , to meet the ball at that instant. What must be his speed so, that he could catch the ball before hitting the ground ?, (a) 2.82 m/s, (b) 2/ 2 m/s, (c) 39.2 m/s, (d) 10 m/s, Q.13 A ball is thrown from ground level so as to just clear a wall, 4 metres high at a distance of 4 metres and falls at a distance, of 14 metres from the wall. The magnitude of velocity of, the ball will be, (a), , 182 m/s, , (b), , 181 m/s, , 185 m/s, (d) 186 m/s, (c), Q.14 A ball is projeced from O with an initial velocity 700 cm/, s in a direction 37° above the horizontal. A ball B, 500 cm, away from O on the line of the initial velocity of A, is, released from rest at the instant A is projected. The height, through which B falls, before it is hit by A and the direction, of the velocity A at the time of impact will respectively be, [given g = 10 m/s2, sin 37° = 0.6 and cos 37° = 8.0], (a) 250 cm, 28° 42', (b) 255 cm, 27° 43', (c) 245 cm, 20° 44', (d) 300 cm, 27° 43', Q.15 A ball is thrown horizontally from a height of 20 m. It hits, the ground with a velocity three times its initial velocity., The initial velocity of ball is, (a) 2 m/s (b) 3 m/s, (c) 5 m/s, (d) 7 m/s, Q.16 A projectile thrown from a height of 10 m with velocity of, 2 m/s, the projectile will fall, from the foot of, projection, at distance-(g = 10 m/s2), , (a) 1 m, (b) 2 m, (c) 3m, (d), 2m, Q.17 Savita throws a ball horizontally with a velocity of 8 m/s, from the top of her building. The ball strikes to her brother, Sudhir playing at 12 m away from the building. What is the, height of the building ?, (a) 11m (b) 10 m, (c) 8 m, (d) 7 m, Q.18 A body is projected downdwards at an angle of 30º to the, horizontal with a velocity of 9.8 m/s from the top of a tower, 29.4 m high. How long will it take before striking the, ground?, (a) 1s, (b) 2s, (c) 3s, (d) 4s, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 06, , 3, , Q.19 A ball is thrown from the top of a tower with an initial velocity, of 10 m/s at an angle of 30º above the horizontal. It hits the, ground at a distance of 17.3 m from the base of the tower., The height of the tower (g = 10 m/s2) will be, (a) 10 m (b) 12 m, (c) 110 m, (d) 100 m, Q.20 A ball 'A' is projected from origin with an initial velocity, v0 = 700 cm/sec in a direction 37º above the horizontal as, shown in fig .Another ball 'B' 300 cm from origin on a line, 37º above the horizontal is released from rest at the instant, A starts. How far will B have fallen when it is hit by A ?, Y, , B, O, , =, , 30, , O, , (a) 9 cm, (c) 0.9 cm, , B, cm A, 0, y-axis, X, , DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, , (b) 90 cm, (d) 900 cm, , 19., , 20., , Velocity at a general point P(x, y) for a horizontal projectile, motion is given by, u, u, , v=, , v 2x + v 2y ; tan a =, , h, , vy, , gt, , vx, , R, , v= u 2 + g 2 t 2, , DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Choose the correct options, (1) A ball is dropped from the window of a moving train, on horizontal rails, the path followed by the ball as, observed by the observer on the ground is parabolic, path., (2) If T be the total time of flight of a current of water, and H be the maximum height attained by it from the, point of projection, then H/T will be (1/4) u sinq, (u = projection velocity and q = projection angle), , RESPONSE GRID, , (3) A hunter aims his gun and fires a bullet directly at a, monkey on a tree. At the instant bullet leaves the gun,, monkey drops, the bullet misses to hit the monkey., (4) If a baseball player can throw a ball at maximum, distance = d over a ground, the maximum vertical, height to which he can throw it, will be d, (Ball have same initial speed in each case), Q.22 A ball projected with speed 'u' at an angle of projection, 15º has range R. The other angle of projection at which the, range will not be same with same initial speed 'u' is, (1) 45º, (2) 35º, (3) 90º, (4) 75º, Q.23 A projectile can have the same range R for two angles of, projections. If t1 and t2 be the times of flight in two cases,, then choose the incorrect relations –, (1) t1t2 µ 1/R2, (2) t1t2 µ R2, (3) t1t2 µ 1/R, (4) t1t2 µ R, , a is angle made by v with horizontal in clockwise direction, Trajectory equation for a horizontal projectile motion is given, by x = vxt = ut, a, , y, , = – (1/2) gt2, , vx, , vy, v, , eliminating t, we get y = – (1/2), , 21., , Space for Rough Work, , 22., , gx 2, u2, , 23.
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t.me/Magazines4all, DPP/ P 06, , 4, Q.24 A ball rolls off top of a stair way with a horizontal velocity, u m/s. If the steps are h m high and b meters wide, the ball, will just hit the edge of nth step if n equals to, (a), , (c), , hu 2, , (b), , gb 2, 2hu 2, , (d), , gb 2, , u 2g, gb 2, 2u 2g, hb 2, , Q.25 An aeroplane is in a level flying at an speed of 144 km/hr, at an altitude of 1000 m. How far horizontally from a given, target should a bomb be released from it to hit the target ?, (a) 571.43 m, (b) 671.43 m, (c) 471.34 m, (d) 371.34 m, Q.26 An aeroplane is flying horizontally with a velocity of 720, km/h at an altitude of 490 m. When it is just vertically above, the target a bomb is dropped from it. How far horizontally it, missed the target?, (a) 1000 m, (b) 2000 m, (c) 100 m, (d) 200 m, , RESPONSE GRID, , 24., , 25., , DIRECTIONS (Q.27-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select, the correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement -1 : Two projectiles are launched from the top, of a cliff with same initial speed with different angles of, projection. They reach the ground with the same speed., Statement -2 : The work done by gravity is same in both the, cases., Q.28 Statement-1 : A man projects a stone with speed u at some, angle. He again projects a stone with same speed such that, time of flight now is different. The horizontal ranges in, both the cases may be same. (Neglect air friction), Statement-2 : The horizontal range is same for two, projectiles projected with same speed if one is projected, at an angle q with the horizontal and other is projected at, an angle (90° – q) with the horizontal. (Neglect air friction), , 26., , 27., , DAILY PRACTICE PROBLEM SHEET 6 - PHYSICS, Total Questions, 28, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 28., , 112, , 42, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 07, , SYLLABUS : MOTION IN A PLANE-2 (Horizontal Circular Motion), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A particle completes 1.5 revolutions in a circular path of, radius 2 cm. The angular displacement of the particle will, be – (in radian), (a) 6 p, (b) 3 p, (c) 2 p, (d) p, Q.2 A particle revolving in a circular path completes first one, third of circumference in 2 sec, while next one third in 1, sec. The average angular velocity of particle will be – (in, rad/sec), (a) 2p/3, (b) p/3, (c) 4p/3, (d) 5p/3, Q.3 The ratio of angular speeds of minute hand and hour hand, of a watch is -, , RESPONSE GRID, , 1., , 2., , (a) 1 : 12, (b) 6 : 1, (c) 12 : 1 (d) 1 : 6, Q.4 The angular displacement of a particle is given by, 1 2, at , where w0 and a are constant and, 2, w0 = 1 rad/sec, a = 1.5 rad/sec2. The angular velocity at, time, t = 2 sec will be (in rad/sec) (a) 1, (b) 5, (c) 3, (d) 4, Q.5 The magnitude of the linear acceleration of the particle, moving in a circle of radius 10 cm with uniform speed, completing the circle in 4 s, will be (a) 5p cm/s2, (b) 2.5p cm/s2, (c) 5p2 cm/s2, (d) 2.5p2 cm/s2, , q = w0t +, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 07, , 2, Q.6 A cane filled with water is revolved in a vertical circle of, radius 4 m and water just does not fall down. The time, period of revolution will be –, (a) 1 s, (b) 10 s, (c) 8 s, (d) 4 s, Q.7 The length of second's hand in a watch is 1 cm. The change, in velocity of its tip in 15 second is (a) 0, (c), , (b), , p, cm/s, 30, , (d), , p, , cm/s, , 30 2, p 2, cm/s, 30, , Q.8 An electron is moving in a circular orbit of radius 5.3 × 10–, 11 metre around the atomic nucleus at a rate of 6.6 × 1015, revolutions per second. The centripetal force acting on the, electron will be (The mass of the electron is 9.1 × 10–31kg), (a) 8.3 × 10–8N, (b) 3.8 × 10–8N, –8, (c) 4.15 × 10 N, (d) 2.07 × 10–8N, Q.9 An air craft executes a horizontal loop of radius 1 km with, a steady speed of 900 km/h. The ratio of centripetal, acceleration to that gravitational acceleration will be(a) 1 : 6.38, (b) 6. 38 : 1, (c) 2.25 : 9.8, (d) 2.5 : 9.8, Q.10 A car driver is negotiating a curve of radius 100 m with a, speed of 18 km/hr. The angle through which he has to lean, from the vertical will be 1, 1, (a) tan–1, (b) tan–1, 4, 40, æ 1 ö÷, æ 1 ö÷, (c) tan –1 çççè ø÷÷, (d) tan –1 çççè ø÷÷, 2, 20, , Q.11 A particle moves in a circle of radius 20cm with a linear, speed of 10m/s. The angular velocity will be (a) 50 rad/s, (b) 100 rad/s, (c) 25 rad/s, (d) 75 rad/s, Q.12 The angular velocity of a particle is given by w = 1.5 t – 3t2 +, 2, the time when its angular acceleration decreases to be, zero will be (a) 25 sec, (b) 0.25 sec, (c) 12 sec, (d) 1.2 sec, , RESPONSE, GRID, , Q.13 A particle is moving in a circular path with velocity varying, with time as v = 1.5t2 + 2t. If the radius of circular path is, 2 cm, the angular acceleration at t = 2 sec will be (a) 4 rad/sec2, (b) 40 rad/sec2, 2, (c) 400 rad/sec, (d) 0.4 rad/sec2, Q.14 A grind stone starts from rest and has a constant-angular, acceleration of 4.0 rad/sec2.The angular displacement and, angular velocity, after 4 sec. will respectively be (a) 32 rad, 16 rad/s, (b) 16 rad, 32 rad/s, (c) 64 rad, 32 rad/s, (d) 32 rad, 64 rad/s, Q.15 The shaft of an electric motor starts from rest and on the, application of a torque, it gains an angular acceleration, given by a = 3t – t2 during the first 2 seconds after it starts, after which a = 0. The angular velocity after 6 sec will be (a) 10/3 rad/sec, (b) 3/10 rad/sec, (c) 30/4 rad/sec, (d) 4/30 rad/sec, Q.16 Using rectangular co-ordinates and the unit vectors i and, j, the vector expression for the acceleration a will be (r is, a position vector) (a) wr2, (b) –w2 r/2, 2, (c) –2wr, (d) –w2r, Q.17 The vertical section of a road over a canal bridge in the, direction of its length is in the form of circle of radius 8.9, metre. Find the greatest speed at which the car can cross, this bridge without losing contact with the road at its highest, point, the center of gravity of the car being at a height h =, 1.1 metre from the ground. (Take g = 10 m/sec2), (a) 5 m/s, (b) 7 m/s, (c) 10 m/s, (d) 13 m/s, Q.18 The maximum speed at which a car can turn round a curve, of 30 metre radius on a level road if the coefficient of, friction between the tyres and the road is 0.4, will be (a) 10.84 m/s, (b) 17.84 m/s, (c) 11.76 m/s, (d) 9.02 m/s, Q.19 The angular speed with which the earth would have to rotate, on its axis so that a person on the equator would weigh, (3/5)th as much as present, will be:, (Take the equatorial radius as 6400 km), (a) 8.7 × 104 rad/sec, (b) 8.7 × 103 rad/sec, 4, (c) 7.8 × 10 rad/sec, (d) 7.8 × 103 rad/sec, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 07, , 3, , Q.20 A smooth table is placed horizontally and a spring of, unstreched length l0 and force constant k has one end fixed, to its centre. To the other end of the spring is attached a, mass m which is making n revolution per second around, the centre. Tension in the spring will be, (a) 4p2 m k l0 n2/ (k – 4p2 m n2), (b) 4p2 m k l0 n2/ (k + 4p2 m n2), (c) 2p2 m k l0 n2/ (k – 4p2 m n2), (d) 2p m k l0 n2/ (k – 4p2 m n2), Q.21 A motor car is travelling at 30 m/s on a circular road of, radius 500 m. It is increasing its speed at the rate of 2 m/, s2. Its net acceleration is (in m/s2) –, (a) 2, (b) 1. 8, (c) 2.7, (d) 0, DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 Three identical particles are connected by three strings as, shown in fig. These particles are revolving in a horizontal, circle. The velocity of outer most particle is v, then choose, correct relation for T1,T2 and T3, (where T1 is tension in the outer most string etc.), O, , (1) T1 =, , l, , m, , 2, mvA, 3l, , l, , m, , l, , m, , (2) T2 =, , 5mv 2A, 9l, , 6mvA2, 5mvA2, (4) T3 =, 9l, 9l, Q.23 A particle describes a horizontal circle on the smooth, surface of an inverted cone. The height of the plane of the, circle above the vertex is 9.8 cm, then choose the correct, options, , (3) T3=, , RESPONSE, GRID, , (1) The speed of the particle will be 0.98 m/s, (2) tan q =, , rg, , (q is semi-apex angle), v2, (3) The speed of the particle will be 98 m/s, rg, (4) tan q =, (q is semiapex angle), v, Q.24 Choose the correct statements, (1) Centripetal force is not a real force. It is only the, requirement for circular motion., (2) Work done by centripetal force may or may not be, zero., (3) Work done by centripetal force is always zero., (4) Centripetal force is a fundamental force., DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, The velocity of the particle changes moving on the curved path,, this change in velocity is brought by a force known as centripetal, force and the acceleration so produced in the body is known as, centripetal acceleration. The direction of centripetal force or, acceleration is always towards the centre of circular path., Q.25 A ball is fixed to the end of a string and is rotated in a, horizontal circle of radius 5 m with a speed of 10 m/sec., The acceleration of the ball will be (a) 20 m/s2, (b) 10 m/s2, 2, (c) 30 m/s, (d) 40 m/s2, Q.26 A body of mass 2 kg lying on a smooth surface is attached, to a string 3 m long and then whirled round in a horizontal, circle making 60 revolution per minute. The centripetal, acceleration will be(a) 118.4 m/s2, (b) 1.18 m/s2, 2, (c) 2.368 m/s, (d) 23.68 m/s2, Q.27 A body of mass 0.1 kg is moving on circular path of diameter, 1.0 m at the rate of 10 revolutions per 31.4 seconds. The, centripetal force acting on the body is (a) 0.2 N, (b) 0.4 N, (c) 2 N, (d) 4 N, , 20., , 21., , 22., , 25., , 26., , 27., , Space for Rough Work, , 23., , 24.
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t.me/Magazines4all, DPP/ P 07, , 4, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason). Each, of these questions has four alternative choices, only one of which is, the correct answer. You have to select the correct choice., (a), (b), (c), (d), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , Q.28 Statement - 1 : In non-uniform circular motion, velocity, vector and acceleration vector are not perpendicular to each, other., Statement - 2 : In non-uniform circular motion, particle has, , RESPONSE GRID, , 28., , 29., , normal as well as tangential acceleration., Q.29 Statement - 1 : A cyclist is cycling on rough horizontal, circular track with increasing speed. Then the frictional force, on cycle is always directed towards centre of the circular, track., Statement - 2 : For a particle moving in a circle, radial, component of net force should be directed towards centre., r, Q.30 Statement - 1 : If net force F acting on a system is, r, changing in direction only, the linear momentum ( p) of, system changes in direction., Statement - 2 : In case of uniform circular motion,, magnitude of linear momentum is constant but direction, of centripetal force changes at every instant., , 30., , DAILY PRACTICE PROBLEM SHEET 7 - PHYSICS, Total Questions, 30, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), , Space for Rough Work, , 120, , 48, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 08, SYLLABUS : MOTION IN A PLANE-3 (Vertical Circular Motion, Relative Motion), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.19) : There are 19 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A man whirls a stone round his head on the end of a string, 4.0 metre long. Can the string be in a horizontal, plane? If, the stone has a mass of 0.4 kg and the string will break, if, the tension in it exceeds 8 N. The smallest angle the string, can make with the horizontal and the speed of the stone, will respectively be (Take g = 10 m/sec2), (a) 30º, 7.7 m/s, (b) 60º, 7.7 m/s, (c) 45º, 8.2 m/s, (d) 60º, 8.7 m/s, Q.2 In figure ABCDE is a channel in the vertical plane, part BCDE, being circular with radius r. A ball is released from A and, slides without friction and without rolling. It will complete, the loop path when, , RESPONSE GRID, , 1., , 2., , A, D, E r, r, , r C, , B, (a) h > 5 r/2, (b) h < 5 r/2, (c) h < 2r/5, (d) h > 2r/5, Q.3 An aircraft loops the loop of radius R = 500 m with a, constant velocity v = 360 km/h. The weight of the flyer of, mass m = 70 kg in the lower, upper and middle points of, the loop will respectively be(a) 210 N, 700 N, 1400 N (b) 1400 N, 700 N, 2100 N, (c) 700 N, 1400 N, 210 N (d) 2100 N, 700 N, 1400 N, , 3., Space for Rough Work
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , DPP/ P 08, , 2, Q.4 A particle of mass 3 kg is moving under the action of a, central force whose potential energy is given by U(r) =, 10r3 joule. For what energy and angular momentum will, the orbit be a circle of radius 10 m ?, (a) 2.5 × 104 J, 3000 kgm2/sec, (b) 3.5 × 104 J, 2000 kgm2/sec, (c) 2.5 × 103 J, 300 kgm2/sec, (d) 3.5 × 103 J, 300 kgm2/sec, Q.5 A string of length 1 m is fixed at one end and carries a, mass of 100 gm at the other end. The string makes 2/p, revolutions per second about a vertical axis through the, fixed end. The angle of inclination of the string with the, vertical, and the linear velocity of the mass will respectively be - (in M.K.S. system), (a) 52º14', 3.16, (b) 50º14', 1.6, (c) 52º14', 1.6, (d) 50º14', 3.16, Q.6 A particle of mass m is moving in a circular path of constant, radius r such that its centripetal acceleration ac is varying, with time t as ac = k2rt2, where k is a constant. The power, delivered to the particle by the force acting on it will be (a) mk2t2r, (b) mk2r2t2 (c) m2k2t2r2 (d) mk2r2t, Q.7 A car is moving in a circular path of radius 100 m with, velocity of 200 m/sec such that in each sec its velocity, increases by 100 m/s, the net acceleration of car will be (in m/sec), (a) 100 17 (b) 10 7 (c) 10 3 (d) 100 3, Q.8 A 4 kg balls is swing in a vertical circle at the end of a cord, 1 m long. The maximum speed at which it can swing if the, cord can sustain maximum tension of 163.6 N will be (a) 6 m/s, (b) 36 m/s (c) 8 m/s (d) 64 m/s, Q.9 The string of a pendulum is horizontal. The mass of the, bob is m. Now the string is released. The tension in the, string in the lowest position is (a) 1 mg, (b) 2 mg, (c) 3 mg, (d) 4 mg, Q.10 A swimmer can swim in still water at a rate 4 km/h. If he, swims in a river flowing at 3 km/h and keeps his direction, (w.r.t. water) perpendicular to the current. Find his velocity, w.r.t. the ground., (a) 3 km/hr, (b) 5 km/hr, (c) 4 km/hr, (d) 7 km/hr, , RESPONSE, GRID, , Q.11 The roadway bridge over a canal is the form of an arc of a, circle of radius 20 m. What is the minimum speed with, which a car can cross the bridge without leaving contact, with the ground at the highest point (g = 9.8 m/s2), (a) 7 m/s, (b) 14 m/s (c) 289 m/s (d) 5 m/s, Q.12 A cane filled with water is revolved in a vertical circle of, radius 0.5 m and the water does not fall down. The, maximum period of revolution must be (a) 1.45, (b) 2.45, (c) 14.15 (d) 4.25, Q.13 A particle of mass m slides down from the vertex of semihemisphere, without any initial velocity. At what height, from horizontal will the particle leave the sphere2, 3, 5, 8, R, (b), R, (c), R, (d), R, 3, 2, 8, 5, Q.14 A body of mass m tied at the end of a string of length l is, , (a), , projected with velocity, , 4lg , at what height will it leave, , the circular path 5, 3, 1, 2, l, (b), l, (c), l, (d), l, 3, 5, 3, 3, Q.15 A string of length L is fixed at one end and carries a mass, M at the other end. The string makes 2/p revolutions per, second around the vertical axis through the fixed end as, shown in the figure, then tension in the string is, , (a), , (a) ML, , S, q, , (b) 2 ML, , L, T, , (c) 4 ML, , M, , (d) 16 ML, , R, , Q.16 A train has to negotiate a curve of radius 400 m. By how, much should the outer rail be raised with respect to inner, rail for a speed of 48 km/hr. The distance between the rails, is 1 m., (a) 12 m, (b) 12 cm, (c) 4.5 cm, (d) 4.5 m, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 08, , 3, , Q.17 A ship is steaming towards east at a speed of 12 ms–1. A, woman runs across the deck at a speed of 5 ms–1 in the, direction at right angles to the direction of motion of the, ship i.e. towards north. What is the velocity of the woman, relative to sea ?, (a) 13 m/s (b) 5 m/s (c) 12 m/s (d) 17 m/s, Q.18 A man is walking on a level road at a speed of 3 km/h., Raindrops fall vertically with a speed of 4 km/h. Find the, velocity of raindrops with respect to the men., (a) 3 km/hr (b) 4 km/hr (c) 5 km/hr (d) 7 km/hr, Q.19 A stone of mass 1 kg tied to a light inextensible string of, , length L = 10 m is whirling in a circular path of radius L, , 3, in a vertical plane. If the ratio of the maximum tension in, the string to the minimum tension in the string is 4 and if g, is taken to be 10m / sec2, the speed of the stone at the, highest point of the circle is, , (a) 20 m/sec, , (b) 10 3m/sec, , (c) 5 2m/sec, , (d) 10 m/sec, , DIRECTIONS (Q.20-Q.22) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.20 Two bodies P and Q are moving along positive x-axis their, r, position-time graph is shown below. If VPQ is velocity of, r, P w.r.t. Q and VQP is velocity of Q w.r.t P, then, (1), (2), (3), (4), , r, r, | VPQ | = | VQP | = constant, r, VPQ towards origin, r, VQP towards origin, r, r, | VPQ | ¹ | VQP | = constant, , RESPONSE, GRID, , 17., , Q.21 A swimmer who can swim in a river with speed mv (with, respect to still water) where v is the velocity of river, current, jumps into the river from one bank to cross the, river. Then, (1) If m < 1 he cannot cross the river, (2) If m £ 1 he cannot reach a point on other bank directly opposite to his starting point., (3) If m > 1 he can reach a point on other bank, (4) He can reach the other bank at some point, whatever, be the value of m., Q.22 Consider two children riding on the merry-go-round Child, 1 sits near the edge, Child 2 sits closer to the centre., Let vl and v2 denote the linear speed of child 1 and child 2,, respectively. Which of the following is/are wrong ?, (1) We cannot determine v 1 & v 2 without more, information, (2) v1 = v2, (3) v1 < v2, (4) v1 > v2, DIRECTIONS (Q.23-Q.25) : Read the passage given below, and answer the questions that follows :, Three of the fundamental constants of physics are the universal, gravitational constant, G = 6.7 × 10–11m3kg–1s–2, the speed of, light, c = 3.0 × 108 m/s, and Planck’s constant, h = 6.6 × 10–34 Js–1., Two particles A and B are projected in the vertical plane with, same initial velocity u0 from part (0, 0) and (l, –h) towards each, other as shown in figure at t = 0., y, u0, g = 10m/s2, , x, , x, , A, (0,0), , P, , l, , Q, , (l, –h) B, , t, , 18., , h, , u0, , 19., , 22., , Space for Rough Work, , 20., , 21.
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t.me/Magazines4all, DPP/ P 08, , 4, Q.23 The path of particle A with respect to particle B will be –, (a) parabola, (b) straight line parallel to x-axis, (c) straight line parallel to y-axis, (d) None of these, Q.24 Minimum distance between particle A and B during motion, will be –, (a) l, (b) h, (d) l + h, l2 + h 2, Q.25 The time when separation between A and B is minimum is, (c), , (a), , x, u 0 cosq, , (b), , l, , 2h, g, , (d) u cosq, 0, , DIRECTIONS (Qs. 26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., , RESPONSE, GRID, , 23., , 24., , (b), (c), (d), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement-1 is False, Statement-2 is True., Statement-1 is True, Statement-2 is False., , Q.26 Statement-1 : The relative velocity between any two, bodies moving in opposite direction is equal to sum of the, velocities of two bodies., Statement-2 : Sometimes relative velocity between two, bodies is equal to difference in velocities of the two., Q.27 Statement-1: A river is flowing from east to west at a, speed of 5m/min. A man on south bank of river, capable of, swimming 10 m/min in still water, wants to swim across, the river in shortest time. He should swim due north., , 2l, , (c) 2u cos q, 0, , (a), , Statement-2 : For the shortest time the man needs to swim, perpendicular to the bank., Q.28 Statement-1 : Rain is falling vertically downwards with, velocity 6 km/h. A man walks with a velocity of 8 km/h., Relative velocity of rain w.r.t. the man is 10 km/h., Statement-2 : Relative velocity is the ratio of two, velocities., , 25., , 26., , 27., , 28., , DAILY PRACTICE PROBLEM SHEET 8 - PHYSICS, Total Questions, 28, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 112, , 44, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 09, SYLLABUS : LAWS OF MOTION-1 (Newton's laws, momentum, pseudo force concept), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A boy standing on a weighing machine observes his weight, as 200 N. When he suddenly jumpes upwards, his friend, notices that the reading increased to 400 N. The, acceleration by which the boy jumped will be(a) 9.8 m/s2, (b) 29.4 m/s2, 2, (c) 4.9 m/s, (d) 14.7 m/s2, Q.2 A force of (6 î + 8 ĵ ) N acted on a body of mass 10 kg. The, displacement after 10 sec, if it starts from rest, will be (a) 50 m along tan –1 4/3 with x axis, (b) 70 m along tan –1 3/4 with x axis, (c) 10 m along tan–1 4/3 with x axis, (d) None, , RESPONSE GRID, , 1., , 2., , Q.3 A boat of mass 1000 kg is moving with a velocity of, 5 m/s. A person of mass 60 kg jumps into the boat. The, velocity of the boat with the person will be (a) 4.71 m/s, (b) 4.71 cm/s, (c) 47.1 m/s, (d) 47.1 cm/s, Q.4 A disc of mass 10 gm is kept horizontally in air by firing, bullets of mass 5 g each at the rate of 10/s. If the bullets, rebound with same speed. The velocity with which the, bullets are fired is (a) 49 cm/s (b) 98 cm/s (c) 147 cm/s (d) 196 cm/s, Q.5 A fire man has to carry an injured person of mass 40 kg, from the top of a building with the help of the rope which, can withstand a load of 100 kg. The acceleration of the, fireman if his mass is 80 kg, will be(a) 8.17 m/s2, (b) 9.8 m/s2, 2, (c) 1.63 m/s, (d) 17.97 m/s2, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 09, , 2, Q.6 A body of mass 0.02 kg falls from a height of 5 metre into, a pile of sand. The body penetrates the sand a distance of 5, cm before stoping. What force has the sand exerted on the, body ?, (a) 1.96 N, (b) –19.6 N, (c) –0.196 N, (d) 0.0196 N, Q.7 The magnitude of the force (in newton) acting on a body, varies with time t (in microsecond) as shown in fig. AB,, BC, and CD are straight line segments. The magnitude of, the total impulse of the force on the body from t = 4 ms to, t = 16 ms is, Force F (N), , (a) 5 × 10–4 N.s, (b) 5 × 10–3 N.s, (c) 5 × 10–5 N.s, , C, , 800, 600, 400, A, 200, , B, F, , 0 2, , E, 4 6 8 10 12 1416D, Time (m s)____, , (d) 5 × 10–2 N.s, Q.8 The total mass of an elevator with a 80 kg man in it is 1000, kg. This elevator moving upward with a speed of 8 m/sec,, is brought to rest over a distance of 16m. The tension T in, the cables supporting the elevator and the force exerted, on the man by the elevator floor will respectively be(a) 7800 N, 624 N, (b) 624 N, 7800 N, (c) 78 N, 624 N, (d) 624 N, 78 N, Q.9 In the arrangement shown in fig. the ends P and Q of an, unstretchable string move downwards with a uniform speed, U. Pulleys A and B are fixed. Mass M moves upwards with, a speed of, B, , A, q q, , Q, , P, , M, , (a) 2U cos q, (b) U cos q, (c) 2U/cos q, (d) U/cos q, Q.10 An engine of mass 5 × 104 kg pulls a coach of mass 4 ×, 104 kg. Suppose that there is a resistance of 1 N per 100, kg acting on both coach and engine, and that the driving, , RESPONSE, GRID, , force of engine is 4500 N. The acceleration of the engine, and tension in the coupling will respectively be(a) 0.04 m/s2, 2000 N, (b) 0.4 m/s2, 200 N, 2, (c) 0.4 m/s , 20 N, (d) 4 m/s2, 200 N, Q.11 A body whose mass 6 kg is acted upon by two forces, ˆ N and (4iˆ + 8j), ˆ N. The acceleration produced will, (8iˆ + 10j), be (in m/s2) –, ˆ, (a) (3iˆ + 2j), , (b) 12iˆ + 18jˆ, , 1 ˆ ˆ, (i + j), (d) 2iˆ + 3jˆ, 3, Q.12 A car of 1000 kg moving with a velocity of 18 km/hr is, stopped by the brake force of 1000 N. The distance covered, by it before coming to rest is (a) 1 m, (b) 162 m (c) 12.5 m (d) 144 m, Q.13 A block of metal weighing 2 kg is resting on a frictionless, plane. It is struck by a jet releasing water at a rate of 1 kg/, s and at a speed of 5 m/s. The initial acceleration of the, block will be –, (a) 2.5 m/s2 (b) 5 m/s2 (c) 0.4 m/s2 (d) 0, Q.14 A man fires the bullets of mass m each with the velocity v, with the help of machine gun, if he fires n bullets every, sec, the reaction force per second on the man will be -, , (c), , m, mv, vn, n, (b) m n v (c), (d), v, n, m, Q.15 A body of mass 15 kg moving with a velocity of 10 m/s is, to be stopped by a resistive force in 15 sec, the force will, be (a) 10 N, (b) 5 N, (c) 100 N, (d) 50 N, Q.16 A cricket ball of mass 250 gm moving with a velocity of, 24 m/s is hit by a bat so that it acquires a velocity of 28 m/, s in the opposite direction. The force acting on the ball, if, the contact time is 1/100 of a second, will be (a) 1300 N in the final direction of ball, (b) 13 N in the initial direction of ball, (c) 130 N in the final direction of ball, (d) 1.3 N in the initial direction of ball, , (a), , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 09, , 3, , Q.17 A force of 2 N is applied on a particle for 2 sec, the change, in momentum will be (a) 2 Ns, (b) 4 Ns, (c) 6 Ns, (d) 3 Ns, Q.18 A body of mass 2 kg is moving along x-direction with a, velocity of 2 m/sec. If a force of 4 N is applied on it along, y-direction for 1 sec, the final velocity of particle will be (a), , 2 2 m/s, , (b), , 2 m/s, , (c) 1/ 2 m/s, (d) 1/ 2 2 m/s, Q.19 A cricket ball of mass 150 g is moving with a velocity of, 12m/sec and is hit by a bat so that the ball is turned back, with a velocity of 20 m/sec, the force on the ball acts for, 0.01 sec, then the average force exerted by the bat on the, ball will be, (a) 48 N, (b) 40 N, (c) 480 N (d) 400 N, Q.20 A body of mass 20 kg moving with a velocity of 3 m/s,, rebounds on a wall with same velocity. The impulse on the, body is (a) 60 Ns, (b) 120 Ns (c) 30 Ns (d) 180 Ns, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 A mass of 60 kg is on the floor of a lift moving down. The, lift moves at first with an acceleration of 3 m/sec2, then, with constant velocity and finally with a retardation of, 3m/sec2 . Choose the correct options related to possible, reactions exerted by the lift on the body in each part of the, motion –, (1) 408 N, (2) 588 N (3) 768 N (4) 508 N, Q.22 A mass of 10 kg is hung to a spring balance in lift. If the, lift is moving with an acceleration g/3 in upward &, downward directions, choose the correct options related, to the reading of the spring balance., (1) 13.3 kg (2) 6.67 kg (3) 32.6 kg (4) 0, , RESPONSE, GRID, , Q.23 Choose the correct options, (1) A reference frame in which Newton’s first law is valid, is called an inertial reference frame., (2) Frame moving at constant velocity relative to a known, inertial frame is also an inertial frame., (3) Idealy, no inertial frame exists in the universe for, practical purpose, a frame of reference may be, considered as inertial if its acceleration is negligible, with respect to the acceleration of the object to be, observed., (4) To measure the acceleration of a falling apple, earth, cannot be considered as an inertial frame., DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, Pseudo force is an imaginary force which is recognised only by a, non-inertial observer to explain the physical situation according, to newton’s laws. Magnitude of pseudo force Fp is equal to the, product of the mass m of the object and the acceleration a of the, frame of reference. The direction of the force is opposite to the, direction of acceleration, Fp = –ma, Q.24 A spring weighing machine inside a stationary lift reads, 50 kg when a man stand on it. What would happen to the, scale reading if the lift is moving upward with (i) constant, velocity (ii) constant acceleratioin ?, æ, 50a ö, (a) 50 kg wt, 50 +, kg wt, çè, g ÷ø, 50g ö, æ, (b) 50 kg wt, ç 50 +, kg wt, è, a ÷ø, , æ 50a ö, (c) 50 kg wt, ç, kg wt, è g ÷ø, æ 50g ö, (d) 50 kg wt, ç, ÷ kg wt, è a ø, Q.25 A 25 kg lift is supported by a cable. The acceleration of, the lift when the tension in the cable is 175 N, will be (a) 2.8 m/s2, (b) 16.8 m/s2, 2, (c) – 9.8 m/s, (d) 14 m/s2, , 17., , 18., , 19., , 20., , 22., , 23., , 24., , 25., , Space for Rough Work, , 21.
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t.me/Magazines4all, DPP/ P 09, , 4, Q.26 A body is suspended by a string from the ceiling of an, elevator. It is observed that the tension in the string is, doubled when the elevator is accelerated. The acceleration, will be (a) 4.9 m/s2, (b) 9.8 m/s2, 2, (c) 19.6 m/s, (d) 2.45 m/s2, DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (b), (c), (d), , Statement-1 is True, Statement-2 is True; Statement-2 is NOT, a correct explanation for Statement-1., Statement-1 is False, Statement-2 is True., Statement-1 is True, Statement-2 is False., , Q.27 Statement-1 : A cloth covers a table. Some dishes are kept, on it. The cloth can be pulled out without dislodging the, dishes from the table., Statement-2 : To every action there is an equal and opposite reaction., Q.28 Statement-1 : If the net external force on the body is zero, then its acceleration is zero., Statement-2 : Acceleration does not depend on force., Q.29 Statement-1 : The slope of momentum versus time graph, give us the acceleration., Statement-2 : Force is given by the rate of change of, momentum., , 28., , 29., , DAILY PRACTICE PROBLEM SHEET 9 - PHYSICS, Total Questions, 29, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), , Space for Rough Work, , 116, , 44, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 10, SYLLABUS : LAWS OF MOTION-2 (Blocks in contact, connected by string, pulley arrangement), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A block of mass M is pulled along a horizontal frictionless, surface by a rope of mass m. If a force P is applied at the, free end of the rope, the force exerted by the rope on the, block will be Pm, MP, mP, (a) P, (b), (c), (d), M+m, M+m, M+m, Q.2 A body of mass 50 kg is pulled by a rope of length 8 m on, a surface by a force of 108N applied at the other end. The, force that is acting on 50 kg mass, if the linear density of, rope is 0.5 kg/m will be (a) 108 N (b) 100 N (c) 116 N, (d) 92 N, Q.3 A rope of length 15 m and linear density 2 kg/m is lying, length wise on a horizontal smooth table. It is pulled by a, , RESPONSE GRID, , 1., , 2., , force of 25 N. The tension in the rope at the point 7 m, away from the point of application, will be (a) 11.67 N, (b) 13.33 N, (c) 36.67 N, (d) None of these, Q.4 A force of 100 N acts in the direction as shown in figure, on a block of mass 10 kg resting on a smooth horizontal, table. The speed acquired by the block after it has .moved, a distance of 10 m, will be (in m/sec) (g = 10 m/sec2), 100N, (a) 17 m/sec, 30°, (b) 13.17 m/sec, 10kg, (c) 1.3 m/sec, (d) 1.7 m/sec, Q.5 In the above example, the velocity after 2 sec will be (in m/sec), (a) 10 3, , 3., Space for Rough Work, , (b) 5 3, , 4., , (c) 10, , 5., , (d) 5
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , DPP/ P 10, , 2, Q.6 Two blocks of mass m = 1 kg and M = 2 kg are in contact on, a frictionless table. A horizontal force F(= 3N) is applied to, m. The force of contact between the blocks, will be(a) 2 N, (b) 1 N, (c) 4 N, (d) 5 N, Q.7 A force produces an acceleration of 5 m/s2 in a body and, same force an acceleration of 15 m/s2 in another body., The acceleration produced by the same force when applied, to the combination of two bodies will be (a) 3.75 m/s2, (b) 20 m/s2, 2, (c) 10 m/s, (d) 0.667 m/s2, Q.8 What is the tension in a rod of length L and mass M at a, distance y from F1 when the rod is acted on by two unequal, forces F1 and F2 (<F1) as shown in fig, y, , B, , C, F2, , T, , (a), , æ, æyö, yö, F1 çç1 , ÷÷ ∗ F2 çç ÷÷, ÷, èç L ø, èç L ø÷, , (c), , æ, æ yö, yö, F1 çç1 ∗ ÷÷ ∗ F2 çç ÷÷, ÷, çè, èç L ø÷, Lø, , A, F1, , T, , L, , (b), (d), , M æç F1 , F2 ö÷, yç, ÷, L çè M ø÷, , M æç F1 ∗ F2 ö÷, yç, ÷, L çè M ø÷, , 1 1 1, , , , ......... (all in m/, 2 3 4, s2 ), applied separetly to n bodies. If these bodies are, combined to form single one, then the acceleration of the, system will be, if same force is taken into account., , Q.9 A force produces acceleration 1 ,, , (a), , n, 2, , (b), , 2, n(n + 1), , (c), , n2, 2, , (d), , n 2 (n + 1), 2, , Q.10 Two blocks of masses 6 kg and 4 kg connected by a rope of, mass 2 kg are resting on frictionless floor as shown in fig. If a, constant force of 60 N is applied to 6 kg block, tension in the, rope at A, B, and C will respectively be 6kg, 2kg 2kg, F=60N, CBA, P, Q, T, , (a) 30 N, 25 N, 20 N, (c) 20 N, 30 N, 25 N, , RESPONSE, GRID, , Q.11 The pulley arrangements of fig (a) and (b) are identical. The, mass of the rope is negligible. In (a) the mass m is lifted up, by attaching a mass 2 m to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a, constant downward force F = 2 mg. Which of the following, is correct?, , A m, B 2m, , m, , F=2mg, , (a), (b), (a) Acceleration in case (b) is 3 times more than that in, case (a), (b) In case (a) acceleration is g, while in case (b) it is 2g, (c) In both the cases, acceleration is same, (d) None of the above, Q.12 Three equal weights of mass m each are hanging on a string, passing over a fixed pulley as shown in fig. The tensions in, the string connecting weights A to B and B to C will, respectively be 2, 2, (a), mg, mg, 3, 3, 2, 4, mg , mg, (b), T1, T1, 3, 3, A, 4, 2, (c), mg, mg, B T2, 3, 3, 3, 3, C, (d), mg, mg, 4, 2, Q.13 In the situation shown in figure, both the pulleys and the, strings are light and all the surfaces are frictionless. The, acceleration of mass M, tension in the string PQ and force, exerted by the clamp on the pulley, will respectively be (a) (2/3)g, (1/3)Mg, ( 2 /3)Mg, (b) (1/3)g, (1/3)Mg, ( 2 /3)Mg, (c) (1/3)g, (2/3)Mg, 3 Mg, (d) 2g, (1/2)g, 2 Mg, , (b) 25 N, 30 N, 20 N, (d) 30 N, 20 N, 25 N, , 6., , 7., , 8., , 11., , 12., , 13., , Space for Rough Work, , 9., , T’, , Q, , T, , B 2M, T, 2Mg, A M, , 10.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 10, , 3, , Q.14 A body of mass 50kg resting on a smooth inclined plane is, connected by a massless inextensible string passing over a, smooth pulley, at the top of the inclined plane have another, mass of 40 kg as shown in the figure. The distance through, which 50 kg mass fall in 4 sec will be (The angle of the inclined plane is 30º), T, (a) 13.04 m, T, (b) 1.63m, C, B, (c) 1.304 m, mgsinq, mgcosq, mg q, (d) 16.3m, 40g, Q.15 A bob is hanging over a pulley inside a car through a string., The second end of the string is in the hand of a person, standing in the car. The car is moving with constant, acceleration 'a' directed horizontally as shown in figure., Other end of the string is pulled with constant acceleration, ‘a’ vertically. The tension in the string is –, (a), , car, , m g2 + a 2, 2, , a, , 2, , (b) m g + a - ma, , a, , m, , (c) m g 2 + a 2 + ma, (d) m (g + a), Q.16 In the fig shown, the velocity of each, particle at the end of 4 sec will be (a) 0.872 m/s, (b) 8.72 m/s, T, T, a, (c) 0.218 m/s, A, a, (d) 2.18 m/s, B, Q.17 In the above example, the height, ascended or descended, as the case may, 11.5g, be, during that time i.e. 4 sec will be (a) 1.744 m, (b) 17.44 m, (c) 0.1744 m, (d) None of these, Q.18 In the above question, if at the end of 4 sec, the string be, cut, the position of each particle in next 2 seconds will, respectively be (a) 17.856 m, 21.344 m, (b) –21.344 m, 17.856 m, (c) –17.856 m, 21.344 m (d) –17.856 m, –21.344 m, , RESPONSE, GRID, , Q.19 Consider th e double Atwood’s, machine as shown in the figure., What is acceleration of the masses ?, (a) g/3, (b) g/2, (c) g, (d) g/4, a, Q.20 In above question, what is the, tension in each string ?, (a) mg/3, (b) 4mg/3 (c) 2mg/3, , T, , T, m, , a, , mg, T, m, , m, , mg, , T, a, , mg, , (d) 5mg/3, , DIRECTIONS (Q.21-Q.22) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, Q.21 Choose the correct options, (1) Inertia µ mass, r, (3) Thrust on rocket F =, , (b) 1 and 2 are correct, (d) 1 and 3 are correct, –, (2) 1 newton = 105 dyne, r, DM r, v - Mg, Dt, (4) Apparent weight of a body in the accelerated lift is, W = m (g + a)., Q.22 Choose the correct statements –, (1) For equilibrium of a body under the action of, ®, ®, ®, ®, concurrent forces F1 + F2 + F3 + ..... Fn = 0, (2) If the downward acceleration of the lift is a = g, then, the body will experience weightlessness., (3) If the downward acceleration of the body is a > g,, then the body will rise up to the ceiling of lift, (4) If the downward acceleration of the lift is a > g, then, the body will experience weightlessness., , DIRECTIONS (Q.23-Q.25) : Read the passage given below, and answer the questions that follows :, A bead of mass m is attached to one end of, a spring of natural length R and spring, , B, , ( 3 + 1) mg, , constant K =, . The other end of, R, the spring is fixed at point A on a smooth, ring of radius R as shown in figure. When, bead is released to move then, , 14., , 15., , 16., , 17., , 19., , 20., , 21., , 22., , Space for Rough Work, , A, , 18., , 30°
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t.me/Magazines4all, DPP/ P 10, , 4, Q.23 Initial elongation in the spring is –, (a) R, (b) 2R, (c), 2R, Q.24 The normal reaction force at B is –, (a), , mg, 2, , (b), , 3mg, , (c) 3 3mg, , (d), , 3R, , (d), , 3 3mg, 2, , Q.25 Tangential acceleration of bead just after it is released., g, 3, g, 2, g, g, (a), (b), (c), (d), 2, 4, 4, 3, , DIRECTIONS (Qs. 26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.26 Statement–1 In fig the ground is smooth and the masses, of both the blocks are different. Net force acting on each, of block is not same., , RESPONSE, GRID, , 23., , 24., , Statement– 2 Acceleration of the both blocks will be, different., F, , B, , A, , Q.27 Statement– 1 Block A is moving on horizontal surface, towards right under the action of force F. All surfaces are, smooth. At the instant shown the force exerted by block A, on block B is equal to net force on block B., Statement– 2 From Newton’s third law of motion, the, force exerted by block A on B is equal in magnitude to, force exerted by block B on A., B, , A, F, , Q.28 Statement– 1 : In the given fig., æ m , m1 ÷ö, ÷g, a < ççç 2, çè m1 ∗ m 2 ÷÷ø, , Statement– 2 : In the given fig.,, m ∗ m2, g, T< 1, 2m1m 2, , 25., , 26., , a, T, m1, , m2, , 27., , 28., , DAILY PRA CTICE PROBLEM SHEET 10 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , a, T, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 11, SYLLABUS : LAWS OF MOTION-3 (Friction), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A body of mass 400 g slides on a rough horizontal surface., If the frictional force is 3.0 N, the angle made by the contact, force on the body with the vertical will be, (a) 37º, (b) 53º, (c) 63º, (d) 27º, Q.2 In the above question, the magnitude of the contact force, is, (g = 10 m/s2), (a) 3.0 N, (b) 4.0 N (c) 5.0 N (d) 7.0 N, Q.3 The coefficient of static friction between a block of mass, m and an inclined plane is ms = 0.3. What can be the, maximum angle q of the inclined plane with the horizontal, so that the block does not slip on the plane?, , RESPONSE GRID, , 1., , 2., , (a) tan–1 (0.1), (b) tan –1 (0.2), –1, (c) tan (0.3), (d) tan –1 (0.4), Q.4 The coefficient of static friction between the two blocks, shown in figure is m and the table is smooth. What maximum, horizontal force F can be applied to the block of mass M, so that the blocks move together?, m, M, , (a) mg (M + m), (c) 2mg (M + m), , 3., Space for Rough Work, , 4., , F, , (b) mg (M – m), (d) mg (M + 2m)
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t.me/Magazines4all, DPP/ P 11, , 2, Q.5 Block A weighs 4 N and block B weighs 8 N. The coefficient of, kinetic friction is 0.25 for all surfaces. Find the force F to slide B at, a constant speed when A rests on B and moves with it., (a) 2N, (b) 3N, (c) 1N, (d) 5N, Q.6 In the above question, find the force F to slide B at a, constant speed when A is held at rest., (a) 2N, (b) 3N, (c) 1N, (d) 4N, Q.7 In the above question, find the force F to slide B at a, constant speed when A and B are connected by a light cord, passing over a smooth Pulley., (a) 2N, (b) 3N, (c) 1N, (d) 5N, Q.8 Find the maximum value of M/m in the situation shown in, figure so that the system remains at rest. Friction, coefficient at both the contacts is m., , é, , (a) Rêê1 +, ë, é, , (c) R êê1 ë, , ù, ú, (m 2 + 1) úû, , (b) R ê1 -, , ù, ú, (m 2 + 1) úû, , (d) R ê1 -, , é, , 1, , ê, ë, , é, , 2, , ê, ë, , ù, ú, (m2 - 1) úû, 1, , ù, ú, (m + 1) úû, 1, 2, , Q.12 A body of mass m is released from the top of a rough, inclined plane as shown in figure. If the frictional force be, F, then body will reach the bottom with a velocity, , m, L, , h, , m, M, , q, , (a), , m, sin q - m cos q, , (c), , m, sin q + m cos q, , (b), , 2m, sin q - m cos q, , (d), , m, cos q - m sin q, , Q.9 A block placed on a horizontal surface is being pushed by, a force F making an angle q with the vertical, if the, coefficient of friction is m, how much force is needed to, get the block just started?, (a), , m, sin q - m cos q, , 2m, , (b) sin q - m cos q, , m, , m, , (c) sin q + m cos q, (d) cos q - m sin q, Q.10 Assuming the length of a chain to be L and coefficient of, static friction m. Compute the maximum length of the chain, which can be held outside a table without sliding., 2m L, , mL, , mL, , 3mL, , (a) 1 + m, (b) 1 - m, (c) 1 + m, (d) 1 + m, Q.11 If the coefficient of friction between an insect and bowl is, m and the radius of the bowl is r, find the maximum height, to which the insect can crawl in the bowl., , RESPONSE, GRID, , 1, (mgh - FL), m, , (a), , 2, (mgh - FL), m, , (b), , (c), , 2, (mgh + FL), m, , (d) None of these, , Q.13 A block of mass 2 kg is placed on the floor. The coefficient, of static friction is 0.4.A force F of 2.5 N is applied on the, block, as shown. Calculate the force of friction between, the block and the floor. (g = 9.8 ms–2), (a) 2.5 N, (b) 25 N, (c) 7.84 N (d) zero, Q.14 A block is kept on a horizontal table. The table is undergoing, simple harmonic motion of frequency 3 Hz in a horizontal, plane. The coefficient of static friction between the block and, the table surface is 0.72. Find the maximum amplitude of the, table at which the block does not slip on the surface (g = 10 ms–, 2), (a) 0.01 m (b) 0.02 m (c) 0.03 m (d) 0.04 m, Q.15 Two cars of unequal masses use similar tyres. If they are, moving at the same initial speed, the minimum stopping, distance (a) is smaller for the heavier car, (b) is smaller for the lighter car, (c) is same for both cars, (d) depends on the volume of the car, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 11, , 3, , Q.16 Consider the situation shown in figure. The wall is smooth, but the surfaces of A and B in contact are rough in, equilibrium the friction on B due to A –, (a) is upward, F, A B, (b) is downward, (c) is zero, (d) the system cannot remain in equilibrium, Q.17 A block is placed on a rough floor and a horizontal force F, is applied on it. The force of friction f by the floor on the, block is measured for different values of F and a graph is, plotted between them –, (i) The graph is a straight line of slope 45°, (ii) The graph is straight line parallel to the F axis, (iii) The graph is a straight line of slope 45º for small F and, a straight line parallel to the F-axis for large F., (iv) There is small kink on the graph, (a) iii, iv, (b) i, iv, (c) i, ii, (d) i, iii, Q.18 The contact force exerted by a body A on another body B, is equal to the normal force between the bodies. We, conclude that (i) the surfaces must be smooth, (ii) force of friction between two bodies may be equal to, zero, (iii) magnitude of normal reaction is equal to that of, friction, (iv) bodies may be rough, (a) ii, iv, (b) i, ii, (c) iii, iv, (d) i, iv, Q.19 It is easier to pull a body than to push, because (a) the coefficient of friction is more in pushing than that, in pulling, (b) the friction force is more in pushing than that in, pulling, (c) the body does not move forward when pushed, (d) None of these, Q.20 A block of metal is lying on the floor of a bus. The, maximum acceleration which can be given to the bus so, that the block may remain at rest, will be (a) µg, (b) µ/g, (c) µ2g, (d) µg2, , RESPONSE, GRID, , DIRECTIONS (Q.21-Q.23) : In the following questions, more, than one of the answers given are correct. Select the correct, answers and mark it according to the following codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Choose the correct statements –, (1) Kinetic friction is lesser than limiting friction., (2) In rolling the surfaces at contact do not rub each other., (3) If a body is at rest and no pulling force is acting on it,, force of friction on it is zero., (4) Kinetic friction is greater than limiting friction., Q.22 Choose the correct statements –, (1) Force of friction is partically independent of, microscopic area of surface in contact and relative, velocity between them. (if it is not high), (2) Normally with increase in smoothness friction, decreases. But if the surface area are made too smooth, by polishing and cleaning the bonding force of, adhesion will increase and so the friction will increase, resulting in 'Cold welding', (3) Friction is a non conservative force, i.e. work done, against friction is path dependent., (4) Force of fricton depends on area, Q.23 Choose the correct options –, (1) Friction always opposes the motion, (2) Friction may opposes the motion, (3) If the applied force is increased the force of static, friction remains constant., (4) If the applied force is increased the force of static, friction also increases upto limiting friction., DIRECTIONS (Q.24-Q.26) : Read the passage given below and, answer the questions that follows :, A block of mass 1 kg is placed on a rough horizontal surface. A, spring is attached to the block whose other end is joined to a rigid, wall, as shown in the figure. A horizontal force is applied on the, block so that it remains at rest while the spring is elongated by x., , 16., , 17., , 18., , 21., , 22., , 23., Space for Rough Work, , 19., , 20.
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t.me/Magazines4all, DPP/ P 11, , 4, mmg, . Let Fmax and Fmin be the maximum and minimum values, k, of force F for which the block remains a equilibrium. For a particular, x,, Fmax – Fmin = 2N., Also shown is the variation of Fmax + Fmin versus x, the elongation of the spring., , ///////////////, , x³, , K, , Fmax+Fmin, 5N, , 1 kg, , //////////////////////////////////////////, , 0.1M, , x, , Q.24 The coefficient of friction between the block and the horizontal surface is –, (a) 0.1, (b) 0.2, (c) 0.3, (d) 0.4, Q.25 The spring constant of the spring is –, (a) 25 N/m, (b) 20 N/m, (c) 2.5 N/m, (d) 50 N/m, Q.26 The value of Fmin, if x = 3 cm. is –, (a) 0, (b) 0.2 N, (c) 5N, (d) 1N, , RESPONSE GRID, , 24., , 25., , DIRECTIONS (Q. 27-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices, only, one of which is the correct answer. You have to select the correct, choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 A solid sphere and a hollow sphere of same mass M and, same radius R are released from the top of a rough, inclined plane. Friction coefficient is same for both the, bodies. If both bodies perform imperfect rolling, then, Statement - 1 : Work done by friction for the motion of, bodies from top of incline to the bottom will be same, for both the bodies., Statement - 2 : Force of friction will be same for both, the bodies., Q.28 Statement - 1 : Maximum value of friction force between, two surfaces is m × normal reaction., where m = coefficient of friction between surfaces., Statement - 2 : Friction force between surfaces of two, bodies is always less than or equal to externally applied, force., , 26., , 27., , 28., , DAILY PRA CTICE PROBLEM SHEET 11 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 12, SYLLABUS : Work, Energy and Power-1 (Work by constant and variable forces,, kinetic and potential energy, work energy theorem), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., r, Q.1 A body is acted upon by a force F = -ˆi + 2jˆ + 3kˆ . The work, done by the force in displacing it from (0,0,0) to (0,0,4m) will, be (a) 12 J, (b) 10 J, (c) 8 J, (d) 6 J, Q.2 The work done in pulling a body of mass 5 kg along an inclined, plane (angle 60º) with coefficient of friction 0.2 through 2, m, will be (a) 98.08 J, (b) 94.08 J, (c) 90.08 J, (d) 91.08 J, , RESPONSE GRID, , 1., , 2., , r, Q.3 A force F = (7 – 2x + 3x2) N is applied on a 2 kg mass, which displaces it from x = 0 to x = 5 m. Work done in, joule is (a) 70, (b) 270, (c) 35, (d) 135, Q.4 An automobile of mass m accelerates from rest. If the engine, supplies a constant power P, the velocity at time t is given by (a) v =, , Pt, m, , (b) v =, , 2Pt, (c), m, , Pt, m, , (d), , 2Pt, m, , Q.5 In the above question, the position (s) at time (t) is given by æ, ö, (a) çè m ÷ø t, 2Pt, , æ 9P ö, , (c) ç ÷, è 8m ø, , 3., Space for Rough Work, , 1/2, , æ 8P ö, , 1/2, , æ 8P ö, , 1/ 2, , (b) çè ÷ø, 9m, t1/2, , 4., , (d) ç ÷, è 9m ø, , 5., , t 3/2, t
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t.me/Magazines4all, DPP/ P 12, , 2, Q.6 A particle moving in a straight line is acted by a force, which, works at a constant rate and changes its velocity from u to, v in passing over a distance x. The time taken will be (a) x =, , æ v+u ö, , v-u, 2, , v +u, , (b) x çè 2 2 ø÷, v +u, , 2, , æ v2 - u 2 ö, 3, æ vö, (x) ç 3 3 ÷, (d) x çè u ÷ø, è v -u ø, 2, Q.7 A chain of linear density 3 kg /m and length 8 m is lying on, the table with 4 m of chain hanging from the edge. The, work done in lifting the chain on the table will be (a) 117.6 J (b) 235.2 J (c) 98 J, (d) 196 J, Q.8 The work done in lifting water from a well of depth 6 m, using a bucket of mass 0.5 kg and volume 2 litre, will be(a) 73.5 J, (b) 147 J (c) 117. 6 J (d) 98 J, Q.9 An object of mass 5 kg falls from rest through a vertical, distance of 20 m and reaches a velocity of 10 m/s. How, much work is done by the push of the air on the object ?, (g = 10 m/s2)., (a) 350 J, (b) 750 J (c) 200 J (d) 300 J, Q.10 A boy pulls a 5 kg block 20 metres along a horizontal surface at a constant speed with a force directed 45° above, the horizontal. If the coefficient of kinetic friction is 0.20,, how much work does the boy do on the block?, (a) 163.32 J, (b) 11.55 J, (c) 150 J, (d) 115 J, Q.11 A uniform chain is held on a frictionless table with onefifth of its length hanging over the edge. If the chain has a, length l and a mass m, how much work is required to pull, the hanging part back on the table ?, (a) mg l / 10, (b) mg l / 5, (c) mg l / 50, (d) mg l / 2, Q.12 A bus of mass 1000 kg has an engine which produces a, constant power of 50 kW. If the resistance to motion,, assumed constant is 1000 N. The maximum speed at which, the bus can travel on level road and the acceleration when, it is travelling at 25 m/s, will respectively be (a) 50 m/s, 1.0 m/s2, (b) 1.0 m/s, 50 m/s2, 2, (c) 5.0 m/s, 10 m/s, (d) 10 m/s, 5m/s2, , (c), , RESPONSE, GRID, , U235, , Q.13 The power output of a 92, reactor if it takes 30 days to, use up 2 kg of fuel and if each fission gives 185 MeV of, energy (Avogadro number = 6 × 1023/mole) will be (a) 58.4 MW, (b) 5.84 MW, (c) 584 W, (d) 5840 MW, Q.14 The stopping distance for a vehicle of mass M moving with, a speed v along a level road, will be (µ is the coefficient of friction between tyres and the road), (a), , v2, mg, , (b), , 2v2, mg, , (c), , v2, 2mg, , (d), , v, mg, , Q.15 The earth circles the sun once a year. How much work, would have to be done on the earth to bring it to rest, relative to the sun, (ignore the rotation of earth about its own axis) Given that mass of the earth is 6 × 1024 kg, and distance between the sun and earth is 1.5 × 108 km(a) 2.7 × 1033, (b) 2.7 × 1024, 23, (c) 1.9 × 10, (d) 1.9 × 1024, Q.16 A particle of mass m is moving in a horizontal circle of, radius r, under a centripetal force equal to (–k/r 2), where k, is a constant. The total energy of the particle is (a) k/2r, (b) – k/2r, (c) kr, (d) –k/r, Q.17 The work done by a person in carrying a box of mass 10 kg, through a vertical height of 10 m is 4900 J. The mass of, the person is (a) 60 kg, (b) 50 kg, (c) 40 kg, (d) 130 kg, Q.18 A uniform rod of length 4 m and mass 20 kg is lying, horizontal on the ground. The work done in keeping it, vertical with one of its ends touching the ground, will be (a) 784 J, (b) 392 J (c) 196 J (d) 98 J, Q.19 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 surface of the earth to a height equal to radius, R of the earth is - [M = mass of earth], (a), , GMm, 2R, , (b), , GM, R, , (c), , GMm, R, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , Space for Rough Work, , (d), , GM, 2R, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 12, , 3, , Q.20 The potential energy between two atoms in a molecule is given, , a, , b, , , where a and b are positive constant and x, x12 x 6, is the distance between the atoms. The atoms is an stable, equilibrium, whenby, U(x) =, , (a), , x=0, , (c), , æ 2a ö, x= ç ÷, è bø, , æ a ö, (b) x = ç ÷, è 2b ø, 1/6, , æ 11a ö, (d) x = ç, è 5b ÷ø, , 20., , In the figure shown, the system is released from rest with both, the springs in unstretched positions. Mass of each block is 1 kg, and force constant of each spring is 10 N/m., , 1/6, , 1/6, , DIRECTIONS (Q.21-Q.22) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 A man pushes a wall and fails to displace it. Choose, incorrect statements related to his work, (1) Negative work, (2) Positive but not maximum work, (3) Maximum work, (4) No work at all, Q.22 Choose the correct options –, (1) The work done by forces may be equal to change in, kinetic energy, (2) The work done by forces may be equal to change in, potential energy, (3) The work done by forces may be equal to change in, total energy, (4) The work done by forces must be equal to change in, potential energy., , RESPONSE, GRID, , DIRECTIONS (Q.23-Q.25) : Read the passage given below, and answer the questions that follows :, , 21., , Q.23 Extension of horizontal spring in equilibrium is:, (a) 0.2 m, (b) 0.4 m, (c) 0.6 m, (d) 0.8 m, Q.24 Extension of vertical spring in equilibrium is, (a) 0.4 m, (b) 0.2 m, (c) 0.6 m, (d) 0.8 m, Q.25 Maximum speed of the block placed horizontally is:, (a) 3.21 m/s, (b) 2.21 m/s, (c) 1.93 m/s, (d) 1.26 m/s, DIRECTIONS (Qs. 26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), (c), (d), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , 22., , 25., , Space for Rough Work, , 23., , 24.
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t.me/Magazines4all, DPP/ P 12, , 4, Q.26 As shown in the figure, a uniform sphere is rolling on a, horizontal surface without slipping, under the action of a, horizontal force F., F, , Statement - 1 : Power developed due to friction force is, zero., Statement - 2 : Power developed by gravity force is nonzero., , RESPONSE GRID, , 26., , 27., , Q.27 Statement - 1 : Sum of work done by the Newton’s 3rd law, pair internal forces, acting between two particles may be, zero., Statement - 2 : If two particles undergo same displacement, then work done by Newton’s 3rd law pair forces on them is, of opposite sign and equal magnitude., Q.28 Statement - 1: A particle moves along a straight line with, constant velocity. Now a constant non-zero force is applied, on the particle in direction opposite to its initial velocity., After the force is applied, the net work done by this force, may be zero in certain time intervals., Statement - 2 : The work done by a force acting on a particle, is zero in any time interval if the force is always, perpendicular to velocity of the particle., , 28., , DAILY PRA CTICE PROBLEM SHEET 12 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 13, SYLLABUS : Work, Energy and Power-2 (Conservation of momentum and energy, collision, rocket case), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., , travelled a distance of 15 m. The coefficient of friction, between the block and the floor will be - (Duration of, impact is very short), , Q.1 A rifle man, who together with his rifle has a mass of 100, kg, stands on a smooth surface fires 10 shots horizontally., Each bullet has a mass 10 gm and a muzzle velocity of 800, m/s. What velocity does rifle man acquire at the end of 10, shots, (a) 0.8 m/s (b) 0.5 m/s (c) 0.3 m/s (d) 1.2 m/s, Q.2 A bullet of mass 10 g travelling horizontally with a velocity of, 300 m/s strikes a block of wood of mass 290 g which rests, on a rough horizontal floor. After impact the block and the, bullet move together and come to rest when the block has, , 2, 1, 1, 3, (c), (d), (b), 3, 3, 2, 4, Q.3 A 20 g bullet pierces through a plate of mass m1 = 1 kg and, then comes to rest inside a second plate of mass m 2 = 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 (a) 20%, (b) 25%, (c) 30%, (d) 45%, , RESPONSE GRID, , 1., , 2., , (a), , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 13, , 2, Q.4 A bullet of mass 20 g hits a block of mass 1.98 kg suspended, from a massless string of length 100 cm and sticks to it., The bullet flies down at an angle of 30º to the horizontal, with a velocity of 200 m/s. Through what height the block, will rise(a) 0.15 m, (b) 0.30 m, , v, (M+m), , (c) 0.45 m, , 30°, , h, , M, (d) 0.75 m, Q.5 A bullet of mass 0.01 kg travelling at a speed of 500 m/s, strikes a block of mass 2 kg, which is suspended by a string, of length 5 m. The centre of gravity of the block is found to, rise a vertical distance of 0.1 m. The speed of the bullet, after it emerges from the block will be -, , (a) 1.4 m/s, (b) 110 m/s, , 1, mv2, (b) mv2, 2, 3, mv2, (d) 2 mv2, (c), 2, Q.12 A body of mass M splits into two parts aM and (1 – a) M, by an internal explosion, which generates kinetic energy T., After explosion if the two parts move in the same direction, as before, their relative speed will be -, , (a), , (c) 220 m/s, (d) 14 m/s, , 0.1m, , u1, m1 m, 2, , v1, v2, , Q.6 The rate of burning of fuel in a rocket is 50 gm/sec. and, comes out with and velocity 4 × 103 m/s. The force exerted, by gas on rocket will be (a) 200 N, (b) 250 N, (c) 2.5 × 106 N, (d) 2.5 × 104 N, Q.7 A body of mass 1 kg strikes elastically with another body at, rest and continues to move in the same direction with one, fourth of its initial velocity. The mass of the other body is (a) 0.6 kg, (b) 2.4 kg (c) 3 kg, (d) 4 kg, Q.8 A ball moving with a speed of 9 m/s strikes with an identical, stationary ball such that after the collision the direction of, each ball makes an angle of 30° with the original line of, motion. Find the speeds of the two balls after the collision., Is the kinetic energy conserved in this collision process ?, (a) 3 3 m/s, no, (c) 6 3 m/s, yes, , RESPONSE, GRID, , Q.9 The mass of a rocket is 500 kg and the relative velocity of, the gases ejecting from it is 250 m/s with respect to the, rocket. The rate of burning of the fuel in order to give the, rocket an initial acceleration 20 m/s2 in the vertically, upward direction (g = 10 m/s2), will be (a) 30 kg/s, (b) 60 kg/s, (c) 45 kg/s, (d) 10 kg/s, Q.10 A slow moving electron collides elastically with a hydrogen, atom at rest. The initial and final motions are along the, same straight line. What fraction of electron's kinetic, energy is transferred to the hydrogen atom? The mass of, hydrogen atom is 1850 times the mass of electron., (a) 0.217 % (b) 2.17 % (c) 0.0217 % (d) 21.7 %, Q.11 A particle of mass 4 m which is at rest explodes into three, fragments, two of the fragments each of mass m are found, to move each with a speed v making an angle 90º with each, other. The total energy relased in this explosion is -, , (a), , T, (1 - a )M, , (b), , 2T, a(1 - a)M, , 2T, T, (d), (1 - a )M, 2(1 - a )M, Q.13 A body of mass 1 kg initially at rest explodes and breaks, into three fragments of masses in the ratio 1 : 1 : 3. The, two pieces of equal mass fly off perpendicular to each other, with a speed of 30 m/sec each. What is the velocity of the, heavier fragment ?, (c), , (b) 3 3 m/s, no, (d) 0, yes, , (a) 10 2 m/s, , (b) 15 2 m/s, , (c) 5 2 m/s, , (d) 20 2 m/s, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 13, , 3, , Q.14 A body of mass m moving with a velocity v1 in the Xdirection collides with another body of mass M moving in, Y-direction with a velocity v2. They coalasce into one body, during collision. The magnitude and direction of the, momentum of the final body, will beæ Mv 2 ö, , (a), , (mv1 ) + (Mv2 ) , tan–1 çè mv ÷ø, 1, , (b), , (mv1 ) + (Mv2 ) , tan–1 çè mv ÷ø, , (c), , æ Mv 2 ö, (mv1 ) 2 + (Mv 2 ) 2 , tan–1 çè mv ÷ø, 1, , (d), , (mv1 ) 2 + (Mv 2 ) 2 , tan–1 çè mv ÷ø, , æ Mv1 ö, 2, , æ Mv1 ö, 2, , Q.15 A ball of mass m hits a wall with a speed v making an angle, q with the normal. If the coefficient of restitution is e, the, direction and magnitude of the velocity of ball after, reflection from the wall will respectively be æ tan q ö, , (a) tan–1 çè e ÷ø , v sin 2 q + e 2 cos 2 q, æ e ö 1, 2, 2, 2, (b) tan–1 çè tan q ÷ø ,, v e sin q + cos q, v, (c) tan–1(e tan q), tan q, e, (d) tan–1 (e tan q), v sin 2 q + e2, Q.16 A tennis ball dropped from a height of 2 m rebounds only 1.5, metre after hitting the ground. What fraction of energy is lost, in the impact?, (a) 1/2, (b) 1/4, (c) 1/8, (d) 1/16, Q.17 A bullet is fired from the gun. The gun recoils, the kinetic, energy of the recoil shall be(a) equal to the kinetic energy of the bullet, (b) less than the kinetic energy of the bullet, (c) greater than the kinetic energy of the bullet, (d) double that of the kinetic energy of the bullet, Q.18 Conservation of linear momentum is equivalent to(a) Newton's second law of motion, (b) Newton's first law of motion, (c) Newton's third law of motion, (d) Conservation of angular momentum., , RESPONSE, GRID, , Q.19 In an inelastic collision(a) momentum is conserved but kinetic energy is not, conserved, (b) momentum is not conserved but kinetic energy is, conserved, (c) neither momentum nor kinetic energy is conserved, (d) both the momentum and kinetic energy are conserved, Q.20 Inelastic collision is the(a) collision of ideal gas molecules with the walls of the, container, (b) collision of electron and positron to an inhilate each, other., (c) collision of two rigid solid spheres lying on a, frictionless table, (d) scattering of a-particles with the nucleus of gold atom, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Which of the following statements is false for collisions(1) Momentum is conserved in elastic collisions but not, in inelastic collisions., (2) Total-kinetic energy is conserved in elastic collisions, but momentum is not conserved., (3) Total kinetic energy and momentum both are conserved, in all types of collisions, (4) Total kinetic energy is not conserved in inelastic, collisions but momentum is conserved, Q.22 Which of the following hold when two particles of masses, m1 and m2 undergo elastic collision?, (1) When m1 = m2 and m2 is stationary, there is maximum, transfer of kinetic energy in head on collision, (2) When m1 = m2 and m2 is stationary, there is maximum, transfer of momentum in head on collision, (3) When m1 >> m2 and m2 is stationary, after head on, collision m2 moves with twice the velocity of m1., (4) When the collision is oblique and m1 = m2 with m2, stationary, after the collision the particle move in, opposite directions., , 14., , 15., , 16., , 17., , 19., , 20., , 21., , 22., , Space for Rough Work, , 18.
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t.me/Magazines4all, DPP/ P 13, , 4, Q.23 Two balls at the same temperature collide inelastically., Which of the following is not conserved?, (1) Kinetic energy, (2) Velocity, (3) Temperature, (4) Momentum, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, A small particle of mass m/10 is moving horizontally at a height, of 3R/2 from ground with velocity 10 m/s. A perfectly inelastic, collision occurs at point P of sphere of mass m placed on smooth, horizontal surface. The radius of sphere is R. (m = 10 kg and R =, 0.1 m) (Assume all surfaces to be smooth)., m/10, , P, , 10 m/s, , DIRECTIONS (Q. 27-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 A particle of mass m strikes a wedge of mass M horizontally, as shown in the figure., m, , R/2, m, , M, , R, B, , Q.24 Speed of particle just after collision is, (a) approx 5.0 m/s, (b) approx 10 m/s, (c) approx. 15.0 m/s, (d) approx 20.0 m/s, Q.25 Speed of sphere just after collision is, (a) 27/43 m/s, (b) 30/43 m/s, (c) 35/43 m/s, (d) 40/43 m/s, Q.26 Angular speed of sphere just after collision is, (a) zero, (b) 2 rad/sec, (c) 2.5 rad/sec, (d) 3 rad/sec, , RESPONSE, GRID, , 23., , 24., , Statement - 1 : If collision is perfectly inelastic then, it, can be concluded that the particle sticks to the wedge., Statement - 2 : In perfectly inelastic collision velocity of, both bodies is same along common normal just after, collision., Q.28 Statement - 1 : In an elastic collision in one dimension, between two bodies, total momentum remains the same, before, during and after the collision., Statement - 2 : In an elastic collision in one dimension, between two bodies, total kinetic energy remains the same, before, during and after the collision., [Assume external forces are absent in both the above, statements]., , 25., , 26., , 27., , 28., , DAILY PRA CTICE PROBLEM SHEET 13 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , SYLLABUS : Centre of mass and its motion, , Max. Marks : 112, , 14, Time : 60 min., , GENERAL INSTRUCTIONS, •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 In the HCl molecule, the separation between the nuclei of, the two atom is about 1.27 A° (1A° = 10 –10 m). The, approximate location of the centre of mass from the, hydrogen atom, assuming the chlorine atom to be about, 35.5 times massive as hydrogen is, (a) 1 Å, (b) 2.5 Å (c) 1.24 Å (d) 1.5 Å, Q.2 A 2 kg body and a 3 kg body are moving along the x-axis. At, a particular instant the 2 kg body has a velocity of 3 ms–1 and the, 3 kg body has the velocity of 2ms–1. The velocity of the, centre of mass at that instant is, (a) 5 ms–1, (b) 1 ms–1, (c) 0, (d) None of these, , RESPONSE GRID, , 1., , 2., , Q.3 The distance between the carbon atom and the oxygen atom, in a carbon monoxide molecule is 1.1 Å. Given, mass of, carbon atom is 12 a.m.u. and mass of oxygen atom is 16, a.m.u., calculate the position of the centre of mass of the, carbon monoxide molecule, (a) 6.3 Å from the carbon atom, (b) 1 Å from the oxygen atom, (c) 0.63 Å from the carbon atom, (d) 0.12 Å from the oxygen atom, Q.4 The velocities of three particles of masses 20g, 30g and, 50g are 10iˆ,10 ˆj and10kˆ respectively. The velocity of the, centre of mass of the three particles is, (a), , 2iˆ + 3 ˆj + 5kˆ, , (c) 20iˆ + 30 ˆj +5 kˆ, , 3., Space for Rough Work, , 4., , (b) 10(iˆ + ˆj + kˆ), (d) 2iˆ + 30 ˆj +50 kˆ
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t.me/Magazines4all, DPP/ P 14, , 2, Q.5 The centre of mass of a triangle shown in figure has, coordinates, h, b, 2, 2, b, h, x= , y=, 2, 2, , (a) x = , y =, (b), , b, 3, h, x= , y=, 3, , (c) x = , y =, (d), , y, , (a) zero, , h, , h, 3, b, 3, , (c), , x, , b, , Q.6 Two bodies of masses 2 kg and 4 kg are moving with, velocities 2 m/s and 10 m/s respectively along same, direction. Then the velocity of their centre of mass will be, (a) 8.1 m/s, (b) 7.3 m/s, (c) 6.4 m/s, (d) 5.3 m/s, Q.7 Four particles of masses m, 2m, 3m and 4m are arranged at, the corners of a parallelogram with each side equal to a, and one of the angle between two adjacent sides is 60°., The parallelogram lies in the x-y plane with mass m at the, origin and 4m on the x-axis. The centre of mass of the, arrangement will be located at, (a), , æ 3, ö, ç 2 a, 0.95a ÷, è, ø, , (c), , æ 3a a ö, çè , ÷ø, 4 2, , (b), , æ, 3 ö, ç 0.95a , 4 a ÷, è, ø, , (d), , æ a 3a ö, çè , ÷ø, 2 4, , Q.8 Three identical metal balls each of radius r are placed touching, each other on a horizontal surface such that an equilateral, triangle is formed, when centres of three balls are joined., The centre of the mass of system is located at, (a) Horizontal surface, (b) Centre of one of the balls, (c) Line joining centres of any two balls, (d) Point of intersection of the medians, Q.9 2 bodies of different masses of 2 kg and 4 kg are moving, with velocities 20 m/s and 10 m/s towards each other due, to mutual gravitational attraction. What is the velocity of, their centre of mass?, (a) 5 m/s, (b) 6 m/s, (c) 8 m/s, (d) Zero, , RESPONSE, GRID, , Q.10 Two particles of masses m1 and m2 initially at rest start, moving towards each other under their mutual force of, attraction. The speed of the centre of mass at any time t,, when they are at a distance r apart, is, , æ m1m2 1 ö, çè G 2 . m ÷ø t, r, 2, , (b), , æ m1m2 1 ö, çè G 2 . m ÷ø t, r, 1, , (d), , æ m1m2, 1 ö, çè G 2 . m + m ÷ø t, r, 1, 2, , Q.11 A 'T' shaped object, dimensions shown in the figure, is, r, , lying on a smooth floor. A force ' F ' 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, (a), , 4, l, 3, , l, , A, , B, , (b) I, (c), , 2, l, 3, , (d), , 3, l, 2, , P, F, , 2l, C, , Q.12 Two spheres of masses 2M and M are initially at rest at a, distance R apart. Due to mutual force of attraction, they, approach each other. When they are at separation R/2, the, acceleration of the centre of mass of spheres would be, (a) 0 m/s2, (b) g m/s2, (c) 3 g m/s2, (d) 12 g m/s2, Q.13 Masses 8 kg, 2 kg, 4 kg and 2 kg are placed at the corners, A, B, C, D respectively of a square ABCD of diagonal 80, cm. The distance of centre of mass from A will be, (a) 20 cm, (b) 30 cm, (c) 40 cm, (d) 60 cm, Q.14 If linear density of a rod of length 3m varies as l = 2 + x,, them the position of the centre of gravity of the rod is, (a), (c), , 7, m, 3, 10, m, 7, , (b), (d), , 12, m, 7, 9, m, 7, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 14, , 3, , Q.15 Four bodies of equal mass start moving with same speed, as shown in the figure. In which of the following, combination the centre of mass will remain at origin?, , Q.19 A cricket bat is cut at the location of its centre of mass as, shown in the fig. Then, , Y, , (a) c and d, , c, , d, , (b) a and b, , X, , X', , (c) a and c, a, , (d) b and d, , b, Y', , Q.16 Three identical spheres, each of mass 1 kg are kept as, shown in figure, touching each other, with their centres on, a straight line. If their centres are marked P, Q, R, respectively, the distance of centre of mass of the system, from P is, PQ + PR + QR, 3, PQ + PR, (b), 3, PQ + QR, (c), 3, PR + QR, (d), 3, , (a), , y, , P, , Q, , R, , (a), (b), (c), (d), , The two pieces will have the same mass, The bottom piece will have larger mass, The handle piece will have larger mass, Mass of handle piece is double the mass of bottom, piece, Q.20 Consider a system of two particles having mass m1 and m2., If the particle of mass m1 is pushed towards the centre of, mass of particles through a distance d, by what distance, would be particle of mass m2 move so as to keep the centre, of mass of particles at the original position?, m1, m, m2, d (b) 1 d, d, (a), (c) d, (d), m1 + m2, m2, m1, , x, , DIRECTIONS (Q.21-Q.22) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, , Q.17 A ladder is leaned against a smooth wall and it is allowed, to slip on a frictionless floor. Which figure represents, trace of motion of its centre of mass, , Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Choose the wrong statements about the centre of mass, (CM) of a system of two particles, (1) The CM lies on the line joining the two particles, midway between them, (2) The CM lies on the line joining them at a point whose, distance from each particle is proportional to the, square of the mass of that particle, (3) The CM is on the line joining them at a point whose, distance from each particle is proportional to the mass, of that particle, (4) The CM lies on the line joining them at a point whose, distance from each particle is inversely proportional, to the mass of that particle, , (a), , (b), Time, , Time, , (c), , (d), Time, , Time, , Q.18 The two particles X and Y, initially at rest, start moving, towards each other under mutual attraction. If at any instant, the velocity of X is V and that of Y is 2V, the velocity of, their centre of mass will be, (a) 0, (b) V, (c) 2V, (d) V/2, , RESPONSE, GRID, , 15., , 16., , 20., , 21., , 17., , Space for Rough Work, , 18., , 19.
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t.me/Magazines4all, DPP/ P 14, , 4, Q.22 Choose the wrong statements about the centre of mass of, a body, (1) It lies always outside the body, (2) It lies always inside the body, (3) It lies always on the surface of the body, (4) It may lie within, outside or on the surface of the body, , Q.25 Maximum extension in the spring after system loses, contact with wall, , DIRECTIONS (Q.23-Q.25) : Read the passage given below, and answer the questions that follows :, , DIRECTIONS (Q. 26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.26 Statement-1 : The centre of mass of a system of n particles, is the weighted average of the position vector of the n, particles making up the system., Statement-2 : The position of the centre of mass of a, system is independent of coordinate system., Q.27 Statement-1 : The centre of mass of a proton and an electron,, released from their respective positions remains at rest., Statement-2 : The centre of mass remains at rest, if no, external force is applied., Q.28 Statement-1 : Position of centre of mass is independent, of the reference frame., Statement-2 : Centre of mass is same for all bodies., , A system consists of block A and B each of mass m connected, by a light spring as shown in the figure with block B in contact, with a wall. The block A compresses the spring by 3mg/k from, natural length of spring and then released from rest. Neglect, friction anywhere., 3mg/k, k, A, , B, , Q.23 Acceleration of centre of mass of system comprising A, and B just after A is released is, (a) 0, (b) 3g/2, (c) 3g, (d) None of these, Q.24 Velocity of centre of mass of system comprising A and B, when block B just loses contact with the wall, 3g, 2, , m, k, , (a) 3 g m, , (b), , (c), , (d) None of these, , k, m, 2g, k, , RESPONSE, GRID, , 22., , 23., , 27., , 28., , (a), (c), , 24., , 3mg, , (b), , 2k, 3mg, 2k, , (d), , 25., , 3mg, 2k, , None of these, , 26., , DAILY PRA CTICE PROBLEM SHEET 14 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 15, , SYLLABUS : Rotational Motion – 1: Basic concepts of rotational motion, moment of a force, torque,, angular momentum and its conservation with application, , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A thin circular ring of mass M and radius r is rotating, about its axis with a constant angular velocity w. Four, objects each of mass m, are kept gently to the opposite, ends of two perpendicular diameters of the ring. The, angular velocity of the ring will be, (a), , Mw, M + 4m, , (b), , ( M + 4m) w, M, , (c), , ( M - 4m) w, M + 4m, , (d), , Mw, 4m, , Q..2 The angular momentum of a system of particles is, conserved, (a) When no external force acts upon the system, (b) When no external torque acts upon the system, , RESPONSE GRID, , 1., , 2., , (c) When no external impulse acts upon the system, (d) When axis of rotation remains same, Q.3 Two rigid bodies A and B rotate with rotational kinetic, energies EA and EB respectively. The moments of inertia, of A and B about the axis of rotation are IA and IB respectively., If IA = IB/4 and EA = 100 EB, the ratio of angular momentum, (LA) of A to the angular momentum (LB) of B is, (a) 25, (b) 5/4, (c) 5, (d) 1/4, Q.4 A uniform heavy disc is rotating at constant angular velocity, w about a vertical axis through its centre and perpendicular, to the plane of the disc. Let L be its angular momentum. A, lump of plasticine is dropped vertically on the disc and, sticks to it. Which of the following will be constant?, (a) w, (b) w and L both, (c) L only, (d) Neither w nor L, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 15, , 2, Q.5 Two discs of moment of inertia I1 and I2 and angular speeds, w1 and w2 are rotating along collinear axes passing through, their centre of mass and perpendicular to their plane. If, the two are made to rotate combindly along the same axis, the rotational KE of system will be, (a), (c), , I1w1 + I 2 w 2, 2 ( I1 + I 2 ), , (b), , ( I1w1 + I2 w 2 ), , ( I1 + I2 )( w1 + w 2 )2, 2, , 2, , I 2w, I1 + I 2, , (b) w, , (c), , I1w, I1 + I 2, , (d), , pMR 3, T, , (b), , MR 2 p, T, , (c), , 2pMR 2, (d), 5T, , ( I1 + I2 ) w, I1, , 4pMR 2, 5T, , Q.10 If the earth is a point mass of 6×1024kg revolving around, the sun at a distance of 1.5×108 km and in time T = 3.14, ×107s. then the angular momentum of the earth around, the sun is, (a) 1.2 × 1018 kg m2/s, (b) 1.8 × 1029 kg m2/s, (c) 1.5 × 1037 kg m2/s, (d) 2.7 × 1040 kg m2/s, , RESPONSE, GRID, , (b) 440 N-m, , (c) 531 N-m, , (d) 628 N-m, , (a), , Q.8 Calculate the angular momentum of a body whose rotational, energy is 10 joule. If the angular momentum vector, coincides with the axis of rotation and its moment of inertia, about this axis is 8×10-7 kg m2, (a) 4 × 10–3 kg m2 /s, (b) 2 × 10–3 kg m2/s, –3, 2, (c) 6 × 10 kg m /s, (d) None of these, Q.9 If the earth is treated as a sphere of radius R and mass M. Its, angular momentum about the axis of rotation with period T, is, (a), , (a) 350 N-m, , Q.12 A constant torque acting on a uniform circular wheel, changes its angular momentum from A0 to 4A0 in 4 seconds., The magnitude of this torque is, , (d) None of these, , 2 ( I1 + I 2 ), , Q.6 A particle performs uniform circular motion with an, angular momentum L. If the frequency of a particle's, motion is doubled and its kinetic energy is halved, the, angular momentum becomes., (a) 2 L, (b) 4 L, (c) L/2, (d) L/4, Q.7 A round disc of moment of inertia I 2 about its axis, perpendicular to its plane and passing through its centre is, placed over another disc of moment of inertia I1 rotating, with an angular velocity w about the same axis. The final, angular velocity of the combination of discs is, (a), , Q.11 An automobile engine develops 100 kW when rotating at a, speed of 1800 rev/min. What torque does it deliver, , 3A 0, 4, , (b) A0, , (c) 4A0, , (d) 12A0, , Q.13 A wheel having moment of inertia 2 kg -m2 about its, vertical axis, rotates at the rate of 60 rpm about this axis., The torque which can stop the wheel's rotation in 1 minute, would be, (a), , 2p, Nm, 15, , (b), , p, Nm, 12, , (c), , p, Nm, 15, , (d), , p, Nm, 18, , Ù Ù, Ù, r, Q.14 Find the torque of a force F = -3 i + j + 5k acting at the, Ù, Ù Ù, r, point r = 7 i + 3 j + k, , Ù, , Ù, , Ù, , Ù, , (a) 14 i - 38 j + 16 k, Ù, , Ù, , Ù, , Ù, , (b) 4 i + 4 j + 6 k, Ù, , (c) -14 i + 38 j - 16 k, , Ù, , Ù, , Ù, , (d) -21 i + 3 j + 5 k, , Q.15 A constant torque of 1000 N -m, turns a wheel of moment, of inertia 200 kg -m2 about an axis passing through the, centre. Angular velocity of the wheel after 3 s will be, (a) 15 rad/s (b) 10 rad/s (c) 5 rad/s (d) 1 rad/s, Q.16 A torque of 30 N-m is applied on a 5 kg wheel whose, moment of inertia is 2kg-m2 for 10 sec. The angle covered, by the wheel in 10 sec will be, (a) 750 rad, , (b) 1500 rad, , (c) 3000 rad, , (d) 6000 rad, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 15, , 3, , Q.17 A horizontal force F is applied such that the block remains, stationary, then which of the following statement is false, a, a, F, , (a) f = mg [where f is the friction force], (b) F = N [ where N is the normal reaction], (c) F will not produce torque, (d) N will not produce torque, Q.18 In a bicycle, the radius of rear wheel is twice the radius of, front wheel. If r F and rr are the radius, vF and vr are speeds, of top most points of wheel, then, (a) vr = 2 vF, (b) vF = 2 vr, (c) vF = vr, (d) vF > vr, Q.19 The wheel of a car is rotating at the rate of 1200 revolutions, per minute. On pressing the accelerator for 10 seconds, it, starts rotating at 4500 revolutions per minute. The angular, acceleration of the wheel is, (a) 30 rad/sec2, (b) 1880 degree/sec2, 2, (c) 40 rad/sec, (d) 1980 degree/sec2, Q.20 A wheel rotates with a constant acceleration of 2.0 radian/, sec 2 . It the wheel starts from rest, the number of, revolutions it makes in the first ten seconds will be, approximately, (a) 8, (b) 16, (c) 24, (d) 32, , Q.21 A child is standing with folded hands at the centre of a, platform rotating about its central axis. The kinetic energy, of the system is K. The child now stretches his arms so, that the moment of inertia of the system doubles. The, kinetic energy of the system now is, (1) less than 2K, (2) equal to K/2, (3) more thanK/4, (4) equal to 4K, Q.22 Two uniforms discs of equal mass but unequal radii are, mounted on fixed horizontal axiles. Light strings are, wrapped on each of the discs. The strings are pulled by, constant equal forces F for same amount of time as shown, in the figure., , Disc I, Disc II, , Angular momenta of discs are L1 and L2 and their kinetic, energies are K1 and K2. Which of the following statements, true –, (1) L1 = L2 (2) L1 < L2 (3) K1 > K2 (4) K1 = K2, DIRECTIONS (Q.23-Q.25) : Read the passage given below, and answer the questions that follows :, Consider a cylinder of mass M = 1kg and radius R = 1 m lying on, a rough horizontal plane. It has a plank lying on its top as shown in, the figure., m = 1kg, , 17., , 18., , 60°, , F, , A, , DIRECTIONS (Q.21-Q.22) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, , RESPONSE, GRID, , F, , F, , M, , R, B, , A force F = 55 N is applied on the plank such that the, plank moves and causes the cylinder to roll. The plank, always remains horizontal. There is no slipping at any, point of contact., , 19., , 22., , Space for Rough Work, , 20., , 21.
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t.me/Magazines4all, DPP/ P 15, , 4, Q.23 Calculate the acceleration of cylinder., (a) 20 m/s2, (b) 10 m/s2, 2, (c) 5 m/s, (d) None of these, Q.24 Find the value of frictional force at A, (a) 7.5 N, (b) 5.0 N, (c) 2.5 N, (d) None of these, Q.25 Find the value of frictional force at B, (a) 7.5 N, (b) 5.0 N, (c) 2.5 N, (d) None of these, , (b), , DIRECTIONS (Q.26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE, GRID, , 23., , 24., , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.26 Statement -1: Torque is equal to rate of change of angular, momentum., Statement -2: Angular momentum depends on moment, of inertia and angular velocity., Q.27 Statement -1: Torque due to force is maximum when angle, r, r, between r and F is 90°., Statement -2: The unit of torque is newton- meter., Q.28 Statement -1: It is harder to open and shut the door if we, apply force near the hinge., Statement -2: Torque is maximum at hinge of the door., , 25., , 26., , 27., , 28., , DAILY PRA CTICE PROBLEM SHEET 15 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 16, , SYLLABUS : Rotational Motion-2 : Moment of inertia, radius of gyration,, (values of moments of inertia simple geometrical objects), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 Five particles of mass 2 kg are attached to the rim of a, circular disc of radius 0.1 m & negligible mass. Moment, of inertia of the system about an axis passing through the, centre of the disc & perpendicular to its plane is, (a) 1 kg-m2 (b) 0.1kg-m2 (c) 2 kg-m2 (d) 0.2 kg-m2, Q.2 Two discs of the same material and thickness have radii 0.2, m and 0.6 m. Their moments of inertia about their axes will, be in the ratio of, (a) 1 : 81, (b) 1: 27 (c) 1 : 9, (d) 1 : 3, Q.3 A cylinder of 500 g and radius 10 cm has moment of inertia, (about its natural axis), , RESPONSE GRID, , 1., , 2., , (a) 2.5 × 10–3 kg–m2, (b) 2 × 10–3 kg–m2, –3, 2, (c) 5 × 10 kg–m, (d) 3.5 × 10–3 kg–m2, Q.4 A constant torque of 31.4 N–m is exerted on a pivoted, wheel. If angular acceleration of wheel is 4 p rad/sec2, then, the moment of inertia of the wheel is, (a) 2.5 kg–m2, (b) 2.5 kg–m2, 2, (c) 4.5 kg–m, (d) 5.5 kg–m2, Q.5 From a uniform wire, two circular loops are made (i) P of, radius r and (ii) Q of radius nr. If the moment of inertia of, loop Q about an axis passing through its centre and, perpendicular to its plane is 8 times that of P about a similar, axis, the value of n is (diameter of the wire is very much, smaller than r or nr), (a) 8, (b) 6, (c) 4, (d) 2, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , DPP/ P 16, , 2, Q.6 The moment of inertia of a sphere of mass M and radius R, , about an axispassing through itscentre is, , 2, MR 2 . The radius, 5, , of gyration of the sphere about a parallel axis to the above, and tangent to the sphere is, (a), , 7, R, 5, , (b), , 3, R, 5, , (c), , æ 7 ö÷, çç ÷ R, çç 5 ÷÷, è ø, , (d), , æ 3 ÷ö, ççç ÷÷ R, çè 5 ÷ø, , Q.7 Four particles each of mass m are placed at the corners of, a square of side length l. The radius of gyration of the, system about an axis perpendicular to the square and, passing through its centre is, (a), , l, 2, , (b), , l, 2, , (c), , l, , 5, Mr 2, 4, , (b), , Mr 2, 4, , (c), , 3, Mr 2, 2, , (d), , Mr 2, 2, , Q.11 Two spheres each of mass M and radius R/2 are connected, with a massless rod of length 2R as shown in the figure., The moment of inertia, of the system about an, axis passing through, the centre of one of the, spheres, and, perpendicular to the, rod will be, (a), , 21 2, Mr, 5, , RESPONSE, GRID, , (b), , 2, Mr 2, 5, , M, , M, , R/2, , R/2, , 2R, , (c), , 5, Mr 2, 2, , (d), , a2, 4, a2, (m1 + m2), 4, , (a) (m2 + m3), , (b) (m1 + m2 + m3) a2, , (c), , (d) (m2 + m3) a2, , Q.13 Three rods each of length L and mass M are placed along, X, Y and Z axis in such a way that one end of each of the, rod is at the origin. The moment of inertia of this system, about Z axis is, , (d) ( 2)l, , Q.8 The radius of gyration of a disc of mass 50 g and radius 2.5, cm, about an axis passing through its centre of gravity and, perpendicular to the plane is, (a) 0.52 cm (b) 1.76 cm (c) 3.54 cm (d) 6.54 cm, Q.9 Moment of inertia of a ring of mass m = 3 gm and radius, r = 1 cm about an axis passing through its edge and parallel, to its natural axis is, (a) 10 gm–cm2, (b) 100 gm–cm2, (c) 6 gm–cm2, (d) 1 gm–cm2, Q.10 A disc is of mass M and radius r. The moment of inertia of, it about an axis tangential to its edge and in plane of the, disc or parallel to its diameter is, (a), , Q.12 Three point masses m1, m2, m3 are located at the vertices, of an equilateral triangle of length 'a'. The moment of, inertia of the system about an axis along the altitude of the, triangle passing through m1, is, , (a), , 2ML2, 3, , (b), , 4ML2, 3, , (c), , 5ML2, 3, , Q.14 ABC is a triangular plate of uniform, thickness. The sides are in the ratio, shown in the figure. IAB, IBC, ICA are, the moments of inertia of the plate, about AB, BC, CA respectively. For, this arrangement which one of the, following relation is correct?, , (d), , ML2, 3, , A, , 5, , 4, , B, , 3, , C, , (a) ICA is maximum, (b) IBC > IAB, (c) IBC > IAB, (d) IAB + IBC = ICA, Q.15 A 1m long rod has a mass of 0.12 kg. The moment of inertia, about an axis passin through the centre and perpendicular, to the length of rod will be, (a) 0.01kg-m2, (b) 0.001 kg-m2, 2, (c) 1 kg-m, (d) 10 kg-m2, Q.16 Two rings of the same radius and mass are placed such that, their centres are at a common point and their planes are, perpendicular to each other. The moment of inertia of the, system about an axis passing through the centre and, perpendicular to the plane of one of the rings is (mass of, the ring = m and radius = r), , 5, Mr 2, 21, , (a), , 1 2, mr, 2, , (b) mr2, , (c), , 3 2, mr, 2, , (d) 2 mr2, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 16, , 3, (a) 3 MR2, , Q.17 One quarter sector is cut from a uniform, circular disc of radius R. This sector has, mass M. It is made to rotate about a line, perpendicular to its plane and passing, through the centre of the original disc., Its moment of inertia about the axis of, rotation is, 1, , 1, , (b), (c), , 90°, , (d), , 1, , MR 2, MR 2, (a), (b) MR2 (c), (d) 2MR 2, 2, 4, 8, Q.18 A thin wire of length L and uniform linear mass density r is, bent into a circular loop with centre at O as shown in figure., The moment of inertia of the loop about the axis XX¢ is, , (a), (b), (c), (d), , rL3, 8p 2, rL3, , X, , 16 p 2, , X¢, 90°, , 5rL3, , O, , 16 p 2, 3rL3, 8p 2, , Q.19 Two disc of same thickness but of different radii are made, of two different materials such that their masses are same., The densities of the materials are in the ratio 1 : 3. The, ratio of the moments of inertia of these discs about the, respective axes passing through their centres and, perpendicular to their planes will be in, (a) 1 : 3, (b) 3 : 1, (c) 1 : 9, (d) 9 : 1, Q.20 A circular disc of radius R and thickness, , R, 6, , has moment, , of inertia I about an axis passing through its centre and, perpendicular to its plane. It is melted and recasted into a, solid sphere. The moment of inertia of the sphere about, one of its diameter as an axis of rotation will be, (a) I, , (b), , 2I, 8, , (c), , I, 5, , (d), , Q.21 Three rings each of mass M and radius R are, arranged as shown in the figure. The moment, of inertia of the system about YY' will be, , RESPONSE, GRID, , I, 10, , Y, , 3, MR 2, 2, 5 MR2, 7, MR 2, 2, , Y¢, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 The density of a rod AB increases linearly from A to B. Its, midpoint is O and its centre of mass is at C. Four axes pass, through A, B, O and C, all perpendicular to the length of, the rod. The moments of inertia of the rod about these, axes are I A, IB, IO and IC respectively then:., (1) IA > IB, (2) IA < IB, (3) IO > IC, (4) IO < IC, Q.23 The moment of inertia of a thin square plate ABCD of, uniform thickness about an axis passing through the centre, O and perpendicular to the plane of the plate is, 4, , (1) I1 + I2, A, (2) I3 + I4, (3) I1 + I3, (4) I1 + I2 + I3 + I4, where I1 , I2 , I3 and I4 are, respectively moments of, D, inertia about axes 1, 2, 3 and, 4 which are in the plane of the, plate., Q.24 Moment of inertia doesn’t depend on, (1) distribution of particles, (2) mass, (3) position of axis of rotation, (4) None of these, , 17., , 18., , 19., , 22., , 23., , 24., , Space for Rough Work, , 20., , 21., , 1, B, , 3, , O, C, , 2
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t.me/Magazines4all, DPP/ P 16, , 4, (a), (b), (c), (d), , DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, Four identical spheres having mass M and, radius R are fixed tightly within a massless ring, such that the centres of all spheres lie in the, plane of ring. The ring is kept on a rough, horizontal table as shown. The string is wrapped, around the ring can roll without slipping., The other end of the string is passed over a massless frictionless, pulley to a block of mass M. A force F is applied horizontally on, the ring, at the same level as the centre, so that the system is in, equilibrium., Q.25 The moment of inertia of the combined ring system about, the centre of ring will be, (a), (b), (c), (d), , 12, MR 2, 5, 48, MR2, 15, 24, MR 2, 5, , F, M, , 48, MR2, 5, , Q.26 The magnitude of F is, (a) Mg, (c), , Mg, 2, , (b) 2Mg, (d) None of these, , Q.27 If the masses of the spheres were doubled keeping their, dimensions same, the force of friction between the ring, and the horizontal surface would, , RESPONSE, GRID, , 25., , 26., , be doubled, increase but be less than double, remain the same, decrease, , DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement-1 is False, Statement-2 is True., (d) Statement-1 is True, Statement-2 is False., Q.28 Statement-1 : Radius of gyration of a body is a constant, quantity., Statement-2 : The radius of gyration of a body about an, axis of rotation may be defined as the root mean square, distance of the particles of the body from the axis of rotation., Q.29 Statement-1 : Moment of inertia of a particle is same,, whatever be the axis of rotation., Statement-2 : Moment of inertia depends on mass and, perpendicular distance of the particle from its axis of, rotation., Q.30 Statement-1 : If earth shrink (without change in mass) to, half of its present size, length of the day would become 6, hours., Statement-2 : When the size of the earth will change, its, moment of inertia will also change., , 27., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 16 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 32, Qualifying Score, 50, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 17, SYLLABUS : Rotational Motion - 3 : Rolling Motion, Parallel and perpendicular theorems and their, applications, Rigid body rotation, equations of rotational motion, , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A disc is rolling (without slipping) on a horizontal surface. C is, its centre and Q and P are two points equidistant from C . Let vP,, vQ and vC be the magnitude of velocities of points P, Q and, C respectively, then, Q, (a) vQ > vC > vP, C, , (b) vQ < vC < vP, (c) vQ = vP , vC =, , vP, 2, , P, , (d) vQ < vC > vP, , RESPONSE GRID, , 1., , Q.2 A uniform rod of length 2L is placed with one end in contact, with the horizontal and is then inclined at an angle a to the, horizontal and allowed to fall without slipping at contact, point. When it becomes horizontal, its angular velocity will, be, 3g sin a, 2L, 2L, (b) w =, 3g sin a, , (a) w =, , 6 g sin a, L, L, (d) w =, g sin a, , (c) w =, , 2., Space for Rough Work
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t.me/Magazines4all, DPP/ P 17, , 2, Q.3 According to the theorem of parallel axes I = I cm + Mx 2 ,, the graph between I and x will be, I, , I, , (a), , Q.8 A solid sphere is rolling on a frictionless surface, shown, in figure with a transnational velocity v m/s. If sphere climbs, up to height h then value of v should be, , h, , (b), O, , v, , O, , x, , x, , I, , I, , (a) ³ 10 gh (b) ³ 2gh (c) 2gh, 7, , (c), x, , x, , O, , Q.4 A solid cylinder of mass M and radius R rolls without, slipping down an inclined plane of length L and height h., What is the speed of its centre of mass when the cylinder, reaches its bottom, 3, gh, 4, , (a), , 4, gh, 3, , (b), , (c), , (d), , 4gh, , (a), , (a), , 2g, (b), 3, , (c), , 5g, 7, , 2 gh, I + mr, , (a), , 1/ 2, , (b), 1/ 2, , é 2mgh ù, ú, ë I + 2 mr 2 û, , (c) ê, , (d), , 2gh, , Q.7 A solid sphere, disc and solid cylinder all of the same mass, and made up of same material are allowed to roll down, (from rest) on an inclined plane, then, (a) Solid sphere reaches the bottom first, (b) Solid sphere reaches the bottom late, (c) Disc will reach the bottom first, (d) All of them reach the bottom at the same time, , RESPONSE, GRID, , (b), , (c), , 3, I, 2, , (d) 2I, , 3, MR 2, 2, , (c) 5MR 2, , 5g, (d), 14, , é 2mgh ù, ê, ú, ë I + mr 2 û, , (b) 3I, , (a) 3MR 2, , 2gh, , Q.6 A cord is wound round the circumference of wheel of radius, r. The axis of the wheel is horizontal and moment of inertia, about it is I. A weight mg is attached to the end of the cord, and falls from the rest. After falling through a distance h,, the angular velocity of the wheel will be, , 5, I, 2, , Q.10 Three rings each of mass M and radius R are arranged as, shown in the figure. The moment of inertia of the system, about YY' will be, Y, , Q.5 An inclined plane makes an angle 30° with the horizontal., A solid sphere rolling down this inclined plane from rest, without slipping has a linear acceleration equal to, g, 3, , 10, gh, 7, , Q.9 Moment of inertia of a disc about its own axis is I. Its, moment of inertia about a tangential axis in its plane is, , (d), O, , (d), , (d), , 7, MR 2, 2, , Y', , Q.11 One circular ring and one circular disc, both are having the, same mass and radius. The ratio of their moments of inertia, about the axes passing through their centres and perpendicular, to their planes, will be, (a) 1 : 1, (b) 2 : 1, (c) 1 : 2, (d) 4 : 1, Q.12 From a disc of radius R, a concentric circular portion of, radius r is cut out so as to leave an annular disc of mass M., The moment of inertia of this annular disc about the axis, perpendicular to its plane and passing through its centre of, gravity is, (a), (c), , 1, M (R2 + r 2 ), 2, 1, M (R4 + r 4 ), 2, , (b), , 1, M (R2 - r 2 ), 2, , (d) 1 M ( R 4 - r 4 ), 2, , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 17, , 3, , Q.13 The moment of inertia of a straight thin rod of mass M and, length l about an axis perpendicular to its length and passing, through its one end, is, (a), , 2, , Ml, 12, , (b), , Ml, 3, , 2, , (c), , Ml, 2, , 2, , (d) Ml2, , 4 2, Ml, 3, , (b), , Ml 2, 3, , (c), , Ml 2, 6, , (d), , 2 2, Ml, 3, , l, , 2MR 2, , l, , l, O, , l, , (b), , 3, MR 2 (c), 2, , 1, MR 2, 2, , (d) MR 2, , Q.16 The moment of inertia of uniform rectangular plate about an, axis passing through its mid-point and parallel to its length, l is (b = breadth of rectangular plate), (a), , Mb2, 4, , (b), , Mb3, 6, , (c), , Mb3, 12, , (d), , Mb2, 12, , Q.17 The moment of inertia of a circular ring about an axis passing, through its centre and normal to its plane is 200 gm × cm2., Then moment of inertia about its diameter is, (a) 400 gm × cm2, (b) 300 gm × cm2, (c) 200 gm × cm2, (d) 100 gm × cm2, Q.18 From a circular disc of radius R and mass 9 M, a small disc of, radius R/3 is removed from the disc. The moment of inertia of, the remaining disc about an axis perpendicular to the plane of, the disc and passing through O is, , RESPONSE, GRID, , (b), , 40, MR 2, 9, , (d), , 2R/3, O, R, , 37, MR 2, 9, , Q.19 The moment of inertia of a thin rod of mass M and length L, about an axis perpendicular to the rod at a distance L/4 from, one end is, (a), , Q.15 The moment of inertia of a uniform circular ring, having a, mass M and a radius R, about an axis tangential to the ring, and perpendicular to its plane, is, (a), , 4MR 2, , (c) 10MR 2, , Q.14 Four thin rods of same mass M and same length l, form a, square as shown in figure. Moment of inertia of this, system about an axis through centre, O and perpendicular to its plane is, (a), , (a), , ML2, 6, , (b), , ML2, 12, , (c), , 7 ML2, 24, , (d), , 7 ML2, 48, , Q.20 A wheel has a speed of 1200 revolutions per minute and is, made to slow down at a rate of 4 radians /s2. The number of, revolutions it makes before coming to rest is, (a) 143, (b) 272, (c) 314, (d) 722, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes:, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 In pure rolling fraction of its total energy associated with, rotation is a for a ring and b for a solid sphere. Then, (1) a = 1 / 2 (2) b = 2 / 7 (3) b = 2 / 5 (4) a = 1 / 4, Q.22 One solid sphere and a disc of same radius are falling along, an inclined plane without slip. One reaches earlier than the, other due to, (1) different size, (2) different radius of gyration, (3) different moment of inertia, (4) different friction, Q.23 A body is rolling down an inclined plane. Its translational, and rotational kinetic energies are equal. The body is not a, (1) solid sphere, (2) hollow sphere, (3) solid cylinder, (4) hollow cylinder, , 13., , 14., , 15., , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., Space for Rough Work
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t.me/Magazines4all, DPP/ P 17, , 4, DIRECTIONS (Q.24-Q.26) : Read the passage given below and, answer the questions that follows :, A uniform solid cylinder of mass 2m and, radius R rolls on a rough inclined plane, 2m, with its axis perpendicular to the line of, R, the greatest slope., m, System is released from rest and as, q, ////////////////////////// ////////////////, the cylinder rolls it winds up a light, string which passes over a light pulley., Q.24 The acceleration of block of mass m is (a), , 2, g (1 - cos q), 7, 2, g (1 - sin q ), 7, , (b), , (c), (d), Q.25 The tension in the string is –, æ 4 + 3sin q ö, ÷ø mg, 7, æ 3 + 4sin q ö, mg, èç, ø÷, 7, , (a) çè, , (b), , 4, g (1 - sin q), 7, 2, g (1 + sin q), 14, æ 3 - 4sin q ö, mg, èç, ø÷, 7, 2, , (1 - sin q) mg, (c), (d), 7, Q.26 The frictional force acting on the cylinder is-, , (a), , 2, (1 - sin q) mg, 7, , (b), , æ 1 + 6 cos q ö, ÷ø mg, 7, , (c) çè, , (d), , æ 6 - sin q ö, çè, ÷ø mg, 7, 1, +, 6, sin, q, æ, ö, çè, ÷ø mg, 7, , DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select, the correct choice., , RESPONSE, GRID, , 24., , 25., , (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : 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, hollow cylinder will reach the bottom of the inclined plane, first., Statement-2 : By the principle of conservation of energy,, the total kinetic energies of both the cylinders are identical, when they reach the bottom of the incline., Q.28 Statement-1: The force of frction in the case of a disc, rolling without slipping down an inclined plane is 1/3 g, sin a., Statement-2: When the disc rolls without slipping, friction, is required because for rolling condition velocity of point, of contact is zero., Q.29 Statement-1: If two different axes are at same distance, from the centre of mass of a rigid body, then moment of, inertia of the given rigid body about both the axes will, always be the same., Statement-2: From parallel axis theorem, I = Icm + md 2,, where all terms have usual meaning., , 26., , 27., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 17 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 18, SYLLABUS : Gravitation - 1 (The Universal law of gravitation, Acceleration due to gravity and its variation, with altitude and depth, Kepler's law of planetary motion), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A mass M splits into two parts m and (M – m), which are, then separated by a certain distance. What ratio of (m / M), maximises the gravitational force between the parts. ?, (a) 2/3, (b) 3/4, (c) 1/2, (d) 1/3, Q.2 What would be the angular speed of earth, so that bodies, lying on equator may experience weightlessness ?, (g = 10m/s2 and radius of earth = 6400 km), (a) 1.25 × 10–3 rad/sec, (b) 1.25 × 10–2 rad/sec, –4, (c) 1.25 × 10 rad/sec, (d) 1.25 × 10–1 rad/sec, , RESPONSE GRID, , 1., , 2., , Q.3 The speed with which the earth have to rotate on its axis so, that a person on the equator would weigh (3/5) th as much, as present will be (Take the equatorial radius as 6400 km.), (a) 3.28 × 10–4 rad/sec, (b) 7.826 × 10–4 rad/sec, (c) 3.28 × 10–3 rad/sec, (d) 7.28 × 10–3 rad/sec, Q.4 On a planet (whose size is the same as that of earth and, mass 4 times to the earth) the energy needed to lift a 2kg, mass vertically upwards through 2m distance on the planet, is (g = 10m/sec2 on the surface of earth) (a) 16 J, (b) 32 J, (c) 160 J, (d) 320 J, Q.5 Two bodies of mass 102 kg and 103 kg are lying 1m apart., The gravitational potential at the mid-point of the line, joining them is (a) 0, (b) –1.47 Joule/kg, (c) 1.47 Joule/kg, (d) –147 × 10–7 Joule /kg, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 18, , 2, Q.6 If g is the acceleration due to gravity on the earth’s surface,, the gain in P.E. of an object of mass m raised from the surface, of the earth to a height of the radius R of the earth is 1, 1, (a) mgR, (b) 2mgR (c), mgR (d), mgR, 2, 4, Q.7 Four particles, each of mass m, are placed at the corners, of square and moving along a circle of radius r under the, influence of mutual gravitational attraction. The speed of, each particle will be (a), , Gm, (2 2 + 1), r, , (b), , Gm æ 2 2 + 1ö, r çè, 4 ÷ø, , (d), , Gm, r, , 2 2Gm, r, Q.8 Three particles of equal mass m are situated at the vertices, of an equilateral triangle of side l. What should be the, velocity of each particle, so that they move on a circular, path without changing l ?, (c), , GM, GM, 2GM, GM, (b), (c), (d), 2l, l, l, 3l, Q.9 What will be the acceleration due to gravity on the surface, of the moon if its radius is 1/4 th the radius of the earth and, its mass is 1/80 th the mass of the earth ?, (a) g/6, (b) g/5, (c) g/7, (d) g/8, Q.10 If the value of 'g' at a height h above the surface of the earth, is the same as at a depth x below it, then (both x and h being, much smaller than the radius of the earth), (a), , h, (d) x = h2, 2, Q.11 At what height above the earth's surface the acceleration, due to gravity will be 1/9 th of its value at the earth’s, surface? Radius of earth is 6400 km., (a) 12800 km, (b) 1280 km, (c) 128000 km, (d) 128 km, Q.12 If the radius of the earth were to shrink by one percent, its, mass remaining the same, the acceleration due to gravity, on the earth’s surface would (a) decrease, (b) remain unchanged, (c) increase, (d) None of these, , (a) x = h, , RESPONSE, GRID, , (b) x = 2h, , Q.13 At what height above the earth’s surface does the force of, gravity decrease by 10% ? Assume radius of earth to be, 6370 km., (a) 350 km., (b) 250 km. (c) 150 km., (d) 300 km., Q.14 A particle is suspended from a spring and it stretches the, spring by 1 cm on the surface of earth. The same particle, will stretches the same spring at a place 800 km above earth, surface by (a) 0.79 cm, (b) 0.1 cm, (c) p / 6 rad/hr., (d) 2p / 7 rad/hr., Q.15 The change in the value of acceleration of earth towards, sun, when the moon comes from the position of solar, eclipse to the position on the other side of earth in line, with sun is (Mass of moon = 7.36 × 1022 kg, the orbital, radius of moon 3.8 × 108m., (a) 6.73 × 10–2 m/s2, (b) 6.73 × 10–3 m/s2, –4, 2, (c) 6.73 × 10 m/s, (d) 6.73 × 10–5 m/s2, Q.16 The radius of the earth is Re and the acceleration due to gravity, at its surface is g. The work required in raising a body of, mass m to a height h form the surface of the earth will be mg, mgh, mgh, mgh, (a), (b), (c), (d), 2, æ, h ö, æ, æ, h ö, h ö, æ, h ö, çè1 + R ÷ø, çè1 + R ÷ø, çè 1 - R ÷ø, çè 1 + R ÷ø, e, e, e, e, Q.17 The masses and the radius of the earth and the moon are, M1, M2 and R1, R2 respectively their centres are at distance, d apart. The minimum speed with which a particle of mass, m should be projected form a point midway between the, two centres so as to escape to infinity will be -, , (c) x =, , (a), , (c), , 2, , G, (M1 + M 2 ), d, , G, (M1 + M 2 ), 2d, , G, (M1 + M 2 ), d, , (b), , (d), , 2, , G M1, d M2, , Q.18 With what velocity must a body be thrown upward form the, surface of the earth so that it reaches a height of 10 Re?, earth’s mass Me = 6 × 1024 kg, radius Re = 6.4 × 106 m and, G = 6.67 × 10–11 N–m2/kg2., (a) 10.7 × 104 m/s, (b) 10.7 × 103 m/s, 5, (c) 10.7 × 10 m/s, (d) 1.07 × 104 m/s, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 18, , 3, , Q.19 Two concentric shells of uniform density having masses M1, and M2 are situated as shown in the figure. The force on the, particle of mass m when it is located at r = b is, , M1, O, r, , m, p, , M2, (a), , b, , GM1m, , (b), , 2, , b, (M1 + M 2 )m, , DIRECTIONS (Q.24-Q.26) : Read the passage given below and, answer the questions that follows :, , GM 2 m, , b2, (M1 - M 2 )m, , (d) G, b2, b2, Q.20 What is the mass of the planet that has a satellite whose, time period is T and orbital radius is r?, (c) G, , (a), , 4p 2 r 3, GT, , 2, , 3p 2r 3, , (b), , GT, , 2, , (c), , 4p 2 r 3, GT, , 3, , (b) 1 and 2 are correct, (d) 1 and 3 are correct, , Sun, , Aphelion, , GT 2, , Q.21 The gravitational force between two point masses m1 and, mm, m2 at separation r is given by F = k 1 2 2, r, The constant k doesn't, (1) depend on medium between masses, (2) depend on the place, (3) depend on time, (4) depend on system of units, Q.22 Which of the following statements about the gravitional, constant are false ?, (1) It is a force, (2) It has no unit, (3) It has same value in all system of units, (4) It doesn’t depend on the value of the masses, , RESPONSE, GRID, , The orbit of Pluto is much more eccentric than the orbits of the, other planets. That is, instead of being nearly circular, the orbit, is noticeably elliptical. The point in the orbit nearest to the sun, is called the perihelion and the point farthest from the sun is, called the aphelion., , 4p2T, , (d), , DIRECTIONS (Q.21-Q.23) : In the following questions, more than, one of the answers given are correct. Select the correct answers, and mark it according to the following codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, , Q.23 Spot the correct statements :, The acceleration due to gravity ‘g’ decreases if, (1) We go down from the surface of the earth towards its, centre, (2) We go up from the surface of the earth, (3) The rotational velocity of the earth is increased, (4) We go from the equator towards the poles on the, surface of the earth, , Perihelion, , Q.24 At perihelion, the gravitational potential energy of Pluto, in its orbit has, (a) its maximum value, (b) its minimum value, (c) the same value as at every other point in the orbit, (d) value which depends on sense of rotation, Q.25 At perihelion, the mechanical energy of Pluto’s orbit has, (a) its maximum value, (b) its minimum value, (c) the same value as at every other point in the orbit, (d) value which depends on sense of rotation, Q.26 As Pluto moves from the perihelion to the aphelion, the, work done by gravitational pull of Sun on Pluto is, (a) is zero, (b) is positive, (c) is negative, (d) depends on sense of rotation, , 19., , 20., , 21., , 24., , 25., , 26., , Space for Rough Work, , 22., , 23.
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DPP/ P 18, , 4, DIRECTIONS (Q. 27-Q.29) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason)., Each of these questions has four alternative choices, only one of, which is the correct answer. You have to select the correct choice., (a), (b), (c), (d), , t.me/Magazines4all, , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , RESPONSE GRID, , 27., , 28., , Q.27 Statement-1 : Gravitational force between two particles is, negligibly small compared to the electrical force., Statement-2 :The electrical force is experienced by charged, particles only., Q.28 Statement-1 :The universal gravitational constant is same, as acceleration due to gravity., Statement-2 :Gravitional constant and acceleration due to, gravity have different dimensional formula., Q.29 Statement-1 :There is no effect of rotation of earth on the, value of acceleration due to gravity at poles., Statement-2 :Rotation of earth is about polar axis., , 29., , DAILY PRA CTICE PROBLEM SHEET 18 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 19, SYLLABUS : Gravitation - 2 (Gravitational potential energy, Gravitational potential,, Escape velocity & Orbital velocity of a satellite, Geo-stationary satellites), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A body of mass 100 kg falls on the earth from infinity., What will be its energy on reaching the earth ? Radius of, the earth is 6400 km and g = 9.8 m/s2. Air friction is, negligible., (a) 6.27 × 109 J, (b) 6.27 × 1010 J, 10, (c) 6.27 × 10 J, (d) 6.27 × 107 J, Q.2 An artificial satellite of the earth is to be established in the, equatorial plane of the earth and to an observer at the equator, it is required that the satellite will move eastward, completing, one round trip per day. The distance of the satellite from the, , RESPONSE GRID, , 1., , 2., , centre of the earth will be- (The mass of the earth is 6.00 ×, 1024 kg and its angular velocity = 7.30 × 10–5 rad./sec.), (a) 2.66 × 103 m., (b) 2.66 × 105 m., 6, (c) 2.66 × 10 m., (d) 2.66 × 107 m., Q.3 Two satellites S1 and S2 revolve round a planet in the same, direction in circular orbits. Their periods of revolutions, are 1 hour and 8 hour respectively. The radius of S1 is 104, km. The velocity of S2 with respect to S1 will be(a) p × 104 km/hr, (b) p/3 × 104 km/hr, 4, (c) 2p × 10 km/hr, (d) p/2 × 104 km/hr, Q.4 In the above example the angular velocity of S2 as actually, observed by an astronaut in S1 is (a) p/3 rad/hr, (b) p/3 rad/sec, (c) p/6 rad/hr, (d) 2p/7 rad/hr, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 19, , 2, Q.5 The moon revolves round the earth 13 times in one year. If, the ratio of sun-earth distance to earth-moon distance is, 392, then the ratio of masses of sun and earth will be (a) 365, (b) 356, (c) 3.56 × 105, (d) 1, Q.6 Two planets of radii in the ratio 2 : 3 are made from the, materials of density in the ratio 3 : 2. Then the ratio of, acceleration due to gravity g1/g2 at the surface of two, planets will be, 4, (a) 1, (b) 2 . 2 5, (c), (d) 0.12, 9, Q.7 A satellite of mass m is revolving in a circular orbit of radius, r. The relation between the angular momentum J of satellite, and mass m of earth will be (a) J = G.Mm 2 r, (b) J = GMm, (c) J =, , (d) J =, , GMmr, , mr, M, , Q.8 A spaceship is launched into a circular orbit close to earth's, surface. What additional velocity has now to be imparted, to the spaceship in the orbit to overcome the gravitational, pull? (Radius of earth = 6400 km, g = 9.8 m/sec2), (a) 3.285 km/sec, (b) 32.85 m/sec, (c) 11.32 km/sec, (d) 7.32 m/sec, Q.9 The ratio of the radius of the Earth to that of the moon is, 10. The ratio of g on earth to the moon is 6. The ratio of, the escape velocity from the earth’s surface to that from, the moon is approximately (a) 10, (b) 8, (c) 4, (d) 2, Q.10 Acceleration due to gravity on a planet is 10 times the value, on the earth. Escape velocity for the planet and the earth, are Vp and Ve respectively. Assuming that the radii of the, planet and the earth are the same, then (a) VP = 10 Ve, (b) VP = 10 Ve, (c) Vp = Ve, , (d) VP =, , 10, , Ve, 10, , Q.11 The Jupiter’s period of revolution round the Sun is 12 times, that of the Earth. Assuming the planetary orbits are circular,, how many times the distance between the Jupiter and Sun, exceeds that between the Earth and the sun., (a) 5.242 (b) 4.242, (c) 3.242 (d) 2.242, , RESPONSE, GRID, , Q.12 The mean distance of mars from sun is 1.524 times the, distance of the earth from the sun. The period of revolution, of mars around sun will be(a) 2.88 earth year, (b) 1.88 earth year, (c) 3.88 earth year, (d) 4.88 earth year, Q.13 The semi-major axes of the orbits of mercury and mars are, respectively 0.387 and 1.524 in astronomical unit. If the, period of Mercury is 0.241 year, what is the period of Mars., (a) 1.2 years, (b) 3.2 years, (c) 3.9 years, (d) 1.9 years, Q.14 If a graph is plotted between T2 and r3 for a planet then its, slope will be (a), , 4p 2, GM, , (b), , GM, 4p 2, , (c) 4p GM, (d) 0, Q.15 The mass and radius of earth and moon are M1, R1 and M2,, R2 respectively. Their centres are d distance apart. With, what velocity should a particle of mass m be projected from, the mid point of their centres so that it may escape out to, infinity (a), , G(M1 + M 2 ), d, , (b), , 2G(M1 + M 2 ), d, , (c), , 4G(M1 + M 2 ), d, , (d), , GM1M 2, d, , Q.16 A satellite has to revolve round the earth in a circular orbit, of radius 8 × 103 km. The velocity of projection of the, satellite in this orbit will be (a) 16 km/sec, (b) 8 km/sec, (c) 3 km/sec, (d) 7.08 km/sec, Q.17 If the satellite is stopped suddenly in its orbit which is at a, distnace = radius of earth from earth’s surface and allowed, to fall freely into the earth, the speed with which it hits the, surface of earth will be (a) 7.919 m/sec, (b) 7.919 km/sec, (c) 11.2 m/sec, (d) 11.2 km/sec, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 19, , 3, , Q.18 A projectile is fired vertically upward from the surface of, earth with a velocity K ve m/s where ve m/s is the escape, velocity and K < 1. Neglecting air resistance, the maximum, height to which it will rise measured from the centre of, the earth is - (where R = radius of earth), 2, 2, R, R, (d) K, (c) 1 - K, (a), (b), 2, R, R, K, 1 - K2, Q.19 A satellite is revolving in an orbit close to the earth’s, surface. Taking the radius of the earth as 6.4 × 106 metre,, the value of the orbital speed and the period of revolution, of the satellite will respectively be (g = 9.8 meter/sec2), (a) 7.2 km/sec., 84.6 minutes, (b) 2.7 km/sec., 8.6 minutes, (c) .72 km/sec., 84.6 minutes, (d) 7.2 km/sec., 8.6 minutes, Q.20 If the period of revolution of an artificial satellite just above, the earth be T second and the density of earth be r, kg/m3, then, (G = 6.67 × 10–11 m3/kg. second2), (a) rT2 is a universal constant, (b) rT2 varies with time, 3p, (c) rT2 =, G, (d) Both (a) and (c), Q.21 Two satellites P and Q of same mass are revolving near the, earth surface in the equitorial plane. The satellite P moves, in the direction of rotation of earth whereas Q moves in, the opposite direction. The ratio of their kinetic energies, with respect to a frame attached to earth will be 2, , æ 7437 ö, æ 8363 ö, (a) ç, (b) ç, è 8363 ÷ø, è 7437 ÷ø, , 2, , æ 8363 ö, æ 7437 ö, (c) çè, ÷ (d) çè, ÷, 7437 ø, 8363 ø, , DIRECTIONS (Q.22-Q.24) : In the following questions, more than, one of the answers given are correct. Select the correct answers, and mark it according to the following codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, , RESPONSE, GRID, , (b) 1 and 2 are correct, (d) 1 and 3 are correct, , 18., , 19., , 23., , 24., , Q.22 Gas escapes from the surface of a planet because it acquires, an escape velocity. The escape velocity will depend on, which of the following factors:, (1) Mass of the planet, (2) Radius of the planet, (3) Mass of the particle escaping, (4) Temperature of the planet, Q.23 ve and vp denotes the escape velocity from the earth and, another planet having twice the radius and the same mean, density as the earth. Then which of the following is (are), wrong ?, (1) ve = vp, (2) ve = 2vp, (3) ve = vp/4, (4) ve = vp / 2, Q.24 Select the wrong statements from the following, (1) The orbital velocity of a satellite increases with the, radius of the orbit, (2) Escape velocity of a particle from the surface of the, earth depends on the speed with which it is fired, (3) The time period of a satellite does not depend on the, radius of the orbit, (4) The orbital velocity is inversely proportional to the, square root of the radius of the orbit, DIRECTIONS (Q.25-Q.27) : Read the passage given below and, answer the questions that follows :, , It can be assumed that orbits of earth and mars are nearly circular, around the sun. It is proposed to launch an artificial planet around, the sun such that its apogee is at the orbit of mars while its perigee, is at the orbit of earth. Let Te and Tm be periods of revolution of, earth and mars. Further the variables are assigned the meanings as, follows., Me ® Mass of earth, Mm ® Mass of mars., Le ® Angular momentum of earth around the sun., Lm ® Angular momentum of mars around the sun., Re ® Semi major axis of earth’s orbit., Rm ® Semi major axis of mars orbit., M ® Mass of the artificial planet., Ee ® Total energy of earth., Em ® Total energy of mars., , 20., , Space for Rough Work, , 21., , 22.
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t.me/Magazines4all, DPP/ P 19, , 4, Q.25 Time period of revolution of the artificial planet about sun, will be (neglect gravitational effects of earth and mars), Te + Tm, TeTm, (a), (b), 2, 3/ 2, , é T 2 / 3 + Tm2 / 3 ù, (d) ê e, ú, 2, êë, úû, Q.26 The total energy of the artificial planet’s orbit will be, , 2TeTm, (c), Te + Tm, , (a), , 2 M æ Re Ee ö, M e çè Re + Rm ÷ø, , (b), , 2M æ Re Ee ö, M m çè Re + Rm ÷ø, , æ, ö, 2 Ee M Re + Rm, ç, ÷, Me ç R2 + R2 ÷, è e, mø, Q.27 Areal velocity of the artificial planet around the sun will, be, (a) less than that of earth, (b) more than that of mars, (c) more than that of earth, (d) same as that of the earth, , (c), , 2 Ee M æ Re + Rm ö, M m çè Rm ÷ø, , RESPONSE, GRID, , 25., , (d), , 26., , DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select, the correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement-1 is False, Statement-2 is True., (d) Statement-1 is True, Statement-2 is False., Q.28 Statement-1 :The speed of revolution of an artificial, satellite revolving very near the earth is 8kms–1., Statement-2 : Orbital velocity of a satellite, become, independent of height of near satellite., Q.29 Statement-1 :If an earth satellite moves to a lower orbit,, there is some dissipation of energy but the speed of, gravitational satellite increases., Statement-2 :The speed of satellite is a constant quantity., Q.30 Statement-1 :Gravitational potential of earth at every place, on it is negative., Statement-2 :Every body on earth is bound by the attraction, of earth., , 27., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 19 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 32, Qualifying Score, 50, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 20, SYLLABUS : Mechanical Properties of solids, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice questions., Each question has 4 choices (a), (b), (c) and (d), out of which ONLY, ONE choice is correct., , Q.1 Length of a wire is doubled, when 20 × 108 N/m2 stress is, applied on it. Its Young’s modulus of elasticity in N/m2 will, be, (a) 20 × 108, (b) 20 × 109, 10, (c) 20 × 10, (d) 10 × 108, Q.2 A steel wire of uniform cross-sectional area 2mm2 is heated, upto 50°C and clamped rigidly at two ends. If the, temperature of wire falls to 30° then change in tension in, the wire will be, if coefficient of linear expansion of steel, is 1.1 × 10–5 /°C and young's modulus of elasticity of steel, is 2 × 1011 N/m2., , RESPONSE GRID, , 1., , 2., , (a) 44 N, , (b) 88 N, , (c) 132 N, , (d) 22 N, , Q.3 The work done in increasing the length of a one metre long, wire of cross-sectional area 1mm2 through 1mm will be, (Y = 2 × 1011 N/m2), (a) 250 J, , (b) 10 J, , (c) 5 J, , (d) 0.1 J, , Q.4 A spring is stretched by 3cm when a load of 5.4 × 10 6 dyne, is suspended from it. Work done will be(a) 8.1 × 106 J, , (b) 8 × 106 J, , (c) 8.0 × 106 erg, , (d) 8.1 ×106 erg, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 20, , 2, 10-8, , m2, , Q.5 A wire of length 1m and area of cross section 4 ×, increases in length by 0.2 cm when a force of 16 N is applied., Value of Y for the material of the wire will be, (a) 2 × 106 N/m2, (c), , Q.12 The compressibility of water is 5 × 10–10 m2/N. If it is, subjected to a pressure of 15 MPa, the fractional decrease in, volume will be -, , (b) 2 × 1011 kg/m2, , 2 × 1011 N/mm2, , (d), , 2 × 1011 N/m2, , Q.6 The volume of a solid rubber ball when it is carried from the, surface to the bottom of a 200 m deep lake decreases by, 0.1%. The value for bulk modulus of elasticity for rubber will, be, (a) 2 × 109 Pa, , (b) 2 × 106 Pa, , (c) 2 × 104 Pa, , (d) 2 × 10–4 Pa, , Q.7 A steel wire is 4.0 m long and 2 mm in diameter. Young’s, modulus of steel is 1.96 × 1011 N/m2. If a mass of 20 kg is, suspended from it the elongation produced will be (a) 2.54 mm, , (b) 1.27 mm, , (c) 0.64 mm, , (d) 0.27 mm, , Q.8 A brass rod is to support a load of 400 N. If its elastic limit is, 4.0 × 108 N/m2 its minimum diameter must be (a) 1.13 mm (b) 2.26 mm, (c) 3.71 mm (d) 4.52 mm, , 4.0 × 107 KN/m2, , (a), , (b), , (c), , 4.0 × 107 N/m2, , (d) None of these, (b) 5 × 10–4, volumetric, , (c) 5 × 10–4, longitudinal, , (d) 5 × 10–3, volumetric, , Q.11 A wire of cross sectional area 3mm2 is just stretched between, two fixed points at a temperature of 20°C. Determine the, tension when the temperature falls to 20°C. Coefficient of, linear expansion a = 10–5 /°C and Y = 2 × 1011 N/m2., (a) 120 KN (b) 20 N, , RESPONSE, GRID, , (c) 120 N, , (c) 7.5 × 10–3, , (d) 1.5 × 10–2, , (a) 5.5 × 104 N/m2, , (b) 1.8 × 106 N/m2, , (c) 2.2 × 108 N/m2, , (d) 2.0 × 1011 N/m2, , Q.14 For a given material, the Young’s modulus is 2.4 times that, of rigidity modulus. It’s poisson’s ratio is, (a) 1.2, , (b) 1.02, , (c) 0.2, , (d) 2, , Q.15 A wire of length 1m is stretched by a force of 10N. The, area of cross-section of the wire is 2 × 10–6 m2 and Y is, 2 × 1011N/m2. Increase in length of the wire will be (a) 2.5 × 10–5 cm, , (b) 2.5 × 10–5 mm, , (c) 2.5 × 10–5 m, , (d) None of these, , Q.16 A stress of 1kg/mm2 is applied on, , a wire. If the modulus of, elasticity of the wire is 1010 dyne/cm2, then the percentage, increase in the length of the wire will be, , Q.10 A copper rod 2m long is stretched by 1mm. Strain will be (a) 10–4, volumetric, , (b) 5.6 × 10–4, , Q.13 The Young’s modulus of steel is 2 × 1011 N/m2 and its, coefficient of linear expansion is 1.1 × 10–5 per deg. The, pressure to be applied to the ends of a steel cylinder to, keep its length constant on raising its temperature by 100ºC,, will be -, , Q.9 A 4.0 m long copper wire of cross sectional area 1.2 cm 2 is, stretched by a force of 4.8 × 103 N stress will be 4.0 × 107 N/mm2, , (a) 3.3 × 10–5, , (a) 0.007, , (b) 0.0098, , (c) 98, , (d) 9.8, , 7800kg/m3, , Q.17 A uniform steel wire of density, is 2.5 m long, and weighs 15.6 × 10–3 kg. It extends by 1.25 mm when, loaded by 8kg. Calculate the value of young’s modulus of, elasticity for steel., , (d) 102 N, , (a) 1.96 × 1011 N/m2, , (b) 19.6 × 1011 N/m2, , (c) 196 × 1011 N/m2, , (d) None of these, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 20, , 3, (2) Reciprocal of bulk modulus of elasticity is called, compressibility, , Q.18 A metallic wire is suspended by suspending weight to it. If S, is longitudinal strain and Y its young's modulus of elasticity, then potential energy per unit volume will be, (a), , 1 2 2, YS, 2, , (b), , (c), , 1 2, YS, 2, , (d) 2YS2, , 1 2, YS, 2, , (4) It is difficult to twist a long rod as compared to small, rod, Q.23 Which statements are false for a metal?, , Q.19 The lengths and radii of two wires of same material are, respectively L, 2L, and 2R, R. Equal weights are applied on, then. If the elongations produced in them are l1 and l2, respectively then their ratio will be, (a) 2 : 1, , (b) 4 : 1, , (c) 8 : 1, , (d) 1 : 8, , (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 4, , (d) 4 : 1, , Q.21 A rod of length l and area of cross-section A is heated from, 0°C to 100°C. The rod is so placed that it is not allowed to, increase in length, then the force developed is proportional to, (b) l –1, , Y<h, , (2), , Y=h, , (3), , Y < 1/ h, , (4), , Y>h, , Q.24 Which of the following relations are false, , (3) K =, , (b) 1 and 2 are correct, , (c), , 2 and 4 are correct, , (d) 1 and 3 are correct, , Q.22 Mark the correct statements, (1) Sliding of molecular layer is much easier than, compression or expansion, , (4) s =, , 0.5Y - h, h, , q, , F, , F, , Q.25 The tensile stress at this plane in terms of F, A and q is, (a), , 1, 2 and 3 are correct, , 9hY, Y+h, , A bar of cross section A is subjected to equal and opposite tensile, forces F at its ends. Consider a plane through the bar making an, angle q with a plane at right angles to the bar as shown in figure., , Codes :, (a), , (2) s = ( 6K + h) Y, , DIRECTIONS (Q.25-Q.27) : Read the passage given below and, answer the questions that follows :, , (d) A–1, , (c) A, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, , RESPONSE, GRID, , (1), , (1) 3Y = K (1 - s ), , Q.20 The ratio of radii of two wires of same material is 2 : 1. If, these wires are stretched by equal forces, then the ratio of, stresses produced in them will be, , (a) l, , (3) Hollow shaft is much stronger than a solid rod of same, length and same mass, , Fcos2 q, A, , (b), , F, A cos2 q, , F, Fsin 2 q, (d), A sin 2 q, A, Q.26 In the above problem, for what value of q is the tensile, stress maximum ?, (c), , (a) Zero, , (b) 90°, , 18., , 19., , 20., , 21., , 23., , 24., , 25., , 26., , Space for Rough Work, , (c) 45°, , 22., , (d) 30°
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t.me/Magazines4all, DPP/ P 20, , 4, Q.27 The shearing stress at the plane, in terms of F, A and q is, (a), , Fcos 2q, 2A, , (b), , Fsin 2q, 2A, , (c), , Fsin q, A, , (d), , F cos q, A, , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , 28., , (d), , Statement -1 is True, Statement-2 is False., Statement -2 : Under given deforming force, steel is, deformed less than rubber., , (a), , 27., , Statement -1 is False, Statement-2 is True., , Q.28 Statement -1 : Steel is more elastic than rubber., , DIRECTIONS (Qs. 28-Q.30) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason). Each, of these questions has four alternative choices, only one of which is, the correct answer. You have to select the correct choice., , RESPONSE GRID, , (c), , Q.29 Statement -1 : Bulk modulus of elasticity (K) represents, incompressibility of the material., Statement -2 : Bulk modulus of elasticity is proportional, to change in pressure., Q.30 Statement -1 :The bridges declared unsafe after a long use., Statement -2 : Elastic strength of bridges losses with time., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 20 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 32, Qualifying Score, 52, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 21, SYLLABUS : Fluid Mechanics, , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The force required to separate two glass plates of area, 10–2m–2 with a film of water 0.05 m thick between them, is, (Surface tension of water is 70 × 10–3 N/m), (a) 28 N, (b) 14 N, (c) 50 N, (d) 38 N, Q.2 A thin metal disc of radius r floats on water surface and bends, the surface downwards along the perimeter making an angle q, with vertical edge of the disc. If the disc displaces a weight of, water W and surface tension of water is T, then the weight of, metal disc is, (a) 2prT + W, (b) 2prT cos q - W, (c) 2prT cos q + W, (d) W - 2prT cos q, , RESPONSE GRID, , 1., , 2., , Q.3 The amount of work done in blowing a soap bubble such that its, diameter increases from d to D is (T = surface tension of the, solution), (a), , 4p( D 2 - d 2 )T, , (b) 8p( D 2 - d 2 )T, , (c) p( D 2 - d 2 )T, (d) 2p( D 2 - d 2 )T, Q.4 A film of water is formed between two straight parallel, wires of length 10 cm each separated by 0.5 cm. If their, separation is increased by 1 mm while still maintaining their, parallelism, how much work will have to be done (Surface, tension of water = 70 × 10–2 N/m), (a) 7.22 × 10–6 Joule, (b) 1.44 × 10–5 Joule, (c) 2.88 × 10–5 Joule, (d) 5.76 × 10–5 Joule, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 21, , 2, Q.5 The liquid meniscus in capillary tube will be convex, if the, angle of contact is, (a) Greater than 90°, (b) Less than 90°, (c) Equal to 90°, (d) Equal to 0°, Q.6 Two soap bubbles of radii r1 and r2 equal to 4 cm and 5 cm, are touching each other over a common surface S1 S2, (shown in figure). Its radius will be, S1, (a) 4 cm, 4 cm, 5 cm, (b) 20 cm, (c) 5 cm, S2, (d) 4.5 cm, Q.7 The radii of two soap bubbles are r1 and r2 . In isothermal, conditions, two meet together in vaccum. Then the radius, R of the resultant bubble is given by, (a) R = (r1 + r2 ) / 2, (b) R = r1 ( r1r2 + r2 ), (c) R 2 = r12 + r22, (d) R = r1 + r2, Q.8 Two parallel glass plates are dipped partly in the liquid of, density ' d ' keeping them vertical. If the distance between, the plates is ' x ' , surface tension for liquids is T and angle, of contact is q , then rise of liquid between the plates due, to capillary will be, 2T cos q, 2T, T cos q, T cos q, (a), (d), (b), (c), xdg cos q, xdg, xdg, xd, , Q.11 Which graph represents the variation of surface tension, with temperature over small temperature ranges for water ?, , (a), , (b), , S.T., , Temp., , Temp., , (c), , S.T., , (d), , S.T., , Temp., , Temp., , Q.12 A solid sphere of density h ( > 1 ) times lighter than water, is suspended in a water tank by a string tied to its base as, shown in fig. If the mass of the sphere is m then the tension, in the string is given by, (a), , æ h - 1ö, çè h ÷ø mg, , (b) hmg, (c), (d), , mg, h -1, ( h - 1)mg, , Q.13 A candle of diameter d is floating on a liquid in a cylindrical, container of diameter D ( D >> d ) as shown in figure. If it, , Q.9 A capillary tube of radius R is immersed in water and water, rises in it to a height H . Mass of water in the capillary, tube is M . If the radius of the tube is doubled, mass of, water that will rise in the capillary tube will now be, (a) M, (b) 2M, (c) M / 2 (d) 4M, Q.10 In a surface tension experiment with a capillary tube water, rises upto 0.1 m. If the same experiment is repeated on an, artificial satellite, which is revolving around the earth, water, will rise in the capillary tube upto a heights of, (a) 0.1 m, (b) 0.2 m, (c) 0.98 m, (d) Full length of the capillary tube, , RESPONSE, GRID, , S.T., , is burning at the rate of 2 cm/ hour then the top of the candle, will, , L, L, d, , D, , (a), (b), (c), (d), , Remain at the same height, Fall at the rate of 1 cm/hour, Fall at the rate of 2 cm/hour, Go up the rate of 1 cm/hour, , 5., , 6., , 7., , 8., , 10., , 11., , 12., , 13., , Space for Rough Work, , 9., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 21, , 3, circular hole of radius R at a depth 4y from the top. When, the tank is completely filled with water the quantities of, water flowing out per second from both the holes are the, same. Then R is equal to, , Q.14 A viscous fluid is flowing through a cylindrical tube. The, velocity distribution of the fluid is best represented by the, diagram, (b), , (a), , (a), , (d) None of these, , (c), , Q.15 When a body falls in air, the resistance of air depends to a, great extent on the shape of the body, 3 different shapes, are given. Identify the combination of air resistances which, truly represents the physical situation. (The cross sectional, areas are the same)., , (a), (c), , R, , R, , R, , W, (1), Disc, , W, (2), Ball, , W, (3), Cigar shaped, , 1< 2 < 3, 3 < 2 <1, , (b), (d), , (a), , 5, d, 4, , d, , H/2, , 2d, , (b), , 4, d, 5, , (d), , 15., , 19., , 20., , (d), , L, 2p, , (b) 50.5 m 2 / s2, , 3m, , (c) 51 m 2 / s 2, 2, , 52.5 cm, , 2, , (d) 52 m / s, Q.19 An incompressible liquid flows through a horizontal tube, as shown in the following fig. Then the velocity v of the, fluid is, v2 = 1.5 m/s, , A, 1.5A, , 3L/4, , 14., , 2p, , (c) L, , (a) 50 m 2 / s 2, , v1 = 3 m/s, , d, 5, , Q.17 A large open tank has two holes in the wall. One is a square, hole of side L at a depth y from the top and the other is a, , RESPONSE, GRID, , L, , Q.18 Water is filled in a cylindrical container to a height of 3m., The ratio of the cross-sectional area of the orifice and the, beaker is 0.1. The square of the speed of the liquid coming, out from the orifice is (g = 10m/s2), , 2 < 3 <1, 3 <1< 2, , (c) d, , (b), , A, , Q.16 A homogeneous solid cylinder of length L( L < H / 2) ., Cross-sectional area A/5 is immersed such that it floats, with its axis vertical at the liquid-liquid interface with length, L/4 in the denser liquid as shown in the fig. The lower, density liquid is open to atmosphere having pressure P0., Then density of solid is given by, H/2, , 2pL, , v, , (a) 3.0 m/s (b) 1.5 m/s, (c) 1.0 m/s, (d) 2.25 m/s, Q.20 Radius of a capillary tube is 2 × 10–3m. A liquid of weight, 6.28 × 10 –4 N may remain in the capillary tube then the, surface tension of liquid will be, (a) 5 × 10–3 N/m, (b) 5 × 10–2 N/m, (c) 5 N/m, (d) 50 N/m, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, , 16., , Space for Rough Work, , 17., , 18.
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t.me/Magazines4all, DPP/ P 21, , 4, Q.21 The temperature at which the surface tension of water is, zero, (1) 370°C, (2) 0°C, (3) Slightly less than 647 K, (4) 277 K, Q.22 Which of the following statements are true in case when, two water drops coalesce and make a bigger drop?, (1) Energy is released., (2) Energy is absorbed., (3) The surface area of the bigger drop is smaller than the, sum of the surface areas of both the drops., (4) The surface area of the bigger drop is greater than the, sum of the surface areas of both the drops., Q.23 An air bubble in a water tank rises from the bottom to the, top. Which of the following statements are true?, (1) Bubble rises upwards because pressure at the bottom, is greater than that at the top., (2) As the bubble rises, its size increases., (3) Bubble rises upwards because pressure at the bottom, is less than that at the top., (4) As the bubble rises, its size decreases., DIRECTIONS (Q.24-Q.25) : Read the passage given below, and answer the questions that follows :, There is a small mercury drop of radius 4.0mm. A surface P of area, 1.0 mm2 is placed at the top of the drop. Atmospheric pressure =, 105 Pa. Surface tension of mercury = 0.465 N/m. Gravity effect is negligible., , RESPONSE, GRID, , Q.24 The force exerted by air on surface P is, (a) 0.1 N, (b) 1.0023 N (c) 105 N, (d) 1.0 N, Q.25 The force exerted by mercury drop on the surface P is, (a) 0.1 N, (b) 1.0023 N, (c) 0.00023 N, (d) 0.10023 N, DIRECTIONS (Qs. 26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.26 Statement-1 : A large soap bubble expands while a small, bubble shrinks, when they are connected to each other by a, capillary tube., Statement-2: The excess pressure inside bubble (or drop), is inverse ly proportional to the radius., Q.27 Statement-1 : Bernoulli’s theorem holds for, incompressible, non-viscous fluids., v2, Statement-2 : The factor, is called velocity head., 2g, Q.28 Statement-1 : The velocity increases, when water flowing, in broader pipe enter a narrow pipe., Statement-2 : According to equation of continuity, product, of area and velocity is constant., , 21., , 22., , 23., , 26., , 27., , 28., , 24., , 25., , DAILY PRA CTICE PROBLEM SHEET 21 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , SYLLABUS : Thermal Expansion, Calorimetry and Change of State, , Max. Marks : 116, , 22, Time : 60 min., , GENERAL INSTRUCTIONS, •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A glass flask is filled up to a mark with 50 cc of mercury at, 18°C. If the flask and contents are heated to 38°C, how much, mercury will be above the mark? (a for glass is 9 × 10–6/°C, and coefficient of real expansion of mercury is 180 × 10–6/, °C), (a) 0.85 cc (b) 0.46 cc (c) 0.153 cc (d) 0.05 cc, Q.2 The coefficient of apparent expansion of mercury in a glass, vessel is 153 × 10–6/°C and in a steel vessel is 144 × 10–6/, °C. If a for steel is 12 × 10–6/°C, then that of glass is, (a) 9 × 10–6/°C, (b) 6 × 10–6/°C, –6, (c) 36 × 10 /°C, (d) 27 × 10–6/°C, , RESPONSE GRID, , 1., , 2., , Q.3 An iron tyre is to be fitted on to a wooden wheel 1m in, diameter. The diameter of tyre is 6mm smaller than that of, wheel. The tyre should be heated so that its temperature, increases by a minimum of (the coefficient of cubical, expansion of iron is 3.6 × 10–5/°C), (a) 167°C (b) 334°C (c) 500°C (d) 1000°C, Q.4 A rod of length 20 cm is made of metal. It expands by, 0.075 cm when its temperature is raised from 0°C to, 100°C. Another rod of a different metal B having the same, length expands by 0.045 cm for the same change in, temperature. A third rod of the same length is composed, of two parts, one of metal A and the other of metal B. This, rod expands by 0.060 cm for the same change in, temperature. The portion made of metal A has the length, (a) 20 cm (b) 10 cm (c) 15 cm (d) 18 cm, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 22, , 2, Q.5 A glass flask of volume one litre at 0°C is filled, level full, of mercury at this temperature. The flask and mercury are, now heated to 100°C. How much mercury will spill out, if, , coefficient of volume expansion of mercury is 1.82 × 10–, 4/°C and, , linear expansion of glass is 0.1 × 10–4/°C?, (a) 21.2 cc (b) 15.2 cc (c) 1.52 cc (d) 2.12 cc, Q.6 The apparent coefficient of expansion of a liquid when, heated in a copper vessel is C and when heated in a silver, vessel is S. If A is the linear coefficient of expansion of, copper, then the linear coefficient of expansion of silver, is, (a), (c), , C + S - 3A, 3, S + 3A - C, 3, , C + 3A - S, 3, C - S + 3A, (d), 3, , (b), , Q.7 The coefficient of volumetric expansion of mercury is 18, × 10–5/°C. A thermometer bulb has a volume 10–6 m3 and, cross section of stem is 0.004 cm2. Assuming that bulb is, filled with mercury at 0°C then the length of the mercury, column at 100°C is, (a) 18.8 mm (b) 9.2 mm (c) 7.4 cm (d) 4.5 cm, Q.8 A piece of metal weight 46 gm in air, when it is immersed, in the liquid of specific gravity 1.24 at 27°C it weighs 30, gm. When the temperature of liquid is raised to 42°C the, metal piece weighs 30.5 gm, specific gravity of the liquid, at 42°C is 1.20, then the linear expansion of the metal will, be, (a) 3.316 × 10–5/°C, (b) 2.316 × 10–5/°C, –5, (c) 4.316 × 10 /°C, (d) None of these, Q.9 2 kg of ice at – 20°C is mixed with 5 kg of water at 20°C, in an insulating vessel having a negligible heat capacity., Calculate the final mass of water remaining in the container., It is given that the specific heats of water and ice are 1, kcal/kg /°C and 0.5 kcal/kg/°C while the latent heat of fusion, of ice is 80 kcal/kg, (a) 7 kg, (b) 6 kg, (c) 4 kg, (d) 2 kg, Q.10 A lead bullet at 27°C just melts when stopped by an, obstacle. Assuming that 25% of heat is absorbed by the, obstacle, then the velocity of the bullet at the time of, striking (M.P. of lead = 327°C, specific heat of lead =, , RESPONSE, GRID, , 0.03 cal/gm°C, latent heat of fusion of lead = 6 cal/gm, and J = 4.2 joule/cal), (a) 410 m/sec, (b) 1230 m/sec, (c) 307.5 m/sec, (d) None of the above, Q.11 The temperature of equal masses of three different liquids, A, B and C are 12°C, 19°C and 28°C respectively. The, temperature when A and B are mixed is 16°C and when B, and C are mixed is 23°C, The temperature when A and C, are mixed is, (a) 18.2°C (b) 22°C, (c) 20.2°C (d) 25.2°C, Q.12 50 gm of copper is heated to increase its temperature by, 10°C. If the same quantity of heat is given to 10 gm of, water, the rise in its temperature is (Specific heat of copper, = 420 Joule-kg–1°C–1), (a) 5°C, (b) 6°C, (c) 7°C, (d) 8°C, Q.13 A beaker contains 200 gm of water. The heat capacity of, the beaker is equal to that of 20 gm of water. The initial, temperature of water in the beaker is 20°C. If 440 gm of, hot water at 92°C is poured in it, the final temperature, (neglecting radiation loss) will be nearest to, (a) 58°C, (b) 68°C, (c) 73°C, (d) 78°C, Q.14 One calorie is defined as the amount of heat required to, raise temperature of 1g of water by 1°C and it is defined, under which of the following condition, (a) From 14.5°C to 15.5°C at 760 mm of Hg, (b) From 98.5°C to 99.5°C at 760 mm of Hg, (c) From 13.5°C to 14.5°C at 76 mm of Hg, (d) From 3.5°C to 4.5°C at 76 mm of Hg, Q.15 A bullet moving with a uniform velocity v, stops suddenly, after hitting the target and the whole mass melts be m,, specific heat S, initial temperature 25°C, melting point, 475°C and the latent heat L. Then v is given by, 1 mv 2, 2 J, , (a) mL = mS (475 - 25) + ., (b) mS (475 - 25) + mL =, , mv 2, 2J, , (c) mS (475 - 25) + mL =, , mv 2, J, , (d) mS (475 - 25) - mL =, , mv 2, 2J, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 22, , 3, , Temperature (°C), , 250, 200, 150, 100, 50, 0, , 1, , 2, , 3 4 5 6, Time (Minute), , (a) 500 cal, 50°C, (c) 1500 cal, 200°C, , RESPONSE, GRID, , 7, , 8, , (b) 1000 cal, 100°C, (d) 1000 cal, 200°C, , DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Heat is supplied to a certain homogenous sample of matter,, at a uniform rate. Its temperature is plotted against time,, as shown. Which of the following conclusions can be, drawn?, Temperature, , Q.16 A stationary object at 4°C and weighing 3.5 kg falls from a, height of 2000 m on a snow mountain at 0°C. If the, temperature of the object just before hitting the snow is, 0°C and the object comes to rest immediately? (g = 10m/, s2 and latent heat of ice = 3.5 × 105 joule/sec), then the, object will melt, (a) 2 kg of ice, (b) 200 gm of ice, (c) 20 gm of ice, (d) 2 gm of ice, Q.17 Density of a substance at 0°C is 10 gm/cc and at 100°C,, its density is 9.7 gm/cc. The coefficient of linear expansion, of the substance will be, (a) 102, (b) 10–2, (c) 10–3, (d) 10–4, Q.18 The real coefficient of volume expansion of glycerine is, 0.000597 per °C and linear coefficient of expansion of, glass is 0.000009 per°C. Then the apparent volume, coefficient of expansion of glycerine is, (a) 0.000558 per°C, (b) 0.00057 per°C, (c) 0.00027 per°C, (d) 0.00066 per°C, Q.19 A constant volume gas thermometer shows pressure, reading of 50 cm and 90 cm of mercury at 0°C and 100°C, respectively. When the pressure reading is 60 cm of, mercury, the temperature is, (a) 25°C, (b) 40°C, (c) 15°C, (d) 12.5°C, Q.20 A student takes 50gm wax (specific heat = 0.6 kcal/kg°C), and heats it till it boils. The graph between temperature and, time is as follows. Heat supplied to the wax per minute and, boiling point are respectively., , Time, , (1) Its specific heat capacity is greater in the liquid state, than in the solid state, (2) Its latent heat of vaporization is greater than its latent, heat of fusion, (3) Its specific heat capacity is greater in the solid state, than in the liquid state, (4) Its latent heat of vaporization is smaller than its latent, heat of fusion, Q.22 A bimetallic strip is formed out of two identical strips,, one of copper and other of brass. The coefficients of linear, expansion of the two metals are αC and α B . On heating,, the temperature of the strip goes up by DT and the strip, bends to form an arc of radius of curvature R. Then R is, (1) inversely proportional to DT, (2) proportional to α B - α C, (3) inversely proportional to α B - αC, (4) proportional to DT, Q.23 A bimetallic strip is heated. Choose wrong statements., (1) does not bend at all, (2) gets twisted in the form of an helix, (3) bends in the form of an arc with the more expandable, metal inside., (4) bend in the form of an arc with the more expandable, metal outside, , 16., , 17., , 18., , 21., , 22., , 23., , Space for Rough Work, , 19., , 20.
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t.me/Magazines4all, DPP/ P 22, , 4, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, In a thermally insulated tube of cross sectional area 4cm2 a liquid, of thermal expansion coefficient 10–3 K–1 is f lowing. Its, velocity at the entrance is 0.1 m/s. At the middle of the tube a, heater of a power of 10kW is heating the liquid. The specific, heat capacity of the liquid is 1.5 kJ/(kg K), and its density is, 1500 kg/m3 at the entrance., Q.24 The rise in temperature of the liquid as it pass through the, tube is, (a), , 1000, °C, 9, , (b), , 1, °C, 9, , (c), , 500, °C, 9, , (d) None, , Q.25 What is the density of liquid at the exit ?, (a) 1450 kg/m3, (b) 1400 kg/m3, 3, (c) 1350 kg/m, (d) None of these, Q.26 How much bigger is the volume rate of flow at the end of, the tube than at the entrance in cubic meters ?, (a) 9 × 10–5, (c), , (b), , 4, × 10–5, 9, , RESPONSE, GRID, , 1, × 10–5, 3, , (d) None, , 24., , 25., , DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select, the correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : Fahrenheit is the smallest unit measuring, temperature., Statement-2 : Fahrenheit was the first temperature scale, used for measuring temperature., Q.28 Statement-1 : A brass disc is just fitted in a hole in a steel, plate. The system must be cooled to loosen the disc from, the hole., Statement-2 : The coefficient of linear expansion for brass, is greater than the coefficient of linear expansion for steel., Q.29 Statement-1 : Laten t heat of fusion of ice is, 336000 J kg–1., Statement-2 : Latent heat refers to change of state without, any change in temperature., , 26., , 27., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 22 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 23, SYLLABUS : Heat transfer & Newton’s law of cooling, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., , Se, , od, , Q.1 Two rods (one semi-circular and other straight) of same, material and of same cross-sectional area are joined as, shown in the figure. The points A and B are maintained at, different temperature. The ratio of the heat transferred, through a cross-section of a semi-circular rod to the heat, transferred through a cross section of the straight rod in a, given time is, circular r, mi, (a) 2 : p, (b) 1 : 2, (c) p : 2, Straight rod, A, B, (d) 3 : 2, , RESPONSE GRID, , 1., , 2., , Q.2 A wall is made up of two layers A and B. The thickness of, the two layers is the same, but materials are different. The, thermal conductivity of A is double than that of B. In thermal, equilibrium the temperature difference between the two, ends is 36°C. Then the difference of temperature at the, two surfaces of A will be, (a) 6°C, (b) 12°C, (c) 18°C, (d) 24°C, Q.3 A room is maintained at 20°C by a heater of resistance 20, ohm connected to 200 volt mains. The temperature is, uniform through out the room and heat is transmitted, through a glass window of area 1m2 and thickness 0.2 cm., What will be the temperature outside? Given that thermal, conductivity K for glass is 0.2 cal/m/°C/sec and J = 4.2 J/, cal, (a) 15.24°C, (b) 15.00°C, (c) 24.15°C, (d) None of these, , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 23, , 2, Q.4 A composite metal bar of uniform section is made up of, length 25 cm of copper, 10 cm of nickel and 15 cm of, aluminium. Each part being in perfect thermal contact with, the adjoining part. The copper end of the composite rod is, maintained at 100°C and the aluminium end at 0°C. The, whole rod is covered with belt so that there is no heat loss, occurs at the sides. If KCu = 2KAl and KAl = 3KNi, then, what will be the temperatures of Cu – Ni and Ni – Al, junctions respectively, Cu, , Ni, , Al, , 100° C, 0° C, (a) 23.33°C and 78.8°C (b) 83.33°C and 20°C, (c) 50°C and 30°C, (d) 30°C and 50°C, Q.5 Three rods of the same dimension have thermal, conductivities 3K, 2K and K. They are arranged as shown, in fig. with their ends at 100°C, 50°C and 20°C. The, temperature of their junction is, 50°C, , (a) 60°C, , 2K, 100°C, , (b) 70°C, , 3K, , (c) 50°C, , K, , Q.9 A cylindrical rod with one end in a steam chamber and the, other end in ice results in melting of 0.1 gm of ice per, second. If the rod is replaced by another with half the length, and double the radius of the first and if the thermal, conductivity of material of second rod is, , the rate at which ice melts in gm/sec will be, (a) 3.2, (b) 1.6, (c) 0.2, (d) 0.1, Q.10 An ice box used for keeping eatable cold has a total wall, area of 1 metre2 and a wall thickness of 5.0 cm. The thermal, conductivity of the ice box is K = 0.01 joule/metre °C. It, is filled with ice at 0° C along with eatables on a day when, the temperature is 30°C. The latent heat of fusion of ice is, 334 × 103 joules/kg. The amount of ice melted in one day, is (1 day = 86,400 seconds), (a) 776 gm, (b) 7760 gm, (c) 11520 gm, (d) 1552 gm, Q.11 A solid copper sphere (density r and specific heat capacity, c) of radius r at an initial temperature 200 K is suspended, inside a chamber whose walls are at almost 0 K. The time, required (in µ s) for the temperature of the sphere to drop, to 100 K is, , 20°C, , (d) 35°C, Q.6 A black body is at a temperature of 2880 K. The energy of, radiation emitted by this object with wavelength between, 499 nm and 500 nm is U1, between 999 nm and 1000 nm, is U2 and between 1499 nm and 1500 nm is U3. The Wein's, constant b = 2.88 × 106 nm K. Then, (a) U1 = 0 (b) U3 = 0 (c) U1 > U2 (d) U2 > U1, Q.7 A body initially at 80° C cools to 64° C in 5 minutes and, to 52° C in 10 minutes. The temperature of the body after, 15 minutes will be, (a) 42.7° C (b) 35° C, (c) 47°, (d) 40° C, Q.8 A 5 cm thick ice block is there on the surface of water in a, lake. The temperature of air is – 10° C; how much time it, will take to double the thickness of the block, (L = 80 cal/g, Kice = 0.004 erg/s-k, dice = 0.92 g cm–3), (a) 1 hour, (b) 191 hours, ., (c) 19.1 hours, (d) 1.91 hours, , RESPONSE, GRID, , 1, that of first,, 4, , (a), , 72 r rc, 7 s, , (b) 7 r rc, 72 s, , (c) 27 rrc, 7 s, , (d) 7 rrc, 27 s, , Q.12 Four rods of identical cross-sectional area and made from, the same metal form the sides of square. The temperature, of two diagonally opposite points are T and 2T, respectively in the steady state. Assuming that only heat, conduction takes place, what will be the temperature, difference between other two points, 2, 2 +1, T, (a), (b), T, 2 +1, 2, (c) 0, (d) None of these, Q.13 Consider two hot bodies B1 and B2 which have temperature, 100° C and 80° C respectively at t = 0. The temperature, of surroundings is 40° C. The ratio of the respective rates, of cooling R1 and R2 of these two bodies at t = 0 will be, (a) R1 : R2 = 3 : 2, (b) R1 : R2 = 5 : 4, (c) R1 : R2 = 2 : 3, (d) R1 : R2 = 4 : 5, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 23, , 3, , Q.14 A body cools from 60° C to 50° C in 10 minutes. If the, room temperature is 25° C and assuming Newton's law of, cooling to hold good, the temperature of the body at the, end of the next 10 minutes will be, (a) 38.5° C (b) 40° C, (c) 42.85° C (d) 45° C, Q.15 The rates of cooling of two different liquids put in exactly, similar calorimeters and kept in identical surroundings are the, same if, (a) The masses of the liquids are equal, (b) Equal masses of the liquids at the same temperature, are taken, (c) Different volumes of the liquids at the same, temperature are taken, (d) Equal volumes of the liquids at the same temperature, are taken, Q.16 For cooking the food, which of the following type of, utensil is most suitable, (a) High specific heat and low conductivity, (b) High specific heat and high conductivity, (c) Low specific heat and low conductivity, (d) Low specific heat and high conductivity, Q.17 Two rods A and B are of equal lengths. Their ends are kept, between the same temperature and their area of crosssections are A1 and A2 and thermal conductivities K1 and, K2. The rate of heat transmission in the two rods will be, equal, if, (a) K1 A2 = K 2 A1, (b) K1 A1 = K2 A2, (c) K1 = K2, (d) K1 A12 = K 2 A22, Q.18 While measuring the thermal conductivity of a liquid, we, keep the upper part hot and lower part cool, so that, (a) Convection may be stopped, (b) Radiation may be stopped, (c) Heat conduction is easier downwards, (d) It is easier and more convenient to do so, Q.19 When fluids are heated from the bottom, convection, currents are produced because, (a) Molecular motion of fluid becomes aligned, (b) Molecular collisions take place within the fluid, (c) Heated fluid becomes more dense than the cold fluid, above it, (d) Heated fluid becomes less dense than the cold fluid, above it, , RESPONSE, GRID, , Q.20 If between wavelength l and l + dl, el and al be the, emissive and absorptive powers of a body and El be the, emissive power of a perfectly black body, then according, to Kirchoff’s law, which is true, (a) el = al = El, (b) el El = al, (c) el = al El, (d) el al El = constant, Q.21 Two thermometers A and B are exposed in sunlight. The, bulb of A is painted black, But that of B is not painted. The, correct statement regarding this case is, (a) Temperature of A will rise faster than B but the final, temperature will be the same in both, (b) Both A and B show equal rise in beginning, (c) Temperature of A will remain more than B, (d) Temperature of B will rise faster, DIRECTIONS (Q.22-Q.24) : In the following questions, more, than one of the answers given are correct. Select the correct, answers and mark it according to the following codes:, Codes :, (a) 1, 2 and 3 are correct (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 Two bodies A and B have thermal emissivities of 0.01 and, 0.81 respectively. The outer surface areas of the two bodies, are the same. The two bodies emit total radiant power at, the same rate. The wavelength λ B corresponding to, maximum spectral radiancy in the radiation from B is, shifted from the wavelength corresponding to maximum, spectral radiancy in the radiation from A, by 1.00 mm. If, the temperature of A is 5802 K, (1) The temperature of B is 1934 K, (2) lB = 1.5 mm, (3) The temperature of B is 11604 K, (4) The temperature of B is 2901 K, Q.23 A cane is taken out from a refrigerator at 0°C. The, atmospheric temperature is 25°C. If t1 is the time taken to, heat from 0°C to 5°C and t2 is the time taken from 10°C, to 15°C, then the wrong statements are, (1) t1 > t2, (2) t1 = t2, (3) There is no relation, (4) t1 < t2, , 14., , 15., , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., , Space for Rough Work
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t.me/Magazines4all, DPP/ P 23, , 4, Q.24 The rate of loss of heat from a body cooling under conditions, of forced convection is proportional to its, (1) surface area, (2) excess of temperature over that of surrounding, (3) heat capacity, (4) absolute temperature, DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, A brass ball of mass 100g is heated to 100°C and then dropped, into 200g of turpentine in a calorimeter at 15°C. The final, temperature is found to be 23°C. Take specific heat of brass as, 0.092 cal/g°C and water equivalent of calorimeter as 4g., Q.25 The specific heat of turpentine is, (a) 0.42 cal/g°c, (b) 0.96 cal/g°c, (c) 0.72 cal/g°c, (d) 0.12 cal/g°c, Q.26 Heat lost by the ball is approximately, (a) 810 cal, (b) 610 cal, (c) 710 cal, (d) 510 cal, Q.27 Heat gained by turpentine and calorimeter is approximately, (a) 810 cal, (b) 610 cal, (c) 710 cal, (d) 510 cal, , RESPONSE, GRID, , 24., , 25., , 29., , 30., , DIRECTIONS (Q.28-Q.30) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason). Each, of these questions has four alternative choices, only one of which is, the correct answer. You have to select the correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.28 Statement-1 : The equivalent thermal conductivity of two, plates of same thickness in contact (series) is less than, the smaller value of thermal conductivity., Statement-2 : For two plates of equal thickness in contact, (series) the equivalent thermal conductivity is given by, 2, 1, 1, =, +, K K1 K 2, , Q.29 Statement-1 : The absorbance of a perfect black body is unity., Statement-2 : A perfect black body when heated emits, radiations of all possible wavelengths at that temperature., Q.30 Statement-1 : As temperature of a black body is raised,, wavelength corresponding to maximum energy reduces., Statement-2 : Higher temperature would mean higher, energy and hence higher wavelength., , 26., , 27., , 28., , DAILY PRA CTICE PROBLEM SHEET 23 - PHYSICS, Total Questions, 30, Total Marks, 120, Attem pted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 24, SYLLABUS : Thermodynamics-1 (Thermal equilibrium, zeroth law of thermodynamics,, concept of temperature, Heat, work and internal energy, Different thermodynamic processes), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice questions., Each question has 4 choices (a), (b), (c) and (d), out of which ONLY, ONE choice is correct., , Q.1 For an isothermal expansion of a perfect gas, the value of, DP, is, P, DV, DV, 1/ 2 DV, 2 DV, (b) (c) -g, (d) -g, (a) -g, V, V, V, V, Q.2 When an ideal gas in a cylinder was compressed, isothermally by a piston, the work done on the gas was, found to be 1.5 × 104 Joule. During this process about, (a) 3.6 × 103 cal of heat flowed out from the gas, (b) 3.6 × 103 cal of heat flowed into the gas, , RESPONSE GRID, , 1., , 2., , (c) 1.5 × 104 cal of heat flowed into the gas, (d) 1.5 × 104 cal of heat flowed out from the gas, Q.3 The latent heat of vaporisation of water is 2240 J/gm. If, the work done in the process of expansion of 1 g of water, is 168 J, then increase in internal energy is, (a) 2408 J, (b) 2240 J, (c) 2072 J, (d) 1904 J, Q.4 One mole of an ideal gas expands at a constant temperature, of 300 K from an initial volume of 10 litres to a final, volume of 20 litres. The work done in expanding the gas is, ( R = 8.31 J/mole - K), (a) 750 Joules, (c) 1500 Joules, , 3., Space for Rough Work, , 4., , (b) 1728 Joules, (d) 3456 Joules
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t.me/Magazines4all, DPP/ P 24, , 2, Q.5 The pressure in the tyre of a car is four times the, atmospheric pressure at 300 K. If this tyre suddenly bursts,, , Q.11 For an adiabatic expansion of a perfect gas, the value of, DP, is equal to, P, , its new temperature will be ( g = 1.4), -0.4 /1.4, , (b) 300 æç 1 ö÷, è4ø, 0.4, /1.4, (d) 300(4)-0.4 /1.4, (c) 300(2), Q.6 A monoatomic gas ( g = 5 / 3) is suddenly compressed to, (a) 300(4)1.4 / 0.4, , 1, of its original volume adiabatically, then the pressure, 8, of the gas will change to, 24, (a), (b) 8, 5, 40, (c), (d) 32 times its initial pressure, 3, 8, Q.7 An ideal gas at 27°C is compressed adiabatically to, of, 27, 5, its original volume. If g = , then the rise in temperature, 3, is, (a) 450 K (b) 375 k (c) 225 K (d) 405 K, Q.8 A given system undergoes a change in which the work done, by the system equals the decrease in its internal energy., The system must have undergone an, (a) Isothermal change, (b) Adiabatic change, (c) Isobaric change, (d) Isochoric change, Q.9 Helium at 27° has a volume of 8 litres. It is suddenly, compressed to a volume of 1 litre. The temperature of the, , gas will be [ g = 5 / 3], (a) 108°C, (b) 9327°C (c) 1200°C (d) 927°C, Q.10 One mole of an ideal gas at an initial temperature of TK, does 6 R joules of work adiabatically. If the ratio of, specific heats of this gas at constant pressure and at, constant volume is 5/3, the final temperature of gas will, be, (a) (T + 2.4) K, , (b) (T - 2.4) K, , (c) (T + 4) K, , (d) (T - 4) K, , RESPONSE, GRID, , DV, DV, (b) V, V, DV, DV, (c) -g, (d) -g 2, V, V, Q.12 If 300 ml of gas at 27°C is cooled to 7°C at constant, pressure, then its final volume will be, (a) 540 ml, (b) 350 ml, (c) 280 ml, (d) 135 ml, , (a), , - g, , Q.13 A sample of gas expands from volume V1 to V2 . The amount, of work done by the gas is greatest when the expansion is, (a) isothermal, (b) isobaric, (c) adiabatic, (d) equal in all cases, Q.14 How much work to be done in decreasing the volume of, and ideal gas by an amount of 2.4 ´ 10-4 m3 at normal, temperature and constant normal pressure of 1 ´ 105 N / m 2, (a) 28 Joule, (b) 27 Joule, (c) 25 Joule, (d) 24 Joule, Q.15 One mole of a perfect gas in a cylinder fitted with a piston has, a pressure P , volume V and temperature T . If the temperature, is increased by 1 K keeping pressure constant, the increase in, volume is, 2V, V, V, (a), (b), (c), (d) V, 273, 91, 273, Q.16 Work done by 0.1 mole of a gas at 27o C to double its, volume at constant pressure is ( R = 2 cal mol -1o K -1 ), (a) 54 cal (b) 600 cal (c) 60 cal (d) 546 cal, Q.17 When an ideal diatomic gas is heated at constant pressure,, the fraction of the heat energy supplied which increases, the internal energy of the gas, is, 2, 3, 3, 5, (a), (b), (c), (d), 5, 5, 7, 7, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 24, , 3, , Q.18 When heat is given to a gas in an isothermal change, the, result will be, (a) external work done, (b) rise in temperature, (c) increase in internal energy, (d) external work done and also rise in temp., Q.19 An ideal gas expands isothermally from a volume V1 to, V2 and then compressed to original volume V1, adiabatically. Initial pressure is P1 and final pressure is, P3 . The total work done is W . Then, (a), , P3 > P1 ,W > 0, , (b), , P3 < P1 , W < 0, , (c) P3 > P1 ,W < 0, (d) P3 = P1 ,W = 0, Q.20 An ideal gas expands in such a manner that its pressure and, volume can be related by equation PV 2 = constant. During, this process, the gas is, (a) heated, (b) cooled, (c) neither heated nor cooled, (d) first heated and then cooled, Q.21 In the following P - V diagram two adiabatics cut two, isothermals at temperatures T1 and T2 (fig.). The value of, Va, will be, Vd, , DIRECTIONS (Q.22 - Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes : (a) 1, 2 and 3 are correct (b) 1 and 2 are correct, (c) 2 and 4 are correct (d) 1 and 3 are correct, Q.22 During the melting of a slab of ice at 273 K and one, atmospheric pressure, (1) Positive work is done on the ice-water system by the, atmosphere, (2) Positive work is done by ice-water system on the, atmosphere, (3) The internal energy of the ice-water system increases, (4) The internal energy of the ice-water system decreases, Q.23 One mole of an ideal monatomic gas is taken from A to C, along the path ABC. The temperature of the gas at A is T0., For the process ABC –, P, , P0, , b, , A, , V0, , B, , 2V0, , (2) Change in internal energy of the gas is, , 11, RT0, 2, , 11, RT0, 2, 13, RT0, (4) Heat absorbed by the gas is, 2, , (3) Heat absorbed by the gas is, , T1, , d, , T0, , (1) Work done by the gas is RT0, , P, , a, , C, , 2P0, , T2, , DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, Va Vd, , (a), , Vb, Vc, , RESPONSE, GRID, , (b), , Vc, Vb, , 18., , Vb Vc, , V, , Vd, (c), Va, , (d) VbVc, , 19., , In the figure n mole of a monoatomic ideal gas undergo the, process ABC as shown in the P-V diagram. The process AB is, isothermal and BC is isochoric. The temperature of the gas at A, is T0. Total heat given to the gas during the process ABC is, measured to be Q., , 20., , 23., , Space for Rough Work, , 21., , 22.
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t.me/Magazines4all, DPP/ P 24, , 4, , P, , A, , DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., , C, , B, 2V, , (a), , 6V, , Q.24 Temperature of the gas at C is equal to, (a) T0, (b) 3 T0, (c) 6 T0, (d) 2 T0, Q.25 Heat absorbed by the gas in the process BC, (a) 3nRT0, (b) nRT0, (c) 2nRT0, (d) 6nRT0, Q.26 The average molar heat capacity of the gas in process ABC, is, (a), , Q, nT0, , (b), , Q, 2nT0, , (c), , Q, 3nT0, , (d), , 2Q, nT0, , RESPONSE, GRID, , 24., , 25., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : The isothermal curves intersect each other, at a certain point., Statement-2 : The isothermal change takes place slowly,, so the isothermal curves have very little slope., Q.28 Statement-1 : In adiabatic compression, the internal, energy and temperature of the system get increased., Statement-2 : The adiabatic compression is a slow process., Q.29 Statement-1 : The specific heat of a gas in an adiabatic, process is zero and in an isothermal process is infinite., Statement-2 : Specific heat of a gas is directly proportional, to change of heat in system and inversely proportional to, change in temperature., , 26., , 27., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 24 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 25, SYLLABUS : Thermodynamics-2 (1st and 2nd laws of thermodynamics, Reversible &, irreversible processes, Carnot engine and its efficiency), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.19) : There are 19 multiple choice questions., Each question has 4 choices (a), (b), (c) and (d), out of which ONLY, ONE choice is correct., , Q.2 An ideal gas is taken from point A to the point B , as, shown in the P - V diagram, keeping the temperature, constant. The work done in the process is, P, , Q.1 Six moles of an ideal gas performs a cycle shown in figure., If the temperature TA = 600 K , TB = 800 K , TC = 2200 K, , A, , PA, , and TD = 1200 K , the work done per cycle is, P, B, , A, , B, , PB, , C, , O, D, , VA, , (a) ( PA - PB )(VB - VA ), T, , (a) 20 kJ, , RESPONSE GRID, , (b) 30 kJ, , 1., , (c) 40 kJ, , (d) 60 kJ, , (c), , 2., Space for Rough Work, , 1, ( PB - PA )(VB - V A ), 2, , VB, , (b), (d), , V, , 1, ( PB - PA )(VB + V A ), 2, 1, ( PB + PA )(VB - V A ), 2
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t.me/Magazines4all, DPP/ P 25, , 2, Q.3 In the diagrams (i) to (iv) of variation of volume with, changing pressure is shown. A gas is taken along the path, ABCD. The change in internal energy of the gas will be, V, , V, , D, , C, , D, , C, , (ii), , (i), A, V, , A, , B, P, , D, , V, , C, , (iii), , B, P, , D, , C, , (iv), A, , A, , B, , B, P, , P, , (a) Positive in all cases (i) to (iv), (b) Positive in cases (i), (ii) and (iii) but zero in case (iv), (c) Negative in cases (i), (ii) and (iii) but zero in case (iv), (d) Zero in all cases, Q.4 A monoatomic ideal gas, initially at temperature T1, is, enclosed in a cylinder fitted with a frictionless piston. The, gas is allowed to expand adiabatically to a temperature T2, by releasing the piston suddenly. If L1and L2 are the lengths, of the gas column before and after expansion respectively,, then T1 / T2 is given by, (a), , æ L1 ö, ç ÷, è L2 ø, , 2/3, , L1, L2, , (b), , (c), , L2, L1, , (d), , æ L2 ö, ç ÷, è L1 ø, , 2/3, , Q.5 A gas mixture consists of 2 moles of oxygen and 4 moles, argon at temperature T. Neglecting all vibrational modes,, the total internal energy of the system is, (a) 4 RT, (b) 15 RT (c) 9 RT, (d) 11 RT, Q.6 Two Carnot engines A and B are operated in succession., The first one, A receives heat from a source at T1 = 800 K, and rejects to sink at T2 K . The second engine B receives, heat rejected by the first engine and rejects to another sink, at T3 = 300 K . If the work outputs of two engines are equal,, then the value of T2 will be, (a) 100K, (b) 300K (c) 550K (d) 700 K, Q.7 A Carnot engine whose low temperature reservoir is at 7°C, has an efficiency of 50%. It is desired to increase the, efficiency to 70%. By how many degrees should the, temperature of the high temperature reservoir be increased, (a) 840K, (b) 280 K (c) 560 K (d) 380K, , RESPONSE, GRID, , Q.8 An ideal heat engine working between temperature T1 and, T2 has an efficiency h, the new efficiency of engine if both, the source and sink temperature are doubled, will be, h, (a), (b) h, (c) 2 h, (d) 3 h, 2, Q.9 Efficiency of a Carnot engine is 50% when temperature, of outlet is 500 K. In order to increase efficiency up to, 60% keeping temperature of intake the same what will be, temperature of outlet, (a) 200 K, (b) 400 K (c) 600 K (d) 800 K, Q.10 A scientist says that the efficiency of his heat engine which, operates at source temperature 120ºC and sink temperature, 27ºC is 26%, then, (a) It is impossible, (b) It is possible but less probable, (c) It is quite probable, (d) Data are incomplete, Q.11 The efficiency of Carnot’s engine operating between, reservoirs, maintained at temperatures 27°C and –123°C,, is, (a) 50%, (b) 24%, (c) 0.75% (d) 0.4%, Q.12 The temperature of sink of Carnot engine is 27°C and, Efficiency of engine is 25%. Then temperature of source, is, (a) 227°C, (b) 327°C (c) 127°C (d) 27°C, Q.13 In changing the state of thermodynamics from A to B state,, the heat required is Q and the work done by the system is, W. The change in its internal energy is, Q -W, (a) Q + W, (b) Q – W (c) Q, (d), 2, Q.14 The first law of thermodynamics is concerned with the, conservation of, (a) Momentum, (b) Energy, (c) Mass, (d) Temperature, Q.15 A system is given 300 calories of heat and it does 600, joules of work. The internal energy of the system change, in this process is, (J = 4.18 Joule/cal), (a) 654 Joule, (b) 156.5 Joule, (c) –300 Joule, (d) – 528.2 Joule, , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 25, , 3, , Q.16 110 J of heat is added to a gaseous system, whose internal, energy change is 40 J, then the amount of external work, done is, (a) 150 J, (b) 70 J, (c) 110 J (d) 40 J, Q.17 For free expansion of the gas which of the following is, true, (a) Q = W = 0 and DEint = 0, (b) Q = 0, W > 0 and DEint = -W, (c) W = 0, Q > 0 , and DEint = Q, (d) W > 0, Q < 0 and DEint = 0, , (2) That the process should be too fast, (3) That the process should be slow so that the working, substance should remain in thermal and mechanical, equilibrium with the surroundings, (4) The loss of energy should be zero and it should be, quasistatic, Q.22 One mole of an ideal gas is taken through the cyclic through, the cyclic process shown in the V-T diagram, where V =, volume and T = absolute temperature of the gas. Which of, the following statements are correct, V, , Q.18 In a given process for 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, Q.19 The specific heat of hydrogen gas at constant pressure is, , V0, , (1), (2), (3), (4), , RESPONSE, GRID, , 2T0, , T, , Heat is given out by the gas, Heat is absorbed by the gas, The magnitude of the work done by the gas is RT0 (ln 2), The magnitude of the work done by the gas is V0T0, , DIRECTIONS (Q.23-Q.25) : Read the passage given below and, answer the questions that follows :, V-T graph of a process of monoatomic ideal gas is as shown in, figure., , DIRECTIONS (Q.20-Q.22) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes : (a) 1, 2 and 3 are correct (b) 1 and 2 are correct, (c) 2 and 4 are correct (d) 1 and 3 are correct, Q.20 Which of the following processes are irreversible?, (1) Transfer of heat by radiation, (2) Electrical heating of a nichrome wire, (3) Transfer of heat by conduction, (4) Isothermal compression, Q.21 For a reversible process, unnecessary conditions are, (1) In the whole cycle of the system, the loss of any type, of heat energy should be zero, , C, , D, T0, , C p = 3.4 ´ 103 cal / kg oC and at constant volume is, , CV = 2.4 ´ 103 cal / kg oC . If one kilogram hydrogen gas is, heated from 10°C to 20°C at constant pressure, the, external work done on the gas to maintain it at constant, pressure is, (a) 105 cal, (b) 104 cal, (c) 103 cal, (d) 5 × 103 cal, , B, , A, , 2V0, , V, a, , b, , d, , c, T, , Q.23 Sum of work done by the gas in process abcd is –, (a) zero, (b) positive, (c) negative, (d) data is insufficient, Q.24 Heat is supplied to the gas in process(s) –, (a) da, ab and bc, (b) da and ab only, (c) da only, (d) ab and bc only, Q.25 Change in internal energy of the gas is zero in process(s) –, (a) da, ab and bc, (b) da and bc only, (c) da only, (d) da and ab only, , 16., , 17., , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , Space for Rough Work
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DPP/ P 25, , 4, DIRECTIONS (Q.26-Q.28) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason). Each, of these questions has four alternative choices, only one of which is, the correct answer. You have to select the correct choice., (a), (b), (c), (d), , t.me/Magazines4all, , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , RESPONSE GRID, , 26., , 27., , Q.26 Statement-1 : It is not possible for a system, unaided by an, external agency to transfer heat from a body at lower, temperature to another body at higher temperature., Statement-2 : According to Clausius statement, “No, process is possible whose sole result is the transfer of, heat from a cooled object to a hotter object., Q.27 Statement-1 : A room can be warmed by opening the door, of a refrigerator in a closed room., Statement-2 : Head flows from lower temperature, (refrigerator) to higher temperature (room)., Q.28 Statement-1 : In isothermal process whole of the heat, energy supplied to the body is converted into internal, energy., Statement-2 : According to the first law of themodynamics, DQ = DU + PDV, , 28., , DAILY PRA CTICE PROBLEM SHEET 25 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 26, SYLLABUS : Kinetic Theory, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice questions., Each question has 4 choices (a), (b), (c) and (d), out of which ONLY, ONE choice is correct., , Q.1 If pressure of a gas contained in a closed vessel is increased, by 0.4% when heated by 1°C, the initial temperature must, be, (a) 250 K, (b) 250°C (c) 2500 K (d) 25°C, Q.2 To double the volume of a given mass of an ideal gas at, 27°C keeping the pressure constant, one must raise the, temperature in degree centigrade to, (a) 54, (b) 270, (c) 327, (d) 600, Q.3 Under which of the following conditions is the law, PV = RT obeyed most closely by a real gas?, (a) High pressure and high temperature, (b) Low pressure and low temperature, , RESPONSE GRID, , 1., , 2., , (c) Low pressure and high temperature, (d) High pressure and low temperature, Q.4 The pressure P, volume V and temperature T of a gas in, the jar A and the other gas in the jar B at pressure 2 P ,, V, and temperature 2T, then the ratio of, the, 4, number of molecules in the jar A and B will be, (a) 1 : 1, (b) 1 : 2, (c) 2 : 1, (d) 4 : 1, Q.5 A flask is filled with 13 gm of an ideal gas at 27°C and its, temperature is raised to 52°C. The mass of the gas that has, to be released to maintain the temperature of the gas in, the flask at 52°C and the pressure remaining the same is, (a) 2.5 g, (b) 2.0 g, (c) 1.5 g, (d) 1.0 g, , volume, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 26, , 2, Q.6 The pressure and temperature of two different gases is, P and T having the volume V for each. They are mixed, keeping the same volume and temperature, the pressure of, the mixture will be, (a) P / 2, (b) P, (c) 2P, (d) 4P, Q.7 The root mean square velocity of a gas molecule of mass, m at a given temperature is proportional to, (a), , mo, , (b) m, , (c), , (d), , m, , 1, m, , Q.8 Which of the following statements is true?, (a) Absolute zero temperature is not zero energy, temperature, (b) Two different gases at the same temperature and, pressure have equal root mean square velocities, (c) The root mean square speed of the molecules of, different ideal gases, maintained at the same, temperature are the same, (d) Given sample of 1 cc of hydrogen and 1 cc of oxygen, both at NTP; oxygen sample has a large number of, molecules, Q.9 At room temperature, the r.m.s. speed of the molecules of, certain diatomic gas is found to be 1930 m/s. The gas is, (a) H 2, (b) F2, (c) O2, (d) Cl2, Q.10 Root mean square velocity of a particle is v at pressure P., If pressure is increased two times, then the r.m.s. velocity, becomes, (a) 2 v, (b) 3 v, (c) 0.5 v, (d) v, Q.11 In the two vessels of same volume, atomic hydrogen and, helium at pressure 1 atm and 2 atm are filled. If temperature, of both the samples is same, then average speed of, hydrogen atoms < CH > will be related to that of helium, < CHe > as, (a) < CH > = 2 < CHe >, , (b) < CH > = < CHe >, , (c) < CH > = 2 < CHe >, , (d) < CH > =, , RESPONSE, GRID, , Q.12 For a gas at a temperature T the root-mean-square velocity, vrms , the most probable speed vmp , and the average speed, v au obey the relationship, (a) vau > vrms > vmp, , (b) vrms > va u > vmp, (d) vmp > vrms > va u, , (c) vmp > vau > vrms, Q.13 One mole of ideal monoatomic gas ( g = 5 / 3) is mixed with, one mole of diatomic gas ( g = 7 / 5) . What is g for the, , Cp ö, æ, mixture? çè g =, ÷, Cv ø, (a) 3/2, (b) 23/15 (c) 35/23 (d) 4/3, Q.14 The value of the gas constant (R) calculated from the, perfect gas equation is 8.32 Joule/gm mol K , whereas its, value calculated from the knowledge of Cp and CV of the, gas is 1.98 cal/gm mole K . From this data, the value of J, is, (a) 4.16 J / cal, (b) 4.18 J / cal, (c) 4.20 J / cal, (d) 4.22 J / cal, Q.15 Gas at a pressure P0 is contained in a vessel. If the masses, of all the molecules are halved and their speeds are, doubled, the resulting pressure P will be equal to, P0, 2, Q.16 The relation between the gas pressure P and average kinetic, energy per unit volume E is, 2, (a) P = 1 E (b) P = E (c) P = 3 E (d) P = E, 3, 2, 2, Q.17 Mean kinetic energy (or average energy) per gm molecule, of a monoatomic gas is given by, 3, 1, 1, 3, RT, KT (c), RT, KT, (a), (b), (d), 2, 2, 2, 2, Q.18 At which of the following temperature would the molecules, of a gas have twice the average kinetic energy they have at, 20°C?, , < CHe >, 2, , (a), , 4P0, , (b) 2P0, , (c) P0, , (a), , 40o C, , (b) 80 o C, , (c) 313o C (d) 586o C, , (d), , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 26, , 3, , Q.19 The kinetic energy of one gram molecule of a gas at normal, temperature and pressure is (R = 8.31 J/mol - K), (a), , (b) 1.3 ´ 10 2 J, , 0.56 ´ 10 4 J, , (c) 2.7 ´ 10 2 J, (d) 3.4 ´ 10 3 J, Q.20 70 calories of heat are required to raise the temperature, o, f, 2 moles of an ideal gas at constant pressure from 30°C to, 35°C. The amount of heat required to raise the temperature, of same gas through the same range (30o C to 35o C) at, constant volume (R = 2 cal/mol K), (a) 30 cal, (b) 50 cal (c) 70 cal (d) 90 cal, Q.21 A vessel contains a mixture of one mole of oxygen and, two moles of nitrogen at 300 K . The ratio of the average, rotational kinetic energy per O2 molecule to that per N2, molecule is, (a) 1 : 1, (b) 1 : 2, (c) 2 : 1, (d) Depends on the moments of inertia of the two, molecules, DIRECTIONS (Q.22-Q.24) : In the following questions, more than, one of the answers given are correct. Select the correct answers, and mark it according to the following codes:, Codes :, , (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, , (b), (d), , 1 and 2 are correct, 1 and 3 are correct, , Q.22 From the following statements, concerning ideal gas at any, given temperature T, select the correct one(s), (1) The coefficient of volume expansion at constant, pressure is same for all ideal gases, (2) In a gaseous mixture, the average translational kinetic, energy of the molecules of each component is same, (3) The mean free path of molecules increases with the, decrease in pressure, (4) The average translational kinetic energy per molecule, of oxygen gas is 3KT (K being Boltzmann constant), , RESPONSE, GRID, , 19., , 20., , 24., , 25., , Q.23 Let v, vrms and v p respectively denote the mean speed,, root mean square speed and most probable speed of the, molecules in an ideal monoatomic gas at absolute, temperature T, the mass of a molecule is m. Then, (1) v p < v < v rms, (2) The average kinetic energy of a molecule is, (3) No molecule can have speed greater than, , 3 2, mv, 4 p, 2vrms, , (4) No molecule can have speed less than v p / 2, Q.24 A gas in container A is in thermal equilibrium with another, gas in container B, both contain equal masses of the two, gases in the respective containers. Which of the following can be true, (2) PA VA = PB VB, (1) PA = PB , VA ¹ VB, (3) PA ¹ PB , VA = VB, , PA, PB, (4) V = V, A, B, , DIRECTIONS (Q.25-Q.27) : Read the passage given below and, answer the questions that follows :, , A diathermic piston divides adiabatic cylinder of volume V0 into, two equal parts as shown in the figure. Both parts contain ideal, monoatomic gases. The initial pressure and temperature of gas, in left compartment are P0 and T0 while that in right compartment, are 2P0 and 2T0. Initially the piston is kept fixed and the system, is allowed to acquire a state of thermal equilibrium., , P0 , T0, , 2 P0 , 2T0, , Q.25 The pressure in left compartment after thermal equilibrium, is achieved is, 3, P0, (a) P0, (b), 2, 4, P0, (c), (d) None of these, 3, , 21., , Space for Rough Work, , 22., , 23.
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t.me/Magazines4all, DPP/ P 26, , 4, Q.26 The heat that flown from right compartment to left, compartment before thermal equilibrium is achieved is, 3, 3, 2, P0V0 (c), P0V0 (d), P0V0, 4, 8, 3, Q.27 If the pin which was keeping the piston fixed is removed, and the piston is allowed to slide slowly such that a state, of mechanical equilibrium is achieved. The volume of left, compartment when piston is in equilibrium is, , (a) P0V0, , (b), , 3, V0, 4, , (b), , (a), , V0, 4, , (c), , V0, 2, , (d), , 2, V0, 3, , DIRECTIONS (Qs. 28-Q.30) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason). Each, of these questions has four alternative choices, only one of which is, the correct answer. You have to select the correct choice., , (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.28 Statement-1 : Internal energy of an ideal gas does not, depend upon volume of the gas., Statement-2 : Internal energy of an ideal gas depends on, temperature of gas., Q.29 Statement-1 : Equal masses of helium and oxygen gases, are given equal quantities of heat. There will be a greater, rise in the temperature of helium compared to that of, oxygen., Statement-2 : The molecular weight of oxygen is more, than the molecular weight of helium., Q.30 Statement-1 : Maxwell speed distribution graph is, asymmetric about most probable speed., Statement-2 : rms speed of ideal gas, depends upon it’s, type (monoatomic, diatomic and polyatomic)., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 26 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 27, SYLLABUS : Oscillations-1 (Periodic motion - period, Frequency, Displacement as a function of time. Periodic, functions, Simple harmonic motion and its equation, Energy in S.H.M. - kinetic and potential energies), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice questions., Each question has 4 choices (a), (b), (c) and (d), out of which ONLY, ONE choice is correct., , Q.1 A simple h armonic motion is represented by, F (t ) = 10sin(20t + 0.5) . The amplitude of the S.H.M. is, (a) a = 30 cm, (b) a = 20 cm, (c) a = 10 cm, (d) a = 5 cm, Q.2 A particle executes a simple harmonic motion of time, period T. Find the time taken by the particle to go directly, from its mean position to half the amplitude, (a) T / 2, (b) T / 4, (c) T / 8, (d) T /12, Q.3 The periodic time of a body executing simple harmonic, motion is 3 sec. After how much time from time t = 0 , its, displacement will be half of its amplitude, , RESPONSE GRID, , 1., , 2., , (a), , 1, sec, 8, , (b), , 1, sec, 6, , (c), , 1, sec, 4, , (d), , 1, sec, 3, , pö, æ, Q.4 If x = a sin ç wt + ÷ and x ' = a cos wt , then what is the, è, 6ø, phase difference between the two waves?, (a) p / 3, (b) p / 6, (c) p / 2, (d) p, Q.5 A body is executing S.H.M. when its displacement from, the mean position is 4 cm and 5 cm, the corresponding, velocity of the body is 10 cm/sec and 8 cm/sec. Then the, time period of the body is, (a) 2p sec, (b) p / 2 sec, (c) p sec, (d) 3p / 2sec, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 27, , 2, Q.6 If a simple pendulum oscillates with an amplitude of 50, mm and time period of 2 sec, then its maximum velocity, is, (a) 0.10 m/s, (b) 0.15 m/s, (c) 0.8 m/s, (d) 0.26 m/s, Q.7 The maximum velocity and the maximum acceleration of, a body moving in a simple harmonic oscillator are 2 m/s, and 4 m/s2. Then angular velocity will be, (a) 3 rad/sec, (b) 0.5 rad/sec, (c) 1 rad/sec, (d) 2 rad/sec, Q.8 The amplitude of a particle executing SHM is 4 cm. At the, mean position the speed of the particle is 16 cm/sec. The, distance of the particle from the mean position at which, , Q.12 The total energy of a particle executing S.H.M. is, proportional to, (a) Displacement from equilibrium position, (b) Frequency of oscillation, (c) Velocity in equilibrium position, (d) Square of amplitude of motion, Q.13 When the displacement is half the amplitude, the ratio of, potential energy to the total energy is, 1, 1, 1, (a), (b), (c) 1, (d), 8, 2, 4, Q.14 A particle is executing simple harmonic motion with, , the speed of the particle becomes 8 3 cm/s, will be, , (a) f / 2, (b) f, (c) 2 f, (d) 4 f, Q.15 A particle executes simple harmonic motion with a, , (a) 2 3 cm, (b), 3 cm, (c) 1cm, (d) 2 cm, Q.9 The amplitude of a particle executing S.H.M. with, frequency of 60 Hz is 0.01 m. The maximum value of the, acceleration of the particle is, (a) 144p 2 m/sec 2, , (b) 144 m/sec 2, , 144, , (d) 288p 2 m/sec 2, m/sec 2, p2, Q.10 A particle executes simple harmonic motion with an, angular velocity and maximum acceleration of 3.5rad/sec, (c), , and 7.5 m /s 2 respectively. The amplitude of oscillation, is, (a) 0.28 m (b) 0.36 m (c) 0.53 m (d) 0.61 m, Q.11 What is the maximum acceleration of the particle doing, é pt, ù, the SHM y = 2sin ê + fú where y is in cm?, 2, ë, û, (a), , p, cm/s 2, 2, , (b), , (c), , p, cm/s 2, 4, , (d), , RESPONSE, GRID, , p2, cm/s 2, 2, p, cm/s 2, 4, , frequency f . The frequency at which its kinetic energy, changes into potential energy is, , frequency f . The frequency with which its kinetic energy, oscillates is, (a) f / 2, (b) f, (c) 2 f, (d) 4 f, Q.16 The kinetic energy of a particle executing S.H.M. is 16 J, when it is in its mean position. If the amplitude of, oscillations is 25 cm and the mass of the particle is 5.12, kg, the time period of its oscillation is, p, sec, (a), (b) 2p sec (c) 20p sec (d) 5p sec, 5, Q.17 The displacement x (in metres) of a particle performing, simple harmonic motion is related to time t (in seconds) as, , pö, æ, x = 0.05cos ç 4pt + ÷ . The frequency of the motion will, 4ø, è, be, (a) 0.5 Hz, (b) 1.0 Hz (c) 1.5 Hz (d) 2.0 Hz, Q.18 A particle executes simple harmonic motion, [amplitude = A ] between x = - A and x = + A . The time, taken for it to go from 0 to A / 2 is T1 and to go from, A / 2 to A is T2 . Then, , (a) T1 < T2, , (b) T1 > T2 (c) T1 = T2 (d) T1 = 2T2, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 27, , 3, , Q.19 A cylindrical piston of mass M slides smoothly inside a, long cylinder closed at one end, enclosing a certain mass, of gas. The cylinder is kept with its axis horizontal. If the, piston is disturbed from its equilibrium position, it, oscillates simple harmonically. The period of oscillation, will be, h, , M, , Gas, P, , A, , (a) T = 2p æç Mh ö÷, è PA ø, , (b) T = 2p æç MA ö÷, è Ph ø, , M ö, (c) T = 2p æç, ÷, PAh, è, ø, , (d) T = 2p MPhA, , Q.20 A particle is performing simple harmonic motion along, x-axis with amplitude 4 cm and time period 1.2 sec. The, minimum time taken by the particle to move from x = 2, cm to x = + 4 cm and back again is given by, (a) 0.6 sec, (b) 0.4 sec, (c) 0.3 sec, (d) 0.2 sec, Q.21 A spring of force constant k is cut into two pieces such, that one piece is double the length of the other. Then the, long piece will have a force constant of, (a) (2/3)k, (b) (3/2)k, (c) 3k, (d) 6k, Q.22 A simple pendulum has time period T 1 . The point of, suspension is now moved upward according to equation, y = kt2 where k =1m/sec2. If new time period is T2 then, ratio, , T12, T22, , will be, , (a) 2/3, (c) 6/5, , RESPONSE, GRID, , (b) 5/6, (d) 3/2, , 19., , 20., , 24., , 25., , DIRECTIONS (Q.23-Q.25) : In the following questions, more than, , one of the answers given are correct. Select the correct, answers and mark it according to the following codes:, Codes : (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.23 A particle constrained to move along the x-axis in a, potential V = kx2, is subjected to an external time dependent, r, force f (t) , here k is a constant, x the distance from the, origin, and t the time. At some time T, when the particle, has zero velocity at x = 0, the external force is removed., Choose the incorrect options –, (1) Particle executes SHM, (2) Particle moves along +x direction, (3) Particle moves along – x direction, (4) Particle remains at rest, Q.24 Three simple harmonic motions in the same direction, having the same amplitude a and same period are, superposed. If each differs in phase from the next by 45°,, then –, (1) The resultant amplitude is (1 + 2) a, (2) The phase of the resultant motion relative to the first is, 90°, (3) The energy associated with the resulting motion is, , (3 + 2 2) times the energy associated with any single, motion, (4) The resulting motion is not simple harmonic, Q.25 For a particle executing simple harmonic motion, which, of the following statements is correct?, (1) The total energy of the particle always remains the same, (2) The restoring force always directed towards a fixed, point, (3) The restoring force is maximum at the extreme, positions, (4) The acceleration of the particle is maximum at the, equilibrium position, , 21., , Space for Rough Work, , 22., , 23.
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t.me/Magazines4all, DPP/ P 27, , 4, DIRECTIONS (Q.26-Q.27) : Read the passage given below and, answer the questions that follows :, , The differential equation of a particle undergoing SHM is given, by a, , d2 x, 2, , + bx = 0 . The particle starts from the extreme position., , dt, Q.26 The ratio of the maximum acceleration to the maximum, velocity of the particle is –, , (a), , b, a, , (b), , a, b, , a, b, (d), b, a, Q.27 The equation of motion may be given by :, , (c), , æ bö, (a) x = Asin ç a ÷ t, è, ø, , æ bö, (b) x = Acos ç a ÷ t, è, ø, æ b, ö, (c) x = A sin ç a t + q÷ where q ¹ p/2, è, ø, (d) None of these, , RESPONSE GRID, , 26., , 27., , DIRECTIONS (Q.28-Q.30) : Each of these questions contains two, statements: Statement-1 (Assertion) and Statement-2 (Reason). Each, of these questions has four alternative choices, only one of which is, the correct answer. You have to select the correct choice., , (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.28 Statement-1 : In S.H.M., the motion is ‘to and fro’ and, periodic., Statement-2, :, Velocity of the particle, (v=, ) w k 2 - x 2 (where x is the displacement and k is, , amplitude), Q.29 Statement-1 : In simple harmonic motion, the velocity is, maximum when acceleration is minimum., Statement-2 : Displacement and velocity of S.H.M. differ, in phase by p / 2, Q.30 Statement-1 : The graph of total energy of a particle in, SHM w.r.t., position is a straight line with zero slope., Statement-2 : Total energy of particle in SHM remains, constant throughout its motion., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 27 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 45, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 28, SYLLABUS : Oscillations-2 (Oscillations of a spring, simple pendulum, free,, forced and damped oscillations, Resonance), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A particle of mass m is, attached to three identical, springs A, B and C each of, force constant k as shown, in figure., If the particle of mass m, is pushed slightly against, the spring A and released, then the time period of, oscillations is, , RESPONSE GRID, , 1., , B, , C, 90°, , O, , m, , m, m, 2m, m, (d) 2p, (b) 2p, (c) 2p, 3k, k, k, 2k, Q.2 Three masses 700g, 500g, and 400g are suspended at the, end of a spring as shown and are in equilibrium., When the 700g mass is removed, the system oscillates, with a period of 3 seconds, when the 500 gm mass is also, removed, it will oscillate with a period of, , (a), , 2p, , (a) 1 s, (b) 2 s, (c) 3 s, (d), , A, , 2., Space for Rough Work, , 12, s, 5, , 700gm, 500gm, 400gm
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t.me/Magazines4all, DPP/ P 28, , 2, Q.3 The bob of a simple pendulum is displaced from its, equilibrium position O to a position Q which is at height h, above O and the bob is then released., Assuming the mass of the bob to be m and time period of, oscillations to be 2.0 sec, the tension in the string when, the bob passes through O is, (a) m( g + p 2 g h), (b), , m ( g + p2 g h ), , (c), , æ, ö, p2, mç g +, g h÷, ç, ÷, 2, è, ø, , Q, h, , æ, 2, ö, (d) m ç g + p g h ÷, ç, è, , 3, , O, , ÷, ø, , Q.4 A spring of force constant k is cut into two pieces such, that one piece is double the length of the other. Then the, long piece will have a force constant of, (a) (2 / 3)k (b) (3 / 2)k (c) 3k, (d) 6k, Q.5 A pendulum suspended from the ceiling of a train has a, period T, when the train is at rest. When the train is, accelerating with a uniform acceleration a, the period of, oscillation will, (a) increase, (b) decrease, (c) remain unaffected, (d) become infinite, Q.6 A simple pendulum is set up in a trolley which moves to the, right with an acceleration a on a horizontal plane. Then the, thread of the pendulum in the mean position makes an angle, q with the vertical is, a, (a) tan -1 in the forward direction, g, a, (b) tan -1 in the backward direction, g, g, in the backward direction, a, -1 g, in the forward direction, (d) tan, a, , (c) tan -1, , RESPONSE, GRID, , Q.7 The time period of a second’s pendulum is 2 sec. The, spherical bob which is empty from inside has a mass of 50, gm. This is now replaced by another solid bob of same, radius but having different mass of 100 gm. The new time, period will be, (a) 4 sec, (b) 1 sec, (c) 2 sec, (d) 8 sec, Q.8 The length of a simple pendulum is increased by 1%. Its, time period will, (a) Increase by 1%, (b) Increase by 0.5%, (c) Decrease by 0.5%, (d) Increase by 2%, Q.9 The bob of a pendulum of length l is pulled aside from its, equilibrium position through an angle q and then released., The bob will then pass through its equilibrium position with, a speed v, where v equals, (a), (b) 2gl (1 + cos q), 2gl (1- sin q), (c) 2gl (1 - cos q), (d) 2gl (1+ sin q), Q.10 A simple pendulum is executing simple harmonic motion, with a time period T. If the length of the pendulum is, increased by 21%, the percentage increase in the time, period of the pendulum of is, (a) 10%, (b) 21%, (c) 30%, (d) 50%, Q.11 A chimpanzee swinging on a swing in a sitting position,, stands up suddenly, the time period will, (a) Become infinite, (b) Remain same, (c) Increase, (d) Decrease, Q.12 A simple pendulum consisting of a ball of mass m tied to a, thread of length l is made to swing on a circular arc of angle q, in a vertical plane. At the end of this arc, another ball of mass, m is placed at rest. The momentum transferred to this ball at, rest by the swinging ball is, (a) Zero, , (b), , mq, , g, l, , (c), , mq, l, , l, g, , (d), , m, l, 2p, l, g, , Q.13 The time period of a simple pendulum of length L as, measured in an elevator descending with acceleration g / 3, is, (a) 2p 3L, g, , (b), , æ 3L ö, p ç ÷, è gø, , (c), , æ 3L ö, 2p ç ÷, è 2g ø, , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., Space for Rough Work, , (d) 2p 2L, 3g, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 28, , 3, is maximum for w = w2, then (where w0 natural frequency, of oscillation of particle), (a) w1 = w0 and w 2 ¹ w 0 (b) w1 = w0 and w 2 = w 0, (c) w1 ¹ w0 and w 2 = w 0 (d) w1 ¹ w0 and w 2 ¹ w 0, , Q.14 A mass m is suspended from the two coupled springs, connected in series. The force constant for springs are k1, and k2. The time period of the suspended mass will be, (a), (c), , æ m ö, T = 2p ç, ÷, è k1 + k 2 ø, æ m(k1 + k2 ) ö, T = 2p ç, ÷, è k1k2, ø, , (b), , æ m ö, T = 2p ç, ÷, è k1 + k 2 ø, , (d), , æ mk k ö, T = 2p ç 1 2 ÷, è k1 + k 2 ø, , c, , Q.15 A spring having a spring constant k is loaded with a mass, m. The spring is cut into two equal parts and one of these is, loaded again with the same mass. The new spring constant is, (a) k / 2, (b) k, (c) 2k, (d) k2, Q.16 A mass m = 100 gm is attached at the end of a light spring, which oscillates on a frictionless horizontal table with an, amplitude equal to 0.16 metre and time period equal to 2, sec. Initially the mass is released from rest at t = 0 and, displacement x = –0.16 metre. The expression for the, displacement of mass at any time t is, (a) x = 0.16 cos(pt ), (b) x = - 0.16 cos(pt ), (c) x = 0.16sin(pt + p), (d) x = - 0.16sin(pt + p ), Q.17 Two masses m1 and m2 are suspended together by a massless, spring of constant k. When the masses are in equilibrium,, m1 is removed without disturbing the system. The amplitude, of oscillations is, (a), (b), (c), , m1g, k, m2 g, k, (m1 + m2 ) g, k, (m1 - m2 ) g, k, , m1, m2, , (d), Q.18 The composition of two simple harmonic motions of equal, periods at right angle to each other and with a phase, difference of p results in the displacement of the particle, along, (a) Straight line, (b) Circle, (c) Ellipse, (d) Figure of 8, Q.19 A particle with restoring force proportional to, displacement and resisting force proportional to velocity, is subjected to a force F sin wt. If the amplitude of the, particle is maximum for w = w1 and the energy of the particle, , RESPONSE, GRID, , 14., 19., , 15., 20., , Q.20 Amplitude of a wave is represented by A =, a +b-c, Then resonance will occur when, (a) b = – c/2, (b) b = 0 & a = c, (c) b = – a/2, (d) None, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Two blocks A and B each of mass m are connected by a, massless spring of natural length L and spring constant k., The blocks are initially resting on a smooth horizontal floor, with the spring at its natural length. A third identical block, C also of mass m moves on the floor with a speed v along, the line joining A and B and collides with A. Then, (1) The kinetic energy of the A – B system at maximum, compression of the spring is mv2/4, (2) The maximum compression of the spring is v m / 2k, (3) The kinetic energy of the A-B system at maximum, compression of the spring is zero, (4) The maximum compression of the spring is v m / k, Q.22 A simple pendulum of length L and mass (bob) M is, oscillating in a plane about a vertical line between angular, limits – f and + f. For an angular displacement θ ( θ < f) ,, the tension in the string and the velocity of the bob are T, and v respectively. The following relations hold good under, the above conditions, (1), , T - Mg cos θ =, , Mv 2, L, , (2) T cosθ = Mg, (3) The magnitude of the tangential acceleration of the bob, a T = g sin θ, , (4) T = Mg cos q, , 16., 21., , Space for Rough Work, , 17., 22., , 18.
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t.me/Magazines4all, DPP/ P 28, , 4, , DIRECTIONS (Q.24-Q.25) : Read the passage given below, and answer the questions that follows :, A particle performs linear SHM such that it is placed on platform & platform along with particles oscillate vertically up and, down with amplitude A = 1cm. If the particle does not loose, contact with platform anywhere and mass of particle is 1 kg,, find :, Q.24 The minimum, possible time period (Take p = g ), (a) 0.1 sec., (b) 0.2 sec., (c) 0.3 sec., (d) 0.4 sec., Q.25 For minimum time period condition average potential, energy between t = 0 to t = 0.05 sec (Take g = 10 m/s2), (a) 0.025 Joule, (b) 0.1 Joule, (c) 0.08 Joule, (d) 0.06 Joule, , RESPONSE, GRID, , 23., , 24., , DIRECTIONS (Q.26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.26 Statement-1 : Consider motion for a mass spring system, under gravity, motion of M is not a simple harmonic motion, unless Mg is negligibly small., Statement-2 : For simple harmonic, k = force constant, X, motion acceleration must be, proportional to displacement and is, directed towards the mean position., M = Mass, Q.27 Statement-1 : The periodic time of a hard spring is less as, compared to that of a soft spring., Statement-2 : The periodic time depends upon the spring, constant, and spring constant is large for hard spring., Q.28 Statement-1 : The percentage change in time period is, 1.5%, if the length of simple pendulum increases by 3%, Statement-2:Time period is directly proportional to length of, pendulum., Ma = kx + Mg, , Q.23 Identify wrong statements among the following, (1) The greater the mass of a pendulum bob, the shorter, is its frequency of oscillation, (2) A simple pendulum with a bob of mass M swings with, an angular amplitude of 40°. When its angular, amplitude is 20°, the tension in the string is less than, Mgcos20°., (3) The fractional change in the time period of a pendulum, on changing the temperature is independent of the, length of the pendulum., (4) As the length of a simple pendulum is increased, the, maximum velocity of its bob during its oscillation will, also decreases., , 25., , 26., , 27., , 28., , DAILY PRA CTICE PROBLEM SHEET 28 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 29, SYLLABUS : Waves-1 (Wave motion, longitudinal and transverse waves, speed of a wave, displacement relation for a, progressive wave, principle of superposition of waves, reflection of waves), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A tuning fork makes 256 vibrations per second in air. When, the velocity of sound is 330 m/s then wavelength of the tone, emitted is, (a) 0.56 m (b) 0.89 m (c) 1.11 m (d) 1.29 m, Q.2 In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero, displacement is 0.170 second. The frequency of the wave, is, (a) 1.47Hz (b) 0.36 Hz (c) 0.73 Hz (d) 2.94 Hz, Q.3 A man is standing between two parallel cliffs and fires a, gun. If he hears first and second echoes after 1.5 s and, , RESPONSE GRID, , 1., , 2., , 3.5s respectively, the distance between the cliffs is, (Velocity of sound in air = 340 ms–1), (a) 1190 m, (b) 850 m, (c) 595 m, (d) 510 m, Q.4 v1 and v2 are the velocities of sound at the same temperature, in two monoatomic gases of densities r 1 and r 2, respectively. If r1 / r 2 =, , 1, then the ratio of velocities v 1, 4, , and v2 will be, (a) 1 : 2, (b) 4 : 1, (c) 2 : 1, (d) 1 : 4, Q.5 A wave of frequency 500 Hz has velocity 360 m/sec. The, distance between two nearest points 60° out of phase, is, (a) 0.6 cm (b) 12 cm (c) 60 cm (d) 120 cm, , 3., Space for Rough Work, , 4., , 5.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 29, , 3, , Q.18 Equation of motion in the same direction is given by, y1 = A sin( wt - kx ), y 2 = A sin(wt - kx - q) .The amplitude, of the medium particle will be, (a), , q, 2, q, 2 A cos, 2, , 2 A cos, , (b) 2 A cos q, , (c), (d) 2A cos q, Q.19 The amplitude of a wave, represented by displacement, 1, 1, sin wt ±, cos wt will be, equation y =, a, b, a+b, a+ b, a± b, a+b, (c), (d), (b), ab, ab, ab, ab, Q.20 The displacement due to a wave moving in the positive, 1, at time t = 0 and by, x -direction is given by y =, (1 + x2 ), 1, y=, at t = 2 seconds, where x and y are in, [1 + ( x - 1)2 ], metres. The velocity of the wave in m/s is, (a) 0.5, (b) 1, (c) 2, (d) 4, , (a), , DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes : (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 P, Q and R are three particles of a medium which lie on the, x-axis. A sine wave of wavelength l is travelling through, the medium in the x-direction. P and Q always have the, same speed, while P and R always have the same velocity., The minimum distance between –, (1) P and Q is l, (2) P and Q is l/2, (3) P and R is l/2, (4) P and R is l, Q.22 A wave represented by the given equation, , (1) A wave travelling in the negative X direction with a, velocity of 1.5 m/sec, (2) A wave travelling in the negative X direction with a, wavelength of 0.2 m, (3) A wave travelling in the positive X direction with a, velocity of 1.5 m/sec., (4) A wave travelling in the positive X direction with a, wavelength of 0.2 m, Q.23 It is usually more convenient to describe a sound wave in, terms of pressure wave as compared to displacement wave, because –, (1) Two waves of same intensity but different frequencies, have different displacement amplitude but same, pressure amplitude, (2) The human ear responds to the change in pressure and, not to the displacement wave., (3) The electronic detector (microphone) does respond, to the change in pressure but not to the displacement., (4) None of the above, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, Sound from a point isotropic source spreads equally in all, directions in homogeneous medium. Therefore its intensity, decreases with square of distance from the source. When, distance between observer and the source changes, apart from, changes in intensity, the observer listens sound of pitch higher, or lower than actual pitch depending upon the fact that the, distance between the observer and source is decreasing or, increasing respectively. An observer O is at a distance 2R from, centre of a circle of radius R. A point isotropic sound source S, moves on the circle with uniform angular velocity w = p/3 rad/, s. Initially observer, source and centre of the circle are in same, line., 2R, R=1m, , πö, æ, Y = A sin ç10πx + 15πt + ÷ , where x is in meter and t is, è, 3ø, in second. The expression represents, , RESPONSE, GRID, , 18., , 19., , O, , 20., , 23., , Space for Rough Work, , S, 1R, , 21., , C, R, , 22.
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t.me/Magazines4all, DPP/ P 29, , 4, Q.24 Starting from initial moment, the source moves through, an angular displacement 180°. Intensity of the sound as, observed by the observer decreases by a factor of –, (a) 2, (b) 3, (c) 4, (d) 9, Q.25 During a complete round trip of star on the circle, the, observer listens a sound, whose –, (a) wavelength first decreases to a maximum value then, increases to the original value, (b) wavelength first increases to a maximum value then, decreases to the original value, (c) During the first half time wavelength increases then, decreases to the original value, (d) None of the above is correct because in Doppler's, effect, it is the pitch of sound which changes and not, its wavelength, irrespective of motion of source or, observer., Q.26 Sound emitted by the source at two successive instants t1, and t 2 has minimum and maximum observed pitch, respectively, then –, (a) t1 = 1s, t2 = 5s, (b) t1 = 5s, t2 = 7s, (c) t1 = 7s, t2 = 11s, (d) t1 = 5s, t2 = 11s, , RESPONSE, GRID, , 24., , 25., , DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : Particle velocity and wave velocity both are, independent of time., Statement-2 : For the propagation of wave motion, the, medium must have the properties of elasticity and inertia., Q.28 Statement-1 : Speed of wave =, , Wavelength, Time period, , Statement-2 : Wavelength is the distance between two, nearest particles vibrating in phase., Q.29 Statement-1 : Transverse waves are not produced in liquids, and gases., Statement-2 : Light waves are transverse waves., , 26., , 27., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 29 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 30, SYLLABUS : Waves-2 (Standing waves in strings and organ pipes, Fundamental mode and, harmonics, Beats, Doppler effect in sound), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A police car moving at 22 m/s, chases a motorcylist. The, police man sounds his horn at 176 Hz, while both of them, move towards a stationary siren of frequency 165 Hz., Calculate the speed of the motorcycle, if it is given that he, does not observes any beats, Motorcycle, , Police Car, , 22 m/s, (176 Hz), , Stationary siren, (165 Hz), , (a) 33 m/s, , (b) 22 m/s, , RESPONSE GRID, , 1., , (c) Zero, , (d) 11 m/s, , 2., , Q.2 A closed organ pipe of length L and an open organ pipe, contain gases of densities r1 and r2 respectively. The, compressibility of gases are equal in both the pipes. Both, the pipes are vibrating in their first overtone with same, frequency. The length of the open organ pipe is, (a), , L, 3, , (b), , 4L, 3, , (c), , 4 L r1, 3 r2, , (d), , 4 L r2, 3 r1, , Q.3 Two whistles A and B produces notes of frequencies 660, Hz and 596 Hz respectively. There is a listener at the midpoint of the line joining them. Now the whistle B and the, listener start moving with speed 30 m/s away from the, whistle A. If speed of sound be 330 m/s, how many beats, will be heard by the listener, (a) 2, (b) 4, (c) 6, (d) 8, , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 30, , 2, 2nd, , Q.4 An open organ pipe is in resonance in its, harmonic, with tuning fork of frequency f1. Now, it is closed at one, end. If the frequency of the tuning fork is increased slowly, from f1 then again a resonance is obtained with a frequency, f2. If in this case the pipe vibrates nth hamonics then, 3, 5, (a) n = 3, f 2 = f1, (b) n = 3, f 2 = f1, 4, 4, 5, 3, (c) n = 5, f 2 = f1, (d) n = 5, f 2 = f1, 4, 4, Q.5 The source producing sound and an observer both are, moving along the direction of propagation of sound waves., If the respective velocities of sound, source and an, observer are v, vs and v0, then the apparent frequency heard, by the observer will be (n = frequency of sound), n(v - v0 ), n (v + v0 ), (a), (b), v - v0, v - vs, n(v + v0 ), n (v - v0 ), (d), (c), v + vs, v + vs, Q.6 A whistle sends out 256 waves in a second. If the whistle, approaches the observer with velocity 1/3 of the velocity, of sound in air, the number of waves per second the, observer will receive, (a) 384, (b) 192, (c) 300, (d) 200, Q.7 A source of sound emitting a note of frequency 200 Hz, moves towards an observer with a velocity v equal to the, velocity of sound. If the observer also moves away from, the source with the same velocity v, the apparent frequency, heard by the observer is, (a) 50 Hz, (b) 100 Hz (c) 150 Hz (d) 200 Hz, Q.8 The speed of sound in air at a given temperature is 350 m/, s. An engine blows whistle at a frequency of 1200 cps. It is, approaching the observer with velocity 50 m/s. The apparent, frequency in cps heard by the observer will be, (a) 600, (b) 1050, (c) 1400, (d) 2400, Q.9 A source of sound of frequency n is moving towards a, stationary observer with a speed S. If the speed of sound in, air is V and the frequency heard by the observer is n1 ,the, value of n1 / n is, (a) (V + S)/V, (b) V/ (V + S), (c) (V – S)/V, (d) V/ (V – S), , RESPONSE, GRID, , Q.10 An observer is moving away from source of sound of, frequency 100 Hz. His speed is 33 m/s. If speed of sound, is 330 m/s, then the observed frequency is, (a) 90 Hz, (b) 100 Hz (c) 91 Hz (d) 110 Hz, Q.11 A whistle giving out 450 Hz approaches a stationary, observer at a speed of 33 m/s. The frequency heard by the, observer in Hz is, (a) 409, (b) 429, (c) 517, (d) 500, Q.12 Two sirens situated one kilometre apart are producing sound, of frequency 330 Hz. An observer starts moving from one, siren to the other with a speed of 2 m/s. If the speed of, sound be 330 m/s, what will be the beat frequency heard, by the observer, (a) 8, (b) 4, (c) 6, (d) 1, Q.13 A small source of sound moves on a circle as shown in the, figure and an observer is standing on O. Let n1, n2 and n3 be, the frequencies heard when the source is at A, B and, C respectively. Then, (a) n1 > n2 > n3, (b) n2 > n3 < n1, (c) n1 = n2 > n3, O, (d) n2 > n1 > n3, Q.14 A person carrying a whistle emitting continuously a note, of 272 Hz is running towards a reflecting surface with a, speed of 18 km/hour. The speed of sound in air is 345ms–, 1. The number of beats heard by him is, (a) 4, (b) 6, (c) 8, (d) 3, Q.15 A source of sound of frequency 256 Hz is moving rapidly, towards a wall with a velocity of 5m/s. The speed of sound, is 330 m/s. If the observer is between the wall and the, source, then the beats heard per second will be, (a) 7.8 Hz, (b) 7.7 Hz, (c) 3.9 Hz, (d) Zero, Q.16 The harmonics which are present in a pipe open at one end, are, (a) odd harmonics, (b) even harmonics, (c) even as well as odd harmonics, (d) None of these, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 30, , 3, , Q.17 A source of sound placed at the open end of a resonance, column sends an acoustic wave of pressure amplitude P0, inside the tube. If the atmospheric pressure is PA, then the, ratio of maximum and minimum pressure at the closed end, of the tube will be, ( PA + P0 ), ( PA + 2 P0 ), (a) ( P - P ), (b) ( P - 2 P ), A, 0, A, 0, 1 ö, æ, çè PA + P0 ÷ø, 2, PA, (d) æ, (c) P, 1 ö, A, çè PA - P0 ÷ø, 2, Q.18 The frequency of fundamental tone in an open organ pipe, of length 0.48 m is 320 Hz. Speed of sound is 320 m/sec., Frequency of fundamental tone in closed organ pipe will, be, (a) 153.8 Hz, (b) 160.0 Hz, (c) 320.0 Hz, (d) 143.2 Hz, Q.19 A standing wave having 3 nodes and 2 antinodes is formed, between two atoms having a distance 1.21Å between them., The wavelength of the standing wave is, (a) 1.21 Å (b) 2.42 Å (c) 6.05 Å (d) 3.63 Å, Q.20 A string on a musical instrument is 50 cm long and its, fundamental frequency is 270 Hz. If the desired frequency, of 1000 Hz, is to be produced, the required length of the, string is, (a) 13.5 cm (b) 2.7 cm (c) 5.4 cm (d) 10.3 cm, Q.21 The loudness and the pitch of a sound depends on, (a) intensity and velocity, (b) frequency and velocity, (c) intensity and frequency, (d) frequency and number of harmonics, Q.22 If in an experiment for determination of velocity of sound, by resonance tube method using a tuning fork of 512 Hz,, first resonance was observed at 30.7 cm and second was, obtainded at 63.2 cm, then maximum possible error in, velocity of sound is (consider actual speed of sound in air, is 332 m/s), (a) 204 cm/sec, (b) 110 cm/sec, (c) 58 cm/sec, (d) 80 cm/sec, , RESPONSE, GRID, , DIRECTIONS (Q.23-Q.25) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.23 You are given four tuning forks, the lowest frequency of, the fork is 300Hz. By striking two tuning forks at a time, any of 1, 2, 3, 5, 7 & 8 Hz beat frequencies are heard. The, possible frequencies of the other three forks –, (1) 301, 302 & 307, (2) 301, 303 & 308, (3) 300, 304 & 307, (4) 305, 307 & 308, Q.24 Doppler shift in frequency depends upon, (1) the frequency of the wave produced, (2) the velocity of the source, (3) the velocity of the observer, (4) distance from the source to the listener, Q.25 The ( x, y ) coordinates of the corners of a square plate are, (0, 0), (L, 0), (L, L) and (0, L). The edges of the plate are, clamped and transverse standing waves are set up in it. If, u ( x, y) denotes the displacement of the plate at the point, ( x, y ) at some instant of time, the possible expression(s), for u is (are) (a = positive constant), px, py, px, 2 py, (1) a sin sin, (2) a sin sin, L, L, L, L, px, py, 2px, py, cos, cos, (3) a cos, (4) a cos, 2L, 2L, L, L, DIRECTIONS (Q.26-Q.27) : Read the passage given below, and answer the questions that follows :, A plate was cut from a quartz crystal and is used to control the, frequency of an oscillating electrical circuit. Longitudinal, standing waves are set up in the plate with displacement antinodes, at opposite faces. The fundamental frequency of vibration is, 2.87 ´ 104, . Here s is thickness of, s, the plate and density of quartz is 2658.76 kg/m3., , given by the equation f0 =, , 17., , 18., , 19., , 20., , 22., , 23., , 24., , 25., , Space for Rough Work, , 21.
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t.me/Magazines4all, DPP/ P 30, , 4, Q.26 Young’s modulus of elasticity for quartz is –, (a) 7 × 1011 N/m2, (b) 8.76 × 1012 N/m2, 12, 2, (c) 2 × 10 N/m, (d) Information insufficient, Q.27 If the quartz plate is vibrating in 3rd harmonic while, measuring the frequency of 1.2 × 106 Hz, then the thickness, of the plateis, (a) 71.75cm, (b) 7.175 cm, (c) 6.02 cm, (d) 0.07 cm, DIRECTIONS (Q.28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., , RESPONSE GRID, , 26., , 27., , Q.28 Statement-1 : Beats cannot be produced by light sources., Statement-2: Light sources have constant phase, difference., Q.29 Statement-1 : In the case of a stationary wave, a person, hear a loud sound at the nodes as compared to the antinodes., Statement-2 : In a stationary wave all the particles of the, medium vibrate in phase., Q.30 Statement-1 : Velocity of particles, while crossing mean, position (in stationary waves) varies from maximum at, antinodes to zero at nodes., Statement-2: 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., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 30 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 31, SYLLABUS : Practical Physics - 1, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.24) : There are 24 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 One cm on the main scale of a vernier callipers is divided, into ten equal parts. If 20 divisions of vernier scale coincide, with 8 small divisions of the main scale. What will be the, least count of callipers ?, (a) 0.05 cm, (b) 0.06 cm, (c) 0.04 cm, (d) 0.01 cm, Q.2 The shape of stress vs strain graph within elastic limit is :, (a) parabolic, (b) curve line, (c) straight line, (d) ellipse, Q.3 In a vernier calliper N divisions of vernier scale coincides, with N – 1 divisions of main scale (in which length of one, division is 1 mm). The least count of the instrument should, be, , RESPONSE GRID, , 1., , 2., , (a) N, (b) N – 1, (c) 1/10 N (d) 1/N – 1, Q.4 The figure shows a situation when the jaws of vernier are, touching each other. Each main scale division is of 1 mm., Find zero correction., (a) – 0.5 mm, main, 0, 1, 2(cm), scale, (b) + 0.5 mm, (c) – 0.4 mm, 10 Vernier, 0, 5, scale, (d) + 0.4 mm, Q.5 In an experiment for measurement of young’s modulus,, following readings are taken. Load = 3.00 kg, length =, 2.820 m, diameter = 0.041 cm and extension = 0.87., Determine the percentage error in the measurement of Y., (a), , ± 5%, , (b) ± 6.5%, , (c) ± 5.5%, , 3., Space for Rough Work, , (d) ± 15%, , 4., , 5.
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t.me/Magazines4all, DPP/ P 31, , 2, Q.6 When the zero of the circular scale of a screw gauge, coincides with the zero of the main scale before A and B, come in contact then the instrument has, (a) positive zero error, (b) negative zero error, AB, (c) no zero error, (d) can't be said anything, Q.7 If h be the elevation or depression of a spherical surface, from the plane glass plate and c be the mean distance, between two consecutive points corresponding to the, impressions made by the three legs of a spherometer then, the radius of curvature is, (a), , c2 h, 6h 2, , (b), , c2 h 2, c2 h, +, +, (c), 6h 2, 6h 2, , (d), , c2 2, +, 6h h, , Q.8 The least count of a spherometer is given by, (a) pitch × no. of circular divisions, (b), , pitch, no. of circular divisions, , (c), , no. of circular divisions, pitch, , (d), , (c) half degree, , pitch, mean distance between two consecutive legs of the spherometer, , Q.9 The specific heat of a solid is determined by the method, known as, (a) the method of fusion, (b) the method of mixture, (c) the method of vaporisation, (d) the method of cooling, Q.10 Which principle is involved in the experiment to determine, the specific heat of a liquid by the method of mixture ?, (a) Heat gained by solid = Heat lost by calorimeter and, liquid., (b) Heat lost by solid = Heat gained by calorimeter and, liquid., (c) Heat lost by solid and liquid = Heat gained by, calorimeter., (d) Heat gained by solid and calorimeter = Heat lost by, liquid., Q.11 Two full turns of the circular scale of a screw gauge cover, a distance of 1mm on its main scale. The total number of, divisions on the circular scale is 50. Further, it is found, , RESPONSE, GRID, , that the screw gauge has a zero error of – 0.03 mm. While, measuring the diameter of a thin wire, a student notes the, main scale reading of 3 mm and the number of circular, scale divisions in line with the main scale as 35. The, diameter of the wire is, (a) 3.32 mm, (b) 3.73 mm, (c) 3.67 mm, (d) 3.38 mm, Q.12 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 instrument is :, (a) half minute, (b) one degree, , (d) one minute, , Q.13 In a screw gauge, the zero of main scale coincides with, fifth division of circular scale in figure (i). The circular, divisions of screw gauge are 50. It moves 0.5 mm on main, scale in one rotation. The diameter of the ball in figure (ii), is, 0, , 0, , 10, 5, 0, , 30, 25, 20, , Figure (ii), , Figure (i), , (a) 2.25 mm, (b) 2.20 mm, (c) 1.20 mm, (d) 1.25 mm, Q.14 A student performs an experiment for determination of, æ 4p 2 l ö, gç=, ÷ . The error in length l is Dl and in time T is DT, è T2 ø, , and n is number of times the reading is taken. The, measurement of g is most accurate for, Dl, DT, n, (a) 5 mm, 0.2 sec, 10, (b) 5 mm, 0.2 sec, 20, (c) 5 mm, 0.1 sec, 10, (d) 1 mm, 0.1 sec, 50, , 6., , 7., , 8., , 9., , 11., , 12., , 13., , 14., , Space for Rough Work, , 10., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 31, , 3, , Q.15 A student performs an experiment to determine the Young’s, modulus of a wire, exactly 2 m long, by Searle’s method. In, a particular reading, the student measures the extension in, the length of the wire to be 0.8 mm with an uncertainty of ±, 0.05 mm at a load of exactly 1.0 kg. The student also, measures the diameter of the wire to be 0.4 mm with an, uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The, Young’s modulus obtained from the reading is, (a) (2.0±0.3)×1011 N/m2 (b) (2.0 ± 0.2) × 1011 N/m2, (c) (2.0±0.1)×1011 N/m2 (d) (2.0 ± 0.05) × 1011 N/m2, Q.16 Students I, II and III perform an experiment for measuring, the acceleration due to gravity (g) using a simple pendulum., They use different lengths of the pendulum and /or record, time for different number of oscillations. The observations, are shown in the table., Least count for length = 0.1 cm, Least count for time = 0.1 s, Time, Total time for, Student Length of the, No. of, pendulum (cm) oscillations (n) oscillations period (s), (s), (n), I, 64.0, 8, 128.0, 16.0, II, 64.0, 4, 64.0, 16.0, III, 20.0, 4, 36.0, 9.0, , (a), (c), , r1, , 3, , A, , RESPONSE, GRID, , B, , (a), , A, , B, , A, , B, , (b), A, , for students I, II and III, respectively, then, , (a) EI = 0, (b) EI is minimum, (c) EI = EII, (d) EII is maximum, Q.17 If the terminal speed of a sphere of gold (density, = 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5, kg/m 3 ), find the terminal speed of a sphere of silver, (density = 10.5 kg/m3) of the same size in the same liquid, (a) 0.4 m/s, (b) 0.133 m/s, (c) 0.1 m/s, (d) 0.2 m/s, Q.18 A spherical solid ball of volume V is made of a material of, density r1. It is falling through a liquid of density r2 (r2<, r1). Assume that the liquid applies a viscous force on the, ball that is proportional to the square of its speed v, i.e.,, Fviscous = –kv2 (k > 0). The terminal speed of the ball is, , (b), , Q.19 A jar is filled with two non-mixing liquids 1 and 2 having densities, r1 and, r2 respectively. A solid ball, made of a material of density, r3 , is dropped in the jar. It comes to equilibrium in the position, shown in the figure.Which of the following is true for r1, r2and, r3?, (a) r3 < r1 < r2, r, (b) r1 > r3 > r2, (c) r1 < r2 < r3, (d) r1 < r3 < r2, Q.20 A capillary tube (A) is dipped in water. Another identical, tube (B) is dipped in a soap-water solution. Which of the, following shows the relative nature of the liquid columns, in the two tubes?, , If EI, EII and EIII are the percentage errors in g, i.e.,, æ Dg, ö, çè g ´ 100÷ø, , Vgr1, k, Vg (r1 – r2 ), (d), k, , Vg (r1 – r2 ), k, Vg r1, k, , B, , (d), , (c), , Q.21 Two wires are made of the same material and have the 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 Dx on applying force F, how much force is, needed to stretch wire 2 by the same amount of energy?, (a) 4 F, (b) 6 F, (c) 9 F, (d) 1 F, Q.22 The vernier constant of two vernier callipers A and B are, 0.01 cm and 0.01 mm respectively. Which one can measure, the length of an object more accurately?, (a) Vernier A, (b) Vernier B, (c) Accuracy in measurement does not depend on vernier, constant, (d) Both A and B are equally accurate., , 15., , 16., , 17., , 20., , 21., , 22., , Space for Rough Work, , 18., , 19.
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , DPP/ P 31, , 4, , (c) l -, , d, 2, , (d) l – d, , Q.24 If x, y, p and q represent the increase in length, the original, length of the experimental wire, load applied to the wire, and area of cross-section of the wire respectively then, Young's modulus of the wire is given by, (a), , xy, pq, , xp, (b) yq, , py, (c) xq, , 80, , 70, , 70, , 60, , (2), , 50, 40, 30, , pq, (d) xy, , DIRECTIONS (Q.25-Q.27) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.25 What is the function of a screw gauge in the experiment of, determining Young's modulus of a wire ?, (1) It measures the extension in the wire., (2) It measures the load applied., (3) It measures the length of the wire., (4) It measures the diameter of the wire., Q.26 Consider the following statements regarding the, experiment to determine the surface tension of water by, capillary rise method. Choose the correct statements., (1) Capillary tube should be clean and liquid should be, free from dirt and grease., , RESPONSE, GRID, , (1), , 80, , Temperature, , d, 2, , 5, , 10, , 80, , (3), , 60, 50, 40, 30, , 0, , 10 20, Time, , 25, , 30, , 35, , 0, , 5, , 0, , 5, , 10, , 80, , 70, , Temperature, , (b) l +, , Temperature, , (a) l + d, , (2) Distilled water should be avoided., (3) Distilled water should be added., (4) Dirty liquid should be used., Q.27 The temperature-time variation graphs, as obtained by four, students 1, 2, 3 and 4 are as shown. The graphs, likely to be, wrong are, Temperature, , Q.23 The acceleration due to gravity at a place can be determined, with the help of a simple pendulum. For this purpose, effective length of the pendulum is considered. If 'l' be, the length of the string and 'd' the diameter of the bob then, the effective length is equal to, , 60, , (4), , 50, 40, , 10, 20, Time, , 25, , 30, , 35, , 70, 60, 50, 40, 30, , 30, 0, , 5, , 10, , 10, 20, Time, , 25, , 30, , 35, , 10, , 10 20, Time, , 25, , 30, , 35, , DIRECTIONS (Q.28-Q.30) : Read the passage given below, and answer the questions that follows :, The internal radius of a 1m long resonance tube is measured as, 3 cm. A tuning fork of frequency 2000 Hz is used. The first, resonating length is measured as 4.6 cm and the second, resonating length is measured as 14.0 cm., Q.28 Calculate the maximum percentage error in measurement, of e., (a) 3.33% (b) 2.23% (c) 4.33% (d) 5.33%, Q.29 Calculate the speed of sound at the room temperature., (a) 275 m/s (b) 376 m/s (c) 356 m/s (d) 330 m/s, Q.30 Calculate the end correction., (a) 0.2 cm (b) 0.3 cm (c) 0.1 cm (d) 0.4 cm, , 23., , 24., , 25., , 28., , 29., , 30., , 26., , 27., , DAILY PRA CTICE PROBLEM SHEET 31 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 32, SYLLABUS : Electrostatics-1 (Coulomb's law, electric field, field lines, Gauss's law), , Max. Marks : 104, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 26 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.18) : There are 18 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A total charge Q is broken in two parts Q1 and Q 2 and they, are placed at a distance R from each other. The maximum, force of repulsion between them will occur, when, Q, Q, , Q1 = Q R, R, Q, 3Q, Q 2 = , Q1 =, 4, 4, , (a) Q 2 =, (c), , Q, 2Q, , Q1 = Q 4, 3, Q, Q, (d) Q1 = , Q2 =, 2, 2, , (b) Q2 =, , Q.2 Two small balls each having the charge + Q are suspended, by insulating threads of length L from a hook. This, arrangement is taken in space where there is no, gravitational effect, then the angle between the two, suspensions and the tension in each thread will be, , RESPONSE GRID, , 1., , 2., , 180o ,, , 1 Q2, 4p Î0 2 L2, , 1, , Q2, , o, (d) 180 , 4p Î 2, 0 L, Q.3 Electric charges of 1mC , -1mC and 2mC are placed in air at, the corners A,B and C respectively of an equilateral triangle, ABC having length of each side 10 cm. The resultant force, on the charge at C is, (a) 0.9 N, (b) 1.8N, (c) 2.7 N, (d) 3.6 N, Q.4 An electron is moving round the nucleus of a hydrogen, atom in a circular orbit of radius r. The coulomb force, ur, 1, ), F between the two is (here K =, , (c), , 4 p Î0, , e, e2 r, e2, e r, (a) - K 3 r$ (b) K 3 r (c) - K 3 r (d) K 2 r$, r, r, r, r, Q.5 Equal charges q are 2placed at the four corners A2, B, C , D of a, 1, Q, 1 Q, o, ,, , force on, square, of length, ofothe, the charge, a . The, (a) 180, (b) 90, 2 magnitude, 2, 2, , 2, , 4 p Î0 (2 L ), , 3., Space for Rough Work, , 4., , 4p Î0 L
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t.me/Magazines4all, DPP/ P 32, , 2, at B will be, (a), , 3q, , 2, , (b), , 4p Î0 a 2, , æ 1+ 2 2 ö, q2, ÷, 2 ø 4p Î0 a 2, , (c) çè, , q, , 2, , p Î0 a 2, æ, , (d) ç 2 +, è, , q2, 1 ö, ÷ø, 2 4 p Î0 a 2, , Q.6 The charges on two spheres are +7mC and –5mC, respectively. They experience a force F. If each of them is, given an additional charge of –2mC, the new force of, attraction will be, (a) F, (b) F/2, (c) F / 3 (d) 2F, Q.7 Electric lines of force about negative point charge are, (a) Circular, anticlockwise (b) Circular, clockwise, (c) Radial, inward, (d) Radial, outward, Q.8 Figure shows the electric lines of force emerging from a, charged body. If the electric field at A and B are E A and, , of –3Q, it is, (a) –E, (b) E/3, (c) –3E, (d) –E/3, Q.12 Charges q, 2q, 3q and 4q are placed at the corners A, B, C, and D of a square as shown in the following figure. The, direction of electric field at the centre of the square is, parallel to side., D, C, (a) AB, 3q, 4q, (b) CB, O, (c) BD, q, 2q, B, A, (d) AC, Q.13 Three infinitely long non-conducting charge sheets are, placed as shown in figure. The electric field at point P is, 2s $, (a), k, Z, e0, , (a), , E A > EB, , (c), , (b) E A < E B, B, A, E, r, (c) E A = B, r, EB, (d) E A = 2, r, Q.9 The magnitude of electric field intensity E is such that,, an electron placed in it would experience an electrical, force equal to its weight is given by, mge, , mg, e, , e2, , e, , (c) mg, (d) 2 g, m, Q.10 A charge particle is free to move in an electric field. It, will travel, (a) Always along a line of force, (b) Along a line of force, if its initial velocity is zero, (c) Along a line of force, if it has some initial velocity in, the direction of an acute angle with the line of force, (d) None of the above, Q.11 Two point charges Q and –3Q are placed at some distance apart., If the electric field at the location of Q is E then at the locality, (a), , RESPONSE, GRID, , (b), , 2s $, k, e0, , (b) –, , EB respectively and if the distance between A and B is r then, , Z = 3a, P, , 4s $, k, e0, , -2 s, , Z=a, x, , 4s $, k, e0, Q.14 Gauss's law is true only if force due to a charge varies as, (a) r–1, (b) r–2, (c) r–3, (d) r–4, Q.15 The electric intensity due to an infinite cylinder of radius, R and having charge q per unit length at a distance r(r > R), from its axis is, (a) Directly proprotional to r2, (b) Directly proprotional to r3, (c) Inversely proprotional to r, (d) Inversely proprotional to r2, Q.16 A sphere of radius R has a uniform distribution of electric, charge in its volume. At a distance x from its centre, for, x < R, the eletric field is directly proportional to, Z=–a, , (d) –, , 1, , 1, (b), (c) x, (d) x2, x, x2, Q.17 A charged ball B hangs from a silk thread S, which makes an, angle q with a large charged conducting sheet P, as shown, , (a), , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 32, , 3, , in the figure. The surface charge density s of the sheet is, proportional to, +, (a) sin q, +, P +, (b) tan q, + q, +, S, +, (c) cos q, +, (d) cot q, B, Q.18 A charge q is placed at the centre of a cube. Then the flux, passing through one face of cube will be, q, q, q, q, (a), (b), (c), (d), 6, Î0, 4, Î, Î0, 2 Î0, 0, DIRECTIONS (Q.19-Q.20) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.19 A solid sphere S1 is connected to a charge reservoir through, a heater H as shown in figure., S2, , S1, Solid, sphere, , H, Heater, , Solid, sphere, , Flux through a closed spherical surface around S1 is given, by f = a t2 where a is a constant and t is time in seconds., If resistance of heater is R then select correct statements, (1) Power consumed by heater will be 4a2e02 Rt2., (2) Electric flux through a closed spherical surface around, S2 will be – a t2., (3) Rate of change of electric flux through a closed, spherical surface around S2 will be –2a t, (4) All of the above are correct, Q.20 A simple pendulum has a time period T. The bob is now, given some positive charge –, (1) If some positive charge is placed at the point of, suspension, T will increases, (2) If some positive charge is placed at the point of, suspension, T will not change, , RESPONSE, GRID, , 17., , 18., , 22., , 23., , (3) If a uniform downward electric field is switched on, T will, increase, (4) If a uniform downward electric field is switched on,, T will decrease, DIRECTIONS (Q.21-Q.23) : Read the passage given below, and answer the questions that follows :, A sphere of radius R contains charge density r (r ) = A( R - r ),, for 0 < r < R. The total electric charge inside the sphere is Q., Q.21 The value of A in terms of Q and R is, (a), , 2Q 2, , (b), , 3Q, , (c), , 3Q 2, , pR 4, pR 4, pR 3, Q.22 The electric field inside the sphere is, , (d), , 3Q, pR, , 2, 12Q2 é 1 æ r ö 1 æ r ö ù, ê ç ÷- ç ÷ ú, R3 êë 3 è R ø 4 è R ø úû, 2, 2, 12 é 1 æ r ö 1 æ r ö ù, 120Q é 1 æ r ö 1 æ r ö ù, (c), ê ç ÷ - ç ÷ ú (d) 2 ê ç ÷ - ç ÷ ú, R Q êë 3 è R ø 4 è R ø úû, 5 Î0 R 2 êë 4 è R ø 3 è R ø úû, , (a), , 2, 3Q é 1 æ r ö 1 æ r ö ù, ê ç ÷- ç ÷ ú, Î0 R 2 êë 3 è R ø 4 è R ø úû, , (b), , 1 ö, Q.23 The electric field outside the sphere is æç k =, 4, p, Î0 ÷ø, è, , (a), , kQ, r, , (b), , kQ, r2, , (c), , kQ, r3, , (d), , kQ 2, r2, , DIRECTIONS (Q. 24-Q.26) : Each of these questions, contains two statements: Statement-1 (Assertion) and, Statement-2 (Reason). Each of these questions has four, alternative choices, only one of which is the correct answer., You have to select the correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.24 Statement-1 : Electric lines of force cross each other., Statement-2 : Electric field at a point superimpose to give, , 19., , Space for Rough Work, , 20., , 21.
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DPP/ P 32, , 4, one resultant electric field., Q.25 Statement-1 : A point charge is brought in an electric field. The, field at a nearby point will increase, whatever be the nature of, the charge., Statement-2 : The electric field is dependent on the nature of, charge., , RESPONSE GRID, , t.me/Magazines4all, , 24., , 25., , Q.26 Statement-1 : Direction of electric field at a point signifies, direction of force experienced by a point charge placed at, that point., Statement-2 : Electric field is a vector quantity., , 26., , DAILY PRA CTICE PROBLEM SHEET 32 - PHYSICS, Total Questions, 26, Total Marks, 104, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 25, Qualifying Score, 40, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 33, SYLLABUS : Electrostatics-2 (Electric potential and potential difference,, equipotential surfaces, electric dipole), , Max. Marks : 104, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 26 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.19) : There are 19 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d),, out of which ONLY ONE choice is correct., Q1. In the electric field of a point charge q , a certain charge is, carried from point A to B, C, D and E. Then the work done, by electric force is, A, (a) least along the path AB, (b) least along the path AD, (c) zero along all the paths AB,, +q, B, E, AC, AD and AE, C, D, (d) least along AE, , RESPONSE GRID, , 1., , 2., , Q2. Four equal charges Q are placed at the four corners of a, square of each side ' a ' . Work done in removing a charge, – Q from its centre to infinity is, , Q2, 2Q2, 2Q2, (a) 0, (b), (c), (d), pÎ0 a, 2pÎ0 a, 4pÎ0 a, Q3. A particle A has charge +q and a particle B has charge +4q, with each of them having the same mass m. When allowed, to fall from rest through the same electric potential, vA, difference, the ratio of their speed, will become, vB, (a) 2 : 1, (b) 1 : 2, (c) 1 : 4, (d) 4 : 1, , 3., Space for Rough Work
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , DPP/ P 33, , 2, Q4. In the figure the charge Q is at the centre of the circle., Work done (by electrostatic force on q) is maximum when, another charge q is taken from point P to (consider both, the charges to be positive), P, , Q, , L, , M, , (a) K, (b) L, (c) M, (d) N, Q5. How much kinetic energy will be gained by an a- particle, in going from a point at 70V to another point at 50V ?, (a) 40 eV (b) 40 keV (c) 40 MeV (d) 0 eV, Q6. Ten electrons are equally spaced and fixed around a circle, of radius R. Relative to V = 0 at infinity, the electrostatic, potential V and the electric field E at the centre C are, ur, ur, (a) V ¹ 0 and E ¹ 0, (b) V ¹ 0 and E = 0, ur, ur, (c) V = 0 and E = 0, (d) V = 0 and E ¹ 0, Q7. The displacement of a charge Q in the electric field, r, ur, E = e1i$ + e2 $j + e3 k$ is r = ai$ + b $j . The work done is, (a) Q ( ae1 + be2 ), , A, (Take 1/ 4pe 0 =1010 Nm2 /C2 ), (a) 2.8 J, (b) 3.5 J, (c) 4.5 J, q, (d) 5.5 J, B, 15 cm, Q11. Electric charges q, q, –2q are placed at the corners of an, equilateral triangle ABC of side l. The magnitude of electric, dipole moment of the system is, q1, , N, , (b) Q, , ( ae1 )2 + (be2 )2, , æ 2 2ö, (d) Q ç e1 + e2 ÷ ( a + b ), è, ø, Q8. As shown in the figure, charges + q and -q are placed at, the vertices B and C of an isosceles triangle. The potential, at the vertex A is, , 5 cm, , K, , Q9. Two electric charges 12µC and – 6µC are placed 20 cm, apart in air. If there will be a point P on the line joining these, charges and outside the region between them, at which the, electric potential is zero, then the distance of P from – 6µC, charge is, (a) 0.10 m (b) 0.15 m (c) 0.20 m (d) 0.25 m, Q10. In the rectangle shown below, the two corners have charges, q1 = –5 mC and q2 = +2.0 mC. The work done by external, agent in moving a charge q = +3.0 mC slowly from B to A is, , 2, , (a), , (d) 4ql, 3ql, Q12. A charge (–q) and another charge (+Q) are kept at two, points A and B respectively. Keeping the charge (+Q) fixed, at B, the charge (–q) at A is moved to another point C such, that ABC forms an equilateral triangle of side l . The net, work done by electrostatic field in moving the charge (–q), is, , (c) Q ( e1 + e2 ) a 2 + b 2, , 1, 2q, ., 4pe0 a 2 + b2, (b) Zero, 1, q, ., (c), 4pe0 a 2 + b2, , A, , (a), , (d), , ( -q ), 1, ., 4pe0 a 2 + b2, , RESPONSE, GRID, , a, B, , b, , b, –q, , +q, , C, , (a), (c), , ql, , (b) 2ql, , 1 Qq, 4pÎ0 l, 1, Qql, 4p Î0, , (c), , 1, , Qq, , (b) 4pÎ 2, 0 l, (d) zero, , Q13. In an hydrogen atom, the electron revolves around the, nucleus in an orbit of radius 0.53 × 10–10m. Then the, electrical potential produced by the nucleus at the position, of the electron is, (a) –13.6V (b) –27.2V (c) 27.2V (d) 13.6V, Q14. Point charge q1 = 2mC and q 2 = -1mC are kept at points x =, 0 and x = 6 respectively. Electrical potential will be zero at, points, (a) x = 2 and x = 9, (b) x = 1 and x = 5, (c) x = 4 and x = 12, (d) x = -2 and x = 2, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., Space for Rough Work
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 33, , 3, , Q15. The distance between H+ and Cl– ions in HCl molecule is 1.28, Å. What will be the potential due to this dipole at a distance of, 12 Å on the axis of dipole, (a) 0.13 V (b) 1.3 V (c) 13 V, (d) 130 V, Q16. Two identical thin rings each of radius R metres are, coaxially placed at a distance R metres apart. If Q1 coulomb, and Q2 coulomb are respectively the charges uniformly, spread on the two rings, the work done by external agent in, moving a charge q slowly from the centre of ring with, charge Q1 to that of other is, q(Q2 - Q1 )( 2 - 1), (a) zero, (b), 2.4 p Î0 R, (c), , q 2(Q1 + Q2 ), 4p Î0 R, , (d), , DIRECTIONS (Q.20-Q.21) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, , Q20. Consider a system of three charges, , q (Q1 + Q2 )( 2 + 1), 4p Î0 R, , (c) ¥, , (b), , q, 8pÎ0 x0 ln 2, , (d), , q ln 2, 4pÎ0 x0, , (1) The electric field at point O is, , 16., , 20., , 21., , y, B, , C, O, , x, , 60°, , q, (3) The potential at point O is, 12pe 0 R, , A, , (4) The magnitude of the force between the charges at C, and B is, , q2, 54pe 0 R 2, , Q21. For spherical symmetrical charge distribution, variation, of electric potential with distance from centre is given in, , y = +1 cm. Then the potentials at the points A, B and C, satisfy, (a) VA < VB (b) VA > VB (c) VA < VC (d) VA > VC, Q19. A point Q lies on the perpendicular bisector of an electrical, dipole of dipole moment p . If the distance of Q from the, dipole is r (much larger than the size of the dipole), then, electric field at Q is proportional to, (a) P–1and r –2, (b) p and r–2, (c) P2 and r–3, (d) p and r–3, , 15., , q, , 8pe 0 R 2, , directed along the negative x-axis, (2) The potential energy of the, system is zero, , diagram. Given that : V =, , Q18. A uniform electric field pointing in positive x - direction, exists in a region. Let A be the origin, B be the point on the, x - axis at x = +1 cm and C be the point on the y - axis at, , RESPONSE, GRID, , q q, 2q, , and placed at, 3 3, 3, , point A, B and C respectively, as shown in the figure. Take, O to be the centre of the circle of radius R and angle, CAB = 60°. Choose the incorrect options, , Q17. Identical point charges, each having + q charge, are fixed, at each of the points x = x0 , x = 3x0 , x = 5 x0 ........., infinite, on the x-axis and a identical point charges, each, having -q charge, ar e fixed at each of the points, x = 2 x0 , x = 4 x0 , x = 6 x0 ... .... i nfinit e. Her e x0 is a, positive constant. Potential at the origin due to the above, system of charges is, (a) 0, , (b) 1 and 2 are correct, (d) 1 and 3 are correct, , q, q, for r £ R0 and V =, 4, pe, 4pe 0 R 0, 0r, V, , for r ³ R0, Then which option (s) are correct :, (1) Total charge within 2R0 is q, r, (2) Total electrostatic energy, r = R0, for r £ R0 is non-zero, (3) At r = R0 electric field is discontinuous, (4) There will be no charge anywhere except at r < R0., , DIRECTIONS (Q.22-Q.23) : Read the passage given below, and answer the questions that follows :, An electric dipole (AB) consisting of two particles of equal and, opposite charge and same mass is released in an electric field., In the figure field lines are without considering effect of field, of dipole., , 17., , Space for Rough Work, , 18., , 19.
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t.me/Magazines4all, DPP/ P 33, , 4, B +q, A, , DIRECTIONS (Q. 24-Q.26) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., , y, , –q, x, , (a), , Q22. The centre of mass of the dipole, (a) Has no acceleration, (b) Has acceleration with positive x and y components, (c) Has acceleration with positive x component and negative y component, (d) Has acceleration with negative x component and positive y component, Q23. Angular acceleration of the dipole, immediately after it is, released, (a) is zero, (b) is clockwise, (c) is anticlockwise, (d) cannot be determined from the given information., , RESPONSE GRID, , 22., , 23., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q24. Statement -1 : A bird perches on a high power line and, nothing happens to the bird., Statement -2 : The level of bird is very high from the, ground., Q25. Statement -1 : Electrons move away from a low potential, to high potential region., Statement- 2 : Because electrons have negative charge, Q26. Statement -1 : Surface of a symmetrical conductor can, be treated as equipotential surface., Statement -2 : Charges can easily flow in a conductor., , 24., , 25., , 26., , DAILY PRA CTICE PROBLEM SHEET 33 - PHYSICS, Total Questions, 26, Total Marks, 104, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 34, SYLLABUS : ELECTROSTATICS -3 (Electrostatic Potential energy, conductors), , Max. Marks : 96, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 24 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.16) : There are 16 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 Three charges Q, + q and + q are placed at the vertices of, right-angled isosceles triangle as shown in the figure. The, net electrostatic energy of the configuration is zero if Q is, equal to, Q, (a), , -q, , (b), , 1+ 2, (c) –2q, , 1., , (b), , 1 2q 2, 4pÎ0 l, , (c), , 1 3q 2, 4pÎ0 l, , (d), , 1 4q 2, 4pÎ0 l, , of side b , then electric potential energy of charge, (+q) which is placed at centre of the cube will be, , +q, , +q, , (a), , 8 2q2, 4pÎ0 b, , (b), , -8 2q 2, p Î0 b, , (c), , -4 2q 2, pÎ0 b, , (d), , -4q2, 3pÎ0 b, , a, , Q.2 Three charges of equal value ‘q’ are placed at the vertices, of an equilateral triangle. What is the net potential energy,, if the side of equilateral Dis l ?, , RESPONSE GRID, , 1 q2, 4pÎ0 l, , Q.3 If identical charges (-q) are placed at each corner of a cube, , -2q, , 2+ 2, (d) + q, , (a), , 2., , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 34, , 2, Q.4 Two charges q1 and q2 are placed 30 cm apart, shown in, the figure. A third charge q3 is moved along the arc of a, circle of radius 40 cm from C to D. The change in the, potential energy of the system is, , q3, k , here k is, 4pÎ0, , (a) 8q2, , q3, C, , (b) 8q1, (c) 6q2, , 40 cm, , (d) 6q1, , q1, , q2, A, , 30 cm, , B, , D, , (a) decreases, (b) increases, (c) remains unchanged, (d) becomes zero, Q.9 Two identical charges are placed at the two corners of an, equilateral triangle. The potential energy of the system is, U. The work done in bringing an identical charge from, infinity to the third vertex is, (a) U, (b) 2U, (c) 3U, (d) zero, Q.10 Potential energy of two equal negative point charges 2mC, held 1 m apart in air is, (a) 2 J, (b) 2eV, (c) 4 J, (d) 0.036 J, Q.11 Four charges + q, –q, + q and –q are put together on four, corners of a square as shown in figure. The work done by, external agent in slowly assembling this configuration is, +q, , –q, , –q, , +q, , Q.5 Three particles, each having a charge of 10mC are placed, at the corners of an equilateral triangle of side 10 cm. The, electrostatic potential energy of the system is (Given, 1, = 9 ´109 N m 2 /C2 ), 4p Î0, , (a) Zero, (b) Infinite (c) 27 J, (d) 100 J, Q.6 Two equal charges q are placed at a distance of 2a and a, third charge -2q is placed at the midpoint. The potential, energy of the system is, (a), , q2, 8p Î0 a, , 3.0 nC, , 6q 2, (b), 8p Î0 a, , a, , 1.0 cm, , 9q 2, 7 q2, (c) –, (d), 8p Î0 a, 8p Î0 a, Q.7 An electric dipole has the magnitude of its charge is q and, its dipole moment is p . It is placed in a uniform electric, field E. If its dipole moment is along the direction of the, field, the force on it and its potential energy are, respectively, (a) 2q.E and minimum, (b) q.E and p.E, , (c) Zero and minimum, (d) q.E and maximum, Q8 In bringing an electron towards another electron,, electrostatic potential energy of the system:, , RESPONSE, GRID, , (a) zero, (b) –2.59kq2/a, (c) +2.59kq2/a, (d) none of these, Q.12 As shown in figure a dust particle with mass m = 5.0 × 10 –9, kg and charge q0 = 2.0 nC starts from rest at point a and, moves in a straight line to point b . What is its speed v at, point b?, b, 1.0 cm, , –3.0 nC, –, 1.0 cm, , (a) 26 ms–1 (b) 34 ms–1 (c) 46 ms–1 (d) 14 ms–1, Q.13 Charges –q, Q and –q are placed at equal distance on a, straight line. If the total potential energy of the system of, three charges is zero, then find the ratio Q/q., –q, , –q, , Q, r, , r, 2r, , (a) 1/2, (b) 1/4, (c) 2/3, (d) 3/4, Q.14 When the separation between two charges is increased,, the electric potential energy of the charges, (a) increases, (b) decreases, (c) remains the same, (d) may increase or decrease, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 34, , 3, , Q.15 A positive charge is moved from a low potential point A to, high potential point B. Then the electric potential energy, of the system, (a) increases, (b) decreases, (c) will remain the same, (d) nothing definite can be predicted, Q.16 If V and u are electric potential and energy density,, respectively, at a distance r from a positive point charge,, then which of the following graph is correct ?, V, , 4, , V, , Q.18 A proton moves a distance d in a uniform electric field, ®, E as shown in the figure. Then which of the following, statements are correct ?, ®, , E, , +, p, d, , (1), (2), (3), (4), , 4, , (b), , (a), u, V, , u, , 4, , V, , DIRECTIONS (Q.19-Q.21) : Read the passage given below, and answer the questions that follows :, , 4, , (d), , (c), u, , u, , DIRECTIONS (Q.17-Q.18) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.17 S is a solid neutral conducting sphere. A point charge q of, 1 × 10–6 C is placed at point A. C is the centre of sphere, and AB is a tangent. BC = 3m and AB = 4m., A, , B, , Three concentric spherical conductors A, B and C of radii R, 2R, and 4R respectively. A and C is shorted and B is uniformly, charged., +, , S, , +, , 15., , 16., , 20., , 21., , B +, A, , C, +, , +, +, +, , 4R, , +, , +, , R, , +, +, , +, , 2R, +, , +, +, , +, , +, , Q.19 Charge on conductor A is, (a) Q/3, (c) 2Q/3, Q.20 Potential at A is, Q, 4pe 0 R, , Q, 20pe0 R, Q.21 Potential at B is, Q, (a), 4 pe 0 R, 5Q, (c), 48pe0 R, , 17., , Space for Rough Work, , (b) – Q/3, (d) None of these, (b), , (c), , (1) The electric potential of the conductor is 1.8 kV, (2) The electric potential of the conductor is 2.25 kV, (3) The electric potential at B due to induced charges on, the sphere is – 0.45 kV, (4) The electric potential at B due to induced charges on, the sphere is 0.45 kV, , +, , +, , (a), , RESPONSE, GRID, , Electric field do a negative work on the proton, Electric potential energy of the proton increases, Electric field do a positive work on the proton, Electric potential energy of the proton decreases, , 18., , Q, 16 pe0 R, , (d) None of these, , (b), , Q, 16 pe0 R, , (d) None of these, , 19.
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t.me/Magazines4all, DPP/ P 34, , 4, DIRECTIONS (Q. 22-Q.24) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.22 Statement-1 : No work is done in taking a small positive, charge from one point to other inside a positively charged, metallic sphere while outside the sphere work is done in, taking the charge towards the sphere. Neglect induction, due to small charge., , Statement-2 : Inside the sphere electric potential is same, at each point, but outside it is different for different points., Q.23 Statement-1 : Electric potential of earth is taken to be, zero as a reference., Statement-2 : The electric field produced by earth in, surrounding space is zero., Q.24 Statement - 1 : The electric potential and the electric field, intensity at the centre of a square having four fixed point, charges at their vertices as shown in figure are zero., +q, , –q, , –q, , +q, , Statement - 2 : If electric potential at a point is zero then, the magnitude of electric field at that point must be zero., , RESPONSE GRID, , 22., , 23., , 24., , DAILY PRA CTICE PROBLEM SHEET 34 - PHYSICS, Total Questions, 24, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 22, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 96, , 39, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 35, SYLLABUS : ELECTROSTATICS-4 (Capacitors, dielectrics), , Max. Marks : 96, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 24 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.15) : There are 15 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A parallel plate capacitor is charged to a potential, difference of 50V. It is discharged through a resistance., After 1 second, the potential difference between plates, becomes 40V. Then, (a) Fraction of stored energy after 1 second is 16/25, (b) Potential difference between the plates after 2, seconds will be 30V, (c) Potential difference between the plates after 2, seconds will be 20V, (d) Fraction of stored energy after 1 second is 4/5, Q.2 Five identical plates each of area A are joined as shown in, the figure. The distance between the plates is d. The plates, are connected to a potential difference of V volts. The, charge on plates 1 and 4 will be respectively, , RESPONSE GRID, , 1., , 2., , (a), (c), (d), , e0 AV 2e0 AV, ,, 2d, 2d, e 0 AV -2e0 AV, ,, d, d, -e 0 AV -2e0 AV, ,, d, d, , (b), , e0 AV 2e0 AV, ,, 2d, 2d, , 1, , 2, , 3, , 4, , –, V, +, , 5, , Q.3 Figure given below shows two identical parallel plate capacitors, connected to a battery with switch S closed. The switch is now opened, and the free space between the plates of capacitors is filled with a, dielectric of dielectric constant 3. What will be the ratio of total, electrostatic energy stored in both capacitors before and after the, introduction of the dielectric?, , (a), (b), (c), (d), , 3., Space for Rough Work, , 3:1, 5:1, 3:5, 5:3, , V, , A, , B
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 35, , 3, , Q.14 A parallel plate capacitor is charged and the charging battery, is then disconnected. If the plates of the capacitor are moved, further apart by means of insulating handles, then, (a) The charge on the capacitor increases, (b) The voltage across the plates decreases, (c) The capacitance increases, (d) The electrostatic energy stored in the capacitor, increases, Q.15 A parallel plate capacitor of plate area A and plates separation, distance d is charged by applying a potential V0 between the, plates. The dielectric constant of the medium between the, plates is K. What is the uniform electric field E between the, plates of the capacitor ?, (a), (c), , E = Î0, E=, , CV0, KA, , (b), , V0, KA, , E=, , (d) E =, , DIRECTIONS (Q.19-Q.21) : Read the passage given below and, answer the questions that follows :, Capacitor C3 in the circuit is variable capacitor (its capacitance, can be varied). Graph is plotted between potential difference V1, (across capacitor C1) versus C3., Electric potential V1 approaches on asymptote of 10 volts as, C3 ® ¥., V1, , V0, Kd, , C1, V, , Kv 0 d, Î0 A, , DIRECTIONS (Q.16-Q.18) : In the following questions, more, than one of the answers given are correct. Select the correct, answers and mark it according to the following codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.16 A parallel plate air condenser is connected with a battery., Its charge, potenital difference, electric field and energy are, Q0 , V0 , E0 , and U0 , respectively. In order to fill the complete, space between the plates a dielectric slab is inserted, the, battery is still connected. Now the corresponding values Q,, V, E and U are in relation with the initially stated as, (1) V > V0 (2) Q > Q0 (3) E > E0 (4) U > U0, Q.17 The false statement are, on increasing the distance between, the plates of a parallel plate condenser,, (1) The electric field intensity between the plates will, decrease, (2) The electric field intensity between the plates will, increase, (3) The P. D. between the plates will decrease, (4) The electric field intensity between the plates will, remain unchanged, , RESPONSE, GRID, , Q.18 The capacitance of a parallel plate condenser depends on, (1) Area of the plates, (2) Medium between the plates, (3) Distance between the plates, (4) Metal of the plates, , C2, , C3, , 10, 8, 6, 4, 2, 2 4, , 6 8, , C3, , C1, will be, C2, (a) 2 / 3, (b) 4 / 3, (c) 3 / 4, (d) 3 / 2, Q.20 The value of C3 for which potential difference across C1, will become 8V, is, (a) 1.5C1, (b) 2.5C1 (c) 3.5 C1 (d) 4.5 C1, Q.21 The ratio of energy stored in capacitor C1 to that of total, energy when C3 ® ¥ is, (a) zero, (b) 1/3, (c) 1, (d) Data insufficient, , Q.19 The ratio of the capacitance, , DIRECTIONS (Q. 22-Q.24) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), (c), (d), , 14., , 15., , 16., , 19., , 20., , 21., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement-1 is False, Statement-2 is True., Statement-1 is True, Statement-2 is False., , Space for Rough Work, , 17., , 18.
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t.me/Magazines4all, DPP/ P 35, , 4, Q.22 Statement-1 : The force with which one plate of a parallel, plate capacitor is attracted towards the other plate is equal, to square of surface density per 2 Î0 per unit area., Statement-2 : The electric field due to one charged plate, of the capacitor at the location of the other is equal to, surface density per 2 Î0., Q.23 Statement-1 : Circuit containing capacitors should be, handled cautiously even when there is no current., , RESPONSE GRID, , 22., , 23., , Statement-2 : The capacitors are very delicate and so quickly, break down., Q.24 Statement-1 : If the distance between parallel plates of a, capacitor is halved and dielectric constant is made three, times, then the capacitance becomes 6 times., Statement-1 : Capacitance of the capacitor does not, depend upon the nature of the material of plates., , 24., , DAILY PRA CTICE PROBLEM SHEET 35 - PHYSICS, Total Questions, 24, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 22, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 96, , 40, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 36, SYLLABUS : CURRENT ELECTRICITY – 1 (Electric Current, drift velocity, Ohm's law, Electrical resistance,, Resistances of different materials, V-I characteristics of Ohm and non-ohmic conductors, electrical energy and power,, Electrical resistivity, Colour code of resistors, Temperature dependance of resistance), , Max. Marks : 92, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 23 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.14) : There are 14 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 In the following fig. the ratio of current in 3W and 1W, resistances is–, 3W, (a) 1/3, 1W, 3A, (b) 2/3, Y, Z, X, (c) 1, 6W, (d) 2, Q.2 The resultant resistance between the points A and B in the, following fig. will be –, , RESPONSE GRID, , 1., , 2., , 1W, 1W, 1W, 1W, (a) 4 W, A, (b) 8 W, 2W, 2W, 2W, (c) 6 W, (d) 2 W, B, Q.3 How will reading in the ammeter, Switch, A be affected if an other, identical bulb Q is connected in Mains, P, parallel to P as shown in the fig., The voltage in the mains is, A, maintained at constant value, (a) the reading will be reduced to one half., (b) the reading will be double of previous one., (c) the reading will not be affected., (d) the reading will increase four fold., , 3., Space for Rough Work, , 1W, , Q
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t.me/Magazines4all, DPP/ P 36, , 2, Q.4 In the circuit shown, the galvanometer G reads zero. If, batteries have negligible internal resistances, the value of, resistance X wil be –, , Q.10 In the fig. shown, Calculate the current through 3 ohm, resistor. The emf of battery is 2 volt and its internal, resistance is 2/3 ohm., , (a) 10 W, (b) 100 W, (c) 200 W, (d) 500 W, Q.5 A cylindrical wire is stretched to increase its length by, 10%. The percentage increase in the resistance of the wire, will be–, (a) 20%, (b) 21%, (c) 22%, (d) 24%, Q.6 In the figure, the equivalent resistance between A and B is–, (a) 2R/3, (b) R/3, (c) R, (d) 3R, Q.7 In the adjoining network of resistors, each is of resistance, r ohm, the equivalent resistance between points A and B is–, , (a) 0.33 amp., (b) 0.44 amp., (c) 1.22 amp., (d) 0.88 amp., Q.11 The current in the given circuit will be, 1, A, (a), 45, i, 1, A, 30 W, (b), 15, 2V, 1, A, (c), 10, 30 W, , 1, A, 5, Q.12 The equivalent resistance of the following infinite network, of resistance is, , (a) 5r, , (d), , (b) 2r/3, (c) r, , 2W, , (d) r/2., Q.8 In the figure a carbon resistor has bands of different, colours on its body as mentioned in the figure. The value, of the resistance is, Silver, (a) 2.2 k W, (b) 3.3 k W, (c) 5.6 k W, White, (d) 9.1 k W, Brown Red, Q.9 Two wires of same material have length L and 2L and crosssectional areas 4A and A respectively. The ratio of their, specific resistance would be, (a) 1 : 2, (b) 8 : 1, (c) 1 : 8, (d) 1 : 1, , RESPONSE, GRID, , 30 W, , 2W, 2W, , 2W, 2W, 2W, , 2W, 2W, 2W, , (a) Less than 4 W, (b) 4W, (c) More than 4W but less than 12W, (d) 12 W, Q.13 A heater coil connected to a supply of a 220 V is dissipating, some power P1. The coil is cut into half and the two halves, are connected in parallel. The heater now dissipates a power, P2. The ratio of power P1 : P2 is, (a) 2 : 1, (b) 1 : 2, (c) 1 : 4, (d) 4 : 1, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 36, , 3, (1), (2), (3), (4), , Q.14 An electric lamp is marked 60 W, 230 V. The cost of a 1, kWh of energy is ` 1.25. The cost of using this lamp 8 hrs, a day for 30 days is (approximately), (a) ` 10, (b) ` 16, (c) ` 18, (d) ` 20, DIRECTIONS (Q.15-Q.17) : In the following questions,, more than one .of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.15 In the fig below the bulbs are identical, The bulbs, light, most brightly are, E, , F, , A, B, , C, D, , P, , Q, , (1) A, (2) B, (3) F, (4) D, Q.16 An electric kettle has two heating coils. When one of the, coils is switched on, the water begins to boil in 6 minutes., When the other is switched on, the boiling begins in 8, minutes. The time when the boiling begin if both coils are, switched on simultaneously is (i) in series (ii) in, parallel, (1) 14 min in series, (2) 3.43 min in parallel, (3) 3.43 min in series, (4) 14 min in parallel, Q.17 For the circuit shown in the figure, , RESPONSE, GRID, , The potential difference across RL is 18 V, The current I through the battery is 7.5 mA, Ratio of powers dissipated in R1 and R2 is 3, If R1 and R2 are interchanged magnitude of the power, dissipated in RL will decrease by a factor of 9, , DIRECTIONS (Q.18-Q.20) : Read the passage given below, and answer the questions that follows :, In the circuit shown in the figure,, 12 V, 1W, , 5W, Q.18 Rate of conversion of chemical energy within the battery, is, (a) 24 W, (b) 20 W (c) 4 W, (d) 14 W, Q.19 Rate of dissipation of electrical energy in battery is, (a) 24 W, (b) 20 W (c) 4 W, (d) 14 W, Q.20 Rate of dissipation of electrical energy in external resistor, is, (a) 4 W, (b) 20 W (c) 14 W (d) 24 W, , DIRECTIONS (Q. 21-Q.23) : Each of these questions contains, two statements': Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.21 Statement-1 : The resistivity of a semiconductor decreases, with temperature., Statement-2 : The atoms of a semiconductor vibrate with, larger amplitude at higher temperatures thereby increasing, its resistivity., , 14., , 15., , 16., , 19., , 20., , 21., , Space for Rough Work, , 17., , 18.
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t.me/Magazines4all, DPP/ P 36, , 4, Q.22 Statement-1 : In a simple battery circuit the point of lowest, potential is negative terminal of the battery., Statement-2 : The current flows towards the point of the, higher potential as it flows in such a circuit from the, negative to the positive terminal., , RESPONSE GRID, , 22., , Q.23 Statement-1 : The temperature coefficient of resistance is, positive for metals and negative for p-type semiconductor., Statement-2 : The effective charge carriers in metals are, negatively charged whereas in p-type semiconductor they, are positively charged., , 23., , DAILY PRA CTICE PROBLEM SHEET 36 - PHYSICS, Total Questions, 23, Total Marks, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 24, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 92, , 40, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 37, SYLLABUS : CURRENT ELECTRICITY – 2 Electrical cell and its internal resistance, Potential difference and E.M.F, of a cell, Combination of cells in series and in parallel, Kirchoff's laws and their applications, RC transient circuit,, Galvanometer, Ammeter, Voltmeter], , Max. Marks : 104, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, , The Daily Practice Problem Sheet contains 26 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.18) : There are 18 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The voltmeter shown in fig, reads 6V across the 60 W, resistor. Then the resistance of the voltmeter is(a) 0 W, (b) ¥ W, (c) 200 W, (d) 300 W, Q.2 If only one hundredth part of total current flowing in the, circuit is to be passed through a galvanometer of resistance, GW, Then the value of shunt resistance required will be-, , RESPONSE GRID, , 1., , 2., , (a) G/10, (b) G/100 (c) G/99, (d) G/999, Q.3 The shunt required for 10% of main current to be sent, through the moving coil galvanometer of resistance 99W, will be(a) 0.9 W, (b) 11 W, (c) 90 W, (d) 9.9 W, Q.4 The reading of voltmeter in the following circuit will be20 W, , 80 W, V, ®, , •, , 80 W, + –, 2V, , (a) 2 volt, , 3., Space for Rough Work, , (b) 0.80 volt (c) 1.33 volt (d) 1.60 volt, , 4.
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t.me/Magazines4all, DPP/ P 37, , 2, Q.5 The figure below shows currents in a part of electric circuit., The current i is, 1 amp, 2 amp, (a) 1.7 amp, 1.3 amp, (b) 3.7 amp, (c) 1.3 amp, 2 amp, (d) 1 amp, i, , Q.6 A voltmeter can measure upto 25 volt and its resistance is, 1000 W. The resistance required to add with voltmeter to, measure upto 250 volt will be(a) 9000 W (b) 1000 W (c) 2500 W (d) 900 W, Q.7 When a Laclanche cell is connected to a 10W resistance, then a current of 0.25 ampere flows in the circuit. If the, resistance is reduced to 4W then current becomes, 0.5ampere. The internal resistance of galvanometer will be(a) 1.5 W, (b) 0.5 W, (c) 1 W, (d) 2 W, Q.8 Consider the circuit shown in the figure. The value of, current I3 is, 28 W, , (a) 5 A, , 54 W, , (b) 3 A, , B, , 6V, , (c) – 3 A, , I3, , (d) – 5/6 A, , 8V, , 10 W, , (b) 10 W, , 5V, , A, 2V, , (b) – 0.4 A, (c) 0.8 A, , (b) Potential adder, (c) Potential substracter, , ( ), Total P.D., , (d) Potential multiplier, , I1, I2, , (d) – 0.8 A, , RESPONSE, GRID, , (a) Potential divider, , X, , Q.10 In the given circuit the current I1 is, (a) 0.4 A, , 5, R, 6, (d) 8 R, Q.14 The arrangement as shown in figure is called as, , B, , (c) 15 W, (d) 20 W, , (a) The reading of ammeter will decrease, (b) The reading of ammeter will increase, (c) The reading of ammeter will remain unchanged, (d) The reading of ammeter will become zero., Q.13 Twelve wires of equal length and same cross-section are, connected in the form of a cube. If the resistance of each, of the wires is R, then the effective resistance between, the two diagonal ends would be, (a) 2 R, (b) 12 R, (c), , 12 V, , Q.9 If VB – VA = 4 V in the given figure, then resistance X will, be, (a) 5 W, , Q.11 To get the maximum current from a parallel combination, of n identical cells each of internal resistance r in an, external resistance R,, (a) R >> r (b) R << r (c) R > r, (d) R = r, Q.12 In the circuit shown below, if the value of R is increased, then what will be the effect on the reading of ammeter if, the internal resistance of cell is negligible-, , 30 W, 40 W, , 40 V I3, , 40 W, , 80 V, , Variable P.D., , Q.15 When a cell of emf E and internal resistance r, is connected, to the ends of a resistance R, then current through, resistance is I. If the same cell is connected to the ends of, a resistance R/2 then the current would be(a) less than I, (b) I, (c) greater then I but less than 2I, (d) greater than 2I, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 37, , 3, , Q.16 The resistance of an ideal voltmeter is, (a) Zero, (b) Very low, (c) Very large, (d) Infinite, Q.17 An ammeter with internal resistance 90 W reads 1.85 A, when connected in a circuit containing a battery and two, resistors 700 W and 410 W in series. Actual current will, be, (a) 1.85 A, (b) Greater than 1.85 A, (c) Less than 1.85 A, (d) None of these, Q.18 The figure shows a network of currents. The magnitude of, currents is shown here. The current I will be, 1A, , I, , 10 A, , Q.21 A microammeter has a resistance of 100W and a full scale, range of 50 µA. It can be used as a voltmeter or a higher, range ammeter provided a resistance is added to it. Pick, the correct range and resistance combination(s)., (1) 10V range with 200 kW resistance in series., (2) 50V range with 10 kW resistance in series., (3) 5 mA range with 1W resistance in parallel., (4) 10 mA range with 1 kW resistance in parallel., DIRECTIONS (Q.22-Q.23) : Read the passage given below, and answer the questions that follows :, A 6V battery of negligible internal resistance is connected across, a uniform wire AB of length 100cm. The positive terminal of, another battery of emf 4V and internal resistance 1W is joined, to the point A as shown in figure. Take the potential at B to be, zero., , 6A, , 6V, , 2A, (b) 9 A, (d) 19 A, , (a) 3 A, (c) 13 A, , A, , DIRECTIONS (Q.19-Q.21) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.19 In the figure,, X, E, r, Y, 15 V, 6V, (1) current may flow from X to Y, (2) current may flow from Y to X, (3) current’s direction depends on E, (4) current’s direction depends on r, Q.20 Kirchoff’s laws are based on conservation of, (1) charge, (2) potential, (3) energy, (4) mass, , RESPONSE, GRID, , D, 4V, , B, C, , 1, , Q.22 What are the potentials at points A and C ?, (a) 6V, 2V, (b) 8V, 4V, (c) 6V, 4V, (d) 8V, 3V, Q.23 If the points C and D are connected by a wire, what will be, the current through it ?, (a) zero, (b) 1A, (c) 2A, (d) 3A, DIRECTIONS (Qs. 24-Q.26) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., , 16., , 17., , 18., , 21., , 22., , 23., , Space for Rough Work, , 19., , 20.
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t.me/Magazines4all, DPP/ P 37, , 4, Q.24 Statement -1 : Voltameter measures current more, accurately than ammeter., Statement -2 : Relative error will be small if measured, from voltameter., Q.25 Statement - 1 :, , Q.26 Statement - 1 :, if, I = 2A., , b, 2, a, 4V, Statement - 2 : Potential difference across the, terminals of a non ideal battery is less than its emf when a, current flows through it., , A larger dry cell has higher emf., , Statement - 2 : The emf of a dry cell is independent of its size., , RESPONSE GRID, , 24., , 25., , In the circuit shown, Vab or Va – Vb = 0,, , 26., , DAILY PRA CTICE PROBLEM SHEET 37 - PHYSICS, Total Questions, 26, Total Marks, 104, Attem pted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 42, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 38, SYLLABUS : CURRENT ELECTRICITY-3 : Wheatstone bridge, Meter bridge, Potentiometer-principle and its, applications., , Max. Marks : 100, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 25 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.16) : There are 16 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A Potentiometer wire of length 1 m is connected in series, with 490 W resistance and 2V battery. If 0.2 mV/cm is the, potential gradient, then resistance of the potentiometer, wire is, (a) 4.9 W, (b) 7.9 W, (c) 5.9 W, (d) 6.9 W, Q.2 Two resistances are connected in two gaps of a metre, bridge. The balance point is 20 cm from the zero end. A, resistance of 15 ohms is connected in series with the, smaller of the two. The null point shifts to 40 cm. The, value of the smaller resistance in ohm is, (a) 3, (b) 6, (c) 9, (d) 12, , RESPONSE GRID, , 1., , 2., , Q.3 In a potentiometer experiment the balancing with a cell is, at length 240 cm. On shunting the cell with a resistance of, 2 W, the balancing length becomes 120 cm. The internal, resistance of the cell is, (a) 4 W, (b) 2 W, (c) 1 W, (d) 0.5 W, Q.4 A potentiometer consists of a wire of length 4 m and, resistance 10 W . It is connected to cell of emf 2 V. The, potential difference per unit length of the wire will be, (a) 0.5 V/m (b) 10 V/m, (c) 2 V/m (d) 5 V/m, Q.5 In given figure, the potentiometer wire AB has a resistance, of 5 W and length 10 m. The balancing length AM for the, emf of 0.4 V is, (a) 0.4 m, (b) 4 m, (c) 0.8 m, (d) 8 m, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 38, , 2, Q.6 In the circuit shown in the figure, the current flowing in, 2W resistance, 2W, 10 W, (a) 1.4 A, 1.4 A, (b) 1.2 A, G, (c) 0.4 A, 25 W, 5W, (d) 1.0 A, Q.7 For the post office box arrangement to determine the value, of unknown resistance the unknown resistance should be, connected between, , (a) 33.3 cm (b) 66.67 cm (c) 25 cm (d) 50 cm, Q.12 A potentiometer has uniform potential gradient across it., Two cells connected in series (i) to support each other, and (ii) to oppose each other are blanced over 6m and 2m, respectively on the potentiometer wire. The e.m.f’s of the, cells are in the ratio of, (a) 1 : 2, (b) 1 : 1, (c) 3 : 1, (d) 2 : 1, Q.13 In a potentiometer experiment two cells of e.m.f E1 and E2, are used in series and in conjunction and the balancing, length is found to be 58 cm of the wire. If the polarity of, E2 is reversed, then the balancing length becomes 29 cm., E1, , The ratio E of the e.m.f. of the two cells is, 2, (a), , (a) B and C (b) C and D, (c) A and D (d) B1and C1, Q.8 The e.m.f. of a standard cell balances across 150 cm length, of a wire of potentiometer. When a resistance of 2W is, connected as a shunt with the cell, the balance point is, obtained at 100 cm. The internal resistance of the cell is, (a) 0.1 W, (b) 1W, (c) 2W, (d) 0.5W, Q.9 Five resistors are connected as shown in the diagram. The, equivalent resistance between A and B is, C, (a) 6 W, 5W, 4W, (b) 9 W, 9W, A, B, (c) 12 W, 8W, 10 W, (d) 15 W, D, Q.10 A potentiometer has uniform potential gradient. The, specific resistance of the material of the potentiometer, wire is 10–7 ohm-meter and the current passing through it, is 0.1 ampere; cross-section of the wire is 10–6m2. The, potential gradient along the potentiometer wire is, (a) 10–4 V/m, (b) 10–6 V/m, –2, (c) 10 V/m, (d) 10–8 V/m, Q.11 Resistance in the two gaps of a meter bridge are 10 ohm, and 30 ohm respectively. If the resistances are interchanged, the balance point shifts by, , RESPONSE, GRID, , 1:1, , (b) 2 : 1, , (c) 3 : 1, , (d) 4 : 1, , Q.14 The resistance of a 10 meter long potentiometer wire is, 1 ohm/metre. A cell of e.m.f. 2.2 volts and a high resistance, box are connected in series with this wire. The value of, resistance taken from resistance box for getting potential, gradient of 2.2 millivolt/metre will be, (a) 790 W, (b) 810 W, (c) 990 W (d) 1000 W, Q.15 In the shown arrangement of the experiment of the meter, bridge if AC correspondin g to null deflection of, galvanometer is x, what would be its value if the radius of, the wire AB is doubled, , R1, , R2, G, , A, , x, , C, , B, , (a) x, (b) x / 4, (c) 4x, (d) 2x, Q.16 In meter bridge or Wheatstone bridge for measurement of, resistance, the known and the unknown resistances are, interchanged. The error so removed is, (a) End correction, (b) Index error, (c) Due to temperature effect, (d) Random error, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 38, , 3, , DIRECTIONS (Q.17-Q.19) : In the following questions,, more than one of the answers given are correct. Select the, correct answ]ers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.17 Which of the following statements are correct?, (1) Voltmeter should have high resistance., (2) Ammeter should have low resistance., (3) Voltmeter is placed in parallel across the conductor, in a circuit., (4) Ammeter is placed in parallel across the conductor, in a circuit., Q.18 Which are correct statements?, (1) The Wheatstone bridge is most sensitive when all the, four resistances are of the same order, (2) Kirchhoff’s first law (for currents meeting at a junction in, an electric circuit) expresses the conservtion of charge., (3) The rheostat can be used as a potential divider., (4) In a balanced Wheatstone bridge, interchanging the, positions of galvanometer and cell affects the balance, of the bridge., Q.19 Figure shows a balanced Wheatstone's bridge, R = 5W, , Q = 50W, , A, , C, S = 10W, , P = 100W, , DIRECTIONS (Q.20-Q.22) : Read the passage given below, and answer the questions that follows :, A battery is connected to a potentiometer and a balance point is, obtained at 84 cm along the wire. When its terminals are, connected by a 5W resistor, the balance point changes to 70 cm, Q.20 Calculate the internal resistance of the cell., (a) 4 W, (b) 2 W, (c) 5 W, (d) 1 W, Q.21 Find the new position of the balance point when 5W, resistance is replaced by 4W resistor., (a) 26.5 cm (b) 52 cm, (c) 67.2 cm (d) 83.3 cm, Q.22 How can we change a galvanometer with Re = 20.0W and, Ifs = 0.00100 A into a voltmeter with a maximum range of, 10.0 V?, (a) By adding a resistance 9980 W in parallel with the, galvanometer, (b) By adding a resistance 9980 W in series with the, galvanometer, (c) By adding a resistance 8890 W in parallel with the, galvanometer, (d) By adding a resistance 8890 W in series with the, galvanometer, DIRECTIONS (Q. 23-Q.25) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , G, (1) If P is slightly increased, the, galvanometer flows from A to C., (2) If P is slightly increased, the, galvanometer flows from C to A., (3) If Q is slightly increased, the, galvanometer flows from C to A., (4) If Q is slightly increased, the, galvanometer flows from A to C., , RESPONSE, GRID, , current in the, current in the, current in the, current in the, , 17., , 18., , 22., , 23., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.23 Statement -1 : In meter bridge experiment, a high, resistance is always connected in series with a, galvanometer., Statement -2 : As resistance increases current through the, circuit increases., , 19., , Space for Rough Work, , 20., , 21.
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DPP/ P 38, , 4, Q.24 Statement -1: A potentiometer of longer length is used, for accurate measurement., Statement -2: The potential gradient for a potentiometer, of longer length with a given source of e.m.f. becomes, small., , RESPONSE GRID, , t.me/Magazines4all, , 24., , Q.25 Statement -1: The e.m.f. of the driver cell in the, potentiometer experiment should be greater than the e.m.f., of the cell to be determined., Statement -2: The fall of potential across the, potentiometer wire should not be less than the e.m.f. of, the cell to be determined., , 25., , DAILY PRA CTICE PROBLEM SHEET 38 - PHYSICS, Total Questions, 25, Total Marks, 100, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 39, SYLLABUS : MAGNETIC EFFECTS OF CURRENT-1 (Magnetic field due to current carrying wires, Biot savart law), , Max. Marks : 108, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 27 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.19) : There are 19 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The magnitude of magnetic field at a point having, perpendicular distance 50 mm from a long straight, conducting wire carrying a current of 3A is, (a) 0.12 G, (b) 1.2 G, (c) 12 G, (d) 0.012 G, Q.2 A circular arc of wire of radius of curvature r subtends an, angle of p/4 radian at its centre. If i current is flowing in, it then the magnetic induction at its centre is m0 i, m i, (a), (b) 0, 8r, 4r, m0 i, (c), (d) 0, 16r, , RESPONSE GRID, , 1., , 2., , Q.3 A current i is flowing in a conductor PQRST shaped as, shown in the figure. The radius of curved part QRS is r and, length of straight portions PQ and ST is very large. The, magnetic field at the centre O of the curved part is R, , r, , Q, , O, , i, , P, , i, , T z, , 90°, S, , y, x, , (a), , µ0 i é 3 p ù ˆ, + 1ú k, 4 pr êë 2, û, , (b), , µ0i é 3p ù ˆ, - 1ú k, 4 pr êë 2, û, , (c), , µ0i é 3p ù ˆ, + 1ú (- k), 4pr êë 2, û, , (d), , µ0i é 3p ù ˆ, - 1ú (- k), 4pr êë 2, û, , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 39, , 2, Q.4 Consider the loop PQRSP, carrying clockwise current i, shown, in the figure. The magnitude of magnetic field at the centre O of, the curved portion is, (a), , m0 i, [p – f + tan f], 2p r, , S, , m0 i, 2p r, (c) 0, , (b), , (d), , O, (, , i, , f f, P Q, , m0 i, [p – f + tan f], 2p r, , i, r, R, , Q.5 A circular coil of 0.2 m diameter has 100 turns and carries a, current of 0.1 ampere. The intensity of magnetic field at the, centre of the coil is (a) 6.28 × 10–4 N/A.m, (b) 62.8 × 10–4 N/A.m, –5, (c) 6.28 × 10 N/A.m, (d) 62.8 × 10–5 N/A.m, Q.6 For the arrangement of two current carrying identical coils, shown in the figure, the magnetic field at the center O is, , (N and a represent number of turns and radius of each coil)y, Coil-2, , m0 NI, , x, , z, , I, I, Coil-1, m 0 NI, , m 0 NI, m NI, (d) 0, 2, 2a, 2a, 2 2a, Q.7 A current is flowing through a conducting hollow pipe, whose area of cross-section is shown in the fig. The value, of magnetic induction will be zero at•R, (a) Point P, Q and R, (b) Point R but not at P and Q, •P, (c) Point Q but not at P and R, •Q, (d) Point P but not at Q and R, Q.8 Dimensional formula of m0 is(a) MLT–2 A–2, (b) MLT–2A–2, –2, 2, (c) MLT A, (d), MLT2 A2, , (a), , RESPONSE, GRID, , (b), , (c), , Q.9 A current of 1.0 ampere is flowing in the sides of an equilateral, triangle of side 4.5 × 10–2 m. Find the magnetic field at the, centroid of the triangle., (Permeability constant m0 = 4p × 10–7 V-s/A-m)., (a) 4.0 × 10–5 weber/m2 (b) 6.0 × 10–8 weber/m2, (c) 2.0 × 10–5 weber/m2 (d) 7.0 × 10–12 weber/m2, Q.10 An air-solenoid has 500 turns of wire in its 40 cm length., If the current in the wire be 1.0 ampere then the magnetic, field on the axis inside the solenoid is (a) 15.7 gauss, (b) 1.57 gauss, (c) 0.157 gauss, (d) 0.0157 gauss, Q.11 A solenoid of length 0.2m has 500 turns on it. If, 8.71 × 10–6 Weber/m2 be the magnetic field at an end of, the solenoid, then the current flowing in the solenoid is –, 174, 17.4, 0.174, 0.0174, (a), A (b), A (c), A (d), A, p, p, p, p, Q.12 A circular current carrying coil has a radius R. The distance, from the centre of the coil on the axis where the magnetic, 1, induction will be th to its value at the centre of the coil,, 8, is, 2, R, R, (a), (b) R 3, (c) 2 3R (d), 3, 3, Q.13 The average radius of an air cored made toroid is 0.1 m, and it has 500 turns. If it carries 0.5 ampere current, then, the magnetic field inside it is :, (a) 5 × 10–4 tesla, (b) 5 × 10–3 tesla, –2, (c) 5 × 10 tesla, (d) 2 × 10–3 tesla, Q.14 The straight long conductors AOB and COD are, perpendicular to each other and carry current i1 and i2. The, magnitude of the magnetic induction at point P at a distance, a from the point O in a direction perpendicular to the plane, ACBD is, B1, μ0, P, i, +, i, B2, (, ), (a), 1 2, 2πa, a, μ0, ( i1 - i2 ), (b), 2πa, A i2, i1, μ 0 2 2 1/ 2, i1 + i2, C, (c), D, 2πa, O, B, μ 0 i1i2, (d) 2πa ( i + i ), 1, 2, , (, , ), , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 39, , 3, , Q.15 A conducting circular loop of radius r carries a constant, ur, current i. It is placed in a uniform magnetic field B , such, ur, that B is perpendicular to the plane of the loop. The, magnetic force acting on the loop is, ur, ur, ur, (a) ir B, (b) 2pri B (c) zero, (d) pri B, Q.16 The radius of a circular loop is r and a current i is flowing, in it. The equivalent magnetic moment will be, (a), , ir, , (b) 2pir, , (c) ipr 2, , 1, , (d), , r2, Q.17 A current of 30 A is flowing in a vertical straight wire. If, the horizontal component of earth's magnetic field is 2 ×, 10–5 tesla then the distance of null point from wire is (a) 0.9 m, (b) 0.3 mm (c) 0.3 cm (d) 0.3 m, Q.18 A charged particle is released from rest in a region of steady, uniform electric and magnetic fields which are parallel to, each other. The particle will move in a, (a) Straight line, (b) Circle, (c) Helix, (d) Cycloid, Q.19 A 6.28m long wire is turned into a coil of diameter 0.2m, and a current of 1 amp is passed in it. The magnetic, induction at its centre will be (a) 6.28 × 10–5 T, (b) 0 T, (c) 6.28 T, (d) 6.28 × 10–3 T, DIRECTIONS (Q.20-Q.21) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.20 Two long straight parallel wires carry, currents I1 and I2 respectively, in the same I, 1, direction (as shown). The distance, between the wires is R. The magnetic, field at the centre of the two wires will, be-, , RESPONSE, GRID, , I2, , m 0 (I1 - I2 ), into the plane of paper (If I1 > I2), pR, m 0 (I 2 - I1 ), (2), out of the plane of paper (if I2 > I1), pR, m 0 (I1 - I 2 ), (3), out of the plane of paper (if I2 > I1), pR 2, m 0 (I 2 - I1 ), into the plane of paper (if I1 > I2), (4), pR 2, Q.21 A wire of length L carrying current I is bent into a circle, of one turn. The field at the center of the coil is B1. A, similar wire of length L carrying current I is bent into a, square of one turn. The field at its center is B2. Then, , (1), , (1), , B1 > B2, , (2) B1 = B2, , (3), , B1, =2, B2, , (4) B1 < B2, , DIRECTIONS (Q.22-Q.24) : Read the passage given below, and answer the questions that follows :, A conducting wire is bent into a loop as shown in the figure. The, segment AOB is parabolic given by the equation y2 = 2x while, segment BA is a straight line parallel to the y-axis., ur, The magnetic field in the region is B = –8k$ and the current in, the wire is 2A., y, A, O, , C, 2m, , D, , x, , B, , Q.22 The torque on the loop will be, (a) 16 2 Nm, , (b) 16 Nm, , (c) 18 2 Nm, (d) Zero, Q.23 The field created by the current in the loop at point C will be, µ, µ0 $, k, (a) – 0 k$, (b) –, 4p, 2p, (c), , R, , –, , µ0 2 $, k, p, , (d) None of these, , 15., , 16., , 17., , 18., , 20., , 21., , 22., , 23., , Space for Rough Work, , 19.
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t.me/Magazines4all, DPP/ P 39, , 4, Q.24 Magnetic field at point D due to segment AO of the loop is, directed parallel to, (a), , k̂, , (b) - k̂, , (c) î, , (d) ĵ, , DIRECTIONS (Q. 25-Q.27) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is] the correct answer. You have to select the, correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 24., , 25., , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is NOT, a correct explanation for Statement-1., (c) Statement-1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.25 Statement -1: Cyclotron does not accelerate electron., Statement-2: Mass of the electron is very small., Q.26 Statement-1: The ion cannot move with a speed beyond a, certain limit in a cyclotron., Statement-2: As velocity increases time taken by ion, increases., Q.27 Statement-1: If an electron, while coming vertically from, outerspace, enter the earth's magnetic field, it is deflected, towards west., Statement-2: Electron has negative charge., , 26., , 27., , DAILY PRA CTICE PROBLEM SHEET 39 - PHYSICS, Total Questions, 27, Total Marks, 108, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 40, SYLLABUS : MAGNETIC EFFECTS OF CURRENT-2 :, (Motion of charge particle in a magnetic field, force between current carrying wires.), , Max. Marks : 104, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 26 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.18) : There are 18 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A proton, a deutron and an a-particle are accelerated, through same potential difference and then they enter a, normal uniform magnetic field. The ratio of their kinetic, energies will be(a) 2 : 1 : 3 (b) 1 : 1 : 2 (c) 1 : 1 : 1 (d) 1 : 2 : 4, Q.2 A proton of energy 8eV is moving in a circular path in a, uniform magnetic field. The energy of an a-particle, moving in the same magnetic field and along the same path, will be(a) 4eV, (b) 2eV, (c) 8eV, (d) 6eV, , RESPONSE GRID, , 1., , 2., , Q.3 An electron is revolving in a circular path of radius, 2×10–10 m with a speed of 3×106 m/s. The magnetic field, at the centre of circular path will be(a) 1.2 T, (b) 2.4 T, (c) 0, (d) 3.6 T, Q.4 An a particle travels at an angle of 30º to a magnetic field, 0.8 T with a velocity of 105 m/s. The magnitude of force, will be(a) 1.28 × 10–14 N, (b) (1.28)Ö3 ×10–4 N, –4, (c) 1.28 × 10 N, (d) (12.8)Ö3 ×10–4 N, Q.5 A beam of protons is moving horizontally towards you. As, it approaches, it passes through a magnetic field directed, downward. The beam deflects× ×, ×, ×, (a) to your left side, v, × ×, F, (b) to your right side, ×, (c) does not deflect, × ×, × ×, (d) nothing can be said, ×, ×, × ×, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 40, , 2, Q.6 If a particle moves in a circular path in clockwise direction, after entering into a downward vertical magnetic field. The, charge on the particle is(a) positive, (b) negative, (c) nothing can be said, (d) neutral, Q.7 In the example above, after how much time, particle comes, to the starting point for the first time. (mass of particle =, m), 2p m, (a), 3qB, 2p m, (b), qB, (c) Never, (d) It will leave the circular path before coming to the, starting point, Q.8 A current of 2.0 amp is flowing through a wire of length, 50 cm. If this wire be placed at an angle of 60º with the, direction of a uniform magnetic field of 5.0 × 10–4 N/Am, the force on the wire will be(a) 4.33 × 10–4 N, (b) 2.50 × 10–4 N, –4, (c) 5.0 × 10 N, (d) 2.33 × 10–4 N, Q.9 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) qb B/m, (b) q(b – a) B/m, (c) qa B/m, (d) q(b + a) B/2m, Q.10 A particle with charge q, moving with a momentum p, enters, a uniform magnetic field normally. The magnetic field has, magnitude B and is confined to a region of width d, where, p, d < Bq , The particle is deflected by an angle q in crossing, × × × ×, the field. Then, Bqd, × × × ×, B, (a) sin q =, p, p, × × × ×, q, p, × × × ×, (b) sin q =, d, Bqd, × × × ×, , RESPONSE, GRID, , (c) sin q =, , Bp, qd, , (d) sin q =, , pd, Bq, , Q.11 An a particle is moving in a magnetic field of (3iˆ + 2ˆj), tesla with a velocity of 5×105 $i m/s. The magnetic force, acting on the particle will be(a) 3.2 × 10–13 dyne, (b) 3.2 × 1013 N, (c) 0, (d) 3.2 × 10–13 N, Q.12 If an a-particle moving with velocity 'v' enters perpendicular, to a magnetic field then the magnetic force acting on it will, be(a) evB, (b) 2evB, (c) 0, (d) 4evB, Q.13 What is the net force on the square coil ?, 10 cm, 2A, , 1A, , 15 cm, , 2 cm, , (a) 25 ´ 10–7 N towards wire, (b) 25 ´ 10–7 N away from wire, (c) 35 ´ 10–7 N towards wire, (d) 35 ´ 10–7 N away from wire, Q.14 A proton is to circulate the earth along the equator with a, speed of 1.0 × 107 m/s. The minimum magnetic field which, should be created at the equator for this purpose., (The mass of proton = 1.7 × 10–27 kg and radius of earth, = 6.37 × 106 m.) will be (in Wb/m2), (a) 1.6 × 10–19, (b) 1.67 × 10–8, –7, (c) 1.0 × 10, (d) 2 × 10–7, Q.15 An a-particle is describing a circle of radius 0.45 m in a, field of magnetic induction 1.2 weber/m2. The potential, difference required to accelerate the particle, (The mass, of a-particle is 6.8 × 10–27 kg and its charge is 3.2 × 10–, 19 coulomb.) will be —, (a) 6 × 106 V, (b) 2.3 × 10–12 V, 6, (c) 7 × 10 V, (d) 3.2 × 10–12 V, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 40, , 3, , Q.16 An electron beam passes through a magnitic field of, 2 × 10–3 weber/m2 and an electric field of 1.0 × 104 volt/, m both acting simultaneously. If the electric field is, removed, what will be the radius of the electron path ?, (a) 1.43 cm. (b) 0.43 cm (c) 2.43 cm. (d) 3.43 cm., Q.17 A straight horizontal copper wire carries a current i = 30, A. The linear mass density of the wire is 45 g/m. What is, the magnitude of the magnetic field needed to balance its, weight?, (a) 147 G (b) 441 G (c) 14.7 G (d) 0 G, Q.18 A 1m long conducting wire is lying at right angles to the, magnetic field. A force of 1 kg. wt is acting on it in a, magnetic field of 0.98 tesla. The current flowing in it will, be(a) 100 A (b) 10 A, (c) 1 A, (d) 0, DIRECTIONS (Q.19-Q.20) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.19 In the fig the two parallel wires PQ and ST are at 30 cm, apart. The currents flowing in the wires are according to, fig. The force acting over a length of 5m of the wires isP, , 10A, , S, , 15A, , 10–4, , Q, , r, A charge particles q enters in a magnetic field B = yiˆ + xjˆ with, r, the velocity v = xiˆ + yjˆ . Neglect any force other than magnetic, force. Now answer the following question., Q.21 When particle arrives at any point P (2, 2) then force acting, on it, will be –, (a) Zero, (b) 4 2q (c) 8q, (d) 2 2q, Q.22 Magnetic force F acting on charge is proportional to –, (a) F µ (x2 – y2), (b) F µ (x2 + y2), x 2 + y2, (d) F does not depend on x or y co-ordinate, Q.23 Which of the following is true for the direction of magnetic, force ?, (a) if x > y then force works along (–z) direction, (b) if x < y then force works along (+z) direction, (c) if x > y then force works along (+z) direction, (d) None of these, , (c), , Fµ, , DIRECTIONS (Qs. 24-Q.26) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , T, , (1) 5 ×, N, (2) attraction, (3) 5 × 10–8 N, (4) repulsion, Q.20 A beam of protons enters a uniform magnetic field of 0.3, tesla with a velocity of 4 × 105 m/s at an angle of 60º to, the field. Then,, (Mass of the proton = 1.7 × 10–27 kg.), (1) the radius of the helical path is 1.226 × 10–2 m, (2) the pitch of the helix is 4.45 × 10–2 m, (3) the radius of the helical path is 1.226 × 10–3 m, (4) the pitch of the helix is 4.45 × 10–4 m, , RESPONSE, GRID, , DIRECTIONS (Q.21-Q.23) : Read the passage given below, and answer the questions that follows :, , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.24 Statement -1 : If two long wires, hanging freely are, connected to a battery in series, they come closer to each, other., Statement -2 : Force of repulsion acts between the two, wires carrying current., , 16., , 17., , 18., , 19., , 21., , 22., , 23., , 24., , Space for Rough Work, , 20.
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DPP/ P 40, , 4, Q.25 Statement - 1 : For a charged particle to pass through a, uniform electro-magnetic field without change in velocity,, its velocity vector must be perpendicular to the magnetic, field., Statement - 2 : Net Lorentz force on the particle is given, r, r r r, by F = q [ E + v ´ B] ., , RESPONSE GRID, , t.me/Magazines4all, , 25., , Q.26 Statement - 1 : If an electron is not deflected while passing, through a certain region of space, then only possibility is, that there is no magnetic region., Statement - 2 : Magnetic force is directly proportional to, the magnetic field applied., , 26., , DAILY PRA CTICE PROBLEM SHEET 40 - PHYSICS, Total Questions, 26, Total Marks, 104, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 41, SYLLABUS : MAGNETIC EFFECTS OF CURRENT-3 (Magnetic dipole, Current carrying loop in, magnetic field,Galvanometer ), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice questions., Each question has 4 choices (a), (b), (c) and (d), out of which ONLY, ONE choice is correct., , Q.1 A circular coil has radius 4 cm and 20 number of turns, carries a current of 3 ampere. It is placed in a magnetic, field of intensity 0.5 weber/m2. The magnetic dipole, 22 ö, æ, moment of the coil is çè Take p = ÷ø, 7, (a) 0.15 ampere m2, (b) 0.3 ampere m2, (c) 0.45 ampere m2, (d) 0.6 ampere m2, Q.2 A circular coil of radius 4 cm has 50 turns. In this coil a, current of 2 A is flowing. It is placed in a magnetic field of, 0.1 weber/m2. The amount of work done in rotating it, through 180° from its equilibrium position will be, (a) 0.1 J, (b) 0.2 J, (c) 0.4 J, (d) 0.8 J, , RESPONSE GRID, , 1., , 2., , Q.3 The deflection in a moving coil galvanometer is, (a) directly proportional to the torsional constant, (b) directly proportional to the number of turns in the coil, (c) inversely proportional to the area of the coil, (d) inversely proportional to the current flowing, Q.4 A moving coil galvanometer has N number of turns in a, coil of effective area A. It carries a current I. The magnetic, field B is radial. The torque acting on the coil is, (a) NA2 B 2 I (b) NABI 2 (c) N 2 ABI (d) NABI, Q.5 A current carrying loop is free to turn in a uniform magnetic, field. The loop will then come into equilibrium when its, plane is inclined at, (a) 0° to the direction of the field, (b) 45° to the direction of the field, (c) 90° to the direction of the field, (d) 135° to the direction of the field, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 41, , 2, Q.6 A 100 turns coil shown in figure carries a current of 2 amp, in a magnetic field B = 0.2Wb/m2. The torque acting on, the coil is, B, 10 cm, , A, N, , S, C, , D, 8 cm, , (a) 0.32 Nm tending to rotate the side AD out of the page, (b) 0.32 Nm tending to rotate the side AD into the page, (c) 0.0032 Nm tending to rotate the side AD out of the, page, (d) 0.0032 Nm tending to rotate the side AD into the page, Q.7 A rectangular coil of size 20 cm × 20 cm has 100 turns, and carries a current of 1 A. It is placed in a uniform, magnetic field B = 0.5 T with the direction of magnetic, field parallel to the plane of the coil. The magnitude of the, torque required to hold this coil in this position is, (a) zero, (b) 200 Nm, (c) 2 Nm, (d) 10 Nm, Q.8 A circular loop of area 0.01m2 carrying a current of 10 A,, is held perpendicular to a magnetic field of intensity 0.1, T. The torque acting on the loop is, (a) zero, (b) 0.01 Nm, (c) 0.001 Nm, (d) 0.8 Nm, Q.9 The magnetic moment of a current carrying circular coil is, (a) directly proportional to the length of the wire, (b) inversely proportional to the length of the wire, (c) directly proportional to the square of the length of, the wire, (d) inversely proportional to the square of the length of, the wire, Q.10 What is the shape of magnet in moving coil galvanometer, to make the radial magnetic field?, (a) Concave cylindrical, (b) Horse shoe magnet, (c) Convex cylindrical, (d) None of these, Q.11 Current i is carried in a wire of length L. If the wire is, turned into a circular coil, the maximum magnitude of, torque in a given magnetic field B will be, , RESPONSE, GRID, , LiB 2, Li 2 B, L2 iB, Li 2 B, (b), (c), (d), 2, 2, 4p, 4p, Q.12 In ballistic galvanometer, the frame on which the coil is, wound is non-metallic. It is, (a) to avoid the production of induced e.m.f., (b) to avoid the production of eddy currents, (c) to increase the production of eddy currents, (d) to increase the production of induced e.m.f., Q.13 A solenoid of length 0.4 m and having 500 turns of wire, carries a current of 3 amp. A thin coil having 10 turns of wire, and of radius 0.01 m carries a current of 0.4 amp. The torque, (in Nm) required to hold the coil in the middle of the solenoid, with its axis perpendicular to the axis of the solenoid is, (m0 = 4p × 10–7 V-s/A-m), (a) 59.2 × 10–6, (b) 5.92 × 10–6, (c) 0.592 × 10–6, (d) 0.592 × 10–4., Q.14 If an electron is moving with velocity v in an orbit of radius, r in a hydrogen atom, then the equivalent magnetic moment is, , (a), , ev, µ0 e, ev´ 10,7, evr, (b) 2, (c), (d), 3, 2r, 2, r, r, Q.15 In a moving coil galvanometer, the deflection of the coil q is, related to the electrical current i by the relation, (a) i µ tan q, (b) i µ q, (a), , (c) i µ q2, (d) i µ q, Q.16 A thin circular wire carrying a current I has a magnetic, moment M. The shape of the wire is changed to a square, and it carries the same current. It will have a magnetic, moment, 4, p, 4, M, M, (a) M, (b) 2 M, (c), (d), 4, p, p, Q.17 A ring of radius R, made of an insulating material carries a, charge Q uniformly distributed on it. If the ring rotates about, the axis passing through its centre and normal to plane of, the ring with constant angular speed w, then the magnitude, of the magnetic moment of the ring is, (a), , QwR 2, , (b), , 1, QwR 2 (c) Qw 2 R, 2, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , (d), , 1, Qw 2 R, 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 41, , 3, , Q.18 The (t - q) graph for a current carrying coil placed in a, uniform magnetic field is, Q, , P, , (b), , (a), 0°, , 90°, , 180°, , q, , (c), , 0°, , 90°, , 180°, , 0°, , 90°, , 180°, , q, , (d), 0°, , 90°, , 180°, , q, , q, , Q.19 A rectangular loop carrying a current i is placed in a, uniform magnetic field B. The area enclosed by the loop, is A. If there are n turns in the loop, the torque acting on, the loop is given by, r r, r r, (a) ni A ´ B, (b) ni A.B, 1 r r, 1 r r, (i A ´ B), (c), (d) (iA.B), n, n, Q.20 The pole pieces of the magnet used in a pivoted coil, galvanometer are, (a) plane surfaces of a bar magnet, (b) plane surfaces of a horse-shoe magnet, (c) cylindrical surfaces of a bar magnet, (d) cylindrical surfaces of a horse-shoe magnet, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Four wires each of length 2.0 metre are bent into four loops, P, Q, R and S and then suspended into uniform magnetic, field. Same current is passed in each loop. Which, statements are incorrect?, , RESPONSE, GRID, , R, , S, , (1) Couple on loop P will be the highest, (2) Couple on loop Q will be the highest, (3) Couple on loop R will be the highest, (4) Couple on loop S will be the highest, Q.22 The sensitivity of a moving coil galvanometer can be, increased by, (1) decreasing the couple per unit twist of the suspension, (2) increasing the number of turns in the coil, (3) decreasing the area of the coil, (4) decreasing the magnetic field, Q.23 A current carrying rectangular coil is placed in a uniform, magnetic field. In which orientation, the coil will tend to, rotate, (1) The magnetic field is parallel to the plane of the coil, (2) The magnetic field is at 45° with the plane of the coil, (3) In any orientation, (4) The magnetic field is perpendicular to the plane of, the coil, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, A wire carrying a 10 A current is, bent to pass through sides of a cube, of side 10 cm as shown in figure. A, r, ˆ is, magnetic field B = (2iˆ - 3jˆ + k)T, , y, 2, , 3, , 1, x, , 4, present in the region. Then, find, z, Q.24 The net force on the loop, r, r, ˆ, (a) Fnet = 0, (b) Fnet = (0.1iˆ - 0.2k)N, r, r, ˆ, ˆ, (c) Fnet = (0.3iˆ + 0.4k)N, (d) Fnet = (0.36k)N, Q.25 The magnetic moment vector of the loop., 2 (b), 2, ˆ, ˆ, (a) (0.1iˆ + 0.05jˆ - 0.05k)Am, (0.1iˆ + 0.05jˆ + 0.05k)Am, 2 (d), 2, ˆ, ˆ, (c) (0.1iˆ - 0.05jˆ + 0.05k)Am, (0.1iˆ - 0.05jˆ - 0.05k)Am, , 18., , 19., , 20., , 23., , 24., , 25., , Space for Rough Work, , 21., , 22.
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t.me/Magazines4all, DPP/ P 41, , 4, Q.26 The net torque on the loop., (b) -0.1iˆ - 0.4kˆ Nm, (a) - 0.1iˆ + 0.4kˆ Nm, (c) 0.1iˆ - 0.4kˆ Nm, , (d) 0.1iˆ - 0.4kˆ Nm, , DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (c) Statement-1 is False, Statement-2 is True., (d) Statement-1 is True, Statement-2 is False., Q.27 Statement-1 : The coil is bound over the metallic frame, in moving coil galvanometer., Statement-2 : The metallic frame help in making steady, deflection without any oscillation., Q.28 Statement-1 : Torque on the coil is maximum, when coil, is suspended in a radial magnetic field., Statement-2 : The torque tends to rotate the coil on its, own axis., Q.29 Statement-1 : A current carr ying loop placed in, equilibrium in a uniform magnetic field starts oscillating, when disturbed from equilibrium., Statement-2 : A system when disturbed slightly from stable, equilibrium oscillates., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 41 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 42, SYLLABUS : MAGNETISM AND MATTER - 1 (Bar magnet as an equivalent solenoid, Magnetic field lines,, Earth's magnetic field and magnetic elements), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 mul1tiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A north pole of strength 50 Am and south pole of strength, 100 Am are separated by a distance of 10 cm in air. Find, the force between them., (a) 50 × 10–3 N, (b) 25 × 10–3 N, (c) 20 × 10–6 N, (d) 30 × 10–18 N, Q.2 Calculate magnetic induction at a distance of 20 cm from, a pole of strength 40 Am in air., (a) 10–4 wb/m2, (b) 10–8 wb/m2, –1, 2, (c) 10 wb/m, (d) 10–12 wb/m2, Q.3 A bar magnet of length 0.2 m and pole strength 5 Am is, kept in a uniform magnetic induction field of strength 15, , RESPONSE GRID, , 1., , 2., , wb/m–2 making an angle of 30° with the field. Find the, couple acting on it., (a) 2.5 Nm (b) 5.5 Nm (c) 7.5 Nm (d) 9.0 Nm, Q.4 The force experienced by a pole of strength 100 Am at a, distance of 0.2 m from a short magnet of length 5 cm and, pole strength of 200 Am on its axial line will be, (a) 2.5 × 10-2 N, (b) 2.5 × 10-3 N, -2, (c) 5.0 × 10 N, (d) 5.0 × 10-3 N, Q.5 A magnet of moment M is lying in a magnetic field of induction, B. W1 is the work done in turning it from 0º to 60º and W2 is, the work done in turning it from 30º to 90º. Then, W1, (a) W2 = W1, (b) W2 =, 2, (c) W2 = 2W1, , 3., Space for Rough Work, , 4., , (d) W2 =, , 5., , 3 W1
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t.me/Magazines4all, DPP/ P 42, , 2, magnet of magnetic moment 4.0 A-m2 is free to rotate, , Q.6 A bar, about a vertical axis through its centre. The magnet is, released from rest from east-west position. Kinetic energy, of the magnet in north-south position will be (Horizontal, component of earth’s magnetic field BH = 25mT), (a) 10-2 J (b) 10-4 J (c) 10-6 J (d) 0, Q.7 The length of a bar magnet is 10 cm and its pole strength is, 10-3 Weber. It is placed in a magnetic field of induction, 4 p × 10-3 Tesla in a direction making an angle 30º with, the field direction. The value of torque acting on the magnet, will be –, (a) 2p × 10-7 N-m, (b) 2p × 10-5 N-m, 2, (c) 0.5 × 10 N-m, (d) None of these, Q.8 At magnetic poles of earth, angle of dip is, (a) zero, (b) 45°, (c) 90 °, (d) 180 °, Q.9 A short bar magnet is placed with its north pole pointing, south. The neutral point is 10 cm away from the centre of, the magnet. If H = 0.4 gauss, calculate magnetic moment, of the magnet., (a) 2 Am2 (b) 1A m2 (c) 0.1 A m2 (d) 0.2 Am2, Q.10 A bar magnet with its poles 25 cm apart and of pole strength, 24.0 A-m rests with its centre on a frictionless pivot. A, force F is applied on the magnet at a distance of 12 cm, from the pivot, so that it is held in equilibrium at an angle, of 30º with respect to a magnetic field of induction 0.25, T. The value of force F is, (a) 65.62 N (b) 2.56 N (c) 6.52 N (d) 6.25 N, Q.11 A small magnet of magnetic moment 4A-m2 is placed on a, deflection magnetometer in tan-B position at a distance, of 20 cm from the compass needle. At what distance from, compass needle should another small magnet of moment, 0.5A-m2 be placed such that the deflection of the needle, remains zero ?, (a) 12 cm (b) 10 cm (c) 20 cm (d) 30 cm, Q.12 The ratio of intensities of magnetic field, at distances x, and 2x from the centre of magnet of length 2cm on its, axis, will be, (a) 4 : 1, (b) 4 : 1 approx, (c) 8 : 1, (d) 8 : 1 approx, Q.13 Two magnets A and B are identical and these are arranged, as shown in the figure. Their length is negligible in, , RESPONSE, GRID, , comparison to the separation between them. A magnetic, needle is placed between the magnets at point P which gets, deflected through an angle q under the influence of, magnets. The ratio of distances d1 and d2 will be, B, , (a) (2 tan q)1/3, (b) (2 tan q)-1/3, (c) (2 cot, , A, , q)1/3, , d1, , (d) (2 cot q)-1/3, , P, q d, 2, , Q.14 The period of oscillation of a freely suspended bar magnet, is 4 second. If it is cut into two equal parts length wise, then the time period of each part will be, (a) 4 sec, (b) 2 sec, (c) 0.5 sec (d) 0.25 sec, Q.15 The length, breadth and mass of two bar magnets are same, but their magnetic moments are 3M and 2M respectively., These are joined pole to pole and are suspended by a string., When oscillated in a magnetic field of strength B, the time, period obtained is 5s. If the poles of either of the magnets, are reverse then the time period of the combination in the, same magnetic field will be –, (b) 2 2 s (c) 5 5 s (d) 1s, (a) 3 3 s, Q.16 A thin magnetic needle oscillates in a horizontal plane with, a period T. It is broken into n equals parts. The time period, of each part will be, T, T, (a) T, (b), (c) Tn2, (d) 2, n, n, Q.17 A bar magnet made of steel has a magnetic moment of 2.5 Am2 and a mass of 6.6 × 103 kg. If the density of steel is, 7.9 × 109 kg/m3, find the intensity of magnetization of the, magnet., (a) 3.0 × 106 A/m, (b) 2.0 × 106 A/m, 6, (c) 5.0 × 10 A/m, (d) 1.2 × 106 A/m, Q.18 A short magnet of length 4 cm is kept at a distance of 20, cm to the east of a compass box such that is axis is, perpendicular to the magnetic meridian. If the deflection, produced is 45°, find the pole strength (H = 30 Am–1), (a) 17.7 Am, (b) 44.2 Am, (c) 27.7 Am, (d) 37.7 Am, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 42, , 3, , Q.19 A 10 cm long bar magnet of magnetic moment 1.34 Am2, is placed in the magnetic meridian with its south pole, pointing geographical south. The neutral point is obtained, at a distance of 15 cm from the centre of the magnet., Calculate the horizontal component of earth’s magnetic, field., (a) 0.12 × 10–4 T, (b) 0.21× 10–4 T, –4, (c) 0.34 × 10 T, (d) 0.87 × 10–7 T, Q.20 A 30 cm long bar magnet is placed in the magnetic meridian, with its north pole pointing south. The neutral point is, obtained at a distance of 40 cm from the centre of the, magnet. Pole strength of the magnet is (The horizontal, component of earth’s magnetic field is 0.34 Gauss), (a) 26.7 Am (b) 16.7 Am (c) 12.7 Am (d) 15.2 Am, Q.21 A long straight horizontal cable carries a current of 2.5 A, in the direction 10° south of west to 10° north of east. The, magnetic meridian of the place happens to be 10° west of, the geographic meridian. The earth’s magnetic field at the, location is 0.33 Gauss, and the angle of dip is zero., Distance of the line of neutral points from the cable is, (Ignore the thickness of the cable)., (a) 1.5 cm (b) 2.5 cm (c) 3.5 cm (d) 2.0 cm, DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 Which of the following is/are not the main difference, between electric lines of force and magnetic lines of force, ?, (1) Electric lines of force are closed curves whereas, magnetic lines of force are open curves., (2) Magnetic lines of force cut each other whereas, electric lines of force do not cut., (3) Electric lines of force cut each other whereas, magnetic lines of force do not cut., (4) Electric lines of force are open curves whereas, magnetic lines of force are closed curves., , RESPONSE, GRID, , Q.23 The correct statements regarding the lines of force of the, magnetic field B are, (1) Magnetic intensity is a measure of lines of force, passing through unit area held normal to it, (2) Magnetic lines of force form a closed curve, (3) Due to a magnet magnetic lines of force never cut, each other, (4) Inside a magnet, its magnetic lines of force move from, north pole of a magnet towards its south pole, Q.24 A short bar magnet of magnetic moment 5.25 × 10–2 JT–1, is placed with its axis perpendicular to the earth’s field, direction. Magnitude of the earth’s field at the place is given, to be 0.42 G. Ignore the length of the magnet in comparison, to the distance involved. Then, (1) the distance from the centre of the magnet on its, normal bisector at which the resultant field is inclined, at 45° with the earth's field is 5 cm, (2) the distance from the centre of the magnet on its axis, at which the resultant field inclined at 45° with the, earth's field is 6.3 cm, (3) the distance from the centre of the magnet on its, normal bisector at which the resultant field inclined, at 45° with the earth's field is 8.3 cm, (4) the distance from the centre of the magnet on its axis, at which the resultant field inclined at 45° with the, earth's field is 8 cm, DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, A telephone cable at a place has four long, straight horizontal wires, carrying a current of 1.0 A in the same direction east to west. The, earth’s magnetic field at the place is 0.39 Gauss, and the angle of, dip is 35°. The magnetic declination is nearly zero. (cos 35° =, 0.82, sin 35° = 0.57), Q.25 The magnetic field produced by four current carrying, straight cable wires at a distance 4 cm is, (a) 0.2 Gauss, (b) 0.3 Gauss, (c) 0.4 Gauss, (d) 0.5 Gauss, Q.26 The resultant magnetic field below at points 4cm and above, the cable are, (a) 0.25, 0.56 Gauss, (b) 0.14, 0.32 Gauss, (c) 0.23, 0.34 Gauss, (d) 0.52, 0.62 Gauss, , 19., , 20., , 21., , 24., , 25., , 26., , Space for Rough Work, , 22., , 23.
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t.me/Magazines4all, DPP/ P 42, , 4, Q.27 The angle that resultant makes with horizontal in case below, and above the cable respectively, are, (a) 30°, 45°, (b) tan–1 1.8, tan–1 0.43, (c) tan–1 2, tan–1 2, , (d) sin–1 0.7, sin–1 0.9, , DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , RESPONSE GRID, , 27., , 28., , (c) Statement-1 is False, Statement-2 is True., (d) Statement-1 is True, Statement-2 is False., Q.28 Statement-1 : Gauss theorem is not applicable in, magnetism., Statement-2 : Mono magnetic pole does not exist., Q.29 Statement-1 : A compass needle when placed on the, magnetic north pole of the earth cannot rotate in vertical, direction., Statement-2 : The earth has only horizontal component, of its magnetic field at the north poles., Q.30 Statement-1 : We cannot think of magnetic field, configuration with three poles., Statement-2 : A bar magnet does not exert a torque on, itself due to its own field., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 42 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 50, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 43, SYLLABUS : MAGNETISM & MATTER-2 (Para, dia and ferro-magnetic substances, magnetic susceptibility and, permeability, Hysteresis, Electromagnets and permanent magnets.), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.24) : There are 24 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 Susceptibility of ferromagnetic substance is, (a) > 1, (b) < 1, (c) 0, (d) 1, Q.2 Among the following properties describing diamagnetism, identify the property that is wrongly stated., (a) Diamagnetic material do not have perman ent, magnetic moment, (b) Diamagnetism is explained in terms of, electromagnetic induction., (c) Diamagnetic materials have a small positive, susceptibility, (d) The magnetic moment of individual electrons, neutralize each other, , RESPONSE GRID, , 1., , 2., , Q.3 If the magnetic dipole moment of an atom of diamagnetic, material, paramagnetic material and ferromagnetic material, denoted by m d , m p , m f respectively then, md , ¹ 0 and m f ¹ 0, , (b) m p = 0 and m f ¹ 0, , (c) m d = 0 and m p ¹ 0, , (d) m d ¹ 0 and m p = 0, , (a), , Q.4 When a piece of a ferromagnetic substance is put in a, uniform magnetic field, the flux density inside it is four, times the flux density away from the piece. The magnetic, permeability of the material is, (a) 1, , (b) 2, , (c) 3, , (d) 4, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 43, , 2, Q.5 The given figure represents a material which is, , Q.13 The relative permeability is represented by µr and the, susceptibility is denoted by c for a magnetic substance., Then for a paramagnetic substance, , (a) Paramagnetic, (b) Diamagnetic, (c) Ferromagnetic, , (a) m r < 1, c < 0, , (b) m r < 1, c > 0, , (d) None of these, , (c) m r > 1, c < 0, , (d) m r > 1, c > 0, , Q.6 Liquid oxygen remains suspended between two pole faces, of a magnet because it is, (a) diamagnetic, (b) paramagnetic, (c) ferromagnetic, (d) antiferromagnetic, Q.7 A superconductor exhibits perfect, (a) ferrimagnetism, (b) ferromagnetism, (c) paramagnetism, (d) diamagnetism, Q.8 Which of the following is the most suitable for the core, of electromagnets?, (a) Soft iron, (b) Steel, (c) Copper-nickel alloy (d) Air, Q.9 The universal property of all substances is, (a) diamagnetism, (b) ferromagnetism, (c) paramagnetism, (d) all of these, Q.10 If a magnetic substance is kept in a magnetic field, then, which of the following substance is thrown out ?, (a) Paramagnetic, (b) Ferromagnetic, (c) Diamagnetic, (d) Antiferromagnetic, Q.11 In the hysteresis cycle, the value of H needed to make the, intensity of magnetisation zero is called, (a) Retentivity, (b) Coercive force, (c) Lorentz force, (d) None of these, Q.12 If a diamagnetic solution is poured into a U-tube and one, arm of this U-tube placed between the poles of a strong, magnet with the meniscus in a line with field, then the level, of the solution will, (a) rise, (b) fall, (c) oscillate slowly, (d) remain as such, , RESPONSE, GRID, , Q.14 The use of study of hysteresis curve for a given material is, to estimate the, (a) voltage loss, (b) hysteresis loss, (c) current loss, (d) all of these, Q.15 The magnetic moment of atomic neon is, (a) zero, (b) mB/2, (c) mB, (d) 3mB/2, Q.16 A ferromagnetic material is heated above its Curie, temperature, then which one is a correct statement ?, (a) Ferromagnetic domains are perfectly arranged, (b) Ferromagnetic domains becomes random, (c) Ferromagnetic domains are not influenced, (d) Ferromagnetic material changes itself into, diamagnetic material, Q.17 If a diamagnetic substance is brought near north or south, pole of a bar magnet, it is, (a) attracted by the poles, (b) repelled by the poles, (c) repelled by the north pole and attracted by the south pole, (d) attracted by the north pole and repelled by the south pole, Q.18 The material of permanent magnet has, (a) high retentivity, low coercivity, (b) low retentivity, high coercivity, (c) low retentivity, low coercivity, (d) high retentivity, high coercivity, Q.19 Diamagnetic substances are, (a) feebly attracted by magnets, (b) strongly attracted by magnets, (c) feebly repelled by magnets, (d) strongly repelled by magnets, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 43, , 3, , Q.20 For an isotropic medium B, µ, H and M are related as (where, B, µ0, H and M have their usual meaning in the context of, magnetic material), (a) (B – M) = µ0H, (b) M = µ0(H + M), (c) H = µ0(H + M), (d) B = µ0(H + M), Q.21 Relative permeability of iron is 5500, then its magnetic, susceptibility will be, (a) 5500 × 107, (b) 5500 × 10–7, (c) 5501, (d) 5499, Q.22 A magnetising field of 2 × 103 amp/m produces a magnetic, flux density of 8p Tesla in an iron rod. The relative, permeability of the rod will be, (a) 102, (b) 100, (c) 104, (d) 101, Q.23 The mass of a specimen of a ferromagnetic material is 0.6, kg. and its density is 7.8 × 103 kg/m 3 . If the area of, hysteresis loop of alternating magnetising field of, frequency 50Hz is 0.722 MKS units then the hysteresis, Y, loss per second will be, Soft iron, , (a) 277.7 × 10-5 Joule, (b) 277.7 × 10-6 Joule, (c) 277.7 ×, , 10-4, , I, Steel, X', , H, , Joule, , X, , Q.25 A magnetising field of 1600 Am–1 produces a magnetic, flux of 2.4 × 10–5 weber in a bar of iron of area of crosssection 0.2 cm2. Then,, (1) the magnetic permeability of th e bar is, 7.5 × 10–4 TA–1m, (2) the susceptibility of the bar is 596.1, (3) the magnetic permeability of the bar is 4.1 Wbm–2, (4) the susceptibility of the bar is 496.1, Q.26 Which of the following statements are correct about, hysteresis?, (1) This effect is common to all ferromagnetic substances, (2) The hysteresis loop area is proportional to the thermal, energy developed per unit volume of the material, (3) The shape of the hysteresis loop is characteristic of, the material, (4) The hysteresis loop area is independent of the thermal, energy developed per unit volume of the material, Q.27 Which of the following statments are false about the, magnetic susceptibility c m of paramagnetic substance?, (1) Value of c m is directly proportional to the absolute, temperature of the sample, (2) c m is negative at all temperature, (3) c m does not depend on the temperature of the sample, , (d) 27.77 × 10-4 Joule, Y', , Q.24 A diamagnetic material in a magnetic field moves, (a) from weaker to the stronger parts of the field, (b) perpendicular to the field, (c) from stronger to the weaker parts of the field, (d) None of these, DIRECTIONS (Q.25-Q.27) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, , RESPONSE, GRID, , (b) 1 and 2 are correct, (d) 1 and 3 are correct, , (4) c m is positive at all temperature, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), (c), (d), , 20., , 21., , 22., , 25., , 26., , 27., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , Space for Rough Work, , 23., , 24.
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DPP/ P 43, , 4, Q.28 Statement-1 : The ferromagnetic substance do not obey, Curie’s law., Statement-2 : At Curie point a ferromagnetic substance, start behaving as a paramagnetic substance., Q.29 Statement-1 : A paramagnetic sample displays greater, magnetisation (for the same magnetising field) when, cooled., , RESPONSE GRID, , t.me/Magazines4all, , 28., , 29., , Statement-2 : The magnetisation does not depend on, temperature., Q.30 Statement-1 : The permeability of a ferromagnetic, material dependent on the magnetic field., Statement-2 : Permeability of a material is a constant, quantity., , 30., , DAILY PRA CTICE PROBLEM SHEET 43 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 44, SYLLABUS : ELECTROMAGNETIC INDUCTION-1 (Magnetic flux, Faraday's law of electromagnetic induction, Lenz's, law, motional e.m.f.), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., r, Q.1 A loop of wire is placed in a magnetic field B = 0.02iˆ T.., Then the flux through the loop if its area vector, r, A = 30iˆ + 16ˆj + 23kˆ cm2 is, (a) 60 mWb, (b) 32 m Wb, (c) 46 m Wb, (d) 138 m Wb, Q.2 The magnetic flux passing perpendicular to the plane of, the coil and directed into the paper is varying according to, the relation f = 3t2 + 2t + 3, where f is in milliweber and, t is in second. Then the magnitude of emf induced in the, loop when t = 2 second is-, , RESPONSE GRID, , 1., , 2., , (a) 31 mV, (b) 19 mV, (c) 14 mV, (d) 6 mV, Q.3 A current carrying solenoid is approaching a conducting, loop as shown in the figure. The direction of induced, current as observed by an observer on the other side of the, loop will be (a) anti-clockwise, (b) clockwise, V, (c) east, (d) west, , 3., Space for Rough Work
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t.me/Magazines4all, DPP/ P 44, , 2, Q.4 Consider the arrangement shown in figure in which the north, pole of a magnet is moved away from a thick conducting, loop containing capacitor. Then excess positive charge will, arrive on, (a) plate a, (b) plate b, (c) both plates a and b, (d) neither a nor b plates, Q.5 An electron moves along the line AB, which lies in the, same plane as a circular loop of conducting wires as shown, in the diagram. What will be the direction of current, induced if any, in the loop?, (a) No current will be induced, (b) The current will be clockwise, A, B, (c) The current will be anticlockwise, (d) The current will change direction as the electron passes by, Q.6 When a small piece of wire passes between the magnetic, poles of a horse-shoe magnet in 0.1 sec, emf of 4 × 10 –3, volt is induced in it. The magnetic flux between the poles, is :, (a) 4 × 10–2 weber, (b) 4 × 10–3 weber, –4, (c) 4 × 10 weber, (d) 4 × 10–6 weber, Q.7 The normal magnetic flux passing through a coil changes, with time according to following equation, f = 10t2 + 5t + 1, where f is in milliweber and t is in second. The value of, induced e.m.f. produced in the coil at t = 5s will be –, (a) zero, (b) 1 V, (c) 2 V, (d) 0.105 V, Q.8 A bicycle wheel of radius 0.5 m has 32 spokes. It is rotating, at the rate of 120 revolutions per minute, perpendicular to, the horizontal component of earth's magnetic field, BH = 4 × 10–5 tesla. The emf induced between the rim and, the centre of the wheel will be(a) 6.28 × 10–5 V, (b) 4.8 × 10–5 V, –5, (c) 6.0 × 10 V, (d) 1.6 × 10–5 V, Q.9 A thin semicircular conducting ring of radius R is falling, with its plane vertical in a horizontal magnetic induction, B. At the position MNQ shown in the fig, the speed of the, ring is V. The potential difference developed across the, semicircular ring is, , RESPONSE, GRID, , (a) Zero, × × × × ×, (b) B vR2/2 and M is at higher potential × B × N × ×, (c) pRBV and Q is at higher potential × × ×V × ×, × × × × ×, (d) 2RBV and Q is at higher potential, M, Q, Q.10 An aeroplane having a distance of 50 metre between the, edges of its wings is flying horizontally with a speed of, 360km/hour. If the vertical component of earth’s magnetic, field is 4 × 10–4 weber/m2, then the induced emf between, the edges of its wings will be –, (a) 2 mV, (b) 2 V, (c) 0.2 V (d) 20 V, Q.11 At certain location in the northern hemisphere, the earth's, magnetic field has a magnitude of 42 mT and points down, ward at 57º to vertical. The flux through a horizontal surface, of area 2.5 m2 will be- (Given cos 33º = 0.839, cos 57º =, 0.545), (a) 42 × 10–6 Wb/m2, (b) 42 × 10–6 Wb/m2, –6, 2, (c) 57 × 10 Wb/m, (d) 57 × 10–6 Wb/m2, Q.12 A square loop of side a is rotating about its diagonal with, angular velocity w in a perpendicular magnetic field as shown, in the figure. If the number of turns in it is × × × × × ×, × × × × × ×, 10 then the magnetic flux linked with the × × × × × ×, × × ×, × ××, loop at any instant will be–, × ×, × × ××, × × w× × × ×, (a) 10Ba2 cos wt (b) 10Ba, ×, × × × ××, 2, 2, × × × × ××, (c) 10Ba, (d) 20Ba, Q.13 Two identical coaxial circular loops carry current i each, circulating in the clockwise direction. If the loops are, approaching each other, then, (a) Current in each loop increases, (b) Current in each loop remains the same, (c) Current in each loop decreases, (d) Current in one-loop increases and in the other it, decreases, Q.14 The distance between the ends of wings of an aeroplane is, 3m. This aeroplane is descending down with a speed of, 300 km/hour. If the horizontal component of earths, magnetic field is 0.4 gauss then the value of e.m.f. induced, in the wings of the plane will be –, (a) 1 V, (b) 2 V, (c) 0.01 V (d) 0.1 V, , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 44, , 3, , Q.15 A gramophone disc of brass of diameter 30 cm rotates, horizontally at the rate of 100/3 revolutions per minute. If, the vertical component of the earth's magnetic field be 0.01, weber/metre2, then the emf induced between the centre, and the rim of the disc will be(a) 7.065 × 10–4 V, (b) 3.9 × 10–4 V, –4, (c) 2.32 × 10 V, (d) None of the above, Q.16 A closed coil consists of 500 turns on a rectangular frame, of area 4.0 cm2 and has a resistance of 50 ohm. The coil is, kept with its plane perpendicular to a uniform magnetic, field of 0.2 weber/meter 2. The amount of charge flowing, through the coil if it is turned over (rotated through 180º), will be (a) 1.6 × 10–19 C, (b) 1.6 × 10–9 C, –3, (c) 1.6 × 10 C, (d) 1.6 × 10–2 C, Q.17 A copper disc of radius 0.1 m rotates about its centre with, 10 revolution per second in 'a uniform magnetic field of, 0.1 T. The emf induced across the radius of the disc is (a) p/10 V (b) 2p/10 V (c) 10p mV (d) 20p mV, Q.18 Two rail tracks, insulated from each other and the ground,, are connected to milli voltmeter. What is the reading of, the milli voltmeter when a train passes at a speed of 180, km/hr along the track ? Given that – the horizontal, component of earth’s magnetic field BH is 0.2 × 10–4 Wb/, m2 and rails are separated by 1 metre., (a) 1 mV, (b) 10 mV (c) 100 mV (d) 1 V, Q.19 The annular disc of copper, with inner radius a and outer, radius b is rotating with a uniform angular speed w, in a, region where a uniform magnetic field B along the axis of, rotation exists. Then, the emf induced between inner side, and the outer rim of the disc is(a) Zero, , (b), , 1, Bwa2, 2, , 1, 1, Bwb2 (d), Bw (b2 – a2), 2, 2, Q.20 A conducting wire in the shape of Y with each side of length, l is moving in a uniform magnetic field B, with a uniform, speed v as shown in fig. The induced emf at the two ends X, and Y of the wire will be-, , (c), , RESPONSE, GRID, , 15., 20., , 16., 21., , (a) zero, (b) 2 Blv, (c) 2 Blv sin (q/2), (d) 2 Blv cos (q/2), DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 A rectangular coil of size 10 cm ×20 cm has 60 turns. It is, rotating about one of its diagonals in magnetic field, 0.5 Wb/m2 with a rate of 1800 revolution per minute. The, induced e.m.f. in the coil can be, (1) 111 V, (2) 112 V (3) 113 V (4) 114 V, Q.22 A closed coil of copper whose area is 1m × 1m is free to, rotate about an axis. The coil is placed perpendicular to a, magnetic field of 0.10 Wb/m2. It is rotated through 180º, in 0.01 second. Then (The resistance of the coil is 2.0W), (1) The induced e.m.f. in the coil is 20 V, (2) The induced current in the coil is 10 A, (3) The induced e.m.f. in the coil is 10 V, (4) The induced current in the coil is 20 A, Q.23 5.5 × 10–4 magnetic flux lines are passing through a coil, of resistance 10 ohm and number of turns 1000. If the, number of flux lines reduces to 5 × 10–5 in 0.1 sec. Then, (1) The electromotive force induced in the coil is 5V, (2) The electromotive force induced in the coil is, 5 × 10–4 V, (3) The current induced in the coil is 0.5 A, (4) The current induced in the coil is 10 A, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, In the figure shown, the rod has a resistance R, the horizontal, rails have negligible friction. Magnetic field of intensity B is, directed perpendicular into the plane of paper. A cell of e.m.f. E, and negligible internal resistance is connected between points a, and b. The rod is initially at rest., , 17., 22., , Space for Rough Work, , 18., 23., , 19.
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t.me/Magazines4all, DPP/ P 44, , 4, ROD, a, , B, , × × ×, , ×, , × × ×, , ×, , E, , L, , b, , × × ×, , ×, , × × ×, , ×, , Q.24 The velocity of the rod as a function of time t (where, t = mR/Bl2) is, E, (1 - e - t / t ), Bl, (c) 3 E (1 - e- t / t ), 2 Bl, , (a), , (b), (d), , E, (1 + e - t / t ), Bl, E, (1 - e - t / t ), 2Bl, , Q.25 After some time the rod will approach a terminal speed., The speed is, (a), , 3 E, 2 Bl, , (b), , 2E, R, , (b), , E, 2Bl, , (c), , E, R, , (c), , E, Bl, , (d), , 3E, 2R, , (d) zero, , 2E, Bl, , Q.26 The current when the rod attains its terminal speed is, (a), , RESPONSE, GRID, , 24., , 25., , DIRECTIONS (Q. 27-Q.29) : Each of these questions, contains two statements: Statement-1 (Assertion) and, Statement-2 (Reason). Each of these questions has four, alternative choices, only one of which is the correct answer., You have to select the correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : The induced e.m.f. and current will be same, in two identical loops of copper and aluminium, when, rotated with same speed in the same magnetic field., Statement-2 : Induced e.m.f. is proportional to rate of, change of magnetic field while induced current depends, on resistance of wire., Q.28 Statement-1 : An aircraft flies along the meridian, the, potential at the ends of its wings will be the same., Statement-2 : Whenever there is change in the magnetic, flux e.m.f. induces., Q.29 Statement-1 : Lenz’s law violates the principle of, conservation of energy., Statement-2 : Induced e.m.f. opposes the change in, magnetic flux responsible for its production., , 26., , 27., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 44 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 45, SYLLABUS : ELECTROMAGNETIC INDUCTION - 2 : Self inductance, mutual inductance,, Growth and decay of current in L.R. circuit, Transformer, Electric motor, Generator, , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A current increases uniformly from zero to one ampere in, 0.01 second, in a coil of inductance 10 mH. The induced, e.m.f. will be (a) 1 V, (b) 2 V, (c) 3 V, (d) 4 V, Q.2 The current in a coil varies with respect to time t as I = 3t2 +, 2t. If the inductance of coil be 10 mH, the value of induced, e.m.f. at t = 2s will be(a) 0.14 V (b) 0.12 V (c) 0.11 V (d) 0.13 V, Q.3 Two circular coils can be arranged in any of the three, situations shown in the figure. Their mutual inductance will, be, , RESPONSE GRID, , 1., , 2., , (A), (B), (C), (a) Maximum in situation (A), (b) Maximum in situation (B), (c) Maximum in situation (C), (d) The same in all situations, Q.4 A current of 10 A in the primary coil of a circuit is reduced, to zero at a uniform rate in 10–3 second. If the coefficient, of mutual inductance is 3H, the induced e.m.f. in the, secondary coil will be(a) 3 kV, (b) 30 kV (c) 2 kV, (d) 20 kV, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 45, , 2, , /\/\/\/, , Q.5 Three inductances are connected as shown below., Assuming no coupling, the resultant inductance will be(L1 = 0.75 H, L2 = L3 = 0.5 H), L2, (a) 0.25 H, L1, (b) 0.75 H, (c) 0.01 H, (d) 1 H, L3, Q.6 A solenoid has an inductance of 50 mH and a resistance of, 0.025 W. If it is connected to a battery, how long will it, take for the current to reach one half of its final equilibrium, value?, (a) 1.34 s (b) 1.38 s (c) 1.38 ms (d) 0.23 s, Q.7 The current in the primary coil of a transformer (assuming, no power loss) as shown in fig. will be –, S, (a) 0.01 A, P, (b) 1.0 A, 220 W, VP = 220 V, (c) 0.1 A, (d) 10–6 A., P, S, Q.8 A current of 5A is flowing at 220V in the primary coil of a, transformer. If the voltage produced in the secondary coil, is 2200V and 50% of power is lost, then the current in the, secondary coil will be –, (a) 2.5A, (b) 5A, (c) 0.25A (d) 0.025A, Q.9 An inductor (L = 100 mH), a resistor (R = 100W) 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 ms after the short circuit is, L, (a) e A, (b) 0.1 A, R, (c) 1 A, A, B, (d) 1/e A, E, Q.10 Which of the following is constructed on the principle of, electromagnetic induction ?, (a) Galvanometer, (b) Electric motor, (c) Generator, (d) Voltmeter, Q.11 In the circuit, E = 10 volt, R1 = 5.0 ohm, R2 = 10 ohm and, L = 5.0 henry. The current just after the switch S is pressed, is., , RESPONSE, GRID, , L, , (a) 2.0A, (b) 3.0A, (c) 5.0A, (d) 6.0A, Q.12 Two inductors L1 and L2 are at a sufficient distance apart., Equivalent inductance when they are connected (i) in series, (ii) in parallel are, L1L 2, L1L2, (a) L1 + L2,, (b) L1 – L2,, L1 + L2, L1 - L2, L1 + L2, (c) L1 L2, L L, (d) None of these, 1 2, Q.13 A small coil of N1 turns, l1 length is tightly wound over the, centre of a long solenoid of length l2, area of cross-section, A and number of turns N2. If a current I flows in the small, coil, then the flux through the long solenoid is, (a) zero, , (b), , m 0 N12 AI, l1, , m 0 N1 N 2 AI, l2, Q.14 If the current in the primary coil is reduced from 3.0, ampere to zero in 0.001 second, the induced e.m.f in the, secondary coil is 1500 volt. The mutual inductance of the, two coils will be(a) 0.5 H, (b) 0.05 H (c) 0.005 H (d) 0.0005 H, Q.15 A 50 Hz a.c. current of crest value 1A flows through the, primary of a transformer. If the mutual inductance between, the primary and secondary be 1.5 H, the crest voltage, induced in secondary is(a) 75 V, (b) 150 V (c) 471 V (d) 300 V, Q.16 In an inductor of inductance L = 100 mH, a current of I =, 10A is flowing. The energy stored in the inductor is, (a) 5 J, (b) 10 J, (c) 100 J, (d) 1000 J, , (c) inifinite, , (d), , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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DPP/ P 45, , 4, DIRECTIONS (Qs. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), (c), (d), , t.me/Magazines4all, , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement-1 is False, Statement-2 is True., Statement-1 is True, Statement-2 is False., , RESPONSE GRID, , 27., , 28., , Q.27 Statement-1 : Soft iron is used as a core of transformer., Statement-2 : Area of hysteresis loop for soft iron is, small., Q.28 Statement-1 : An electric motor will have maximum, efficiency when back e.m.f. is equal to half of the applied, e.m.f., Statement-2 : Efficiency of electric motor depends only, on magnitude of back e.m.f., Q.29 Statement-1 : A transformer cannot work on dc supply., Statement-2 : dc changes neither in magnitude nor in, direction., , 29., , DAILY PRA CTICE PROBLEM SHEET 45 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 46, SYLLABUS : ALTERNATING CURRENT - 1 (Alternating currents, peak and rms value of, alternating current/voltage; reactance and impedance, Pure circuits, LR, CR ac circuits.), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The resistance of a coil for dc is 5 ohm. In ac, the resistance, will, (a) remain same, (b) increase, (c) decrease, (d) be zero, Q.2 If instantaneous current is given by i = 4 cos(wt + f), amperes, then the r.m.s. value of current is, (a) 4 ampere, , (b) 2 2 ampere, , (c) 4 2 ampere, , (d) zero ampere, , Q.3 In an ac circuit I = 100 sin 200pt. The time required for, the current to achieve its peak value will be, 1, 1, 1, 1, sec (b), sec (c), sec (d), sec, (a), 100, 200, 300, 400, Q.4 The frequency of ac mains in India is, (a) 30 c/s or Hz, (b) 50 c/s or Hz, (c) 60 c/s or Hz, (d) 120 c/s or Hz, Q.5 The peak value of an alternating e.m.f. E given by, E = E0 cos wt is 10 volts and its frequency is 50 Hz. At, time t =, , 1, sec, the instantaneous e.m.f. is, 600, , (a) 10 V, , RESPONSE GRID, , 1., , 2., , 3., Space for Rough Work, , (b) 5 3 V, , 4., , (c) 5 V, , 5., , (d) 1 V
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t.me/Magazines4all, DPP/ P 46, , 2, Q.6 An alternating current is given by the equation, i = i1 cos wt + i2 sin wt . The r.m.s. current is given by, (a), , 1, 2, 1, , (i1 + i2 ), , (b), , 1, 2, , (i1 + i2 ) 2, , 1 2 2 1/ 2, (i1 + i2 ), 2, 2, Q.7 In a circuit, the value of alternating current is measured by, hot wire ammeter as 10 ampere. Its peak value will be, (a) 10 A, (b) 20 A, (c) 14.14 A (d) 7.07 A, Q.8 The frequency of an alternating voltage is 50 cycles/sec, and its amplitude is 120V. Then the r.m.s. value of voltage, is, (a) 101.3V (b) 84.8V (c) 70.7V (d) 56.5V, Q.9 A resistance of 20W is connected to a source of an, , (c), , (i12 + i22 )1/ 2, , (d), , alternating potential V = 220sin(100pt ) . The time taken, by the current to change from its peak value to r.m.s. value, is, (a) 0.2 sec, (b) 0.25 sec, (c) 25 × 10–3 sec, (d) 2.5 × 10–3 sec, Q.10 An alternating current of frequency f is flowing in a circuit, containing a resistor of resistance R and a choke of, inductance L in series. The impedance of this circuit is, (a), , R + 2pfL, , (b), , R 2 + 4p2 f 2 L2, , (d) R 2 + 2pfL, R 2 + L2, Q.11 An alternating voltage is connected in series with a, resistance R and an inductance L. If the potential drop across, the resistance is 200 V and across the inductance is 150 V,, then the applied voltage is, (a) 350 V, (b) 250 V, (c) 500 V, (d) 300 V, Q.12 An inductive circuit contains resistance of 10W and an, inductance of 20 H. If an ac voltage of 120 V and frequency, 60 Hz is applied to this circuit, the current would be nearly, (c), , RESPONSE, GRID, , (a) 0.32 A, (b) 0.016 A, (c) 0.48 A, (d) 0.80 A, Q.13 A 20 volt ac is applied to a circuit consisting of a resistance, and a coil with negligible resistance. If the voltage across, the resistance is 12 V , the voltage across the coil is, (a) 16 volt, (b) 10 volt, (c) 8 volt, (d) 6 volt, Q.14 An alternating voltage E = 200 2 sin(100t ) is connected, to a 1 microfarad capacitor through an ac ammeter. The, reading of the ammeter will be, (a) 10 mA, (b) 20 mA, (c) 40 mA, (d) 80 mA, Q.15 A resistor and a capacitor are connected in series with an, a.c. source. If the potential drop across the capacitor is 5, V and that across resistor is 12 V, applied voltage is, (a) 13 V, (b) 17 V, (c) 5 V, (d) 12 V, Q.16 A 120 volt ac source is connected across a pure inductor, of inductance 0.70 henry. If the frequency of the ac source, is 60 Hz, the current passing through the inductor is, (a) 4.55 amp, (b) 0.355 amp, (c) 0.455 amp, (d) 3.55 amp, Q.17 The instantaneous value of current in an A.C. circuit is, I = 2 sin (100 pt + p/3)A. The current will be maximum, for the first time at, 1, 1, s, s, (a) t =, (b) t =, 100, 200, 1, 1, s, s, (c) t =, (d) t =, 400, 600, , æ 0.4 ö, Q.18 In an L – R circuit, the value of L is ç, ÷ henry and the, è p ø, value of R is 30 ohm. If in the circuit, an alternating e.m.f., of 200 volt at 50 cycles per sec is connected, the, impedance of the circuit and current will be, (a) 11.4W,17.5A, (b) 30.7 W,6.5A, (c) 40.4W,5A, (d) 50W, 4A, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 46, , 3, , Q.19 The voltage across a pure inductor is represented by the, following diagram. Which one of the following diagrams, will represent the current?, V, , t, , Ai, , Ai, , (a), , (b), , t, , Ai, , t, , Ai, , (c), , t, , (d), , t, , Q.20 One 10 V, 60 W bulb is to be connected to 100 V line. The, required induction coil has self-inductance of value, ( f = 50 Hz), (a) 0.052 H, (b) 2.42 H, (c) 16.2 mH, (d) 1.62 mH, 1, Q.21 A resistance of 300W and an inductance of, henry are, p, connected in series to a ac voltage of 20 volt and 200 Hz, frequency. The phase angle between the voltage and current, is, -1 4, -1 3, (a) tan, (b) tan, 3, 4, 3, 2, -1, -1, (c) tan, (d) tan, 2, 5, , RESPONSE, GRID, , DIRECTIONS (Q.22-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 If an alternating voltage is represented as E = 141 sin, (628t), then, (1) the rms voltage is 141V, (2) the rms voltage is 100V, (3) the frequency is 50 Hz, (4) the frequency is 100 Hz, Q.23 The r.m.s. value of an ac of 50 Hz is 10 A., (1) The time taken by the alternating current in reaching, from zero to maximum value is 5×10–3 sec, (2) The time taken by the alternating current in reaching, from zero to maximum value is 2×10–3 sec, (3) The peak current is 14.14 A, (4) The peak current is 7.07 A, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, If the voltage in an ac circuit is represented by the equation,, V = 220 2 sin (314t – f), then, Q.24 RMS value of the voltage is, (a) 220 V, (b) 314 V, (c) 220 2 V, (d) 200 / 2 V, Q.25 Average voltage is, (a) 220 V, (b) 622/pV, (c) 220 2 V, Q.26 Frequency of ac is, (a) 50 Hz, (c) 50 2 Hz, , 19., , 20., , 21., , 24., , 25., , 26., , Space for Rough Work, , 22., , (d) 200 / 2 V, (b) 50 2 Hz, (d) 75 Hz, , 23.
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DPP/ P 46, , 4, DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), (c), (d), , t.me/Magazines4all, , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , RESPONSE GRID, , 27., , 28., , Q27 Statement-1: The alternating current lags behind the em.f., by a phase angle of p/2, when ac flows through an inductor., Statement-2 : The inductive reactance increases as the, frequency of ac source decreases., Q.28 Statement-1 : An alternating current does not show any, magnetic effect., Statement-2: Alternating current varies with time., Q.29 Statement-1 : A capacitor of suitable capacitance can be, used in an ac circuit in place of the choke coil., Statement-2 : A capacitor blocks dc and allows ac only., , 29., , DAILY PRA CTICE PROBLEM SHEET 46 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 47, SYLLABUS : ALTERNATING CURRENT - 2 (LCR series circuit, resonance, quality factor, power in AC circuits,, wattless and power current), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 In a series LCR circuit capacitance is changed from C to 2, C. For the resonant frequency to remain unchanged, the, inductance would be changed from L to, (a) L/2, (b) 2 L, (c) 4 L, (d) L/4, Q.2 The power factor of LCR circuit at resonance is, (a) 0.707, (b) 1, (c) Zero, (d) 0.5, Q.3 An alternating current source of frequency 100 Hz is, joined to a combination of a resistance, a capacitance and, a inductance in series. The potential difference across the, inductance, the resistance and the capacitor is 46, 8 and, , RESPONSE GRID, , 1., , 2., , 40 volt respectively. The electromotive force of alternating, current source in volt is, (a) 94, (b) 14, (c) 10, (d) 76, Q.4 A 10 ohm resistance, 5 mH inductance coil and 10 mF, capacitor are joined in series. When a suitable frequency, alternating current source is joined to this combination,, the circuit resonates. If the resistance is halved, the, resonance frequency, (a) is halved, (b) is doubled, (c) remains unchanged, (d) is quadrupled, Q.5 The phase difference between the current and voltage of, LCR circuit in series combination at resonance is, (a) 0°, (b) p/2, (c) p, (d) – p, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 47, , 2, Q.6 The coefficient of induction of a choke coil is 0.1H and, resistance is 12W. If it is connected to an alternating current, source of frequen cy 60 Hz, then power factor is, approximately, (a) 0.4, (b) 0.30, (c) 0.2, (d) 0.1, Q.7 The resonant frequency of a circuit is f. If the capacitance, is made 4 times the initial values, then the resonant, frequency will become, (a) f /2, (b) 2f, (c) f, (d) f/4, Q.8 In the non-resonant circuit, what will be the nature of the, circuit for frequencies higher than the resonant frequency?, (a) Resistive, (b) Capacitive, (c) Inductive, (d) None of the above, Q.9 In a series LCR circuit, resistance R = 10W and the impedance, Z = 20W . The phase difference between the current and the, voltage is, (a) 30°, (b) 45°, (c) 60°, (d) 90°, 1 ö, æ, Q.10 An alternating e.m.f. of frequency n ç =, is, è 2 p LC ÷ø, applied to a series LCR circuit. For this frequency of the, applied e.m.f., (a) The circuit is at resonance and its impedance is made, up only of a reactive part, (b) The current in the circuit is not in phase with the, applied e.m.f. and the voltage across R equals this, applied emf, (c) The sum of the p.d.’s across the inductance and, capacitance equals the applied e.m.f. which is 180°, ahead of phase of the current in the circuit, (d) The quality factor of the circuit is wL / R or 1/ wCR, and this is a measure of the voltage magnification, (produced by the circuit at resonance) as well as the, sharpness of resonance of the circuit, Q.11 In a circuit L,C and R are connected in series with an, alternating voltage source of frequency f. The current leads, the voltage by 45°. The value of C is, 1, 1, (a) 2pf (2pfL + R ), (b), pf (2pfL + R ), 1, 1, (c), (d), 2pf (2pfL - R ), pf (2pfL - R), , RESPONSE, GRID, , Q.12 For the series LCR circuit shown in the figure, what is the, resonance frequency and the amplitude of the current at, the resonating frequency, 8 mH, , 220 V, , 20µF, , 44 W, , (a) 2500 rad s -1 and 5 2A (b) 2500rad s -1 and 5A, 5, A (d) 25rad s -1 and 5 2A, (c) 2500rad s -1 and, 2, Q.13 In an ac circuit, V and I are given by V = 100sin(100 t ) volt,, , pö, æ, I = sin ç 100t + ÷ mA . The average power dissipated in, è, 3ø, circuit is, (a) 104 watt, (b) 10 watt, (c) 0.025 watt, (d) 2.5 watt, Q.14 For a series LCR circuit R = XL = 2XC. The impedance of, the circuit and phase difference between V and I, respectively will be, (a), , 5R, , tan -1 (2), 2, , (b), , 5R, , tan -1 (1 / 2), 2, , (c), , 5X C , tan -1 (2), , (d), , 5R, tan -1 (1/ 2), , pö, æ, Q.15 If a current I given by I0 sin ç wt - ÷ flows in an ac circuit, 2ø, è, across which an ac potential of E = E0 sin wt has been, applied, then the average power consumption P in the circuit, will be, E I, (a) P = 0 0, (b) P = 2 E0l0, 2, (c) P = E0 I 0, (d) P = 0, 2, Q.16 An ac supply gives 30V r.m.s. which passes through a 10 W, resistance. The power dissipated in it is, (a) 90 2 W (b) 90 W, , (c) 45 2 W (d) 45 W, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 47, , 3, , Q.17 The figure shows variation of R,XL and X C with frequency, f in a series L, C , R circuit. Then for what frequency, point, the circuit is inductive, XC, , XL, , R, , A B, , f, , C, , (a) A, (b) B, (c) C, (d) All points, Q.18 An alternating e.m.f. of angular frequency w is applied, across an inductance. The instantaneous power developed, in the circuit has an angular frequency, w, w, (a), (b), (c) w, (d) 2w, 4, 2, Q.19 In the circuit shown in figure neglecting source resistance, the voltmeter and ammeter reading will respectively, be, V, , A, , R = 30W, , X L = 25W, , X C = 25W, , DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, , 240 V, , (a), , (b) 150V ,3 A, (c) 150V , 6 A, (d) 0V ,8 A, Q.20 In an LCR circuit, the sharpness of resonance depends on, (a) Inductance (L), (b) Capacitance (C), (c) Resistance (R), (d) All of these, 0V ,3 A, , DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, , RESPONSE, GRID, , Q.21 For series LCR circuit, correct statements are, (1) Applied e.m.f. and potential difference across, resistance may be in phase, (2) Applied e.m.f. and potential difference at inductor coil, have phase difference of p/2, (3) Potential difference across resitance and capacitor, have phase difference of p/2, (4) Potential difference at capacitor and inductor have, phase difference of p/2, Q.22 An ac source is connected to a resistive circuits. Which, of the following statements are false?, (1) Current leads the voltage, (2) Current lags behind the voltage, (3) Any of (1) or (2) may be true depending upon the value, of resistance, (4) Current and voltage are in same phase, Q.23 A series LCR arrangement with XL = 80 W, XC = 50 W, R = 40 W, is applied across a.c. source of 200 V. Choose the correct, options., (1) Wattless current = 3.2 A, (2) Power current = 3.2 A, (3) Power factor = 0.6, (4) Impedance of circuit = 50 W, , A student constructs a series RLC circuit. While operating the, circuit at a frequency f she uses an AC voltmeter and measures, the potential difference across each device as (DVR) = 8.8 V,, (DVL) = 2.6V and (DVC) = 7.4V., Q.24 The circuit is constructed so that the inductor is next to, the capacitor. What result should the student expect for, a measurement of the combined potential difference, (DVL + DVC) across the inductor and capacitor ?, (a) 10.0 V, (b) 7.8 V, (c) 7.4 V, (d) 4.8 V, Q.25 What result should the student expect for a measurement of, the amplitude Em of the potential difference across the power, supply ?, (a) 18.8 V (b) 13.6 V (c) 10.0 V (d) 4.0 V, , 17., , 18., , 19., , 20., , 22., , 23., , 24., , 25., , Space for Rough Work, , 21.
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t.me/Magazines4all, DPP/ P 47, , 4, Q.26 What will happen to the value of (DVL) if the frequency is, adjusted to increase the current through the circuit?, (a) (DVL) will increase., (b) (DVL) will decrease., (c) (DVL) will remain the same regardless of any changes, to f., (d) There is not enough information to answer the, question., DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : For an electric lamp connected in series, with a variable capacitor and ac source, its brightness, increases with increase in capacitance., Statement-2 : Capacitive reactance decreases with, increase in capacitance of capacitor., Q.28 Statement-1 : When capacitive reactance is smaller than, the inductive reactance in LCR current, e.m.f. leads the, current., Statement-2 : The phase angle is the angle between the, alternating e.m.f. and alternating current of the circuit., Q.29 Statement-1 : Choke coil is preferred over a resistor to, adjust current in an ac circuit., Statement-2 : Power factor for inductance is zero., , 27., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 47 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 48, SYLLABUS : EM Waves, , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 Light is an electromagnetic wave. Its speed in vacuum is, given by the expression, (a), , m o eo, , mo, eo, , (b), , (c), , eo, mo, , (d), , Q.2 The range of wavelength of the visible light is, (a) 10 Å to 100 Å, (b) 4,000 Å to 8,000 Å, (c) 8,000 Å to 10,000 Å, (d) 10,000 Å to 15000 Å, , RESPONSE GRID, , 1., , 2., , 1, m o eo, , Q.3 Which of the following radiations has the least wavelength?, (a) g-rays, (b) b-rays, (c) a-rays, (d) X -rays, Q.4 A parallel plate capacitor with plate area A and seperation, between the plates d, is charged by a constant current i., Consider a plane surface of area A/4 parallel to the plates, and drawn symetrically between the plates, what is the, displacement current through this area?, (a) i, (b) 2 i, (c) i /4, (d) i / 2, Q.5 The charging current for a capacitor is 1 A, then the, displacement current is, (a) 1 A, (b) 0.5 A, (c) 0, (d) 2 A, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 48, , 2, Q.6, , Q.7, , Q.8, , Q.9, , r, r, If E and B be the electric and magnetic field of E.M., wave then the direction of propogation of E.M. wave is, along the direction., r, r, (a) E, (b) B, r, r, r, r, (c) E × B, (d) B × E, Which of the following pairs of space and time varying E, and B fields would generate a plane electromagnetic wave, travelling in (–Z) direction ?, (a) Ex, By, (b) Ey, Bx, (c) Ez, By, (d) Ey, Bz, Choose the wrong statement for E.M. wave. They(a) are transverse, (b) travel in vacuum with the speed of light, (c) are produced by accelerated charges, (d) travel with same speed in all medium, The intensity of light from a source is 500/p W/m2. Find, the amplitude of electric field in this wave(a), (c), , 3 × 102 N/C, , 3, × 102 N/C, 2, , (b) 2 3 × 102 N/C, , (c), , (d) 2 3 ×101 N/C, , 10–4, , (b), , 30, , 3 × 10–4, , (d), , 3 × 10–2, , (a) 2.25, , (b) 1.25, , (c) 3.25, , (d) 0.25, , Q.13 A magnetic field of a plane electromagnetic wave is given, by By = 2 × 10–7 sin(0.5 × 103x + 1.5 × 1011t) T. Fequency, of the wave is, (a) 23.9 Hz, , (b) 13.9 Hz, , (c) 33.9 Hz, , (d) 12.9 Hz, , Q.14 The electric field of a plane electromagnetic wave, in vacuum is represented by, Ex = 0, Ey = 0.5 cos[2p × 108 (t – x/c)] and Ez = 0., Determine the wavelength of the wave., (a) 4 m, , (b) 5 m, , (c) 3 m, , (d) 6 m, , Q.15 A light beam travelling in the X-direction is described by, the electric field Ey = (300 V/m)sinw(t – x/c). An electron, is constrained to move along the Y-direction with a speed, of 2.0 × 107 m/s. Find the maximum magnetic force (in N), on the electron., (a) 3.2 × 10–18, , (b) 5.1 × 10–16, , 10–11, , (d) 7.8 × 10–12, , Q.16 Which of the following waves have minimum frequency ?, , Q.11 In a EM wave the amplitude of electric field is 10 V/m., The frequency of wave is 5 × 10 4 Hz. The wave is, propagating along Z-axis. Then the average energy density, of magnetic field is(a) 2.21 × 10–10 J/m3, (b) 2.21 × 10–8 J/m3, (c) 2 × 10–8 J/m3, (d) 2 × 10–10 J/m3, , RESPONSE, GRID, , 2 × 108 m/s. The relative permeability of the medium is 1., What is the relative permittivity of the medium ?, , (c) 6.5 ×, , Q.10 A point source of 2 watt is radiating uniformly in all, direction in vacuum. Find the amplitude of electric field, at a distance 2m from it(a) 3 ×, , Q.12 Elecromagnetic waves travel in a medium with a speed of, , (a) Microwaves, , (b) Audible waves, , (c) Ultrasonic waves, , (d) Radiowaves, , Q.17 Electromagnetic waves travel in a medium which has, relative permeability 1.3 and relative permittivity 2.14., Then the speed of the electromagnetic wave in the medium, will be, (a) 13.6 × 106 m/s, , (b) 1.8 × 102 m/s, , (c) 3.6 × 108 m/s, , (d) 1.8 × 108 m/s, , Q.18 If lv, lx and lm represent the wavelength of visible light, x-rays and microwaves respectively, then, (a) lm > lx > lv, , (b) lv > lm > lx, , (c) lm > lv > lx, , (d) lv > lx > lm, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 48, , 3, , Q.19 Light wave is travelling along + y-direction. If the, r, corresponding E vector at that time is along + x-direction,, r, B vector must be directed along., y, (a) y-axis, (b) x-axis, , x, , (c) + z-axis, , z, (d) – z axis, Q.20 A wave is propagating in a medium of dielectric constant 2, and relative magnetic permeability 50. The wave impedance, of such a medium is, (a) 5W, (b) 376.6W, (c) 1883 W, , (d) 3776 W, , DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Which of the following statements are true ?, (1) Photographic plates are sensitive to ultraviolet rays., (2) Photographic plates can be made sensitive to infrared, rays., (3) Infrared rays are emitted by hot objects., (4) Infrared photon has more energy than the photon of, visible light., Q.22 Which of the following are electromagnetic waves ?, (1) Cosmic rays, (2) Gamma rays, (3) X-rays, (4) b- rays, , RESPONSE, GRID, , Q.23 An electromagnetic wave of frequency n = 3.0 MHz passes, from vacuum into a dielectric medium with permitivity, e = 4.0. Then the wrong statements are, (1) Wavelength is doubled and the frequency remains, unchanged, (2) Wavelength is doubled and frequency becomes half, (3) Wavelength and frequency both remain unchanged, (4) Wavelength is halved and frequency remains, unchanged, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, The electron density of a layer of ionosphere at a height 150, km from the earth surface is 9 ×1010 per m3. For the sky wave, transmission from this layer upto a range of 250 km, find, Q.24 The critical frequency of the layer, (a) 2.7 × 106 Hz, (b) 2.7 × 105 Hz, (c) 4.7 × 106 Hz, (d) 4.8 × 105Hz, Q.25 The maximum usuable frequency, (a) 3.17 × 108 Hz, (b) 3.17 × 106 Hz, (c) 4.57 × 106 Hz, (d) 4.57 × 106Hz, Q.26 The angle of incidence of this layer, (a) 34.5°, (b) 25.2°, (c) 31.6°, (d) 40°, DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., , 19., , 20., , 21., , 24., , 25., , 26., , Space for Rough Work, , 22., , 23.
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DPP/ P 48, , 4, Q.27 Statement-1: The electromagnetic waves of shorter, wavelength can travel longer distances on earth's surface, than those of longer wavelengths., Statement-2: Shorter the wavelength, the larger is the, velocity of wave propagation., , RESPONSE GRID, , t.me/Magazines4all, , 27., , 28., , Q.28 Statement-1: Ultraviolet radiation are of higher frequency, waves and are dangerous to human being., Statement-2: Ultraviolet radiation are absorbed by the, atmosphere., Q.29 Statement-1: Radio waves can be polarised., Statement-2: Sound waves in air are longitudinal in nature., , 29., , DAILY PRA CTICE PROBLEM SHEET 48 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 49, SYLLABUS : RAY OPTICS-1 (Reflection on plane mirrors and curved mirrors), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 Find the number of images formed by two mutually, perpendicular mirrors –, (a) 3, , (b) 4, , (c) 1, , (d) 2, , (a) 3, , (b) 45º £ q £ 72º, , (c) 60º £ q £ 72º, , (d) 15º £ q £ 72º, , Q.3 Two mirrors are inclined at an angle of 50º. Then what is, the number of images formed for an object placed in between the mirrors ?, , RESPONSE GRID, , 1., , 2., , (c) 1, , (d) 7, , Q.4 Two plane mirrors are inclined at an angle q. A ray of light, is incident on one mirror at an angle of incidence i. The, ray is reflected from this mirror, falls on the second mirror from where it is reflected parallel to the first mirror., What is the value of i, the angle of incidence in terms q ?, , Q.2 The angle q between two plane mirrors producing five, images of a given object is given by., (a) 30º £ q £ 72º, , (b) 6, , (a) 2q - 90º, , (b) 4q - 90º, , (c) q - 90º, , (d) 3q - 90º, , Q.5 A girl stands at a distance 30 cm from the mirror. She is, able to see her erect image but of 1/5 height of actual, height. The mirror will be :, (a) plane mirror, , (b) concave mirror, , (c) convex mirror, , (d) plane convex mirror, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 49, , 2, , (c) 12.5 cm (d) 100cm, , Q.7 Two plane mirrors are inclined at an angle of 30º. Then the, first four images of an object O placed between the two, mirrors are correctly represented by, M1, , I1, , I4, , M2, , (a), , I4, , I3, , M1, , (b), , M2, , I2, , I3, , I1, , I3, , I2, , I1 I3, , M1, M2, , I2, , (d), , I2 I4, , Q.8 The plane of a mirror makes an angle of 30º with horizontal., If a vertical ray is incident on a mirror, then what is the, angle between mirror and reflected ray ?, (b) 90º, , æ R ö, (a) ç, .v, è 2u - R ÷ø 0, , (c) 45º, , æ 2R ö, (c) ç, .v, è 2u - R ÷ø 0, , (b) 60º, , (d) 30º, , (c) 75º, , (d) 90º, , Q.10 A 0.2 cm high object is placed 15 cm from a concave mirror, of focal length 5 cm. Find position and size of the image., (a) 7.5 cm, 0.1 cm., , (b) 7.5 cm, 0.4 cm., , (c) 10.0 cm, 0.5 cm., , (d) 7.5 cm, 0.4 cm., , RESPONSE, GRID, , (d) None of these, , Q.14 A short linear object of length b lies along the axis of a, concave mirror of focal length f at a distance u from the, pole of the mirror. Find the approximate size of the image., , Q.9 Two plane mirrors are placed at an angle a so that a ray parallel to one mirror gets reflected parallel to the second mirror, after two consecutive reflections. The value of a will be, (a) 30º, , 2, , æ R ö, (b) ç, .v, è 2u + R ÷ø 0, , 2, , M2, , I4, , (a) 60º, , 2, , M1, , I1, , (c), , //// ////, , (b) 25 cm, , ////, , (a) 50 cm, , Q.11 A 0.5 cm high object is placed at 30 cm from a convex, mirror whose focal length is 20 cm. Find the position and, size of the image., (a) 12 cm, 0.2 cm, (b) 18 cm, 0.2 cm, (c) 6 cm, 0.5 cm, (d) 5 cm, 0.1 cm, Q.12 There is a convex mirror of radius 50 cm. The image of a, point at a distance 50 cm from the pole of mirror on its, axis will be formed at :, //, ////, (a) infinity, O·, ·C, 50 cm, 50 cm, (b) pole, (c) focus, (d) 16.67 cm behind the mirror, Q.13 A particle is moving at a constant speed v0 from a large, distance towards a concave mirror of radius R along its, principle axis. Find the speed of the image formed by the, mirror as a function of the distance u of the particles from, the mirror., /// ////, , Q.6 An object is placed at a distance of 50 cm from a convex, mirror. A plane mirror is placed in front of the convex, mirror in such a way that it convers half of the convex, mirror. If the distance between object and plane mirror is, 30 cm then there is no parallax between the images formed, by two mirrors , the radius of curvature of convex mirror, will be :, , æ f ö, (a) b ç, è u - f ÷ø, , 2, , æ 2f ö, (c) b ç, è u - f ÷ø, , 2, , æ f ö, (b) b ç, è u + f ÷ø, , 2, , (d) None of these, , Q.15 The relation between the linear magnification m, the object, distance u and the focal length f for a spherical mirror is, (a) m =, , f -u, f, , (b) m = f - u, , (c) m =, , f +u, f, , (d) m = f + u, , f, f, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 49, , 3, , Q.16 An object of length 1 cm is placed at a distance of 15 cm, from a concave mirror of focal length 10 cm. The nature, and size of the image are, (a) real, inverted, 1.0 cm, (b) real, inverted, 2.0 cm, (c) virtual, erect, 0.5 cm, (d) virtual, erect, 1.0 cm, Q.17 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 its 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, (b) f< x< 2f (c) x = 2f (d) x > 2f, Q.18 Two plane mirrors are inclined to each other at some angle., A ray of light incident at 30º on one, after reflection from, the other retraces its path. The angle between the mirrors, is, (a) 30º, (b) 45º, (c) 60º, (d) 90º, Q.19 A convex mirror is used to form the image of an object., Which of the following statements is wrong ?, (a) The image lies between the pole and the focus, (b) The image is diminished in size, (c) The image is erect, (d) The image is real, Q.20 A point source of light B is placed at a distance L in front, of the centre of a mirror of width 'd' hung vertically on a, wall. A man walks in front of the mirror along a line parallel, to the mirror at a distance 2L from it as shown in fig. The, greatest distance over which he can see the image of the, light source in the mirror is, B, , Q.21 A concave mirror of focal length f0(in magnitude) produces, a real image n time the size of the object. What is the, distance of the object from the mirror?, (a), , (a) 20 cm (only), , (b) 40 cm (only), , (c) 30 cm (only), , (d) 20 cm or 40 cm, , Codes :, (a), , 1, 2 and 3 are correct, , (b) 1 and 2 are correct, , (c), , 2 and 4 are correct, , (d) 1 and 3 are correct, , Q.23 A plane mirror reflecting a ray of incident light is rotated, through an angle q about an axis through the point of, incidence in the plane of the mirror perpendicular to the, plane of incidence, then, (1) The reflected ray rotates through an angle 2q, (2) The incident ray is fixed, (3) The reflected ray does not rotate, (4) The reflected ray rotates through an angle q, Q.24 The light reflected by a plane mirror will not form a real, image, (1) If the rays incident on the mirror are diverging, (2) Under no circumstances, (3) If the object is real, , 2L, , RESPONSE, GRID, , -(n2 +1), (n - 1), f0 (d), f0, n, n, , DIRECTIONS (Q.23-Q.25) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, , L, , (b) d, , (c), , Q.22 The focal length of a concave mirror is 30 cm. Find the, position of the object in front of the mirror, so that the, image is three times the size of the object., , d, , (a) d/2, , -(n +1), (n +1), f0 (b), f0, n, n, , (c) 2d, , (4) If the rays incident on the mirror are converging, , (d) 3d, , 16., , 17., , 18., , 19., , 21., , 22., , 23., , 24., , Space for Rough Work, , 20.
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t.me/Magazines4all, DPP/ P 49, , 4, Q.25 Which of the following form(s) a virtual and erect image, for all positions of the object?, (1) Convex lens, , (2) Concave lens, , (3) Convex mirror, , (4) Concave mirror, , DIRECTIONS (Q.26-Q.27) : Read the passage given below, and answer the questions that follows :, A plane mirror (M1) and a concave mirror, (M2) of focal length 10 cm are arranged as, shown in figure. An object is kept at origin., Answer the following questions. (Consider, image formed by single reflection in all, cases), , Y, , M2, , 10 cm, , 20 cm, , 45°, O, , X, M1, , Q.26 The co-ordinates of image formed by plane mirror are, (a) (–20 cm, 0), , (b), , (10 cm, – 60 cm), , (c) (10 cm, –10 cm), , (d) (10 cm, 10 cm), , Q.27 The co-ordinates of image formed by concave mirror are, (a) (10 cm, – 40 cm), , (b) (10 cm, – 60 cm), , (c) (10 cm, 8 cm), , (d) None of these, , RESPONSE, GRID, , 25., , 26., , DIRECTIONS (Qs. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.28 Statement-1: The mirrors used in search lights are, parabolic and not concave spherical., Statement-2: In a concave spherical mirror the image, formed is always virtual., Q.29 Statement-1: When an object is placed between two plane, parallel mirors, then all the images found are of different, intensity., Statement-2: In case of plane parallel mirrors, only two, images are possible., Q.30 Statement-1: The size of the mirror doesn't affect the, nature of the image., Statement-2: Small mirror always forms a virtual image., , 27., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 49 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 50, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 50, SYLLABUS : RAY OPTICS - II (Refraction on plane surface, total internal reflection, prism), , Max. Marks : 108, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 27 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A thin prism of angle A = 6° produces a deviation d = 3°., Find the refractive index of the material of prism., (a) 1.5, (b) 1.0, (c) 2.5, (d) 0.5, Q.2 A ray of light is incident at an angle of 60° on one face of, a prism which has an angle of 30°. The ray emerging out of, the prism makes an angle of 30° with the incident ray., Calculate the refractive index of the material of the prism, (a) 1, , (b), , (c), , (d) 2, , 3, , RESPONSE GRID, , 1., , Q.3 Light of wavelength 6000Å enters from air into water, (a medium of refractive index 4/3). Find the speed and, wavelength [c = 3 × 108 m/s], (a) 2.25 × 108 m/s, 4500Å (b) 1.25 × 108 m/s, 2500Å, (c) 3.15 × 108 m/s, 3500Å (d) 3.45 × 108 m/s, 5500Å, Q.4 A ray of light is incident on a transparent glass-slab of, refractive index 1.5. If the reflected and refracted rays are, mutually perpendicular, what is the angle of incidence ?, (a) 30°, , 2, , -1, (c) tan, , 2., , 3., Space for Rough Work, , 2, 3, , 4., , -1, (b) sin, , 2, 3, , -1, (d) tan, , 3, 2
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t.me/Magazines4all, DPP/ P 50, , 2, Q.5 An optical fibre consists of core of m1 surrounded by a, cladding of m2 < m1. A beam of light enters from air at an, angle a with axis of fibre. The highest a for which ray can, be travelled through fibre is, (a) cos -1 m 22 - m12, (b) sin, , -1, , m12, , - m 22, , m2, a, , Q.7, , Q.8, , Q.9, , m1, , -1, m12 - m 22, (d) sec, A glass plate 4 mm thick is viewed from the above through, a microscope. The microscope must be lowered 2.58 mm, as the operator shifts from viewing the top surface to, viewing the bottom surface through the glass. What is the, index of refraction of the glass ?, (a) 1.61, (b) 1.55, (c) 3.24, (d) 1.21, A vertical microscope is focussed on a point at the bottom, of an empty tank. Water (m = 4/3) is then poured into the, tank. The height of the water column is 4cm. Another lighter, liquid, which does not mix with water and which has, refractive index 3/2 is then poured over the water.The height, of liquid column is 2cm. What is the vertical distance, through which the microscope must be moved to bring the, object in focus again ?, (a) 2.61 m (b) 1.55 m (c) 3.12 m (d) 1.67 m, Light from a sodium lamp (l0 = 589 nm) passes through a, tank of glycerin (refractive index 1.47) 20m long in a time, t1. If it takes a time t2 to traverse the same tank when filled, with carbon disulphide (index 1.63), then the difference t2 –, t1 is, (a) 6.67 × 10-8 sec, (b) 1.09 × 10-7 sec, -7, (c) 2.07 × 10 sec, (d) 1.07 × 10-8 sec, A light beam is travelling from Region I to Region IV (Refer, Figure). The refractive index in Regions I, II, III and IV are, n n, n, n0, 0 , 0 and 0 , respectively. The angle of incidence q, 2 6, 8, for which the beam just misses entering Region IV is, , RESPONSE, GRID, , q, n0, , n0, 2, 0, , (c) tan -1 m12 - m 22, , Q.6, , Region I Region II Region III Region IV, n0, 6, , n0, 8, , 0.2 m, , 0.6 m, æ 3ö, -1 æ 1 ö, (a) sin -1 ç ÷, (b) sin ç ÷, 4, è8ø, è ø, 1, -1 æ ö, -1 æ 1 ö, (c) sin ç ÷, (d) sin ç ÷, è4ø, è3ø, Q.10 The refractive index of the material of a prism is 2 and, its prism angle is 30º. One of its refracting faces is polished., The incident beam of light will return back for the angle of, incidence, (a) 60º, (b) 45º, (c) 30º, (d) 0º, Q.11 A ray of light incident on a prism surface at an angle of 50º, in the minimum deviation position. If the angle of prism is, 60º then the values of dm and m will be respectively –, (sin 50º = 0.766), (a) 40º and 1.532, (b) 60º and 1.532, (c) 90º and 1.532, (d) 0º and 1.532, Q.12 A glass prism of refractive index 1.5 and angle of prism 6º, is put in contact with another prism of refractive index 1.6, when a ray of light is made incident on this combination, normally then it emerges out undeviated. The angle of, second prism will be –, (a) 6º, (b) 5º, (c) 4º, (d) 3º, Q.13 A crown glass prism of angle 5° is to be combined with a, flint glass prism in such a way that the mean ray passes, undeviated. Find the angle of the flint glass prism needed, and the angular dispersion produced by the combination, when white light goes through it. Refractive indices for, red, yellow and violet light are 1.514, 1.517 and 1.523, respectively for crown glass and 1.613, 1.620 and 1.632, for flint glass., (a) 4.2°, 0.0348°, (b) 4.2°, 0.0138°, (c) 1.2°, 0.0348°, (d) 4.4°, 0.0218°, Q.14 Calculate the dispersive power for crown glass from the, (given data : mv= 1.5230, m r =1.5145), (a) 0.0163 (b) 0.0183 (c) 0.0142 (d) 0.0112, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 50, , 3, , Q.15 A prism of dispersive power 0.021 and refractive index, 1.53 form an achromatic combination with prism of angle, 4.2° and dispersive power 0.045 having refractive index, 1.65. Find the resultant deviation., (a) 1.12°, (b) 2.16°, (c) 3.12°, (d) 4.18°, Q.16 A ray of light fall normally on a refracting face of a prism, of refractive index 1.5. Find the angle of the prism if the, ray just fails to emerge from the prism., (a) 55°, (b) 22°, (c) 12°, (d) 42°, Q.17 The refractive indices of material of a prism for blue and, red colours are 1.532 and 1.514 respectively. Calculate, angular dispersion produced by the prism if angle of prism, is 8°., (a) 0.144° (b) 0.122° (c) 0.133° (d) 0.111°, Q.18 A ball is dropped from a height of 20 m above the surface, of water in a lake. The refractive index of water is 4/3. A, fish inside the lake, in the line of fall of the ball, is looking, at the ball. At an instant, when the ball is 12.8 m above the, water surface, the fish sees the speed of ball as [g = 10 m/, s2], (a) 9 m/s, (b) 12 m/s (c) 16 m/s (d) 21.33 m/s, Q.19 The dispersive powers of crown and flint glasses are 0.03, and 0.05 respectively. The refractive indices for yellow, light for these glasses are 1.517 and 1.621 respectively ., It is desired to form an achromatic combination of prisms, of crown and flint glasses which can produce a deviation, of 1° in the yellow ray. The refracting angle of the flint, glass prism is, (a) 2.4°, (b) 1.4°, (c) 3.4°, (d) 5.2°, Q.20 A glass prism (m = 1.5) is dipped in water ( m = 4/3) as, shown in figure. A light ray is incident normally on the, surface AB. It reaches the surface BC after T.I.R, B, , q, , Q.21 A prism having an apex angle 4° and refraction index 1.5, is located in front of a vertical plane mirror as shown in, figure. Through what total angle is the ray deviated after, reflection from the mirror, 90° 4°, , (a) 176°, (c) 178°, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 A ray of monochromatic light is incident on the plane, surface of separation between two media x and y with angle, of incidence i in the medium x and angle of refraction r in, the medium y. The graph shows the relation between sin i, and sin r., , sin r, , 30°, sin i, (1) The speed of light in the medium y is 3 times than in, medium x, 1, (2) The speed of light in the medium y is, times than in, 3, medium x., (3) The total internal reflection can take place when the, incidence is in x., (4) The total internal reflection can take place when the, incidence is in y, , A, , C, , (a) sin q ³ 8/9, (c) sin q £ 2/3, , RESPONSE, GRID, , (b) 4°, (d) 2°, , (b) 2/3 < sin q < 8/9, (d) It is not possible, , 15., , 16., , 17., , 20., , 21., , 22., Space for Rough Work, , 18., , 19.
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t.me/Magazines4all, DPP/ P 50, , 4, Q.23 Dispersive power does not depend upon, (1) The shape of prism, (2) Angle of prism, (3) Height of the prism, (4) Material of prism, Q.24 The wrong statements are, (1) The order of colours in the primary and the secondary, rainbows is the same, (2) The intensity of colours in the primary and the, secondary rainbows is the same, (3) The intensity of light in the primary rainbow is greater, and the order of colours is the same than the secondary, rainbow, (4) The intensity of light for different colours in primary, rainbow is greater and the order of colours is reverse, as that in the secondary rainbow, DIRECTIONS (Qs. 25-Q.27) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , (c), , Statement -1 is False, Statement-2 is True., , (d), , Statement -1 is True, Statement-2 is False., , Q.25 Statement-1: There is no dispersion of light refracted, through a rectangular glass slab., Statement-2: Dispersion of light is the phenomenon of, splitting of a beam of white light into its constituent, colours., Q.26 Statement-1: Dispersion of light occurs because velocity, of light in a material depends upon its colour., Statement-2: The dispersive power depends only upon the, material of the prism, not upon the refracting angle of the, prism., Q.27 Statement-1: If a plane glass slab is placed on the letters, of different colours all the letters appear to be raised up to, the same height., Statement-2: Different colours h ave differen t, wavelengths., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 23., , 24., , 25., , 26., , 27., , DAILY PRA CTICE PROBLEM SHEET 50 - PHYSICS, Total Questions, 27, Total Marks, 108, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 51, SYLLABUS : RAY OPTICS - 3 (Refraction on curved surface lens, Optical instrument), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 A small point object is placed at O, at a distance of 0.60, metre in air from a convex spherical surface of refractive, index 1.5. If the radius of the curvature is 25 cm, then what, is the position of the image on the principal axis ?, (a) 4.5 m, (b) 2.5 m, (c) 1.5 m, (d) 5.5 m, Q.2 The radius of a glass ball is 5 cm. There is an air bubble at, 1cm from the centre of the ball and refractive index of, glass is 1.5. The position of image viewed from surface, near the bubble is., (a) 3.63 cm, (b) 4.63 cm, (c) 2.12 cm, (d) 5.12 cm, , RESPONSE GRID, , 1., , 2., , Q.3 In case of thin lens of focal length f an object is placed at, a distance x1 from first focus and its image is formed at a, distance x2 from the second focus, find x1 x2, (a) f, (b) f 3, (c) f 2, (d) 1/f, Q.4 What is the refractive index of material of a plano-convex, lens , if the radius of curvature of the convex surface is, 10 cm and focal length of the lens is 30 cm ?, (a) 1/3, (b) 4/3, (c) 2/3, (d) 1/4, Q.5 A convex lens of focal length 10.0 cm is placed in contact, with a convex lens of 15.0 cm focal length. What is the, focal length of the combination ?, (a) 6 cm, (b) 12 cm, (c) 8 cm, (d) 4 cm, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 51, , 2, Q.6 A convex lens of focal length 20 cm is placed in contact, with a diverging lens of unknown focal length. The lens, combination acts as a converging lens and has a focal length, of 30 cm. What is the focal length of the diverging lens ?, (a) – 10 cm (b) – 30 cm (c) – 60 cm (d) – 90 cm, Q.7 A pin is placed 10 cm in front of a convex lens of focal, length 20 cm, made of material having refractive index 1.5., The surface of the lens farther away from the pin is silvered, and has a radius of curvature 22 cm. Determine the position, of the final image., (a) 11 cm in front, (b) 21 cm in front, (c) 15 cm in front, (d) 31 cm in front, Q.8 An image is formed on the screen by a convex lens. When, upper half part of lens is covered with black paper, then :, (a) half image is formed, (b) full image is formed, (c) intensity of image will be enhanced, (d) None of these, Q.9 A convex lens is made out of a substance of 1.2 refractive, index. The two surfaces of lens are convex. If this lens is, placed in water whose refractive index is 1.33, it will, behave as :, (a) convergent lens, (b) divergent lens, (c) plane glass plate, (d) like a prism, Q.10 An equiconvex lens has a power of 5 diopter. If it is made, of glass of refractive index 1.5 then the radius of the, curvature of each surface will be, (a) 20 cm (b) 10 cm (c) 5 cm, (d) zero, Q.11 A convex lens when placed in the first position forms a, real image of an object on a fixed screen. The distance, between the object and the screen is 75 cm. On displacing, the lens from first position by 25 cm to the second position,, again a real image is formed on the screen. Then the focal, length of the lens is, D = 75 cm, x = 25 cm, , I2, , Object, , First position Second position, , RESPONSE, GRID, , I1, , (a) 25.0 cm (b) 16.7 cm (c) 50.3 cm (d) 33.3 cm, Q.12 A lens is placed between a source of light and a wall. It, forms images of area A1 and A2 on the wall for its two, different positions. The area of the source of light is, (a), , é 1, 1 ù, (b) ê +, ú, ë A1 A 2 û, , A1 +A 2, 2, , -1, , 2, , é A1 + A 2 ù, ú, (c), A1A 2, (d) ê, 2, ëê, ûú, Q.13 A convex lens of power 4D is kept in contact with a concave, lens of power 3D, the effective power of combination will, be, (a) 7D, (b) 4D/3, (c) 1D, (d) 3D/4, Q.14 The power of a plano-convex lens is P. If this lens is cut, longitudinally along its principal axis into two equal parts, and then they are joined as given in the figure. The power, of combination will be :, (a) P, (b) 2P, (c) P/2, (d) zero, , Q.15 The plane surface of a planoconvex lens is silvered. If, radius of curved surfaceisR and refractiveindex is m, then, the system behaves like a concave mirror whose radius, will be, R, (a), (b) Rm, m, R, (c), (d) R (m – 1), m -1, Q.16 A slide projector lens has a focal length 10 cm. It throws, an image of a 2cm × 2 cm slide on a screen 5 m from the, lens. Find the size of the picture on the screen., (a) ( 98 × 98) cm2, (b) (88 × 88) cm2, 2, (c) (64 × 64) cm, (d) (78 × 78) cm2, Q.17 If the focal length of a magnifier is 5 cm calculate the, power of the lens., (a) 20D, (b) 10D, (c) 5D, (d) 15D, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 51, , 3, , Q.18 In the above question, find the magnifying power of the, lens for relaxed and strained eye., (a) 2×, 3×, (b) 5×, 6×, (c) 4×, 2×, (d) 1×, 2×, Q.19 A 35 mm film is to be projected on a 20 m wide screen, situated at a distance of 40 m from the film-projector., Calculate the focal length of projection lens., (a) 70 mm, (b) 35 mm, (c) 40 mm, (d) 20 mm, Q.20 In a compound microscope the objective and the eye- piece, have focal lengths of 0.95 cm and 5 cm respectively, and, are kept at a distance of 20 cm. The last image is formed at, a distance of 25 cm from the eye- piece. Calculate the, total magnification., (a) 94, (b) 84, (c) 75, (d) 88, Q.21 A Galilean telescope consists of an objective of focal, length 12 cm and eye- piece of focal length 4 cm. What, should be the separation of the two lenses when the virtual, image of a distant object is formed at a distance of 24 cm, from the eye- piece?, (a) 7.2 cm, (b) 8.2 cm., (c) 12.4 cm., (d) 2.8cm., , Q.23 Resolving power of a microscope doesn't depend upon, (1) Velocity of light used, (2) Frequency of light used, (3) Focal length of objective, (4) Wavelength of light used, Q.24 The light gathering power of a camera lens, doesn't depend on, (1) Ratio of focal length and diameter, (2) Product of focal length and diameter, (3) Wavelength of light used, (4) Its diameter, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, , A cylindrical tube filled with water (µw = 4/3) is closed at its, both ends by two silvered plano convex lenses as shown in the, figure. Refractiveindex of lenses L1 and L2 are 2.0 and 1.5 while, their radii of curvature are 5 cm and 9 cm respectively. A point, object is placed somewhere at a point O on the axis of cylindrical, tube. It is found that the object and image coincide each other., Q.25 The position of object w.r.t lens L1 is, (a) 8 cm, (b) 10 cm, (c) 12 cm, (d) 14 cm, Q.26 The position of object w.r.t lens L2 is, (a) 8 cm, (b) 10 cm, (c) 12 cm, (d) 14 cm, Q.27 The length of the cylindrical tube is, (a) 16 cm, (b) 18 cm, (c) 20 cm, (d) 22 cm, , Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 An astronomical telescope has an angular magnification, of magnitude 5 for distant objects. The separation between, the objective and eye- piece is 36 cm and the final image, is formed at infinity., (1) the focal length of objective is 30 cm, (2) the focal length of objective is 25 cm, (3) the focal length of eye piece is 6 cm, (4) the focal length of eye piece is 12 cm, , RESPONSE, GRID, , DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, , m1, , mw, O, , L1, , m2, L2, , 18., , 19., , 20., , 21., , 22., , 23., , 24., , 25., , 26., , 27., , Space for Rough Work
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t.me/Magazines4all, DPP/ P 51, , 4, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., , Q.28 Statement-1: A double convex lens (m = 1.5) has focal, length 10 cm. When the lens is immersed in water (m = 4/, 3) its focal length becomes 40 cm., , (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , (c), , Statement -1 is False, Statement-2 is True., , (d), , Statement -1 is True, Statement-2 is False., , Q.29 Statement-1: The focal length of lens changes when red, light is replaced by blue light., Statement-2: The focal length of lens does not depend on, colour of light used., Q.30 Statement-1: By increasing the diameter of the objective, of telescope, we can increase its range., Statement-2: The range of a telescope tells us how far, away a star of some standard brightness can be spotted by, telescope., , RESPONSE GRID, , 28., , 29., , Statement-2:, , 1 ml - mm æ 1, 1ö, =, - ÷, ç, f, m m è R1 R2 ø, , 30., , DAILY PRA CTICE PROBLEM SHEET 51 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 52, SYLLABUS : WAVE OPTICS - I (Interference of Light), , Max. Marks : 116, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 29 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The intensity ratio of two waves is 9 : 1. These waves, produce the event of interference. The ratio of maximum, to minimum intensity will be, (a) 1 : 9, (b) 9 : 1, (c) 1 : 4, (d) 4 : 1, Q.2 The equation of two light waves are y1 = 6cos wt,, y2 = 8cos(wt + f). The ratio of maximum to minimum, intensities produced by the superposition of these waves, will be, (a) 49 : 1, (b) 1 : 49, (c) 1 : 7, (d) 7 : 1, Q.3 In a Young’s double slit experiment, the separation between, the slits is 0.10 mm, the wavelength of light used is 600, , RESPONSE GRID, , 1., , 2., , nm and the interference pattern is observed on a screen, 1.0 m away. Find the separation between the successive, bright fringes., (a) 6.6 mm (b) 6.0 mm (c) 6 m, (d) 6 cm., Q.4 In Young’s double slit experiment the two slits are, illuminated by light of wavelength 5890 Å and the angular, separation between the fringes obtained on the screen is, 0.2°. If the whole apparatus is immersed in water then the, angular fringe width will be, if the refractive index of water, is 4/3?, (a) 0.30°, (b) 0.15°, (c) 15°, (d) 30°, Q.5 The intensities of two light sources are I and 9I respectively., If the phase difference between the waves emitted by them, is p then the resultant intensity at the point of observation, will be –, (a) 3I, (b) 4I, (c) 10I, (d) 82I, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 52, , 2, Q.6 In Fresnel’s biprism experiment the width of 10 fringes is, 2cm which are formed at a distance of 2 meter from the, slit. If the wavelength of light is 5100Å then the distance, between two coherent sources will be, (a) 5.1 × 10–4 m, (b) 5.1 × 104 cm, –4, (c) 5.1 × 10 mm, (d) 10.1 × 10–4cm, Q.7 Two coherent sources of intensity ratio 1 : 4 produce an, interference pattern. The fringe visibility will be (a) 1, (b) 0.8, (c) 0.4, (d) 0.6, Q.8 When a mica sheet (m = 1.6) of thickness 7 microns is, placed in the path of one of interfering beams in the biprism, experiment then the central fringe gets shifted at the, position of seventh bright fringe. The wavelength of light, used will be –, (a) 4000Å (b) 5000Å (c) 6000 Å (d) 7000Å, Q.9 In Young’s double slit experiment, the distance between, two slits is made three times then the fringe width will, become –, (a) 9 times (b) 1/9 times (c) 3 times (d) 1/3 times, Q.10 In the given diagram, CP represents a wavefront and AO &, BP, the corresponding two rays. Find the condition on q, for constructive interference at P between the ray BP and, reflected ray OP, O, , Q, , R, , qq, C, , A, , d, , P, B, , (a) cos q = 3l/2d, (b) cos q = l/4d, (c) sec q – cosq = l/d, (d) sec q – cosl = 4l/d, Q.11 In Young’s double slit experiment 10th order maximum is, obtained at the point of observation in the interference, pattern for l = 7000 Å. If the source is replaced by another, one of wavelength 5000 Å then the order of maximum at, the same point will be–, , RESPONSE, GRID, , (a) 12 th, (b) 14 th, (c) 16 th, (d) 18 th, Q.12 In Young’s double slit experiment, white light is used. The, separation between the slits is b. The screen is at a distance, d (d > > b) from the slits. Some wavelengths are missing, exactly in front of one slit. One of these wavelengths is, b2, 2b 2, b2, 2b 2, (b) l =, (c) l =, (d) l =, 6d, d, 3d, 3d, th, Q.13 In Fresnel’s biprism experiment distance of m bright, fringe from zeroth order fringe will be –, lD, mDl, (a) (2m – 1), (b), 2d, d, , (a) l =, , lD, md, (d) (2m + 1), lD, 2d, Q.14 Consider interference between waves from two sources, of Intensites I & 4I. Find intensities at points where the, , (c), , p, ., 2, (a) I, (b) 5 I, (c) 4 I, (d) 3 I, Q.15 The width of one of the two slits in a Young’s double slit, experiment is double of the other slit. Assuming that the, amplitude of the light coming from a slit is proportional, to slit-width. Find the ratio of the maximum to the, minimum intensity in the interference pattern., (a) 34 : 1, (b) 9 : 1, (c) 4 : 1, (d) 16 : 1, Q.16 The intensity of the light coming from one of the slits in a, young’s double slit experiment is double the intensity from, the other slit. Find the ratio of the maximum intensity to, the minimum intensity in the interference fringe pattern, observed., (a) 9 : 1, (b) 34 : 1, (c) 4 : 1, (d) 16 : 1, Q.17 Two waves originating from source S1 and S2 having zero, phase difference and common wavelength l will show, completely destructive interference at a point P if, (S1 P – S2 P) is(a) 5l, (b) 3l/4, (c) 2l, (d) 11l/2, , phase difference is, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 52, , 3, , Q.18 In an interference pattern, at a point we observe the 16th, order maximum for l1= 6000Å. What order will be visible, here if the source is replaced by light of wavelength ?, l2 = 4800 Å., (a) 40, (b) 20, (c) 10, (d) 80, Q.19 In Young’s experiment the wavelength of red light is, 7.5 × 10–5 cm. and that of blue light 5.0 × 10–5 cm. The, value of n for which (n + 1)th blue bright band coincides, with nth red bright band is(a) 8, (b) 4, (c) 2, (d) 1, Q.20 In Young’s double slit experiment, carried out with light of, wavelength l = 5000Å, the distance between the slits is, 0.2mm and the screen is at 200 cm from the slits. The, central maximum is at x = 0. The third maximum will be at, x equal to., (a) 1.67 cm, (b) 1.5 cm, (c) 0.5 cm, (d) 5.0 cm, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 The Young’s double slit experiment, the ratio of intensities, of bright and dark fringes is 9. This means that, (1) The intensities of individual sources are 5 and 4 units, respectively, (2) The intensities of individual sources are 4 and 1 units, respectively, (3) The ratio of the their amplitudes is 3, (4) The ratio of their amplitude is 2, Q.22 In an experiment similar to Young’s experiment,, interference is observed using waves associated with, electrons. The electrons are being produced in an electron, gun. In order to decrease the fringe width, , RESPONSE, GRID, , 18., , 19., , 23., , 24., , (1) electron gun voltage be increased., (2) the slits be moved away from each other., (3) the screen be moved closer to interfering slits., (4) electron gun voltage be decreased., Q.23 Interference fringes were produced in Young’s double slit, experiment using light of wave length 5000 Å. When a film, of material 2.5 × 10–3 cm thick was placed over one of the, slits, the fringe pettern shifted by a distance equal to 20, fringe width. The refractive index of the material of the, film cannot be, (1) 1.25, (2) 1.33, (3) 1.5, (4) 1.4, DIRECTIONS (Q.24-Q.26) : Read the passage given below, and answer the questions that follows :, In a Young’s double slit experiment a monochromatic light whose, wavelength is l strikes on the slits, separated by distance d, as, shown in the figure. Refractive index of the medium between, slits and screen varies with time t as n = n0 + kt. Here n0 and k, are positive constants. Position of any point P on screen is, measure by its y-coordinate as shown., , P, AIR, 2, , d, , n = (n0+ kt), , O, , f, l, , y, , S1, D, , Q.24 The y co-ordinate of central maxima at any time t is, D cos f, D sin f, (a), (b), n0 + kt, n0 + kt, (c), , 20., , Space for Rough Work, , D sin f, (n0 + kt ), , (d), , 2, , 21., , D cos f, (n0 + kt )2, , 22.
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t.me/Magazines4all, DPP/ P 52, , 4, Q.25 The velocity of central maxima at any time t as a function, of time t is, –2kD sin f, – kD sin f, (a), (b), 2, (n0 + kt ), (n0 + kt )2, – kD sin f, –2kD sin f, (d), (n0 + kt ), (n0 + kt ), Q.26 If a glass plate of small thickness b is placed in front of S1., How should its refractive index vary with time so that, central maxima is formed at O., 2 d sin f, 2d sin f, (a) n0 + kt +, (b) n0 + kt –, b, b, d sin f, d sin f, (c) n0 + kt –, (d) n0 + kt +, b, b, , (c), , DIRECTIONS (Q. 27-Q.29) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.27 Statement-1 : No interference pattern is detected when, two coherent sources are infinitely close to each other., Statement-2 : The fringe width is directly proportional to, the distance between the two slits., Q.28 Statement-1 : In Young’s experiment, the fringe width for, dark fringes is same as that for white fringes., Statement-2 : In Young’s double slit experiment performed, with a source of white light, only black and bright fringes, are observed., Q.29 Statement-1 : In Young’s double slit experiment, the fringes, become indistinct if one of the slits is covered with, cellophane paper., Statement-2 : The cellophane paper decreases the, wavelength of light., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 25., , 26., , 27., , 28., , 29., , DAILY PRA CTICE PROBLEM SHEET 52 - PHYSICS, Total Questions, 29, Total Marks, 116, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 53, SYLLABUS : WAVE OPTICS - II (Diffraction and polarisation of light), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The first diffraction minima due to a single slit diffraction, is at q = 30° for a light of wavelength 5000Å. The width of, the slit is(a) 5 × 10–5 cm, (b) 1.0 × 10–4 cm, –5, (c) 2.5 × 10 cm, (d) 1.25 × 10–5 cm, Q.2 Two spectral line of sodium D1 & D2 have wavelengths of, approximately 5890Å and 5896Å. A sodium lamp sends, incident plane wave on to a slit of width 2 micrometre. A, screen is located at 2m from the slit. Find the spacing, between the first maxima of two sodium lines as measured, on the screen., (a) 10–4 m, (b) 9 × 10–4 m, 4, (c) 9 × 10 m, (d) None, , RESPONSE GRID, , 1., , 2., , Q.3 Width of slit is 0.3mm. Fraunhoffer diffraction is observed, in focus plane of lense of a lense of focal length 1 m. If, third minima is at 5 mm distance from central maxima,, then wavelength of light is(a) 7000Å (b) 6500Å (c) 6000Å (d) 5000Å, Q.4 When a wave of wavelength 0.2 cm is made incident, normally on a slit of width 0.004m, then the semi-angular, width of central maximum of diffraction pattern will be(a) 60°, (b) 30°, (c) 90°, (d) 0°, Q.5 A parallel beam of monochromatic light is incident on a, narrow rectangular slit of width 1mm. When the diffraction, pattern is seen on a screen placed at a distance of 2m. the, width of principal maxima is found to be 2.5 mm. The wave, length of light is(a) 6250 nm, (b) 6200 nm, (c) 5890 nm, (d) 6000 nm, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 53, , 2, Q.6 Light of wavelength 6328Å is incident normally on slit, having a width of 0.2 mm. The width of the central, maximum measured from minimum to minimum of, diffraction pattern on a screen 9.0 meters away will be, about (a) 0.36°, , (b) 0.18°, , (c) 0.72°, (d) 0.09°, Q.7 A screen is placed 2m away from the single narrow slit., Calculate the slit width if the first minimum lies 5mm on, either side of the central maximum. Incident plane waves, have a wavelenght of 5000Å., (a) 2 × 10–4 m, (b) 2 × 10–3 cm, –4, (c) 2 × 10 cm, (d) None, Q.8 Red light of wavelength 6500Å from a distant source falls, on a slit 0.5 mm wide. What is the distance between two, dark bands on each side of central bright band of diffraction, pattern observed on a screen placed 1.8 m from the slit., (a) 4.68 × 10–3 cm, (b) 4.68 × 10–3 mm, –3, (c) 4.68 × 10 nm, (d) 4.68 × 10–3 m, Q.9 Fraunhoffer diffraction pattern is observed at a distance, of 2m on screen, when a plane-wavefront of 6000Å is, incident perpendicularly on 0.2 mm wide slit.Width of, central maxima is:, (a) 10 mm, (b) 6 mm, (c) 12 mm, (d) None of these, Q.10 A diffraction pattern is produced by a single slit of width, 0.5mm with the help of a convex lens of focal length 40cm., If the wavelength of light used is 5896Å. then the distance, of first dark fringe from the axis will be(a) 0.047 cm, (b) 0.047 m, (c) 0.047 mm, (d) 47 cm, Q.11 What should be the size of the aperture of the objective of, telescope which can just resolve the two stars of angular, width of 10–3 degree by light of wavelength 5000Å?, (a) 3.5 cm, (b) 3.5 mm, (c) 3.5 m, (d) 3.5 km, , RESPONSE, GRID, , Q.12 Image of sun formed due to reflection at air water interface, is found to be very highly polarised. Refractive index of, water being m = 4/3, find the angle of sun above the horizon., (a) 36.9º, , (b) 26.9º, , (c) 16.9º, , (d) 46.9º, , Q.13 When light of a certain wavelength is incident on a plane, surface of a material at a glancing angle 30º, the reflected, light is found to be completely plane polarised. Determine, refractive index of given material –, (a), , (b), , 3, , (c) 1 / 2, , 2, , (d) 2, , Q.14 Two polaroids are oriented with their planes perpendicular, to incident light and transmission axis making an angle of, 30º with each other. What fraction of incident unpolarised, light is transmitted ?, (a) 57.5 %, , (b) 17.5 %, , (c) 27.5 %, , (d) 37.5 %, , Q.15 Unpolarised light of intensity 32 Wm–2 passes through, three polarisers such that the transmission axis of the last, polariser is crossed with the first. If the intensity of the, emerging light is 3 Wm –2 . At what angle will the, transmitted intensity be maximum ?, (a) 45º, , (b) 15º, , (c) 35º, , (d) 75º, , Q.16 V0 and VE represent the velocities, m0 and mE the refractive, indices of ordinary and extraordinary rays for a doubly, refracting crystal. Then, (a) V0 ³ VE, m0 £ mE if the crystal is calcite, (b) V0 £ VE, m0 £ mE if the crystal is quartz, (c) V0 £ VE, m0 ³ mE if the crystal is calcite, (d) V0 ³ VE, m0 ³ mE if the crystal is quartz, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 53, , 3, , Q.17 A ray of light is incident on the surface of a glass plate at, an angle of incidence equal to Brewster’s angle f. If m, represents the refractive index of glass with respect to air., then the angle between reflected and refracted rays is, (a) 90° + f, (b) sin–1 (m cosf), (c) 90°, (d) 90° – sin–1 ( cosf/m), Q.18 A light has amplitude A and angle between analyser and, polariser is 60°. Light transmitted by analyser has, amplitude, (a), , A 2, , (b) A / 2, , (c), , 3 A / 2 (d) A/2, , Q.19 A slit of size 0.15 cm is placed at 2.1 m from a screen. On, illuminating it by a light of wavelength 5×10–5 cm, the width, of central maxima will be, (a) 70 mm, , DIRECTIONS (Q.23-Q.25) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, , (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.23 Plane polarised light is passed through a polaroid. On, viewing through the polaroid we find that when the polariod, is given one complete rotation about the direction of the, light, which of the following is not observed ?, (1) The intensity of light gradually decreases to zero and, remains at zero, (2) The intensity of light gradually increases to a, maximum and remains at maximum, (3) There is no change in intensity, , (b) 0.14 mm (c) 1.4 mm (d) 0.14 cm, , Q.20 What will be the angle of diffraction for the first minimum, due to Fraunhoffer diffraction with sources of light of wave, lenght 550 nm and slit width 0.55 mm ?, (a) 0.001 rad, , (b) 0.01 rad, , (c) 1 rad, , (d) 0.1 rad, , (4) The intensity of light is twice maximum and twice, zero, Q.24 Out of the following statements which are correct ?, (1) Nicol’s prism works on the principle of double, refraction and total internal reflection, , Q.21 In Fresnel diffraction, if the distance between the disc and, the screen is decreased, the intensity of central bright spot, will, (a) increase, , (b) decrease, , (c) remain constant, , (d) none of these, , 1, I0, 4, , (b), , 1, I0, 2, , (c) I 0, , (2) Nicol’s prism can be used to produce and analyse, polarised light, (3) Calcite and Quartz are both doubly refracting crystals, (4) When unpolarised light passes through a Nicol’s, prism, the emergent light is elliptically polarised, , 22. When an unpolarized light of intensity I 0 is incident on a, polarizing sheet, the intensity of the light which does not, get transmitted is, (a), , (b) 1 and 2 are correct, , Q.25 Which statements are incorrect for a zone plate and a lens?, (1) Zone plate has one focus whereas lens has multiple, focii, (2) Both zone plate and lens have multi focii, (3) Zone plate has one focus whereas a lens has infinite, , (d) zero, , (4) Zone plate has multi focii whereas lens has one, , RESPONSE, GRID, , 17., , 18., , 19., , 20., , 22., , 23., , 24., , 25., , Space for Rough Work, , 21.
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t.me/Magazines4all, DPP/ P 53, , 4, DIRECTIONS (Q.26-Q.27) : Read the passage given below, and answer the questions that follows :, Angular width of central maximum in the Fraunhoffer-diffraction, pattern of a slit is measured. The slit is illuminated by light of, wavelength 6000 Å. When the slit is illuminated by light of, another wavelength, the angular width decreases by 30%., Q.26 The wavelength of the light is, (a) 4200Å (b) 3500Å (c) 5000 Å (d) 5200Å, Q.27 The same decrease in the angular width of central maximum, is obtained when the original apparatus is immersed in a, liquid. Find refractive index of the liquid., (a) 1.23, (b) 1.43, (c) 2.2, (d) 2.43, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (b), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement-1 is False, Statement-2 is True., (d) Statement-1 is True, Statement-2 is False., Q.28 Statement-1 : The unpolarised light and polarised light, can be distinguished from each other by using polaroid., Statement-2 : A polaroid is capable of producing plane, polarised beams of light., Q.29 Statement-1 : Nicol prism is used to produce and analyse, plane polarised light., Statement-2 : Nicol prism reduces the intensity of light, to zero., Q.30 Statement-1 : The cloud in sky generally appear to be, whitish., Statement-2 : Diffraction due to clouds is efficient in equal, measure at all wavelengths., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 53 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 54, SYLLABUS : DUAL NATURE OF MATTER & RADIATION (Matter Waves, Photon, Photoelectric effect, X-ray), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 Energy of a a-particle, having de broglie wavelength of, 0.004 Å is approximately., (a) 1275 eV, , (b) 1200 KeV, , (c) 1200 MeV, , (d) 1200 GeV, , Q.2 Velocity of a proton is c/20. Associated de-Broglie, wavelength is (Take h = 6.626 × 10–34 J–s), (a) 2.64 × 10–24 mm, (c) 2.64 ×, , 10–14, , RESPONSE GRID, , Å, , 1., , Q.3 One electron & one proton is accelerated by equal potential., Ratio of their de-Broglie wavelengths is(a), , (d) 2.64 ×, , 2., , me, , (b), , me, mp, , (c), , mp, me, , (d), , 1, , Q.4 de-Broglie wavelength of an electron is 10 Å then velocity, will be(a) 7.2 × 107 m/s, (b) 7.2 × 106 m/s, 5, (c) 7.2 × 10 m/s, (d) 7.2 × 104 m/s, Q.5 One electron & one proton have equal energies then ratio, of associated de-Broglie wavelength will be-, , (b) 2.64 × 10–24 cm, 10–14, , mp, , (a) 1 : (1836)2, (c) 1836 : 1, , m, , 3., Space for Rough Work, , 4., , (b), 1836 :1, (d) (1836)2 : 1, , 5.
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t.me/Magazines4all, DPP/ P 54, , 2, Q.6 The ratio of wavelength of deutron & proton accelerated, by an equal potential is, (a), , 1, , (b), , 2, , Q.13 A photon and an electron have equal energy E. lphoton/, lelectron is proportional to, , 2, 1, , (a), , 1, 2, (d), 2, 1, Q.7 In photoelectric effect if intensity of light is doubled then, maximum kinetic energy of photoelectrons will become, (a) Double, (b) Half, (c) Four time, (d) No change, Q.8 Quantum nature of light is explained by which of the, following phenomenon?, (a) Huygen wave theory, (b) Photoelectric effect, (c) Maxwell electromagnetic theory, (d) de- Broglie theory, Q.9 From rest an electron is accelerated between two such, points which has potential 20 & 40 volts respectively., Associated de-Broglie wavelength of electron is(a) 0.75 Å, (b) 7.5 Å, (c) 2.75 Å, (d) 2.75 m, Q.10 An electron microscope uses 40 keV electrons. Find its, resolving limit on the assumption that it is equal to the, wavelength of the electron(a) 0.61 Å, (b) 0.6 Å, (c) 0.06 Å, (d) 0.061 Å, Q.11 A hydrogen atom moving at a speed v absorbs a photon of, wavelength 122 nm and stops. Find the value of v., (Mass of hydrogen atom = 1.67 × 10–27 kg), (a) 3.5 m/s, (b) 32.5 m/s, (c) 3.05 m/s, (d) 3.25 m/s, Q.12 The de-Broglie wavelength of an electron is 0.2 Å. Calculate, the potential difference (approximate) required to retard, it to rest(a) 3.76 × 10–3 V, (b) 3.76 × 103 V, 3, (c) 3.76 × 10 eV, (d) 376.5 V, , (c), , RESPONSE, GRID, , 1, E, , (c), , 1, , (b), , E, , E, , (d) Does not depend upon E., , Q.14 In a photoemissive cell with exciting wavelength l, the, fastest electron has speed v. If the exciting wavelength is, changed to 3l/4, the speed of the fastest emitted electron, will be, (a) v (3/4)1/2, (b) v (4/3)1/2, (c) Less than v (4/3)1/2, (d) Greater than v (4/3)1/2, Q.15 Which of the following figure repesents variation of, particle momentum and the associated de-Broglie, wavelength?, p, , p, , (a), , (b), , l, , l, p, , p, , (c), , (d), l, , l, , Q.16 The work function for the surface of aluminium is 4.2 eV., What will be the wavelength of that incident light for which, the stopping potential will be zero., (h » 6.6 × 10–34 J–s e » 1.6 × 10–19 C), (a) 2496 Å, (b) 2946 × 10–7 m, (c) 2649 Å, (d) 2946 Å, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 54, , 3, , Q.17 Slope of V0 – n curve is(where V0 = Stopping potential and n = frequency), h, (a) e, (b), (c) f0, (d) h, e, Q.18 A radio station is transmitting waves of wavelength 300 m., If diffracting power of transmitter is 10 kW, then numbers, of photons diffracted per second is(a) 1.5 × 1035, (b) 1.5 × 1031, 29, (c) 1.5 × 10, (d) 1.5 × 1033, Q.19 Light of wavelength 3320 Å is incident on metal surface, (work function = 1.07 eV). To stop emission of photo, electron, retarding potential required to be, (Take hc » 12420 eV – Å), (a) 3.74 V (b) 2.67 V (c) 1.07 V (d) 4.81 V, Q.20 The figure shows the variation of photocurrent with anode, potential for a photo-sensitive surface for three different, radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc, be the frequencies for the curves a, b and c respectively., Then, (a) fa = fb and Ia ¹ Ib, (b) fa = fc and Ia = Ic, , c, , b, a, , DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, , (c) fa = fb and Ia = Ib, (d) fa = fb and Ia = Ic, , O, , of frequency 3 × 1015 cycles, , Q.21 An electromagnetic radiation, per second falls on a photo electric surface whose work, function is 4.0 eV. Find out the maximum velocity of the, photo electrons emitted by the surface(a) 13.4 × 10–19 m/s, (b) 19.8 × 10–19m/s, 6, (c) 1.73 × 10 m/s, (d) None, DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, , RESPONSE, GRID, , Q.22 Ultraviolet light of wavelength 280 nm is used in an, experiment on photo electric effect with lithium (f = 2.5, eV) cathode., (1) The maximum kinetic energy is 1 .9 eV, (2) The stopping potential is1.9 V, (3) The maximum kinetic energy is 4.4 V, (4) The stopping potential is 4.4 eV, Q.23 The separation between Bragg’s planes in a crystal is, 10 Å. Then the wavelength of those X-rays which can be, diffracted by this crystal is(1) 5 Å, (2) 10 Å, (3) 20 Å, (4) 25 Å, Q.24 Electrons are accelerated in television tubes through, potential difference of about 10 KV., (1) The lowest wavelength of the emitted X-rays is 12.4Å, (2) The lowest wavelength of the emitted X-rays is 1.24Å, (3) The highest frequency of the emitted X-rays is, 2.4 × 108 Hz, (4) The highest frequency of the emitted X-rays is, 2.4 × 1018 Hz, , (b) 1 and 2 are correct, (d) 1 and 3 are correct, , A physicist wishes to eject electrons by shining light on a metal, surface. The light source emits light of wavelength of 450 nm., The table lists the only available metals and their work functions., Metal, W0 (eV), Barium, 2.5, Lithium, 2.3, Tantalum, 4.2, Tungsten, 4.5, Q.25 Which metal(s) can be used to produce electrons by the, photoelectric effect from given source of light ?, (a) Barium only, (b) Barium or lithium, (c) Lithium, tantalum or tungsten, (d) Tungsten or tantalum, , 17., , 18., , 19., , 20., , 22., , 23., , 24., , 25., , Space for Rough Work, , 21.
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t.me/Magazines4all, DPP/ P 54, , 4, Q.26 Which option correctly identifies the metal that will, produce the most energetic electrons and their energies ?, (a) Lithium, 0.45 eV, (b) Tungsten, 1.75 eV, (c) Lithium, 2.30 eV, (d) Tungsten, 2.75 eV, Q.27 Suppose photoelectric experiment is done separately with, these metals with light of wavelength 450 nm. The, maximum magnitude of stopping potential amongst all the, metals is(a) 2.75 volt, (b) 4.5 volt, (c) 0.45 volt, (d) 0.25 volt, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select, the correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (b), (c), (d), , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , Q.28 Statement -1 : Mass of moving photon varies directly as, the wavelength., Statement -2 : Energy of the particle = Mass × (Speed of, light)2, Q.29 Statement -1 : Photosensitivity of a metal is large if its, work function is small., Statement -2 : Work function = hf 0 where f 0 is the, threshold frequency., Q.30 Statement -1 : The de-Broglie wavelength of a molecule, varies inversely as the square root of temperature., Statement -2 : The root mean square velocity of the, molecule is proportional to square root of absolute, temperature., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 54 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 50, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 55, SYLLABUS : Atoms, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 In nature there may not be an element for which the, principal quantum number n > 4 , then the total possible, number of elements will be, (a) 60, (b) 32, (c) 4, (d) 64, Q.2 In the following atoms and molecule for the transition, from n = 2 to n = 1 , the spectral line of minimum, wavelength will be produced by, (a) Hydrogen atom, (b) Deuterium atom, (c) Uni-ionized helium, (d) Di-ionized lithium, Q.3 The Lyman series of hydrogen sperctum lies in the region, (a) Infrared, (b) Visible, (c) Ultraviolet, (d) X - rays, , RESPONSE GRID, , 1., , 2., , Q.4 The energy levels of the hydrogen spectrum is shown in, figure. There are some transitions. A,B,C,D and E., Transition A, B and C respectively represent, (a) First spectral line of, Lyman series, third, spectral line of Balmer, series and the second, spectral line of Paschen, series., (b) Ionization potential of hydrogen, second spectral line, of Balmer series and third spectral line of Paschen, series, (c) Series limit of Lyman series, third spectral line of, Balmer series and second spectral line of Paschen, series, (d) Series limit of Lyman series, second spectral line of, Balmer series and third spectral line of Paschen series, , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 55, , 2, Q.5 Energy levels A, B, C of a certain atom corresponding to, increasing values of energy i.e. E A < EB < EC . If, l1 , l 2 , l3 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, , l1, l2, , l3, , B, , Q.11 According to Bohr’s theory the moment of momentum of, an electron revolving in second orbit of hydrogen atom will, be, p, h, (d), h, p, Q.12 In the Bohr model of a hydrogen atom, the centripetal force, is furnished by the coulomb attraction between the proton, , (a), , m is the mass, e is the charge on the electron and e0 is the, vacuum permittivity, the speed of the electron is, e, (a) 0, (b), e a m, , l1l 2, (b) l3 = l + l, 1, 2, , (d) l32 = l12 + l 22, Q.6 If m is mass of electron, v its velocity, r the radius of, stationary circular orbit around a nucleus with charge Ze,, then from Bohr’s first postulate, the kinetic energy, 1, K = mv2 of the electron in C.G.S . system is equal to, 2, Ze, 1 Ze 2, 1 Ze 2, Ze2, (a), (b), (c), (d), 2, 2, r2, 2 r, r, r, Q.7 When a hydrogen atom is raised from the ground state to, an excited state, (a) P. E. increases and K. E. decreases, (b) P. E. decreases and K. E. increases, (c) Both kinetic energy and potential energy increase, (d) Both K. E. and P. E. decrease, Q.8 The value of the kinetic energy divided by the total energy, of an electron in a Bohr orbit is, (a) – 1, (b) 2, (c) 0.5, (d) None of these, Q.9 The ratio of the frequencies of the long wavelength limits, of Lyman and Balmer series of hydrogen spectrum is, (a) 27 : 5, (b) 5 : 27, (c) 4 : 1, (d) 1 : 4, Q.10 Ratio of the wavelengths of first line of Lyman series and, first line of Balmer series is, (a) 1 : 3 (b) 27 : 5, (c) 5 : 27, (d) 4 : 9, (c) l1 + l 2 + l3 = 0, , RESPONSE, GRID, , (c), , and the electron. If a0 is the radius of the ground state orbit,, A, , (a) l3 = l1 + l 2, , (b) ph, , 2ph, , 0 0, , e, , 4pe0 a0 m, (d), 4pe0 a0 m, e, Q.13 Which of the following transitions in hydrogen atoms emit, photons of highest frequency?, (a) n = 1 to n = 2, (b) n = 2 to n = 6, (c) n = 6 to n = 2, (d) n = 2 to n = 1, Q.14 As per Bohr model, the minimum energy (in eV ) required, to remove an electron from the ground state of doubly, , (c), , ionized Li atom ( Z = 3) is, (a) 1.51, (b) 13.6, (c) 40.8, (d) 122.4, Q.15 The third line of Balmer series of an ion equivalent to, hydrogen atom has wavelength of 108.5 nm. The ground, state energy of an electron of this ion will be, (a) 3.4 eV, (b) 13.6 eV, (c) 54.4 eV, (d) 122.4 eV, Q.16 The wavelength of radiation emitted is l0 when an electron, jumps from the third to the second orbit of hydrogen atom., For the electron jump from the fourth to the second orbit of, the hydrogen atom, the wavelength of radiation emitted will, be, 20, 16, 27, 25, l0, l 0 (b), l0, l0, (a), (c), (d), 25, 27, 20, 16, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 55, , 3, , Q.17 The energy of electron in the nth orbit of hydrogen atom, -13.6, eV . The shortest and longest, is expressed as En =, n2, wavelength of Lyman series will be, (a) 910 Å, 1213 Å, (b) 5463 Å, 7858 Å, (c) 1315 Å , 1530 Å, (d) None of these, Q.18 Consider a hydrogen like atom whose energy in nth exicited, 13.6 Z 2, state is given by En = when this excited atom, n2, makes a transition from excited state to ground state, most, energetic photons have energy Emax = 52.224 eV and least, energetic photons have energy Emin = 1.224 eV . The, atomic number of atom is, (a) 2, (b) 5, (c) 4, (d) None of these, Q.19 In the Bohr model of the hydrogen atom, let R, v and E, represent the radius of the orbit, the speed of electron and, the total energy of the electron respectively. Which of the, following quantity is proportional to the quantum number, n, (a) R / E, (b) E / v, RE, (c), (d) vR, Q.20 An a - particle of 5 MeV energy strikes with a nucleus, of uranium at stationary at an scattering angle of 180°. The, nearest distance upto which a - particle reaches the, nucleus will be closest to, (a) 1 Å, (b) 10–10cm, –12, (c) 10 cm, (d) 10–15cm, Q.21 In a hypothetical Bohr hydrogen, the mass of the electron, is doubled. The energy E0 and the radius r0 of the first, orbit will be ( a0 is the Bohr radius), (a), , E0 = -27.2eV ; r0 = a0 /2, , (b) E0 = -27.2eV ; r0 = a0, (c) E0 = -13.6eV ; r0 = a0 /2, (d) E0 = -13.6eV ; r0 = a0, , RESPONSE, GRID, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select, the correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 The electron in a hydrogen atom makes a transition n 1 ®, n2, where n 1 and n2 are the principal quantum numbers of, two states. Assume the Bohr model to be valid. The time, period of the electron in the initial state is eight times that, in the final state. Then, (1) n1 = 4 (2) n2 = 2, (3) n2 = 5, (4) n1 = 5, Q.23 A free hydrogen atom in ground state is at rest. A neutron, of kinetic energy K collides with the hydrogen atom., After collision hydrogen atom emits two photons in, succession one of which has energy 2.55eV. Assume that, the hydrogen atom and neutron has same mass., (1) Minimum value of K is 25.5 eV, (2) Minimum value of K is 12.75 eV, (3) The other photon has energy 10.2eV, (4) The upper energy level is of excitation energy 12.5, eV, Q.24 Which of the series of hydrogen spectrum are not in the, visible region?, (1) Lyman series, (2) Paschen series, (3) Bracket series, (4) Balmer series, DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, A gas of identical hydrogen like atoms has some atoms in ground, state and some atoms in a particular excited state and there are, no atoms in any other energy level. The atoms of the gas make, transition to a higher energy state by absorbing monochromatic, light of wavelength 304Å. Subsequently, the atoms emit radiation of only six different photon energies. Some of emitted photons have wavelength 304 Å, some have wavelength more and, some have less than 304Å (Take hc = 12420 eV–Å), Q.25 Find the principal quantum number of the initially excited, state., (a) 1, (b) 2, (c) 3, (d) 4, , 17., , 18., , 19., , 20., , 22., , 23., , 24., , 25., , Space for Rough Work, , 21.
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t.me/Magazines4all, DPP/ P 55, , 4, Q.26 Identify the gas (Z = ?), (a) 1, (b) 2, (c) 3, (d) 4, Q.27 Find the maximum and minimum energies of emitted, photons (in eV), (a) 20.4, 10.6, (b) 10.4, 3.6, (c) 40.8, 10.6, (d) None of these, DIRECTIONS (Q.28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select, the correct choice., (a), (b), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.28 Statement-1 : Bohr postulated that the electrons in, stationary orbits around the nucleus do not radiate energy., Statement-2 : According to classical physics all moving, electrons radiate energy., Q.29 Statement-1 : The force of repulsion between atomic, nucleus and a-particle varies with distance according to, inverse square law., Statement-2 : Rutherford did a-particle scattering, experiment., Q.30 Statement-1 : Hydrogen atom consists of only one, electron but its emission spectrum has many lines., Statement-2 : Only Lyman series is found in the absorption, spectrum of hydrogen atom whereas in the emission, spectrum, all the series are found., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 55 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 50, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 56, SYLLABUS : Nuclei, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 The energy released per fission of uranium 235 is about, 200 MeV. A reactor using U-235 as fuel is producing 1000, kilowatt power. The number of U-235 nuclei undergoing, fission per sec is, approximately(a) 106, (b) 2 × 108 (c) 3 × 1016 (d) 931, Q.2 Power output of 92U235 reactor if it takes 30 days to use up, 2kg of fuel, and if each fission gives 185 MeV of useable, energy is(a) 5.846 kW, (b) 58.46 MW, (c) .5846 kW, (d) None, Q.3 How many electrons, protons and neutrons are there in a, 6 gm of 6C12., , RESPONSE GRID, , 1., , 2., , (a) 6 × 1023, 6 × 1023 , 6 × 1023, (b) 36 × 1023 , 36 × 1023 , 36 × 1023, (c) 12 × 1023, 12 × 1023, 12 × 1023, (d) 18 × 1023, 18 × 1023, 18 × 1023, Q.4 Nuclear radius of 8O16 is 3 × 10–15 m. Find the density of, nuclear matter., (a) 7.5 × 1017 kg m–3, (b) 5.7 × 1017 kg m–3, 17, –3, (c) 2.3 × 10 kg m, (d) 1.66 × 1017 kg m–3, Q.5 Consider the decay of radium-226 atom into an alpha, particle and radon-222. Then, what is the mass defect of the, reactionMass of radium -226 atom = 226.0256 u, Mass of radon - 222 atom = 222.0715 u, Mass of helium - 4 atom = 4.0026 u, (a) 0.0053 u, (b) 0.0083 u, (c) 0.083 u, (d) None, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , DPP/ P 56, , 2, Q.6 If mass equivalent to one mass of proton is completely, converted into energy then determine the energy produced?, , Q.12 If the binding energy of deuterium is 2.23 MeV, then the, mass defect will be- (in a.m.u.), , (a) 931.49 MeV, , (b) 731.49 MeV, , (a) 0.0024, , (b) – 0.0024, , (c) 911.49 MeV, , (d) 431.49 MeV, , (c) – 0.0012, , (d) 0.0012, , Q.7 If mass equivalent to one mass of electron is completely, converted into energy then determine the energy liberated., , Q.13 The ratio of the radii of the nuclei, , (b) 0.51 MeV, , (a) 6 : 10, , (c) 3.12 MeV, , (d) 2.12 MeV, , (c) 40 : 177, , (a) 1.25, , (b) 125 × 104, , (c) 1.25 × 108, , (d) 1.25 × 106, , (d) 14 : 73, Zn64, , nucleus is nearly (in fm)-, , (a) 1.2, , (b) 2.4, , (c) 3.7, , (d) 4.8, , (a) 11, 12, 13, , (b) 11, 11, 13, , (c) 12, 11, 13, , (d) 11, 13, 12, , Q.16 Energy of each photon obtained in the pair production, process will be, if the mass of electron or positron is, , (a) 3.8 days, , (b) 16.5 days, , (c) 33 days, , (d) 76 days, , 1/2000 a.m.u-, , Q.10 In the nuclear reaction,, A, 4, ZTh + 2He , the values of A and Z are-, , (a) A = 234, Z = 94, , (b) A = 234, Z = 90, , (c) A = 238, Z = 94, , (d) A = 238, Z = 90, , Q.11 The mass of helium nucleus is less than that of its, constituent particles by 0.03 a.m.u. The binding energy per, nucleon of 2He4 nucleus will be-, , (a) 0.213 MeV, , (b) 0.123 MeV, , (c) 0.321 MeV, , (d) 0.465 MeV, , Q.17 Deuterium is an isotope of hydrogen having a mass of, 2.01470 amu. Find binding energy in MeV of this isotope, (a) 2.741 MeV, , (b) 2.174 MeV, , (c) 1.741 MeV, , (d) 0.741 MeV, , Q.18 The binding energy per nucleon for 3Li7 will be, if the mass, of 3Li7 is 7.0163 a.m.u., , (a) 7 MeV, , (b) 14 MeV, , (a) 5.6 MeV, , (b) 39.25 MeV, , (c) 3.5 MeV, , (d) 21 MeV, , (c) 1 MeV, , (d) zero, , RESPONSE, GRID, , is, , Q.15 How many electrons, protons, and neutrons are there in a, nucleus of atomic number 11 and mass number 24?, , (Given log 10e = 0.4343), , 92, , e125, 52Te, , (b) 13 : 52, , Q.14 The radius of the 30, , Q.9 The half life of radioactive Radon is 3.8 days. The time at, the end of which 1/20th of the Radon sample will remain, undecayed is, , U238 ®, , and, , approximately -, , (a) 1.51 MeV, , Q.8 If the mass defect in the formation of helium from, hydrogen is 0.5%, then the energy obtained, in kWH, in, forming helium from 1 kg of hydrogen will be-, , 27, 13 Al, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 56, , 3, (1) atoms disintegrated per second in r eactor is, 3.125 ×1016, , Q.19 Sun radiates energy in all direction. The average energy, recieved at earth is 1.4 kW/m2. The average distance, between the earth and the sun is 1.5 × 10 11 m. If this energy, is released by conservation of mass into energy, then the, mass lost per day by the sun is approximately, , (2) atoms disintegrated per second in r eactor is, 3.125 ×1018, (3) decay in mass per hour is 4 × 10–8 kg, , (Use 1day = 86400 sec), , (4) decay in mass per hour is 4 × 10–6 kg, , (a) 4.4 × 109 kg, , (b) 7.6 × 1014 kg, , (c) 3.8 × 1012 kg, , (d) 3.8 × 1014 kg, , Q.23 Which of the following are not examples of nuclear fusion?, (1) Formation of Ba and Kr from U235, , Q.20 Fission of nuclei is possible because the binding energy, per nucleon in them, , (2) Formation of Pu – 235 from U–235, (3) Formation of water from hydrogen and oxygen, , (a) increases with mass number at high mass number, (b) decreases with mass number at high mass number, , (4) Formation of He from H, Q.24 Which of the following are mode of radioactive decay?, , (c) increases with mass number at low mass numbers, , (1) Positron emission, , (2) Electron capture, , (d) decreases with mass number at low mass numbers, , (3) Alpha decay, , (4) Fusion, , Q.21 Half life of Bi210 is 5 days. If we start with 50,000 atoms, of this isotope, the number of atoms left over after 10 days, is, (a) 5,000, , (b) 25,000, , (c) 12,500, , (d) 20,000, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a), , 1, 2 and 3 are correct, , (b) 1 and 2 are correct, , (c), , 2 and 4 are correct, , (d) 1 and 3 are correct, , DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, In a living organism, the quantity of C14 is the same as in the, atmosphere. But in organisms which are dead, no exchange takes, place with the atmosphere and by measuring the decay rate of, 14C in the old bones or wood, the time taken for the activity to, reduce to this level can be calculated. This gives the age of the, wood or bone., Given : T1/2 for 14 C is 5370 years and the ratio of, 14 C/ 12 C is 1.3 × 10 –12 ., Q.25 The decay rate of 14C in 1g of carbon in a living organism, is, , Q.22 On disintegration of one atom of U235 the amount of energy, obtained is 200 MeV. The power obtained in a reactor is, 1000 KW. Then, , RESPONSE, GRID, , 19., , 20., , 24., , 25., , (a) 25 Bq, , (b) 2.5 Bq, , (c) 0.25 Bq, , (d) 5 Bq, , 21., , Space for Rough Work, , 22., , 23.
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t.me/Magazines4all, DPP/ P 56, , 4, Q.26 If in an old sample of wood of 10g the decay rate is 30, decays per minute, the age of the wood is, , Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , (a) 50 years, , (b) 1000 years, , (c), , Statement -1 is False, Statement-2 is True., , (c) 13310 years, , (d) 15300 years, , (d), , Statement -1 is True, Statement-2 is False., , Q.27 The decay rate in another piece is found to be 0.30 Bq per, gm then we can conclude, (a) the sample is very recent, (b) the observed decay is not that of 14C alone, (c) there is a statistical error, (d) all of these, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , (b), , Q.28 Statement-1 : Amongst alpha, beta and gamma rays, g-has, maximum penetrating power., Statement-2 : The alpha particle is heavier than beta and, gamma rays., Q.29 Statement-1 : The mass of b-particles when they are, emitted is higher than the mass of electrons obtained by, other means., Statement-2 : b-particle and electron, both are similar, particles., Q.30 Statement-1 : Electron capture occurs more often than, positron emission in heavy elements., Statement-2 : Heavy elements exhibit radioactivity., , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 56 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 57, SYLLABUS : SEMICONDUCTOR ELECTRONICS - 1 (Semiconductors, LED, Photodiode, Zener diode), , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 When a semiconductor is heated, its resistance, (a) decreases, (b) increases, (c) reamins unchanged, (d) nothing is definite, Q.2 The energy band gap of Si is, (a) 0.70 eV, (b) 1.1 eV, (c) between 0.70 eV to 1.1eV, (d) 5 eV, Q.3 The forbidden energy ban d gap in conductors,, semiconductors and insulators are EG1, EG2 and EG3, respectively. The relation among them is, (a) EG1 = EG2 = EG3, (b) EG1 < EG2 < EG3, (c) EG1 > EG2 > EG3, (d) EG1 < EG2 > EG3, , RESPONSE GRID, , 1., , 2., , Q.4 Let nh and ne be the number of holes and conduction, electrons respectively in a semiconductor. Then, (a) nh > nein an intrinsic semiconductor, (b) nh = nein an extrinsic semiconductor, (c) nh = ne in an intrinsic semiconductor, (d) ne > nh in an intrinsic semiconductor, Q.5 Which statement is correct?, (a) N-type germanium is negatively charged and P-type, germanium is positively charged, (b) Both N-type and P-type germanium are neutral, (c) N-type germanium is positively charged and P-type, germanium is negatively charged, (d) Both N-type and P-type germanium are negatively, charged, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 57, , 2, Q.6 Wires P and Q have the same resistance at ordinary (room), temperature. When heated, resistance of P increases and, that of Q decreases. We conclude that, (a) P and Q are conductors of different materials, (b) P is n-type semiconductor and Q is p-type, semiconductor, (c) P is semiconductor and Q is conductor, (d) P is conductor and Q is semiconductor, Q.7 In extrinsic P and N-type, semiconductor materials, the, ratio of the impurity atoms to the pure semiconductor, atoms is about, (a) 1, (b) 10–1, (c) 10–4, (d) 10–7, Q.8 At zero Kelvin a piece of germanium, (a) becomes semiconductor, (b) becomes good conductor, (c) becomes bad conductor, (d) has maximum conductivity, Q.9 Electronic configuration of germanium is 2, 8, 18 and 4,, To make it extrinsic semiconductor small quantity of, antimony is added, (a) The material obtained will be N-type germanium in, which electrons and holes are equal in number, (b) The material obtained will be P-type germanium, (c) The material obtanied will be N-type germanium, which has more electrons than holes at room, temperature, (d) The material obtained will be N-type germanium, which h as less electrons than holes at room, temperature, Q.10 The intrinsic semiconductor becomes an insulator at, (a) 0°C, (b) –100°C, (c) 300 K, (d) 0 K, Q.11 Energy bands in solids are a consequence of, (a) Ohm’s Law, (b) Pauli’s exclusion principle, (c) Bohr’s theory, (d) Heisenberg’s uncertainty principle, , RESPONSE, GRID, , Q.12 The energy gap for diamond is nearly, (a) 1 ev, (b) 2 ev, (c) 4 ev, (d) 6 ev, Q.13 The valence band and conduction band of a solid overlap at, low temperature, the solid may be, (a) metal, (b) semiconductor, (c) insulator, (d) None of these, Q.14 Choose the correct statement, (a) When we heat a semiconductor its resistance, increases, (b) When we heat a semiconductor its resistance, decreases, (c) When we cool a semiconductor to 0 K then it becomes, super conductor, (d) Resistance of a semiconductor is independent of, temperature, Q.15 If ne and vd be the number of electrons and drift velocity, in a semiconductor. When the temperature is increased, (a) ne increases and vd decreases, (b) ne decreases and vd increases, (c) Both ne and vd increases, (d) Both ne and vd decreases, Q.16 The reverse biasing in a PN junction diode, (a) decreases the potential barrier, (b) increases the potential barrier, (c) increases the number of minority charge carriers, (d) increases the number of majority charge carriers, Q.17 Two PN-junctions can be connected in series by three, different methods as shown in the figure. If the potential, difference in the junctions is the same, then the correct, connections will be, P N, , +, , (a), (b), (c), (d), , N P, , –, , P N P N, , +, , 1, , 2, , –, , In the circuit (1) and (2), In the circuit (2) and (3), In the circuit (1) and (3), Only in the circuit (1), , N P, , +, 3, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , N P, , –, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 57, , 3, , Q.18 The approximate ratio of resistances in the forward and, reverse bias of the PN-junction diode is, (a) 102 : 1, (b) 10–2 : 1, –4, (c) 1 : 10, (d) 1 : 104, Q.19 The dominant mechanisms for motion of charge carriers, in forward and reverse biased silicon P-N junctions are, (a) Drift in forward bias, diffusion in reverse bias, (b) Diffusion in forward bias, drift in reverse bias, (c) Diffusion in both forward and reverse bias, (d) Drift in both forward and reverse bias, Q.20 In a triclinic crystal system, (a) a ¹ b ¹ c, a ¹ b ¹ g, (b) a = b = c, a ¹ b ¹ g, (c) a ¹ b ¹ c, a ¹ b = g, (d) a = b = c, a = b = g, Q.21 The correct cymbol for zener diode is, (a), (b), (c), , (d), , +, , –, , –, , +, , +, , –, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, , RESPONSE, GRID, , Q.22 In the given figure, which of the diodes are forward biased?, +10V, , (2) –12V, , (1), +5V, , R, , R, , –5V, R, , (3), , R, , (4) –10V, , –10V, , Q.23 Which of the following materials are crystalline?, (1) Copper, (2) Sodium chloride, (3) Diamond, (4) Wood, Q.24 A piece of copper and the other of germanium are cooled, from the room temperature to 80 K, then which of the, following would be wrong statements?, (1) Resistance of each increases, (2) Resistance of each decreases, (3) Resistance of copper increases while that of, germanium decreases, (4) Resistan ce of copper decreases while that of, germanium increases, DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, A student performs an experiment for drawing the static, characteristic curve of a triode valve in the laboratory. The, following data were obtained from the linear portion of the, curves:, Grid voltage Vg (volt), –2.0, –3.5, –2.0, Plate voltage Vp (volt), 180, 180, 120, Plate current IP (mA), 15, 7, 10, Q.25 Calculate the plate resistance rp of the triode valve?, (a) 0.12 × 104 ohm, (b) 1.2 × 10 4 ohm, 4, (c) 1.3 × 10 ohm, (d) 1.4 × 10 4 ohm, , 18., , 19., , 20., , 23., , 24., , 25., , Space for Rough Work, , 21., , 22.
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t.me/Magazines4all, DPP/ P 57, , 4, Q.26 Calculate the mutual conductance gm of the triode valve?, (a) 5.33 × 10–3 ohm–1, (b) 53.3 × 10–3 ohm–1, –3, –1, (c) 4.32 × 10 ohm, (d) 5.00 × 10–3 ohm–1, Q.27 Calculate the amplification factor µ, of the triode valve?, (a) 64, (b) 52, (c) 54, (d) 62, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a) Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., , RESPONSE GRID, , 26., , 27., , (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.28 Statement-1 : The number of electrons in a P-type silicon, semiconductor is less than the number of electrons in a, pure silicon semiconductor at room temperature., Statement-2: It is due to law of mass action., Q.29 Statement-1 : The resistivity of a semiconductor decreases, with temperature., Statement-2 : The atoms of a semiconductor vibrate with, larger amplitude at higher temperature there by increasing, its resistivity., Q.30 Statement-1 : We can measure the potential barrier of a, PN junction by putting a sensitive voltmeter across its, terminals., Statement-2 : The current through the PN junction is not, same in forward and reversed bias., , 28., , 29., , 30., , DAILY PRA CTICE PROBLEM SHEET 57 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 28, Qualifying Score, 48, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 58, SYLLABUS : SEMICONDUCTOR ELECTRONICS-2 (Junction transistor, transistor action, characteristics of a, transistor, transistor as an amplifier, logic gates), , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.20) : There are 20 multiple choice, questions. Each] question has 4 choices (a), (b), (c) and (d),, out of which ONLY ONE choice is correct., Q.1 A NPN transistor conducts when, (a) both collector and emitter are positive with respect, to the base, (b) collector is positive and emitter is negative with, respect to the base, (c) collector is positive and emitter is at same potential, as the base, (d) both collector and emitter are negative with respect, to the base, Q.2 In the case of constants a and b of a transistor, (a) a = b, (b) b < 1, a > 1, (c) a = b2, (d) b > 1, a < 1, , RESPONSE GRID, , 1., , 2., , Q.3 In an NPN transistor 1010 electrons enter the emitter in, 10–6 s and 2% electrons recombine with holes in base,, then a and b respectively are, (a) a = 0.98, b = 49, (b) a = 49, b = 0.98, (c) a = 0.49, b = 98, (d) a = 98, b = 0.49, Q.4 If l1, l2 , l3 are the lengths of the emitter, base and collector, of a transistor then, (a) l1 = l2 = l3, (b) l3 < l2 > l1, (c) l3 < l1 < l2, (d) l3 > l1 > l2, Q.5 In an NPN transistor circuit, the collector current is 10, mA. If 90% of the electrons emitted reach the collector,, the emitter current (iE) and base current (iB) are given by, (a) iE = -1mA, iB = 9mA (b) iE = 9mA, iB = -1mA, (c) iE = 1mA, iB = 11mA, , 3., Space for Rough Work, , 4., , (d) iE = 11mA, iB = 1mA, , 5.
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t.me/Magazines4all, DPP/ P 58, , 2, Q.6 The transfer ratio of a transistor is 50. The input resistance, of the transistor when used in the common-emitter, configuration is 1 kW. The peak value for an A.C input, voltage of 0.01 V peak is, (a) 100 μA, (b) 0.01 mA, (c) 0.25 mA, (d) 500 μA, Q.7 For transistor, the current amplification factor is 0.8. The, transistor is connected in common emitter configuration., The change in the collector current when the base current, changes by 6 mA is, (a) 6 mA, (b) 4.8 mA, (c) 24 mA, (d) 8 mA, Q.8 In NPN transistor the collector current is 10 mA. If 90%, of electrons emitted reach the collector, then, (a) emitter current will be 9 mA, (b) emitter current will be 11.1 mA, (c) base current will be 0.1 mA, (d) base current will be 0.01 mA, Q.9 In a transistor in CE configuration, the ratio of power gain, to voltage gain is, (a) a, (b) b / a, (c) ba, (d) b, Q.10 The following truth table corresponds to the logic gate, A 0 0 1 1, B 0 1 0 1, X, , Q.12 For the given combination of gates, if the logic states of, inputs A, B, C are as follows A = B = C = 0 and, A = B = 1,, C = 0 then the logic states of output D are, , (a) 0, 0, (b) 0, 1, (c) 1, 0, (d) 1, 1, Q.13 Correct statement for ‘NOR’ gate is that, it gives, (a) high output when both the inputs are low, (b) low output when both the inputs are low, (c) high output when both the inputs are high, (d) None of these, Q.14 A gate has the following truth table, P 1 1 0 0, Q 1 0 1 0, R 1 0 0 0, , The gate is, (a) NOR, (b) OR, (c) NAND (d) AND, Q.15 What will be the input of A and B for the Boolean, expression ( A + B).( A.B) = 1, (a) 0, 0, (b) 0, 1, (c) 1, 0, (d) 1, 1, Q.16 To get an output 1 from the circuit shown in the figure, the, input can be, , 0 1 1 1, , (a) NAND, (b) OR, (c) AND, (d) XOR, Q.11 The truth table shown in figure is for, , (a), , (a) XOR, (c) XNOR, , RESPONSE, GRID, , (b) A = 1, B = 0, C = 0, , (c) A = 1, B = 0, C = 1, (d) A = 1, B = 1, C = 0, Q.17 The truth-table given below is for which gate?, A 0 0 1 1, B 0 1 0 1, C 1 1 1 0, (a) XOR, (b) OR, (c) AND, (d) NAND, , A 0 0 1 1, B 0 1 0 1, Y, , A = 0, B = 1, C = 0, , 1 0 0 1, , (b) AND, (d) OR, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 58, , 3, , Q.18 The combination of gates shown below produces, , (a) AND gate, (b) XOR gate, (c) NOR gate, (d) NAND gate, Q.19 Figure gives a system of logic gates. From the study of, truth table it can be found that to produce a high output (1), at R, we must have, , Q.22 Given below are symbols for some logic gates. The XOR, gate and NOR gate are respectively, (1), , (2), , (3), , (4), , Q.23 Given below are four logic gates symbol (figure). Those, for OR, NOR and NAND are respectively, y, A, A, y, (2) B, (1), B, A, y, A, y, (4) B, (3) B, DIRECTIONS (Q.24-Q.25) : Read the passage given below, and answer the questions that follows :, , (a) X = 0, Y = 1, (b) X = 1, Y = 1, (c) X = 1, Y = 0, (d) X = 0, Y = 0, Q.20 In the case of a common emitter transistor as/an amplifier,, the ratio Ic /Ie is 0.96, then the current gain (b) of the amplifier, is, (a) 6, (b) 48, (c) 24, (d) 12, DIRECTIONS (Q.21-Q.23) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.21 Which of the following are false?, (1) Common base transistor is commonly used because, current gain is maximum, (2) Common collector is commonly used because current, gain is maximum, (3) Common emitter is the least used transistor, (4) Common emitter is commonly used because current, gain is maximum, , RESPONSE, GRID, , Doping changes the fermi energy of a semiconductor. Consider, silicon, with a gap of 1.11 eV between the top of the valence, bond and the bottom of the conduction band. At 300K the Fermi, level of the pure material is nearly at the mid-point of the gap., Suppose that silicon is doped with donor atoms, each of which, has a state 0.15 eV below the bottom of the silicon conduction, band, and suppose further that doping raises the Fermi level to, 0.11 eV below the bottom of that band., Conduction Band, , 1.11eV, , Fermi, level, , Donor, level, , Valence Band, , Q.24 For both pure and doped silicon, calculate the probability, that a state at the bottom of the silicon conduction band is, occupied? (e4.524 = 70.38), (a) 5.20 × 10–2, (b) 1.40 × 10–2, –2, (c) 10.5 × 10, (d) 14 × 10–2, Q.25 Calculate the probability that a donor state in the doped, material is occupied? e–1.547 = 0.212, (a) 0.824, (b) 0.08, (c) 0.008, (d) 8.2, , 18., , 19., , 20., , 23., , 24., , 25., , Space for Rough Work, , 21., , 22.
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t.me/Magazines4all, DPP/ P 58, , 4, DIRECTIONS (Q. 26-Q.28) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., (b) Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., (c) Statement -1 is False, Statement-2 is True., (d) Statement -1 is True, Statement-2 is False., Q.26 Statement -1 : The logic gate NOT cannot be built by using, diode., Statement -2 : The output voltage and the input voltage of, the diode have 180° phase difference., , RESPONSE GRID, , 26., , 27., , Q.27 Statement -1 : The following circuit represents ‘OR’ gate, , Statement-2 : For the above circuit Y = X = A + B = A + B, Q.28 Statement -1 : De-morgan’s theorem A + B = A.B may be, explained by the following circuit, , Statement -2 : In the following circuit, for output 1 inputs, A,B,C are 1, 0, 1., , 28., , DAILY PRA CTICE PROBLEM SHEET 58 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 46, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 59, SYLLABUS : Communication Systems, Laser, , Max. Marks : 120, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 In short wave communication which of the following, frequencies will be reflected back by the ionospheric layer,, having electron density 1011 per m3., (a) 2.84 MHz, (b) 10.42 MHz, (c) 12.24 MHz, (d) 18.1 MHz, Q.2 In an amplitude modulated wave for audio frequency of, 500 cycle/second, the appropriate carrier frequency will, be, (a) 50 cycles/sec, (b) 100 cycles/sec, (c) 500 cycles/sec, (d) 50, 000 cycles/sec, , RESPONSE GRID, , 1., , 2., , Q.3 Range of frequencies allotted for commercial FM radio, broadcast is, (a) 88 to 108 MHz, , (b) 88 to 108 kHz, , (c) 8 to 88 MHz, , (d) 88 to 108 GHz, , Q.4 The process of superimposing signal frequency (i.e. audio, wave) on the carrier wave is known as, (a) Transmission, , (b) Reception, , (c) Modulation, , (d) Detection, , Q.5 The characteristic impedance of a coaxial cable is of the, order of, (a) 50W, , (b) 200W, , (c) 270W, , (d) None of these, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, DPP/ P 59, , 2, Q.6 If m1 and m 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) m1 > m2, , (b) m1 < m 2, , (c) m1 = m2, (d) None of these, Q.7 Maximum usable frequency (MUF) in F-region layer is x,, when the critical frequency is 60 MHz and the angle of, incidence is 70o . Then x is (cos 70° = 0.34), (a) 150 MHz, (b) 170 MHz, (c) 175 MHz, (d) 190 MHz, Q.8 A laser is a coherent source because it contains, (a) many wavelengths, (b) uncoordinated wave of a particular wavelength, (c) coordinated wave of many wavelengths, (d) coordinated waves of a particular wavelength, Q.9 A laser beam is used for carrying out surgery because it, (a) is highly monochromatic, (b) is highly coherent, (c) is highly directional, (d) can be sharply focussed, Q.10 Laser beams are used to measure long distances because, (a) they are monochromatic, (b) they are highly polarised, (c) they are coherent, (d) they have high degree of parallelism, Q.11 An oscillator is producing FM waves of frequency, 2 kHz with a variation of 10 kHz . What is the modulating, index?, (a) 0.20, (b) 5.0, (c) 0.67, (d) 1.5, Q.12 If f a and ff represent the carrier wave frequencies for, amplitude and frequency modulations respectively, then, (a), , fa > f f, , (b), , fa < f f, , (c), , fa = f f, , (d), , fa ³ f f, , Q.13 An antenna is a device, (a) that converts electromagnetic energy into radio, frequency signal, , RESPONSE, GRID, , (b) that converts radio frequency signal into, electromagnetic energy, (c) that converts guided electromagnetic waves into free, space electromagnetic waves and vice-versa, (d) None of these, Q.14 While tuning in a certain broadcast station with a receiver,, we are actually, (a) varying the local oscillator frequency, (b) varying the frequency of the radio signal to be picked, up, (c) tuning the antenna, (d) None of these, Q.15 In a communication system, noise is most likely to affect, the signal, (a) At the transmitter, (b) In the channel or in the transmission line, (c) In the information source, (d) At the receiver, Q.16 In an FM system a 7 kHz signal modulates 108 MHz, carrier so that frequency deviation is 50 kHz . The carrier, swing is, (a) 7.143, (b) 8, (c) 0.71, (d) 350, Q.17 The phenomenon by which light travels in an optical fibres, is, (a) Reflection, (b) Refraction, (c) Total internal reflection, (d) Transmission, Q.18 In frequency modulation, (a) The amplitude of modulated wave varies as frequency, of carrier wave, (b) The frequency of modulated wave varies as amplitude, of modulating wave, (c) The amplitude of modulated wave varies as amplitude, of carrier wave, (d) The frequency of modulated wave varies as frequency, of modulating wave, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 59, , 3, , Q.19 Audio signal cannot be transmitted because, (a) The signal has more noise, (b) The signal cannot be amplified for distance, communication, (c) The transmitting antenna length is very small to design, (d) The transmitting antenna length is very large and, impracticable, Q.20 For sky wave propagation of a 10 MHz signal, what should, be the minimum electron density in ionosphere, (a), , ~ 1.2 ´ 1012 m -3, , Q.23 What type of modulation is not employed in India for radio, transmission?, (1) A mixture of both frequency and pulse modulation., (2) Pulse modulation, (3) Frequency modulation, (4) Amplitude modulation, Q.24 Which of the following are the characteristics of Laser, beams?, (1) They are monochromatic, , (b) ~ 106 m -3, , (2) They are coherent, , (c) ~ 1014 m-3, , (3) They have high degree of parallelism, , ~ 10 22 m -3, Q.21 What should be the maximum acceptance angle at the, , (d), , aircore interface of an optical fibre if n1 and n2 are the, refractive indices of the core and the cladding, respectively, (a) sin -1 (n2 / n1 ), , (b) sin -1 n12 - n22, , é -1 n2 ù, (c) ê tan n ú, ë, 1û, , é -1 n1 ù, (d) ê tan n ú, ë, 2û, , DIRECTIONS (Q.25-Q.27) : Read the passage given below, and answer the questions that follows :, The electron density of a layer of ionosphere at a height 150 km, from the earth's surface is 9 × 10 9 per m 3 . For the sky, transmission from this layer up to a range of 250 km,, Q.25 The critical frequency of the layer is, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (b) 1 and 2 are correct, (c) 2 and 4 are correct, (d) 1 and 3 are correct, Q.22 In which of the following remote sensing technique is used?, (1) Forest density, (2) Pollution, (3) Wetland mapping, (4) Medical treatment, , RESPONSE, GRID, , (4) They are not monochromatic, , (a) 2 Hz, , (b) 2.7 Hz, , (c) 2.78 kHz, , (d) 2.7 MHz, , Q.26 Maximum usable frequency is, (a) 3.17 Hz, (b) 3.17 × 106 HZ, (c) 3.17×10 3 Hz, (d) 3.17 × 1010Hz, Q.27 Angle of incidence of this layer is, (a) 0°, , (b) sec–1 (1.51), , (c) sec–1 (1.17), , (d) 37°, , 19., , 20., , 21., , 22., , 24., , 25., , 26., , 27., , Space for Rough Work, , 23.
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DPP/ P 59, , 4, DIRECTIONS (Qs. 28-Q.30) : Each of these questions contains, two statements: Statement-1 (Assertion) and Statement-2, (Reason). Each of these questions has four alternative choices,, only one of which is the correct answer. You have to select the, correct choice., (a), (b), (c), (d), , t.me/Magazines4all, , Statement-1 is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement-1., Statement-1 is True, Statement-2 is True; Statement-2 is, NOT a correct explanation for Statement-1., Statement -1 is False, Statement-2 is True., Statement -1 is True, Statement-2 is False., , RESPONSE GRID, , 28., , 29., , Q.28 Statement-1 : Television signals are not received through, sky-wave propagation., Statement-2 : The ionosphere reflects electromagnetic, waves of frequencies greater than a certain critical, frequency., Q.29 Statement-1 : The electromagnetic waves of shorter, wavelength can travel longer distances on earth’s surface, than those of longer wavelengths., Statement-2 : Shorter the wavelength, the larger is the, velocity of wave propagation., Q.30 Statement-1 : A dish antenna is highly directional., Statement-2 : This is because a dipole antenna is omni, directional., , 30., , DAILY PRA CTICE PROBLEM SHEET 59 - PHYSICS, Total Questions, 30, Total Marks, 120, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 30, Qualifying Score, 50, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Name :, , Date :, , Start Time :, , End Time :, , 60, SYLLABUS : Practical Physics - 2, , Max. Marks : 112, , Time : 60 min., GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains 28 MCQ's. For each question only one option is correct. Darken the correct, circle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/, deducted if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min., The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets., After completing the sheet check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation., , DIRECTIONS (Q.1-Q.26) : There are 26 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d), out, of which ONLY ONE choice is correct., Q.1 In making an Ohm’s law circuit, which of the following, connection is correct?, (a) Voltmeter in series and ammeter in parallel, (b) Voltmeter in parallel and ammeter in series, (c) Voltmeter and ammeter both are in parallel, (d) Voltmeter and ammeter both are in series, Q.2 To calculate an unknown resistance with the help of a meter, bridge why is it advised to change the gap with the known, and unknown resistance?, (a) To eliminate the resistance of the connecting wire and, copper strip, , RESPONSE GRID, , 1., , 2., , (b) To include the resistance of the connecting wire and, copper strip, (c) To balance the known and unknown resistance., (d) To eliminate the resistance of the gap., Q.3 Potential gradient of a potentiometer is equal to, (a) e.m.f per unit length, (b) potential drop per unit length, (c) current per unit length, (d) resistance per unit length, Q.4 The refractive index of the material of a prism does not, depend on which of the following factor?, (a) Nature of the material, (b) Wavelength or colour of light, (c) Temperature, (d) Angle of the prism., , 3., Space for Rough Work, , 4.
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t.me/Magazines4all, DPP/ P 60, , 2, Q.5 A meter-bridge is based on the principle of, (a) Wheatstone bridge, (b) Variation of resistance with temperature, (c) Galvanometer, (d) None of these, Q.6 A potentiometer works on the principle that, (a) when a current flows through a wire of uniform, thickness and material, potential difference between, its two points is directly proportional to the length of, the wire between the two points, (b) when a current flows through a wire of uniform, thickness and material, potential difference between, its two points is inversely proportional to the length, of the wire between the points, (c) when a current flows through a wire of uniform, thickness and material, potential difference between, its two points doesn't depend on the length of the wire, between the points, (d) none of these, Q.7 Which of the following statement is wrong regarding a, p–n junction diode?, (a) When the p-type section is connected to the positive, terminal and the n-type section to the negative terminal, of the battery the diode is called forward biased, (b) When the p-type section is connected to the negative, terminal and the n-type section to the positive terminal, of the battery the diode is said to be reverse biased, (c) When the diode is in reverse biased mode a forward, current flows, (d) When the diode is in forward biased mode a forward, current flows., Q.8 A Zener diode operates on which of the following bias?, (a) Forward bias, (b) Reverse bias, (c) Both forward and reverse bias., (d) No biasing is required for it., Q.9 The transfer characteristics of a transistor means a plot of, (a) input voltage versus input current, (b) output voltage versus output current., (c) output voltage versus input voltage, (d) input current versus output current., , RESPONSE, GRID, , Q.10 Current gain is maximum in which of the following, configuration of a transistor ?, (a) common emitter configuration, (b) common base configuration, (c) common collector configuration, (d) equal in both common emitter and common base, configuration, Q.11 Which of the following operations will not increase the, sensitivity of a potentiometer?, (a) Increase in the number of wires of the potentiometer., (b) Reducing the potential gradient., (c) Increasing the current through the potentiometer, (d) Increasing the sensitivity of the galvanometer., Q.12 Which two circuit components are connected in parallel, in the following circuit diagram ?, (a) Rheostat and voltmeter, (b) Voltmeter and ammeter, (c) Voltmeter and resistor, (d) Ammeter and resistor, Q.13 A current of 4A produces a deflection of 30° in the, galvanometer. The figure of merit is, (a) 6.5 A/rad, (b) 7.6 A/rad, (c) 7.5 A/rad, (d) 8.0 A/rad, Q.14 Two potentiometers A and B having 4 wires and 10 wires,, each having 100 cm in length are used to compare e.m.f., of 2 cells. Which one will give a larger balancing length?, (a) Balancing length doesn't depend on the total length, of the wire., (b) Both A and B will give same balancing length, (c) Potentiometer B, (d) Potentiometer A, Q.15 An LED operates under which biasing condition?, (a) Forward bias, (b) Reverse bias, (c) Can operate both in forward and reverse bias, (d) No biasing is required., Q.16 How are the currents flowing in the emitter, base and the, collector related to each other?, (a) Ic =Ib + Ie, (b) Ie =Ib + Ic, (c) Ib =Ie + Ic, (d) Ie =Ic – Ib, , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , 16., Space for Rough Work, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 60, , 3, , Q.17 The potential gradient of a potentiometer can be increased, by which of the following operation?, (a) By increasing the area of cross-section of the, potentiometer wire., (b) By decreasing the area of cross-section of the, potentiometer wire., (c) By decreasing the current through it., (d) By using a wire of material of low specific resistance., when A is decreased, k will increase., Q.18 Of the diodes shown in the following diagrams, which one, is reverse biased ?, +10 V, , (a), , R, , (b), , +5 V, , –12 V, , R, –5 V, +5 V, , (c), , R, , (d), , R, , –10 V, , Q.19 To determine the equivalent resistance of two resistors, when connected in series, a student arranged the circuit, components as shown in the diagram. But he did not, succeed to achieve the objective., , Which of the following mistakes has been committed by, him in setting up the circuit ?, (a) Position of voltmeter is incorrect, (b) Position of ammeter is incorrect, (c) Terminals of voltmeter are wrongly connected, (d) Terminals of ammeter are wrongly connected, Q.20 In the circuit shown, voltmeter is ideal and its least count, is 0.1 V. The least count of ammeter is 1 mA. Let reading, of the voltmeter be 30.0 V and the reading of ammeter is, 0.020 A. Calculate the value of resistance R within error, limits., , RESPONSE, GRID, , (a) (1.5 ± 0.05)kW, R, E, (b) (1.2 ± 0.05) k W, V, R, (c) (1.2 ± 0.08) kW, A, (d) (1.5 ± 0.08) kW, Q.21 In an experiment to measure the focal length of a concave, mirror, it was found that for an object distance of 0.30 m,, the image distance come out to be 0.60 m. Determine the, focal length., (a) (0.2 ± 0.01) m, (b) (0.1 ± 0.01) m, (c) (0.2 ± 0.02) m, (d) (0.1 ± 0.02) m, Q.22 In an experiment to determine an unknown resistance, a, 100 cm long resistance wire is used. The unknown, resistance is kept in the left gap and a known resistance is, put into the right gap. The scale used to measure length has, a least count 1 mm. The null point B is obtained at 40.0 cm, from the left gap. Determine the percentage error in the, computation of unknown resistance., (a) 0.24%, (b) 0.28% (c) 0.50% (d) 0.42%, Q.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 its 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, (b) f < x < 2f, (c) x = 2f, (d) x > 2f, H, , æ 1ö, , æ 1ö, , Q.24 For a convex spherical mirror, the graph of çè v ÷ø verses çè u ÷ø, is, (b) 1/ v, (a) 1/ v, , (c), , O, , (d), , 1/ v, , 1/ u, , O, , 17., , 18., , 19., , 22., , 23., , 24., , Space for Rough Work, , O, , 1/ u, , 20., , 1/ u, , 1/ v, , O, , 21., , 1/ u
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t.me/Magazines4all, DPP/ P 60, , 4, Q.25 If the wire in the experiment to determine the resistivity, of a material using meter bridge is replaced by copper or, hollow wire the balance point i.e. null point shifts to, (a) right, (b) left, (c) at same point, (d) none of these, Q.26 Which device is used to measure the potential difference, between two points of a conductor in the laboratory ?, (a) Voltameter, (b) Ammeter, (c) Potentiometer, (d) Galvanometer, DIRECTIONS (Q.27-Q.28) : Read the passage given below, and answer the questions that follows :, Consider a block of conducting material of resistivity ‘r’ shown, in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We, apply superposition principle to find voltage ‘DV’ developed, between ‘B’ and ‘C’. The calculation is done in the following, steps:, (i) Take current ‘I’ entering from ‘A’ and assume it to spread, over a hemispherical surface in the block., , RESPONSE GRID, , 25., , 26., , (ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s, law E = r j, where j is the current per unit area at ‘r’., I, I, DV, (iii) From the ‘r’ dependence of E(r),, obtain the potential V(r) at r., a, a, b, (iv) Repeat (i), (ii) and (iii) for current, C, A, B, D, ‘I’ leaving ‘D’ and superpose, results for ‘A’ and ‘D’., Q.27 DV measured between B and C is, rI, rI, rI, rI, –, –, (a), (b), pa p(a + b), a (a + b), rI, rI, rI, –, (c), (d), 2p(a - b), 2pa 2p(a + b), Q.28 For current entering at A, the electric field at a distance ‘r’, from A is, (a), , 27., , rI, 8pr, , 2, , (b), , rI, r, , 2, , (c), , rI, 2pr, , 2, , (d), , 28., , DAILY PRA CTICE PROBLEM SHEET 60 - PHYSICS, Total Questions, 28, Total Marks, 112, Attempted, Correct, Incorrect, N et Score, Cut-off Score, 26, Qualifying Score, 44, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , rI, 4pr 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 01, , 2, , ur ur ur ur, ur ur, P.Q = 0 Þ (A + B).(A – B) = 0, ur ur, Þ A2 – B2 = 0 Þ | A |=| B |, 20., , 27., , QP ^ Q, (a) By substituting dimension of each quantitity in R.H.S., of option (a) we get, , 28., 29., , é mg ù é M ´ LT -2 ù, -1, ú = [ LT ]., ê hr ú = ê -1 -1, ë, û ëê ML T ´ L ûú, 21., , [a ] = [ P ][V ]2 = [ ML-1T -2 ] ´ [ L6 ] = [ ML5T -2 ], , 30., , 1, 2, , Þ x + y = 0, - 2 y = -1 or y =, 1, 2, 23. (a) Try out the given alternatives., When x = 1, y = –1, z = 1, , Therefore, x = –, , P x S y C z = P1S-1C1 =, , =, , [ML–1T –2 ] [LT –1 ], 2 –2, , [ML T, , 2, , / L T], , = [M 0 L0 T 0 ], , 2 -2, Dimensions of work, [ W ] = éë ML T ùû, 2 -2, Dimensions of torque, [ t ] = éë ML T ùû, 2 -2, Dimensions of energy, [ E ] = éë ML T ùû, , Dimensions of Young¢s modulus,, , [ Y] = éëML-1T-2 ùû, Dimensions of light year = [L], Dimension of wavelength = [L], 25. (d), 26. (b), , c, , 5, , =, , kgm 2s-1 ´ m3kg -1s-2, 5, , 5, , m /s, , T = 2p g / l ., (a) Unit of quantity (L/R) is Henry/ohm., As Henry = ohm × sec,, hence unit of L/R is sec, i.e. [L/R] = [T]., Similarly, unit of product CR is farad × ohm, , or, , 2 -1, 24. (c) Dimensions of angular momentum, [ L] = éë ML T ùû, , hG, , = 3.0 eV, (a) Both statement -1 and statement -2 are correct and, statement -1 follows from statement -2, (c) Let us write the dimension of various quantities on, two sides of the given relation., L.H.S. = T = [T],, , or, , PC, S, , =, , s2 = s, , Putting the values of h, G and c in above relation, Planck time = 1.3 × 10– 43 s., , hc 6.6 ´10 -34 ´ 3 ´108, =, = 4.95 ´10-19 J, l, 4 ´10-7, , R.H.S. = 2p g / l =, , é a ù, ê 2 ú = [ P], ëV û, , 22. (c) f = c mx ky;, Spring constant k = force/length., [M0L0T–1] = [Mx (MT–2)y] =[ Mx + y T–2y], , E=, , LT -2, = [T -1 ], L, (\ 2p has no dimension)., As dimension of L.H.S. is not equal to dimension of, R.H.S. therefore according to principle of homogeneity, the relation., , This option gives the dimension of velocity., (a) By principle of dimensional homogeneity, , \, , (a), , Coulomb Volt, ×, Volt, Amp, Sec×Amp, = second, Amp, , i.e. [CR] = [T], therefore, [L/R] and [CR] both have the same, dimension., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 02, , 3, , 1., , (d) Density r =, , Þ, , M, M, = 2, V pr L, , 10. (c), , Dr DM, Dr DL, =, +2 +, r, M, r, L, , 2., , 3., , (a), (c), (c), , 7., , (d), , 8., , (c), , l, 4p 2 l, Þ g= 2, g, T, , Here % error in l =, , 1mm, 0.1, ´ 100 =, ´ 100 = 0.1%, 100cm, 100, , Dr, ´ 100 = 0.04 ´100 = 4%, Percentage error =, r, , and % error in T =, , 0.1, ´ 100 = 0.05%, 2 ´ 100, , 1 ö, æ, = ç 2 ´ 1 + 3 ´ 3 + 1´ 2 + ´ 2 ÷ % = 14%, 2 ø, è, Percentage error in X = aa+ bb + cg, Errors in A and B will be added., Given, L = 2.331 cm, = 2.33 (correct upto two decimal places), and B = 2.1 cm = 2.10 cm, \ L + B = 2.33 + 2.10 = 4.43 cm. = 4.4 cm, Since minimum significant figure is 2., The number of significant figures in all of the given, number is 4., Y=, , =, , \, , 11., , b a cb, , d g ed, So, maximum error in a is given by, , Db, Dc, æ Da, ö, ´ 100 ÷, = a. ´ 100 + b. ´100, ç, b, c, è a, ømax, + g., , = (ab1+ bc1+ gd1+ de1)%, , Dd, De, ´ 100 + d. ´ 100, d, e, , = % error in l + 2(% error in T), = 0.1 + 2 × 0.05 = 0.2%, (c) Mean time period T = 2.00 sec, & Mean absolute error DT = 0.05 sec., To express maximum estimate of error, the time period, should be written as (2.00 ± 0.05) sec, , = 3 × % error in radius, = 3 × 1 = 3%, 13. (a) Weight in air = (5.00 ± 0.05)N, Weight in water = (4.00 ± 0.05)N, Loss of weight in water = (1.00 ± 0.1)N, Now relative density =, , weight in air, weight loss in water, , 5.00 ± 0.05, 1.00 ± 0.1, Now relative density with max permissible error, , i.e. R.D =, , æ DM Dg DL 2DD Dl ö, DY, ´ 100 = ç, +, +, +, + ÷ ´ 100, Y, g, L, D, l ø, è M, , (d) a =, , % error in g, , 12. (b) V = 4 pr 3, 3, \ % error in volume, , 4MgL so maximum permissible error in Y, pD 2l, , 1, 1, 1, 1 1 ö, = æç, +, +, + 2 ´ + ÷ ´ 100, 41 87 ø, è 300 981 2820, = 0.065 × 100 = 6.5%, 9., , l, g, , =, , (a) In division (or multiplication), the number of significant, digits in the answer is equal to the number of, significant digits which is the minimum in the given, numbers., (d) Percentage error in A, , 4., 5., 6., , T = 2p, , Þ T 2 = 4p 2, , 0.003, 0.005 0.06, + 2´, +, 0.3, 0.5, 6, = 0.01 + 0.02 + 0.01 = 0.04, , \, , 02, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , =, , 5.00 æ 0.05 0.1 ö, ±ç, ±, ÷ ´100 = 5.0 ± (1 + 10)%, 1.00 è 5.00 1.00 ø, , = 5.0 ± 11%, DV, Dl, æ DR, ö, ´100 ÷, =, ´100 + ´ 100, 14. (b) \ ç, V, l, è R, ømax, 5, 0.2, ´ 100 +, ´100 = (5 + 2)% = 7%, 100, 10, (c) Volume of cylinder V = pr2l, Percentage error in volume =, , =, , 15., , DV, 2Dr, Dl, ´ 100 =, ´ 100 + ´ 100, V, r, l, , 0.1, æ 0.01, ö, ´100 +, ´ 100 ÷ = (1 + 2)% = 3%, =ç 2´, 2.0, 5.0, è, ø
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t.me/Magazines4all, DPP/ P 02, , 4, 24., DV, 2Dr, Dl, ´ 100 =, ´ 100 + ´ 100, V, r, l, , (d) Here, s = (13.8 ± 0.2) m, t = (4.0 ± 0.3) s, s 13.8, =, = 3.45 ms–1 = 3.5 ms–1, t 4.0, (rounding off to two significant figures), , velocity, v =, , 0.1, æ 0.01, ö, ´100 +, ´ 100 ÷ = (1 + 2)% = 3%, = ç 2´, 2.0, 5.0, è, ø, 16., , 17., 18., , DH, æ 2D I D R D t ö, ´ 100 = ç, +, +, ÷ ´ 100, H, R, t ø, è I, = (2 × 3 + 4 + 6)% = 16%, (c) Quantity C has maximum power. So it brings maximum, error in P., , \, , DE, v '2 - v 2, ´ 100 =, ´ 100 = [(1.5)2 – 1] × 100, E, v2, , \, , DE, ´100 = 125%, E, , (b) Required random error =, , 20., , (b) \ E =, , 23., , Dv, 4.94, =±, = ±0.0895, v, 13.8 ´ 4.0, D v = ± 0.0895 × v = ± 0.0895 × 3.45 = ± 0.3087 = ± 0.31, (rounding off to two significant fig.), Hence, v = (3.5 ± 0.31) ms–1, , Þ, , 1, (d) Kinetic energy E = mv 2, 2, , 19., , 21., 22., , Dv, ( 0.8 + 4.14), æ Ds Dt ö, æ 0.2 0.3 ö, = ±ç + ÷ = ±ç, +, ÷=±, v, t ø, 13.8 ´ 4.0, è s, è 13.8 4.0 ø, , (b) H = I2Rt, , x, 4, , 1 2, mv, 2, \ % Error in K.E., = % error in mass + 2 × % error in velocity, = 2 + 2 × 3 = 8%, , (c), (d) Since for 50.14 cm, significant number = 4 and for, 0.00025, significant numbers = 2, (a) Since percentage increase in length = 2%, Hence, percentage increase in area of square sheet, = 2 × 2% = 4%, , % age error in velocity =, , Dv, ´ 100 = ± 0.0895 × 100 = ± 8.95, v, , % = ± 9%, 25. (a) Maximum percentage error in measurement of e, as given, by Reyleigh’s formula., (Given error is measurement of radius is 0.1 cm), De = 0.6 DR = 0.6 × 0.1 = 0.06 cm., De, 0.06, ´ 100 =, ´ 100 = 3.33%, e, 0.6 ´ 3, 26. (b) Speed of sound at the room temperature., l1 = 4.6 cm, l2 = 14.0 cm.,, l = 2 (l2 – l1) = 2 (14.0 – 4.6) = 18.8 cm., , Percentage error is, , 18.8, = 376 m / s, 100, 27. (c) End correction obtained in the experiment., , v = f l = 2000 ´, , l 2 - 3l1 14.0 - 3 ´ 4.6, =, = 0.1 cm., 2, 2, (d) Since zeros placed to the left of the number are never, significant, but zeros placed to right of the number are, significant., (b) The last number is most accurate because it has, greatest significant figure (3)., (a) As the distance of star increases, the parallex angle, decreases, and great degree of accuracy is required for, its measurement. Keeping in view the practical, limitation in measuring the parallex angle, the maximum, distance of a star we can measure is limited to 100 light, year., e=, , 28., , 29., 30., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 03, , 7, , 1 2, 2v, at = vt Þ t =, 2, a, (b) x = 4(t – 2) + a(t – 2)2, At t = 0, x = – 8 + 4a = 4a – 8, , 19. (a), 20., , dx, = 4 + 2 a (t - 2), dt, At t = 0, v = 4 – 4a = 4(1 – a), v=, , But acceleration, a =, 21. (d), , x = ae, , -at, , d2x, dt 2, , dx d, = (ae -at + bebt ), dt dt, , -at, bt, Acceleration = -aae (-a ) + bbe .b, , = aa 2 e-at + bb2 eb t, Acceleration is positive so velocity goes on increasing, with time., 22. (d) (1) a = 3 sin 4t, , Þ, , dv, = 3sin 4t, dt, , \, , ò dv = ò 2sin pt dt, , 2, 2, or v = (1 - cos pt), p, p, Velocity is always non-negative, hence particle always moves, along positive x-direction., \ Distance from time t = 0 to t = t is, , 0, , -3, cos 4t + c, 4, For initial velocity, t = 0, , 2, or v = - cos pt + C, p, , At t = 0, v = 0 \ C =, , t, , Þ v=, , 3, , v0 = 0, 4, Therefore, initial velocity may or may not be zero., (2) Acceleration = 0, Þ a = 3 sin4t = 0 Þ sin 4t = 0, , Þ, , t=, , np, 4, , where n = 0, 1, 2, ................., Therefore, the acceleration of the particle becomes zero, p, second., 4, (3) As acceleration is sinusoidal function of time, so, particle repeats its path periodically. Thus, the particle, comes to its initial position after sometime (period of, function)., (4) The particle moves in a straight line path as it, performs S.H.M., Since (1) & (3) are correct, hence correct answer is (d)., , after each interval of, , 2t 2, sin pt, p p2, , 2, meters., p, 25. (a), 26. (b), 27. (b), 28. (d) Negative slope of position time graph represents that, the body is moving towards the negative direction and, if the slope of the graph decrease with time then it, represents the decrease in speed i.e. retardation in, motion., 29. (c) As per definition, acceleration is the rate of change of, velocity,, r, r dv, i.e. a = ., dt, If velocity is constant, r, dv, r, = 0, \ a = 0, dt, Therefore, if a body has constant velocity it cannot, have non zero acceleration., 30. (d) The displacement is the shortest distance between, initial and final position. When final position of a body, coincides with its initial position, displacement is zero,, but the distance travelled is not zero., , Distance from time t = t to t = 1s =, , At particular value of C =, , t, , 2, 2æ 1, 2, 2, ö, (1 - cos pt) dt = ç t - sin pt ÷ = t - 2 sin pt, ø0 p, p, pè p, p, , Also displacement from time t = 0 to t =, , 3, +C, 4, , Þ 4t = np, , Total displacement, Total time, Q It comes back to its initial position, \ Total displacement is zero., Hence, average velocity is zero., Sol. For Qs. 25-27., a = sin pt, , S=ò, , ò dv = ò 3sin 4t dt + c, , v0 = -, , 2v1v 2, Total distance, 2x, =, =, x, x, Total time, v1 + v 2, +, v1 v 2, , Average velocity =, , bt, , = a.e-at (-a) + bebt (b) = -aae-at + bbebt, , Þ, , =, , = 2a, , + be, , Velocity v =, , 23. (c) For an inertial frame of reference, its acceleration should, be zero. As reference frame attached to the earth i.e. a, rotating or revolving frame is accelerating, therefore, it, will be non-inertial., Thus (2) & (4) are correct, so correct answer is (c)., 24. (c) Average speed
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t.me/Magazines4all, DPP/ P 04, , 8, , DAILY PRACTICE, PROBLEMS, 1., , (c) If a stone is dropped from height h, t2 =, , 1 2, then h = gt, ......... (i), 2, If a stone is thrown upward with velocity u then, 1 2, gt1, ......... (ii), 2, If a stone is thrown downward with velocity u then, 1 2, gt2, 2, From (i), (ii) and (iii) we get, -ut1 +, , 1 2 1 2, gt1 = gt, 2, 2, , 1 2 1 2, gt2 = gt, 2, 2, Dividing (iv) and (v) we get, , ut2 +, , \, , 4., ......... (iii), , ......... (iv), , 5., , t, t2 - t2, or - 1 = 2 12, t2 t - t2, (c) Since direction of v is opposite to the direction of g, and h so from equation of motion, h = -nt +, , 3., , gt 2 - 2nt - 2h = 0, , Þ, , t=, , 2n ± 4n2 + 8 gh, 2g, , Þ, , t=, , 2 gh ù, né, ê1 + 1 + 2 ú, gë, n û, , (c), , 6., , 7., , 8., , 1 2, gt [As u = 0 for it downward motion], 2, (d) In the positive region the velocity decreases linearly, (during rise) and in the negative region velocity, increases linearly (during fall) and the direction is, opposite to each other during rise and fall, hence fall is, shown in the negative region., (a) For the given condition initial height h = d and velocity, of the ball is zero. When the ball moves downward its, velocity increases and it will be maximum when the ball, hits the ground & just after the collision it becomes, half and in opposite direction. As the ball moves, upward its velocity again decreases and becomes zero, at height d/2. This explanation match with graph (a)., (c) Acceleration of body along AB is g cos q, , From DABC, AB = 2R cosq; 2 R cos q =, , v 2 = u 2 + 2 gh Þ n 2 = (0) 2 + 2 g ´ 1 Þ n = 2 g, , t2 =, , For second 1 meter distance, 1 2, gt2 Þ gt22 + 2 2 gt2 - 2 = 0, 2, , 1 2, gt, 2, , 1, 2, Distance travelled in time t sec = AB = ( g cos q)t, 2, , 1 2, 1, gt Þ 1 = 0 ´ t1 + gt12 Þ t1 = 2 / g, 2, 2, Velocity after travelling 1m distance, , h = ut +, , 1 = 2 g ´ t2 +, , for u = 19.6 First ball will just strike the ground(in sky), Second ball will be at highest point (in sky), Third ball will be at point of projection or at ground, (not in sky), (a) The distance covered by the ball during the last t, seconds of its upward motion = Distance covered by it, in first t seconds of its downward motion, , h=, , 1 2, gt, 2, , Þ, , (d) Interval of ball throw = 2 sec., If we want that minimum three (more than two) ball, remain in air then time of flight of first ball must be, greater than 4 sec., T > 4 sec., , From h = ut +, , By solving t = t1t2, 2., , 2/ g, t1, 1, =, =, and so on., t2 (2 - 2) / g, 2 -1, , 2u, > 4sec Þ u > 19.6m / s, g, , ......... (v), , 1, g (t 2 - t12 ), -ut1 2, \, =, 1, ut2, g (t 2 - t22 ), 2, , -2 2 g ± 8 g + 8 g - 2 ± 2, =, 2g, g, , Taking + ve sign t2 = (2 - 2) / g, , h = -ut1 +, , h = ut2 +, , 04, , PHYSICS, SOLUTIONS, , 9., 10., , R, 4R, or t = 2, g, g, , (b) It has lesser initial upward velocity., (b) At maximum height velocity n = 0, , 1, g cos qt 2, 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., DPP/ P 04, , 9, , uuur, Velocity of car w.r.t. train (nct ) is towards, West – North, 21. (a) As the trains are moving in the same direction. So the, initial relative speed (v1 – v2 ) and by applying, retardation final relative speed becomes zero., , We know that n = u + at, hence, 0 = u - gT Þ u = gT, u, When n = , then, 2, , u, u, gT, T, = u - gt Þ gt = Þ gt =, Þt =, 2, 2, 2, 2, , Hence at t =, 11., , (b), , T, u, , it acquires velocity ., 2, 2, , t, h, 2h, Þ 1 = 1, g, t2, h2, , t=, , 12. (c) Speed of the object at reaching the ground n = 2 gh, If heights are equal then velocity will also be equal., 13. (b), , S, , S, , 10, = 10 + (2 ´ 3 - 1) = 35m, 2, , 3rd, , 2, , = 10 +, , nd, , From v = u - at Þ 0 = (v1 - v2 ) - at Þ t = v1 - v2, a, r, r, 22. (d) Let v A and v B be the respective velocities of the, particles at A and B. The relative velocity of particle at, A. w.r.t. to that at B is given by, r, r, r, r, v A - v B = v A + (- v B ), , B, , A, , vA=25m/s, , 300m, , S rd, 10, 7, (2 ´ 2 - 1) = 25 Þ 3 =, S nd 5, 2, 2, , Sn µ (2n - 1) . In equal time interval of 2 seconds, Ratio of distance = 1 : 3 : 5, 15. (c) Net acceleration of a body when thrown upward, = acceleration of body – acceleration due to gravity, =a–g, 16. (d) The initial velocity of aeroplane is horizontal, then the, vertical component of velocity of packet will be zero., , 14. (c), , So t =, , t.me/Magazines4all, , 2h, g, , Totallength, 50 + 50 100, =, =, = 4sec, Relative velocity 10 + 15 25, 18. (d) Relative velocity, = 10 + 5 = 15 m/sec, , (see figure). From triangle law of velocities if OP and, r, r, PQ represent v A and - v B , then the required relative, r, velocity v R is given by OQ ., r, | vR | = 252 + 202 = 625 + 400 = 32.02 m / s, If Ð PQO = q, then tan q =, , vB=20m/s, , 25, æ 5ö, Þ q = tan -1 ç ÷, è 4ø, 20, , 25, , vA, , O, , 17. (b) Time =, , 150, = 10 sec, 15, (a) When two particles moves towards each other then, v1 + v2 = 4, ....... (i), When these particles moves in the same direction then, v1 – v2 = 4, ....... (ii), By solving v1 = 5 and v2 = 1 m/s, uuur uur uur, (b) nct = nc - nt, uuur uur, uur, n ct = n c + (-nt ), vc, \t =, , 19., , 20., , vct, , P, 20, , Q, Thus, the particle at A, appears to approach B, in a direction, making an angle of tan–1 (5/4) with its direction of motion., Let us draw a line from A, as AC, such that Ð BCA is equal, to q ., A, , 300m, , B, , M, C, , Thus, to B, A appears to move along AC. From B, draw a, perpendicular to AC as BM., BM is the shortest distance between them., , \ BM = ABcos q = 300 ´, , 45°, -vt, , vt, , 4, 41, , = 187.41 m, , Also, AM = AB sin q = 234 .26 m, \ time taken to cover a distance
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t.me/Magazines4all, DPP/ P 04, , 10, AB = 234.26 m with a velocity of 32.02 m/s, 234.26, = 7.32 sec., =, 32.02, 23. (d) Since the wind is blowing toward the east, the plane, must head west of north as shown in figure. The velocity, of the plane relative to the ground vr pg will be the sum of, the velocity of the plane relative to the air vr pa and the, velocity of the air relative to the ground vr ag., (i) 1. The velocity of the plane relative to the ground is, given by equation :, r, r, r, v pg = v pa + v ag, 2. The sine of the angle q between the velocity of, the plane and north equals the ratio of vag and, vpa ., sin q =, (ii), , v ag, v pa, , =, , vpg =, , 2, - v ag, , (b) Vplane = 100 m/s, , vplane, , vwind, , =, , =, , 26., , (c), , =, , 1 2, (i) -40 = 10T - gT, 2, , or -40 = 10T - 5T, , u=10m/s, u, 40m, , 27., , 10 + 10 - 4 ´ 5(-40) 10 + 100 + 800, =, T=, 2´5, 10, =, , 2 ´ 10, = 2 sec., g, (iii) v = 10 + g × 2 = 30 m/s, , vwind, , (d), , vtotal, , 20, 100, , cos f =, , f, , vplane, , vplane, v total, , \ v total =, , h, , 2, , v wind, vplane, , æ 20 ö, \ f = tan -1 ç, è 100 ÷ø, , v, or 5T 2 - 10T - 40 = 0, , or, , tan f =, , (200 km / h )2 - (90 km / h )2, , r r, 1r, S = uT + aT 2, 2, , 2, , æ 1 ö, 400 + 10000 + 2 ´ 20 ´ 100 ´ ç ÷, è, 2ø, , = 87 m/s, , = 179 km/h., 24. (a) Using,, , vtotal, , Vtotal = (20)2 + (100)2 + 2 ´ 20 ´ 100 ´ cos135°, , 90 km / h, = 0.45 \ q = 26.74, 200 km / h, , Since vag and vpg are perpendicular, we can use the, Pythagorean theorem to find the magnitude of vr pg., v2pg = v2ag + v2pg, , v 2pa, , 25., , v plane, cos f, , =, , 100, m /s, cos f, , 28. (a) If components of velocities of boat relative to river is, same normal to river flow (as shown in figure) are same,, both boats reach other bank simultaneously., , 10 + 30, = 4 sec., 10, , Boat 2, , v, v, , (ii) t =, , 29., 30., , q q, , Boat 1, River, , (a) Both statement - 1 & statement - 2 are correct and, statement - 2 is correct explanation of statement - 2, (d) Statement - 1 is true but statement - 2 is false., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 05, , 11, , 1., , 5., , æ 90° ö, (d) Dv = 2v sin ç, è 2 ÷ø, , = 2v sin 45° = 2v ´, , 2, , = 2 ´ rw = 2 ´ 1 ´, 2., , 3., , (b) Time taken to cross the river along shortest possible, path is given by, t=, , 1, , 05, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , = 2v, , 15, 1, =, 2, 60, 5 - u2, Þ u = 3 km/h, \, , 10, æ qö, (b) Dv = 2v sin ç ÷ = 2 ´ 5 ´ sin 45° =, 2, è2ø, , So work done from (0, 0) to (a, 0) is given by, ur r, W = F . r = - Ka $j . ai$ = 0, For motion (a, 0) to (a, a), ur, F = - K (ai$ + a $j ) and displacement, r, r = (aiˆ + ajˆ) - (aiˆ + 0 ˆj ) = a $j, , v - u2, , v = velocity of boat in still water, u = velocity of river water, d = width of river, , 2p, 2p, =, cm/s, 60, 30, , 10 / 2, 1, Dv, =, =, m/s2, \ a=, 10, Dt, 2, (c) For motion of the particle from (0, 0) to (a, 0), ur, ur, F = - K (0iˆ + ajˆ) Þ F = - Ka $j, r, Displacement r = (aiˆ + 0 ˆj ) - (0iˆ + 0 ˆj ) = ai$, , d, 2, , 6., , (d) Here d = 320 m =, t = 4 min, v=, , 5, u, 3, , Putting values in t =, 7., , (c), , T, , v - u2, , , u = 60 m/min, , 1.93, R, =, sin q1 sin150°, , Þ R=, , 1.93 ´ sin150° 1.93 ´ 0.5, =, =1, sin q1, 0.9659, , 150°, , So total work done = - Ka 2, (d), , d, 2, , P, Q, R, =, =, sin q1 sin q2 sin150°, Þ, , So work done from (a, 0) to (a, a), ur r, W = F . r = - K (ai$ + a $j ). a $j = - Ka 2, , 4., , 320, km, 1000, , P, , T cos q, , q2, , Q, , q1, R, , q, T sin q, , P, , 8., , (b), , T, , T cos 30°, , W, As the metal sphere is in equilibrium under the effect, of the three forces therefore, ur ur uur, T + P +W = 0, From the figure, ................... (i), T cos q = W, ................... (ii), T sin q = P, From equation (i) and (ii) we get, P = W tan q and T 2 = P 2 + W 2, , 30°, T sin 30°, , 30 N, , W
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 05, , 13, , (AB)2 + (BC) 2 = (400)2 + (300) 2 = 500 m, Distance = AB+BC =400+300=700 m, r, r, r, ˆ B = ˆi - 3jˆ + 5k,, ˆ C = 2iˆ - ˆj + 4kˆ, 23. (a) A = 3iˆ - 2ˆj + k,, , æ - ˆj + kˆ ö, r, r, Unit vector perpendicular to A and B is ç, ÷., 2 ø, è, , r, A = 32 + ( -2) 2 + 12 = 9 + 4 + 1 = 14, , r, 2iˆ + ˆj + kˆ, (2iˆ + ˆj + kˆ) is parallel to A so, unit vector, 6, r, is parallel to A ., , AC =, , Any vector whose magnitude is k (constant) times, , r, B = 12 + ( -3) 2 + 52 = 1 + 9 + 25 = 35, , uv, C = 22 + 12 + ( -4) 2 = 4 + 1 + 16 = 21, , As B = A 2 + C2 therefore ABC will be right angled, triangle., r r, 24. (a) A ´ B = 0 \ sin q = 0 \q = 0°, Two vectors will be parallel to each other., 25. (a), 26 (b), 27. (b), iˆ ˆj kˆ, r r, A´ B = 2 1 1, 1 1 1, , = - ĵ + kˆ, , = iˆ(1 - 1) - ˆj(2 - 1) + kˆ(2 - 1), , 28. (b), , r r, r r, A+ B = A- B, , Þ A2 + B2 + 2AB cosq = A2 + B2 + 2AB cosq, Hence cosq = 0 which gives q = 90°, Also vector addition is commutative., r r r r, Hence A + B = B + A, r, r, 29. (a) Let P and Q are two vectors in opposite direction,, r, r, r r, then their sum P + (-Q) = P - Q, r r, If P = Q then sum equal to zero., , 30. (d) The resultant of two vectors of unequal magnitude, given by R = A2 + B 2 + 2 AB cos q cannot be zero, for any value of q.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 06, , 15, (b) Let at the time of impact, vx and vy be the horizontal, and vertical velocities respectively, then, vx = 700 × cos 37° = 700 × 0.8 = 560 cm/s, and vy = – 700 × sin 37° + 1000 × (5/7), = – 700 × 0.6 + (5000/7) = – 420 + 714.3, = + 294.3 cm/sec (downward), , 2u sin q 2´ 19.6 sin 45º, 4, =, =, = 2.82 sec, 9.8, g, 2, , t=, , 28 . 2 m, = 10 m/s, 2 . 82 sec, (a) Referring to (fig.) let P be a point on the trajectory, whose co-ordinates are (4, 4). As the ball strikes the, ground at a distance 14 metre from the wall, the range is, 4 + 14 = 18 metre. The equation of trajectory is, , Velocity of man =, , 13., , y = x tan q -(1/2) g, , B, 50, , x2, 70, , u 2 cos 2 q, , A, O, , Y, , /, cm, , 0c, , m, , h, vx, , s, , 0, 37º, , v, , vy, C, , 40 0cm, , Velocity of the ball at the time of collision, , usinq, , v sina, , ( v 2x + v 2y ), , v=, , v, ·, , u P v cosa, h, q, , A ucosq, , \v=, B X, , Again tan q =, , gx, é, ù, or y = x tan q ê1 ú, 2, 2, ë 2 u cos q. tan q û, , 15., , é 2u 2, ù, x, or y = x tan q ê1 - g sin q cos q ú, ëê, ûú, é xù, = x tan q ê1 - ú, ë Rû, Here x = 4, y = 4 and R = 18, , [(560 ) 2 + ( 294 .3) 2 ] = 632.6 cm/sec, , vy, vx, , =, , 294 .3, 560, , æ 294 .3 ö, ÷ = 27° 43', or q = tan–1 ç, è 560 ø, (d) Initial velocity is constant let the ball touches the, ground at an angle q and velocity 3u, , Hence 3u cos q = u or cos q = 1/3 or sin q = 8 /3, The vertical component of velocity at the ground, , ... (1), , = 3u sin q =, , u, , 4ù, é, æ7ö, \ 4 = 4 tan q ê1 - ú = 4 tan q ç ÷, ë 18 û, è9ø, , 3 8, =, 3, , 8u, , 20 m, , or tan q = 9/7, sin q = 9/ 130 and, cos q = 7/ 130, Again R = (2/g) u2 sin q cos q, , q, , = (2/9.8) × u2 × (9/ 130 ) × (7/ 130 ), , For a freely falling body it covers 20 m to acquire veloc-, , 18´ 9.8 ´ 130 ´ 130 98 ´ 13, u2 =, =, = 182,, 2´ 9 ´ 7, 7, u = 182 metre per second., 14. (b) The situation is shown in fig., (a) Let the ball collide after t sec, From fig. OC = OB cos 37° = 500 cos 37°, = 500 × 0.8 = 400 cm, ...(1), Horizontal velocity = 700 × cos 37º, \ OC = 700 × cos 37° × t, = 700 × 0.8 × t = 560 t, ... (2), From eqs. (1) and (2) 560 t = 400, or t = (5/7) sec., Now h = (1/2) g t2 = (1/2) × 1000 × (5/7)2 = 255.1 cm, , u, , ity, , 16., , \ ( 8 u)2 – 0 = 2 × 9.8 × 20 or u = 7 m/s, (b) The horizontal range of the projectile on the ground, R= u, , 17., 18., , 8u, , 2h, g ÞR=, , 2, , 2 ´ 10, = 2 . 2 = 2m, 10, , (a) R = ut Þ t = R/u = 12/8, Now h = (1/2) gt2 = (1/2) × 9.8 × (12/8)2 = 11 m, (b) The situation is shown in the adjoining figure., The time taken by the body is equal to the time taken, by the freely falling body from the height 29.4 m. Initial, velocity of body
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t.me/Magazines4all, DPP/ P 06, , 16, 30º, v, 29.4 m, , 19., , B, A, u sin q = 9.8 sin 30º = 4.9 m/s, From the relation, h = u sin q t + (1/2) gt2,, we get 29.4 = 4.9 t + (1/2) × 9.8 t2 Þ t = 2 sec, (b) The horizontal and vertical velocities of the bomb are, independent to each other. The time taken by the bomb, to hit the target can be calculated by its vertical motion. Let this time be t. Putting h = 490 m and g = 9.8 m/, s2 in the formula h = 1/2 gt2,, we have 490 = (1/2) × 9.8 × t2,, 2 ´ 490, = 10 sec, 9.8, The bomb will hit the target after 10 sec of its dropping., The horizontal velocity of the bomb is 60 km/hr which, is constant. Hence the horizontal distance travelled by, the bomb in 10 sec (horizontal velocity × time), = 60 km/hr × 10 sec, = 60 km/hr × 10/(60 × 60) hr = 1/6 km, Hence the distance of aeroplane from the enemy post, is 1/6 km = 1000/6 m = 500/3 meter., The trajectory of the bomb as seen by an observer on, the ground is parabola. Since the horizontal velocity of, the bomb is the same as that of the aeroplane, the falling bomb will always remain below the aeroplane. Hence, the person sitting inside the aeroplane will observe the, bomb falling vertically downward., (a) The angle of projection of the ball is q0 ( = 30º) and the, velocity of projection is u ( = 10 m/s). Resolving u in, horizontal and vertical components,, we have horizontal component,, ux = u cos q0 = 10 cos 30º = 8.65 m/s, and vertical component (upward),, uy = u sin 30º = 5.0 m/s, If the ball hit the ground after t sec of projection, then, the horizontal range is R = ux × t = 8.65 t meter, , \t=, , 20., , R, 17 . 3m, \t=, =, = 2.0 s, 8 .65, 8 .65 m / s, If h be the height of the tower,, then h = u'y t + (1/2) g t2,, where uy' is the vertical component, (downward) of the velocity of the ball., Y, , (2) Total time of flight = T =, , 2u sin q, ,, g, , Maximum height attained H =, , u 2 sin 2 q, 2g, , H u sin q, =, T, 4, (3) Initially the height of the monkey = MB = y = x tan q, Let the monkey drop to along line MA and the bullet, reach along the parabolic path OA. If both reach at A, simultaneously, the monkey is hit by the bullet., , Now, , AB = x tan q –, , gx 2, 2u 2 cos 2 q, , ,, , \ MA = MB – AB, MA = x tan q – x tan q +, , =, , gx 2, 2u 2 cos 2 q, , gx 2, , ....(i), , 2u 2 cos 2 q, M, A, , v, , u, , uy, ux, h, , Here uy = – uy' = – 5.0 m/s and t = 2.0 s, \ h = (–5.0) × 2.0 + 1/2 × 10 × (2.0)2, = – 10 + 20 = 10 meter, 21. (b) Let the ball B hits the ball A after t sec, The X-component of velocity of A is, v0 cos 37º = 700 cos 37º, The X-compoment of position of B is 300 cos 37º, The collision will take place when the X-coordinate of, A is the same as that of B., As the collision takes place at a time t, hence, 700 cos 37º × t = 300 cos 37º, or t = (300/700) = (3/7) sec, In this time the ball B has fallen through a distance, y = – 1/2 gt2 (Free fall of body B), = – 1/2 × 980 × (3 / 7)2 = – 90 cm, Hence the ball B falls a distance 90 cm, 22. (b), (1) Because force is constant hence acceleration will be, constant. When force is in oblique direction with initial, velocity, the resultant path is parabolic path., , X, , O, Tower, , q, , B, , Time taken by the bullet to reach point A,, X, R=17.3m, , t=, , x, u cos q, , ....(ii), , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 06, , 17, , Hence from (1), MA = (1/2) gt2, The monkey drops through distance (1/2) gt2 in the, same time. So the monkey is hit by the bullet., (4) The range R =, , (c) If the ball hits the n th step, the horizontal, and vertical distances traversed are nb and nh, respectively., , u 2 sin 2q, g, , u, , h, , u2, \ Maximum range Rmax = d =, g, Height H =, , 25., , b, , ....(iii), , u 2 sin 2 q, 2g, , R, , u2, Hmax=, 2g, , Let t be the time taken by the ball for these horizontal, and vertical displacement. Then velocity along, horizontal direction remains constant = u ; initial vertical, velocity is zero, \ nb = ut, ....(1), nh = 0 + (1/2) gt2, ....(2), From (1) & (2) we get, nh = (1/2) g (nb/u)2, , ....(iv), , From (iii) & (iv), Hmax = d/2, , u 2 sin 2q, 23. (a) Range of projectile, R =, g, The range is same for two angle q1 and q2 provided, q2 = 90º – q1, , Þ n=, , 2, , u sin 2q1, At an angle q1, range R1 =, g, , 26., , At an angle of projection q2,, , =, , 27., , (b), , 28., 29., , (a), (a), , 30., , (a), , Þ R1 = R2, \ other angle = 90º - q1 = 90º – 15º = 75º, , t2 =, , 2u sin q, g, , 2u sin(90º -q) 2u cos q, =, g, g, , t1 t2 =, , 2, 2 u 2 sin 2q, = .R, g, g, g, , where R is the range, Hence t1t2 µ R, , 2hu 2, gb 2, , (eleminating t), , (a) y = (1/2) gt2 (downward), Þ 1000 = (1/2) × 10 × t2 Þ t = 14.15 sec, , u 2 sin 2q 2, g, , u 2 sin 2 ( 90 º - q1 ) u 2 sin 2q1, =, g, g, , 24. (a) t1 =, , 2, , nth step, , \ Maximum height, , Range R2 =, , 1, , æ 144´10 3 ö, ç, ÷, x = ut = ç 60´60 ÷ × 14.15= 571.43 m, è, ø, Horizontal component of velocity, = 720 × 5/8 = 200 m/s, Let t be the time taken for a freely falling body from 490., Then y = (1/2) gt2, Þ 490 = (1/2) × 9.8 × t2 Þ t = 10 second, Now horizontal distance = Velocity × time, = 200 × 10 =2000m, Hence the bomb missed the target by 2000 m, Since W = D K implies that the final speed will be same., The time of flight depends only on the vertical, component of velocity which remains unchanged in, collision with a vertical wall., In statement-2, if speed of both projectiles are same,, horizontal ranges will be same. Hence statement-2 is, correct explanation of statement-1.
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t.me/Magazines4all, DPP/ P 07, , 18, , (1), , (b) We have angular displacement, =, , (6), , linear displacement, radius of path, , (d) We know that, Time period =, , DS, r, Here, DS = n (2pr) = 1.5 (2p × 2 × 10–2) = 6p × 10–2, , Þ Dq =, , \ Dq =, , (2), , 6 p ´ 10 -2, 2 ´ 10 -2, ®, , (a) We have w av, , = 3p radian, , =, , (7), , 2pr / 3 2 p, =, rad, r, 3, Total angular displacement,, q = q1 + q2 = 4p/3 rad, Total time = 2 + 1 = 3 sec, , (8), , (4), , (5), , (9), , 2p, w, 12, rad/sec Þ 2 =, 60, w1, 1, , 1 2, dq, at Þ, = w0 + at, 2, dt, This is angular velocity at time t., Now angular velocity at t = 2 sec will be, , 2p ´ 1, p, =, cm/s, 60, 30, æ pö, æ pö, çè 30 ÷ø + çè 30 ÷ø, , 2, , p, 2 cm/s, 30, (a) Let the radius of the orbit be r and the number of, revolutions per second be n. Then the velocity of electron, is given by v = 2pnr,, , v2, 4p 2 r 2 n 2, =, = 4 p2 r n2, r, r, Substituting the given values, we have, a = 4 × (3.14)2 × (5.3 × 10–11) (6.6 × 1015)2, = 9.1 × 1022 m/s2 towards the nucleus., The centripetal force is, FC = ma = (9.1 × 10–31) (9.1 × 1022), = 8.3 × 10–8 N towards the nucleus., (b) Given that radius of horizontal loop, r = 1 km = 1000 m, 9000 ´ 5, = 250 m/s, 18, , Centripetal acceleration ac=, , 250 ´ 250, v2, = 62.5 m/s2, =, 1000, r, , Centripetal acceleration, ac, 62.5, \ Gravitational acceleration =, =, = 6.38 : 1, g, 9.8, , (10), , 2 pr, 2p ´ 10, =, = 5p cm/s, t, 4, The linear acceleration is,, , The linear speed is v =, , v2, (5p )2, =, = 2.5 p2 cm/s2, a=, r, 10, This acceleration is directed towards the centre of the circle, , 2 pr, Circumference, =, 60, Time of revolution, , Speed v = 900 km/h =, , (d) We have q = w0t +, , æ dq ö, w = çè dt ÷ø, = w0 + 2a = 1 + 2 × 1.5 = 4 rad/sec, t =2sec, (d) The distance covered in completing the circle is, 2pr = 2p × 10 cm, , = 4 sec, , \ Acceleration a =, , 4p / 3, 4p, 2p, rad/s =, =, rad/s, 3, 6, 3, (c) Angular speed of hour hand,, , w2 =, , 7 ´ 10 ´ 4, , =, , ®, , 2p, Dq, = 12 ´ 60 rad/sec, Dt, Angular speed of minute hand,, , 2 ´ 22 ´ 4, , Change in velocity Dv =, , \ w av =, , w1 =, , 2pr, Circumference, =, Criticalspeed, gr, , 2, , S1, 2 pr / 3, angular displacement, q1 =, =, r, r, For second one third part of circle,, , q2 =, , (b) Velocity =, =, , Total angular displacement, =, Total time, , For first one third part of circle,, , (3), , 07, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 5ö, æ, 18 ´ ÷, 2, ç, è, v, 18 ø, (b) We know that, tan q =, =, rg, 100 ´ 10, , =, (11), , 1, 1, Þ q = tan–1, 40, 40, , (a) The angular velocity is w =, , v, r, , Hence, v = 10 m/s, r = 20 cm = 0.2 m \ w = 50 rad/s, , 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 07, (12), , 19, , (b) Given that w = 1.5t – 3t2 + 2, , (18), , dw, = 1.5 – 6t, dt, When, a = 0, Þ 1.5 – 6t = 0, , a=, , (13), , (a) Let W = Mg be the weight of the car. Friction, force = 0.4 W, M v2 W v2, =, Centripetal force =, r, gr, , 1.5, Þt=, = 0.25 sec, 6, (c) Given v = 1.5 t2 + 2t, Linear acceleration a = dv/dt = 3t + 2, This is the linear acceleration at time t, Now angular acceleration at time t, , 0.4 W =, , (19), , 3t + 2, a, Þa=, 2 ´ 10 -2, r, Angular acceleration at t = 2 sec, , a=, , W v2, gr, , Þ v2 = 0.4 × g × r = 0.4 × 9.8 × 30 = 117.6, Þ v = 10.84 m/sec, (c) Let v be the speed of earth's rotation., We know that W = mg, Hence, , 3, m v2, W = mg –, 5, r, , 8, × 102, =, 2, , or, , 3, m v2, mg = mg –, 5, r, , \, , (14), , = 4 × 102 = 400 rad/sec2, (a) Angular displacement after 4 sec is, , 2, 2g r, m v2, mg =, or v2 =, 3, 5, r, , 2 ´ 9.8 ´ (6400 ´ 103 ), 5, Solving, we get v = 5 × 109 m/sec,, , (15), , 1 2, 1 2 1, at =, at =, × 4 × 42 = 32 rad, 2, 2, 2, Angular velocity after 4 sec, w = w0 + at = 0 + 4 × 4 = 16 rad/sec, (a) Given a = 3t – t2, , (a)at t = 2sec =, , 3 ´ 2+2, 2 ´ 10 -2, , Now v2 =, , q = w0t +, , Þ, , dw, = 3t – t2 Þ dw = (3t – t2)dt, dt, , w=, (20), , 3t 2 t 3, - +c, 2, 3, At t = 0, w = 0, , Þ w=, , 3t 2 t 3, 2, 3, Angular velocity at t = 2 sec, (w)t = 2 sec, , \ c = 0, w =, , \ Centripetal force =, , (d) ˆix + yj, ˆ , x = r cos q,, y = r sin q where q = wt, r = î ( r cos wt) + ĵ (r sin wt), , (17), , k l0, , Þ r=, , ........(3), (k - 4 p 2 n 2 m), Substituting the value of r in eqn. (1) we have, , v = dr/dt = – î (wr sin wt) – ĵ (w r cos wt), a = d2 r/dt2 = – w2 r, (c) Let R be the normal reaction exerted by the road on the, car. At the highest point, we have, , m v2, = mg – R, R should not be negative., (r + h), Therefore v2 £ (r + a)g = (8.9 + 1.1) × 10, or v2 £ 10 × 10v £ 10 m/sec, \ vmax = 10 m/sec, , (a) Let T be the tension produced in the stretched string., The centripetal force required for the mass m to move in a, circle is provided by the tension T. The stretched length of, the spring is r (radius of the circle). Now,, Elongation produced in the spring = (r – l0), Tension produced in the spring,, T = k (r – l0), ........ (1), Where k is the force constant, Linear velocity of the motion v = 2p r n, mv2, m(2prn)2, =, r, r, = 4p2 r n2 m, ........ (2), Equating equation. (1) and (2), we get, k (r – l0) = 4p2 r n2 m, (\ T = mv2/r), 2, 2, Þ kr – k ll0 = 4 p r n m, r (k – 4p2 n2 m) = k l0, , 3, 8 10, = (4) - =, rad/sec, 2, 3, 3, Since there is no angular acceleration after 2 sec, \ The angular velocity after 6 sec remains the same., , (16), , æ 2g ö, çè ÷ø = 7.8 × 104 radian/sec., 5r, , ù, kl 0, 4p 2 n 2 ml 0 k, - l 0 ú or T =, 2 2, (k - 4 p2 n 2 m), ë (k - 4p n m), û, é, , T=k ê, (21), , (c) Two types of acceleration are experienced by the car, (i) Radial acceleration due to circular path,, ar =, , v2, (30)2, =, = 1.8 m/s2, r, 500
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t.me/Magazines4all, DPP/ P 07, , 20, (ii) A tangential acceleration due to increase of tangential, speed given by, at = Dv/Dt = 2 m/s2, Radial and tangential acceleration are perpendicular to each, other., Net acceleration of car, 2, a r2 + a 2t = (1.8) 2 + (2) 2 = 2.7 m/s, , a=, (22), , (a) For A :, , Þ, , mv A2, 3l, (net force towards centre = T1), This will provide required centripetal force, , Required centripetal force =, , \ T1 =, , particle at A,, , mv A2, 3l, , But tan q =, Þ v=, , Þ T2 =, , 2mv 2A, 9l, , + T1 =, , 5mv 2A, 9l, , (a) We know, a =, , (26), , (10)2, = 20 m/s2, 5, (a) Given that the mass of the particle, m = 2 kg, Radius of circle = 3 m, Angular velocity = 60 rev/minute, , 60 ´ 2p, rad/sec = 2p rad/sec, 60, Because the angle described during 1 revolution is 2p radian, The linear velocity v = rw = 2p × 3 m/s = 6p m/s, , =, , 2mv 2A, 9l, , v2, (6p )2, =, m/s2, r, 3, = 118.4 m/s2, , The centripetal acceleration =, (Putting value of T1), , 2, mvC2, mvA, =, 3l, 9l, Net force towards centre = T3 – T2, , Centripetal force., , T3 =, (23), , 2, 2, mvA, mvA, Þ T3 =, + T2, 9l, 9l, , 6mvA2, (on putting value of T2), 9l, , (b) N cos q =, , v2, r, , Hence v = 10 m/s, r = 5 m \ a =, , For C :, , \ T3 – T 2 =, , 9.8 ´ 9.8 ´ 10 -2 = 0.98 m/s, , (25), , 2mv 2A, 9l, , Net force towards the centre T2 – T1 =, , hg =, , (d) (1) Centripetal force is not a real force. It is only the, requirement for circular motion. It is not a new kind of, force. Any of the forces found in nature such as gravitational, force, electric friction force, tension in string, reaction force, may act as centripetal force., (3) Work done by centripetal force is always zero., , m(vB2 ), 2l, Remember w i.e. angular velocity, of all the particles is same, , Thus for B, centripetal force =, , rg, r, r, \, = 2, h, h, v, , (24), , For B : Required centripetal force =, , vA, vB, vC, =, =, \ w=, 3l, 2l, l, When a system of particles rotates about an axis, the, angular velocity of all the particles will be same, but their, linear velocity will be different, because of different, distances from axis of rotation i.e. v = rw., , rg, N sin q, mg, =, Þ tan q = 2, 2, N cos q, v, mv / r, , mv2, and N sin q = mg, r, , (27), , mv2, = mrw2, r, Here m = 0.10 kg, r = 0.5 m, , (a) F =, , 2pn, 2 ´ 3.14 ´ 10, = 2 rad/s, =, t, 31.4, F = 0.10 × 0.5 × (2)2 = 0.2, (28) (a) In non-uniform circular motion acceleration vector, makes some angle with radius hence it is not, perpendicular to velocity vector., (29) (c) If speed is increasing there is a tangential acceleration., Net acceleration is not pointing towards centre., (30) (b) Both statements are true but statement-2 is not correct, explanation for statement-1., , and w =, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 08, , 21, , (1), , O, , (a)., , q, , ·A, , T, , Tcosq, q, , m v2D, ³ mg, r, , · Tsinq ·, mg, , mg, , From equation (2) v 2D = 2g (h – 2r),, 5, r, 2, (d). See fig, Here v = 360 km/hr = 100 m/sec, , ...... (1), , (3), , 2, , 2, , =, , mv, l sin q, , ...... (2), , N, mg, , mg, Form eq. (1), T =, cosq, When the string is horizontal, q must be 90º i.e.,cos 90º = 0, , ·, , mg, , At lower point, N – mg =, , 1, , q = 60º, 2, The angle with horizontal = 90º – 60º = 30º, , N = 70 × 10 +, , (2), , 4 sin 60o, , N=, , 32sin 2 60º, = 80 sin2 60º, 0.4, , 1, m vB2, 2, The ball now rises to a point D, where its potential energy, is mg(h – 2r). If vD be the velocity of the ball at D, then,, , Then, mgh =, , 1, m v 2D, ......(2), 2, Now to complete the circular path, it is necessary that the, , m g (h – 2 r) =, , mv2, R, , 70 ´ (10000), = 2100 N, 500, , At upper point, N + mg =, , 0.4 ´ v 2, , Þ v = 80 sin 60º = 7.7 m/sec, (a). Let m be the mass of the ball. When the ball comes, down to B, its potential energy mgh which is converted, into kinetic energy. Let vB, be the velocity of the ball at B., , mv2, ,, R, , N = weight of the flyer = mg +, , cos q = (4/8) =, , From equation (2), 8 sin 60º =, , N, , N, , mg, \T=, =¥, 0, Thus the tension must be infinite which is impossible, so, the string can not be in horizontal plane., The maximum angle q is given by the breaking tension of, the string in the equation T cos q = m.g, Here T (Maximum) = 8 N and m = 0.4 kg, \ 8 cos q = 0.4 × g = 0.4 × 10 = 4, , v2 =, , v 2D ³ r g, , or, , \ 2g (h – 2r) ³ r g Þ h ³, , Form figure, T cos q = mg, mv, r, , centrifugal force acting upward at point D, should be equal, or greater than the force mg acting downward at point D, should be equal or greater than the force mg acting, downward. Therefore, , l, T, , T sin q =, , 08, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , mv2, ,, R, , mv2, – mg = 1400 – 700 = 700 N, R, , mv2, = 1400 N, R, (a). Given that U(r) = 10r3, So the force F acting on the particle is given by,, , At middle point, N =, , (4), , ¶U, ¶, =–, (10 r3) = –10 × 3 r2 = –30 r2, ¶r, ¶r, For circular motion of the particle,, , F=–, , m v2, = 30 r2, r, Substituting the given values, we have,, , F=, , 3 ´ v2, = 30 × (10)2 or v = 100 m/s, 10
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t.me/Magazines4all, DPP/ P 08, , 22, (8), , The total energy in circular motion, E = K.E. + P.E. =, , 1, mv2 + U(r), 2, , (b). Suppose v be the velocity of particle at the lowest, position B., According to conservation of energy, (K.E. + P.E.) at A = (K.E. + P.E.) at B, , 1, × 3 × (100)2 + 10 + (10)3 = 2.5 × 104 joule, 2, Angular momentum, = mvr = 3 × 100 × 10 = 3000 kg–m2/sec, , O, , 2pr 2 ´ p ´ 10 p, Also time period T =, =, =, sec, v, 5, 100, (a). Let T be the tension, q the angle made by the string, with the vertical through the point of suspension., , B, , =, , (5), , Therefore w =, , 1, g, h, =4Þ, =, h, g 16, , mg, 1, Þ 0 + mgl = mv2 + 0 Þ v =, 2, q, , h, g, cos q =, =, l, 16, = 0.6125 Þ q = 52º 14', Linear velocity, = (l sin q)w =1 × sin 52º 14' × 4, = 3.16 m/s, , (6), , l, , h, 1, =, = p/2, g, frequency, , The time period t = 2p, , A, , l, , (9), , T, , h, , (a). Maximum tension T =, \, , ·, mg, , mv2, r, , 2gl, , mv2, + mg, r, , = T – mg, , mv2, = 163.6 – 4 × 9.8 Þ v = 6 m/s, r, (c). The situation is shown in fig. Let v be the velocity of, the bob at the lowest position. In this position the P.E. of, bob is converted into K.E. hence -, , or, , (10), , v2, (d). Centripetal acceleration, ac =, = k2 rt2, r, , \ Variable velocity v =, , k2r 2 t2 = k r t, The force causing the velocity to varies, , dv, =mkr, dt, The power delivered by the force is,, P = Fv = mkr × krt = mk2r2t, (a). We know centripetal acceleration, , F= m, , (7), , ac =, , 1, mv2 Þ v2 = 2gl, 2, If T be the tension in the string,, , mgl =, , (tangential velocity) 2, (200)2, =, = 400 m/sec2, radius, 100, , mv 2, l, From (1) & (2), T = 3 mg, , then T – mg =, , at, , (11), , ac, O, , =, , 2, , ac + a t, , (400) 2 + (100) 2 = 100 17 m/s2, , 2, , ....(2), , r, (b). The velocity of the swimmer w.r.t. water vSR = 4.0 km/, h in the direction perpendicular to the river. The velocity of, r, river w.r.t. the ground is vRG = 3.0 km/h along the length, of river., Y, , Tangential acceleration, at = 100 m/sec2 (given), 2, 2, o, \ anet = a c + a t + 2a c a t cos 90 =, , ....(1), , VSR, , VSG, , X, VRG, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 08, , 23, , r, The velocity of the swimmer w.r.t. the ground is vSG where, ®, , ®, , mv2, = T + mg cos a, l, On leaving the circular path, T=0, , ®, , V SG = V SR + V RG, 2, 2, VSG = VSR, + VRG, = 4 2 + 32, , (12), , = 16 + 9 = 25 = 5 km / hr, (b). The minimum speed at highest point of a vertical circle, , (13), , is given by vc = rg = 20 ´ 9.8 = 14 m/s, (a). The speed at highest point must be, v>, \ r, , 2p, >, T, 2pr, , 2p, T, , 1, mv2 + mgh, 2, Þ v2 = 2g (2l – h), , Þ 0 + 2mgl =, , rg, , rg, , < 2p, , A, h, , Also, cos a =, (16), , q, mg, , h-l, l, , S, , T sin q = M w 2 R, T sin q = Mw 2 L sin q, From (i) and (ii), , N, q, , (d), , B, v, , mv, = mg cos q, R, Þ v2 = gR cos q, ....(1), According to law of conservation of energy, (K.E. + P.E.) at A = (K.E. + P.E.) at B, , 1, mv2 + mgh, 2, Þ v2 = 2g (R – h), , (17), , R, , (a). v = 60 km/hr =, , 50, m/s, 3, , r = 0.1 km = 100m, \ tan q =, , v2, = 0.283, rg, , \ q = tan–1 (0.283), ....(2), (18), , 2, R, 3, , 2, 3, (a). Let the body will have the circular path at height h, above the bottom of circle from figure, , Also cos q=, , v2, ..... (1), rg, Let h be the relative raising of outer rail with respect to, inner rail. Then, , (c). We know that tan q =, , tan q =, , h, l, , v, B, , A, , 2, , mw R, , = 16 ML, , Þ 0 + mgR =, , O, , L, , æ 2ö, = M4p2 ç ÷ L, è pø, , 2, , h, , T, , q, 2, , mv2, = mg cos q – N, R, When the particle leaves the sphere i.e. N = 0, , (15), , q, , T = Mw 2 L, = M4 p 2 n 2 L, , From (1) & (2), h =, , ....(2), , 5, l, 3, , From (1) & (2) h =, , r, 0.5, < 2p, < 1.4 sec, g, 9.8, Maximum period of revolution = 1.4 sec, (a). Let the particles leaves the sphere at height h,, T<, , (14), , gr , v = rw = r, , mv2, = mg cos a, l, Þ v2 = g l cosa, ....(1), According to law of conservation of energy, (K.E. + P.E.) at A = (K.E. + P.E.) at B, , \, , From (1) & (2) , h =, , ...... (2), (l = separation between rails), , v2, ×l, rg, , Hence v = 48 km/hr =, , mg, \ h=, , 120, m/s, (r = 400 m, l = 1m),, 9, , (120 / 9)2 ´ 1, = 0.045 m = 4.5 cm, 400 ´ 9.8
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t.me/Magazines4all, DPP/ P 08, , 24, (19), , (a). The woman has two velocities simultaneously while, running on the deck, one velocity is equal to the velocity, of ship i.e. 12 m/s due east and other velocity is 5 m/s due, north., , \, , mv2B, TB, v 2 + gL 4, 4, L, =, = or B, =, TT mvT2, 1, v 2T - gL 1, - mg, L, , or, , v 2B + gL = 4v 2T - 4gL but v 2B = v 2T + 4gL, , \, , vT2 + 4gL + gL = 4v 2T - 4gL Þ 3v 2T = 9gL, , N, 5 m/s, , 13 m, , /s, , 12m/s, , E, , The resultant velocity of woman, =, (20), , (12)2 + (5) 2 = 13 m/s, , (c). If we consider velocity of rain with respect to the man, is V km/h., , vmg = 3km/h, , Relative velocity of man w.r.t. ground, ®, , ®, , v mg = vm - v g, , ........(1), , ®, , (c). He can only reach the opposite point if he can cancel, up the velocity of river by his component of velocity., (24) (a). v = Rw, v1 > v2, (25) (b), (26) (b), (27) (c)., The path of a projectile as observed by other projectile is a, straight line., u0sinq, , Velocity of rain w.r.t. ground, ®, , q, , ®, , v rg = vr - v g, , vT2 = 3 ´ g ´ L = 3 ´ 10 ´, , (23), , Road, , ®, , (22), , Rain, , vrg = 4km/h, , 10, or v T = 10m / sec, 3, r, r, r, (d). Use definition of relative velocity VPQ = VP - VQ, r, r, VP = const. ; VQ = const., r, r, r, r, \ | VPQ | = | VQP | = const. ; | VP | > | VQ |, r, r, \ VPQ ® + ve ; VQP = - ve i.e. towards origin., , \, , ........(2), , u0cosq, , u0sinq, , u0cosq q, u0cosq, , vmg = 3 km/hr, , B, , v ba, , u0sinq, , vrm, , v A = u cos q iˆ + (u sin q - gt) ˆj.vAB = (2u cos q) ˆi, , vrm= 4 km/hr, , -vmg, ®, , ®, , ®, , Velocity of rain w.r.t. man v rm = vr - v m, On subtracting eqn. 1 from eqn. 2, ®, , ®, , v B = -u cos q ˆi + (u sin q - gt) ˆj ; a BA = g - g = 0, The vertical component u0 sin q will get cancelled., The relative velocity will only be horizontal which is equal, to 2u0 cos q, Hence B will travel horiozontally towards left w.r.t A with, constant speed 2u0 cos q and minimum distance will be h., Srel, l, =, Vrel 2u 0 cos q, , ®, , v rm = v rg - v mg, , | vrm |= vrg 2 + vmg 2 = 42 + 32 = 5 km / hr, (21), , (d) Since the maximum tension TB in the string moving in, the vertical circle is at the bottom and minimum tension TT, is at the top., , \, , TB =, , mv2B, mv2T, + mg and TT =, - mg, L, L, , (28) (a) When two bodies are moving in opposite direction,, relative velocity between them is equal to sum of the, velocity of bodies. But if the bodies are moving in same, direction their relative velocity is equal to difference in, velocity of the bodies., (29) (b) Time taken is shortest when one aims perpendicular to, the flow., (30) (d), , 2, v r / m = v r2 + v m, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 09, , 25, , (1), , (a) Force causing the acceleration = 400 – 200 = 200N, mass of the boy = 200/9.8, hence acceleration = F/m =, , (2), , (8), , 200, × 9.8 = 9.8 m/s2, 200, , 4, with x-axis, 3, (a) From the law of conservation of momentum, 1000 × 5 + 0 = (1000 + 60) v, , So the displacement is 50 m along tan–1, , 1000 ´ 5, Þv=, = 4.71 m/s, 1060, , (4), , (a) (a) The elevator having an initial upward speed of 8 m/, sec is brought to rest within a distance of 16 m, Hence,, 0 = (8)2 + 2a (16) (Q v2 = u2 + 2as),, a= -, , r, F 6iˆ + 8jˆ, in the direction of force, (a) Acceleration =, =, m, 10, and displacement, , ˆ, ˆ, r r 1r, 1 æ 6i + 8j ö, S = ut + at 2 = 0 +, ç 10 ÷ 100 = 30 î + 40 ĵ, ø, 2, 2 è, , (3), , 09, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , (9), , Resultant upward force on elevator = T – mg. According to, Newton's law., T – mg = ma, or T = mg + ma = m (g + a) = 1000 (9.8 – 2) = 7800 N, (b) Let P be the upward force exerted on the man by the, elevator floor. If m' be the mass of the man, then, weight of, the man acting downward = m' g ,, Upward force on the man = P – m'g, According to Newton's law. P–m' g = m' a or, P = m' (a + g) = (– 2 + 9.8) = 624 N, (d) As P and Q move down, the length l decreases at the, rate of U m/s, , A, , 10, (b) Weight of disc =, kg ,, 1000, Let speed of the bullet = v, So rate of change of momentum of the bullets, , =, , 8´8, = – 2 m/sec2, 2 ´ 16, , b, , 2 ´ 10 ´ 5, 1000, , 2 ´ 10 ´ 5, 10 ´ g, × v=, 1000, 1000, Þ v = 0.98 m/s2 = 98 cm/s2, (c) Total mass = 80 + 40 = 120 kg, The rope cannot with stand this load so the fire man should, slide down the rope with some acceleration, \ The maximum tension = 100 × 9.8 N, m (g – a) = tension ,, 120 (9.8 – a) = 100 × 9.8 Þ a = 1.63 m/s2, (b) Suppose the velocity of the body at the instant when it, reaches the pile of sand be v. Then, v2 = 0 + 2 (9.8) × (5 metre) = 98(Q v2 = u2 + 2as), , a=–, , (7), , 98, = – 980 m/sec2, 2 ´ (0.05), , Now, retarding force, F = mass × acceleration= 0.02 kg × (– 980 m/sec2) = –19.6 N, (b) Impulse = F . t = Area under F-t curve from 4 ms to 16, ms = Area under BCDFB, = Area of trapizium BCEF + area of DCDE, 1, 1, (200+800) (2×10–6) + ×10 × 10–6 × 800, 2, 2, = 10 × 10–4 + 40 × 10–4 N–s = 50 × 10–4, = 5.0 × 10–3 N-s, , =, , Q, , P, , M, From figure, l2 = b2 + y2, Differentiating with respect to time, , Now, , (6), , B, , q q, , v = applied force on the disc, , (5), , b, y, , 2l, , dl, dy, = 2y, dt, dt, , (Q b is constant), , dy l dl, U, 1 dl, = ., ., =, =, dt y dt, cosq, cos q dt, (a) The engine, coach, coupling and resistance are, shown, in figure., \, , (10), , COACH, , ENGINE, T, , DRIVING, FORCE, , Driving force = 4500 N, (5 + 4)104, = 900 N, 100, Resultant force = 4500 – 900 = 3600 N, Mass of engine and coach = 9 × 104 kg, According to Newton's law, F = ma, \ 3600 = 9 × 104 a, or a = (3600) / (9 × 104) = 0.04 m/sec2, So acceleration of the train = 0.04 m/sec2, , Opposing force (Resistance) =
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t.me/Magazines4all, DPP/ P 09, , 26, Now considering the equilibrium of the coach only, we, have (T – R) = 4 × 104 × 0.04 (Q F = ma), , (17), , to the impulse =, , 4 ´ 104, or T –, = 4 × 104 × 0.04,, 100, T = 4 × 104 × 0.04 + 4 × 102 = 1600 + 400 = 2000 N, , (11), , ®, , (12), , | a | = 2 2 + 32 =, (c) From the relation, , 4+9 =, , 13 m/sec2, , (18), , (19), , F, 1000, =, = 1 m/s2, m 1000, As the force is brake force, acceleration is – 1 m/s2 using, relation v2 = u2 + 2as, we obtain, , F = ma Þ a =, , 2, , (13), , 5ö, æ, çè18 ´ ÷ø, 2, 18, u, 2as = u2 Þ s =, =, = 12.5 m, 2, 2a, (a) The water jet striking the block at the rate of 1 kg/s at a, speed of 5 m/s will exert a force on the block, , F=v, , (14), , (16), , impulse, 13, =, = 1300 N, time, 1/ 100, in the direction of the ball., r dpr, r, r, (b). We know F =, Þ Fdt = dp, dt, r, r, Þ 2 × 2 = dp Þ 4 = dp, Therefore change in momentum = 4 Ns, r dpr, F, (a) We know =, dt, r, r, r, r, r, r, Þ Fdt = dp = p2 - p1 = mv 2 - mv1, , r, Þ 4 ĵ . 1 = 2 . v2 – 2(2 î ), r, ˆ = 2.vr - 2 (2i), ˆ = 4jˆ + 4iˆ, Þ 2v2 4j.1, 2, r, Þ v2 = 2iˆ + 2ˆj, r, Þ | v2 | = 2 2 m/s, (c) Initial momentum of the ball, =, , 150, × 12 = 1.8 kg.m/sec, 1000, , 150, × 20= – 3.0 kg m/sec, 1000, Change in momentum = 4.8 kg m/sec, , 2m(v + u), t, , dp, Þ F dt = dp = p2 – p1, (b) F =, dt, Þ F × 1 = mnv – 0, Þ F = mnv, (Total mass of the bullets fired in 1 sec = mn), (a) The initial momentum = 15 × 10 = 150 kgm/s and, , change in momentum, 0 - 150, Force =, = –10 N, =, 15, time, A constant force of 10 N must be acting in opposite, direction to the motion of body., , 4.8, = 480 N, .01, (b) Initial momentum of the body = mu = 20 × 3 = 60, and final momentum of the body = – mu = –20 × 3 = – 60, The change in momentum of body in initial direction, = – 60 – 60 = –120, The change in momemtum imparted to the body in opposite, direction = 120, \ The impulse imparted to the body = 120 Ns, (a) (1) Since the lift is moving down with an acceleration of, 3 m/sec2, then the inertial force F = ma, acts upwards on, the body, , Average force exerted = Impulse/ time =, , a, , Dv, m[(v + u) - {-(v + u)}], =, Dt, t, , =, , æ 250, ö, çè - 1000 ´ 24÷ø = 13 Ns, , Final momentum of the ball =–, , And under the action of this force of 5 N, the block of mass, 2 kg will move with an acceleration given by,, F = ma Þ a = F/m = 5/2 = 2.5 m/s2, (a) Relative speed of the ball = (v + u), Speed after rebouncing = – (v + u), So, F = m, , (15), , (20), , dm, =5×1=5N, dt, , 2 kg, , 2.50, × 28 –, 1000, , and force =, , ®, , ®, ˆ and F = (4iˆ + 8j), ˆ, (d) Given that F1 = (8iˆ + 10j), 2, r, Then the total force F = 12iˆ + 18jˆ, r, r F 12iˆ + 18jˆ, = 2iˆ + 3jˆ m/sec2, So acceleration a = =, m, 6, Net acceleration, , (a) The change in momentum in the final direction is equal, , (21), , (22), , R, , 2, , F=ma, , a=3 m/s, , mg, Now, R + F= mg, or R = mg – F = mg – ma = m (g – a) = 60 (9.8 – 3) = 408 N, (2) When the lift is moving down with constant velocity, a = 0 and hence, R = mg = 60 × 9.8 = 588 N, (3) The lift is now moving down with a retardation of, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 09, , (23), , 27, , 3 m/sec2., The retardation is 3 m/sec2 in the downward direction is, equivalent to an acceleration of 3 m/sec2 upwards., Hence the direction of fictitious force is downwards., Now, R = mg + ma = m (g + a) = 60 (12.8) = 768 N, (b) When the lift is moving up m (g + a) = force, , m(g + a), The scale reading =, =, g, , gö, æ, 10 ç g + ÷, è, 3ø, = 13.3 kg, g, , When lift is moving down the scale reading, , m(g - a), =, =, g, (24), , (25), , gö, æ, 10 ç g - ÷, è, 3ø, = 6.67 kg, g, , (a), (1) A reference frame in which Newton’s first law is valid is, called an inertial reference frame., (2) Frame moving at constant velocity relative to a known, inertial frame is also an inertial frame., (3) Idealy, no inertial frame exists in the universe for, practical purpose, a frame of reference may be considered, as Inertial if its acceleration is negligible with respect to, the acceleration of the object to be observed., (4) To measure the acceleration of a falling apple, earth can, be considered as an inertial frame., (a), (i) In the case of constant velocity of lift, there is no reaction,, therefore the apparent weight = actual weight. Hence the, reading of machine is 50 kg wt., (ii) In this case the acceleration is upward the reaction R =, ma acts downward, therefore apparent weight is more than, actual weight ., , i.e. W' = W + R = m (g + a), Hence, scale show a reading of, 50g ö, æ, m (g + a) Newton = çè 50 +, kg wt, a ÷ø, , (26), , (27), , (a) Tension = m (g + a), when lift moving up, putting the, values, we get, 175 = 25 (9.8 + a) Þ a = 2.8 m/s2, [negative sign shows that lift is moving downward], (b) Apparent tension, T = 2T0, , æ a0 ö, So, T = 2T0 = T0 ç 1 + ÷, gø, è, a, or 2 = 1 + 0 Þ a0 = g = 9.8 m/s2, g, (28) (b) Cloth can be pulled out without dislodging the dishes, from the table because of inertia. Therefore,, statement- 1 is true., This is Newton's third law and hence true. But, statement 2 is not a correct explanation of statement 1., (29) (d) According to Newton’s second law, Force, i.e. if net external force on the, Mass, body is zero then acceleration will be zero., , Acceleration =, , (30) (c), , F=, , dp, = Slope of momentum -time graph, dt, , i.e. Rate of change of meomentum = Slope of, momentum-time graph = force.
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t.me/Magazines4all, DPP/ P 10, , 28, , (1), , (c) Force on the block, = Mass of the block × acceleration of the system, , (5), , (a) We have acceleration, , Fcos q 50 3, =, = 5 3 m /sec2, m, 10, The velocity after 2 sec, v = u + at, , a=, , P, =M×, M+m, , 1, = 4 kg, 2, Total mass = 50 + 4 = 54 kg, , (2), , (b) Mass of the rope = 8 ×, , (3), , F 108, =, = 2 m/s2, m 54, Force utilised in pulling the rope =4×2 = 8 N, Force applied on mass = 108 – 8 = 100 N, (b) Mass of the rope = 15 × 2 = 30 kg, , \, , 10, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , (6), , Þ v = 0 + 5 3 × 2 = 10 3 m/sec, (a) All the forces acting on the two blocks are shown in, fig. As the blocks are rigid under the action of a force F,, both will move together with same acceleration., R1, , a=, , R1, F, , F 25 5, =, = m/s2, m 30 6, At the point 7 m away from point of application the, mass of first part of rope = 14 kg, , acceleration =, , 5, = 11.67 N, 6, The remaining force = (25 –11.67)N = 13.33 N, (b) The various forces acting are shown in fig., The force of 100N has, , \, , (4), , (7), , (8), , 500 3 =, , 1, × 10 × v2 Þ v2 = 100 3 Þ v = 13.17 m/sec, 2, , 5 ´ 15, a1a 2, =, = 3.75 m/s2, 5 + 15, a1 + a 2, (a) As net force on the rod = F1 – F2 and its mass is M so, acceleration of the rod will be, a = (F1 – F2)/M, ...(i), Now considering the motion of part AB of the rod,, which has mass (M/L)y,, Acceleration a given by, (i) Assuming that tension at B is T, , F1 – T =, , W=10×10N, , 50 3 × 10 = 500 3 Joule., If v is the speed acquired by the block, the work done, must be equal to the kinetic energy of the block., Therefore, we have, , MF, 2´3, =, = 2N, M + m 2 +1, (a) As the same force is applied to the combined, mass, we have, , 1 1, 1, = +, a a1 a 2, , 50 3N, , Since the block is always in contact with the table, the, net vertical force, R = mg + F sin q = (10 × 10 + 50) N = 150N, When the block moves along the table, work is done, by the horizontal component of the force. Since the, distance moves is 10 m, the work done is, , Mg, , f = Ma =, , 30°, , 50N, , A, , a = F/(m+M) = 3/(1+2) = 1 m/s2, Now as the mass of larger block is m and its acceleration, a so force of contact i.e. action on it., , R, 100, N, , R, m, mg, , Force used in pulling 14 kg = 14 ×, , (i) horizontal component of 100 cos 30º = 50 3 N, and (ii) A vertical component = 100 sin 30º = 50N, , M, , Þ F1 – T =, , (9), , or, , a=, , M, y×a, L, , (from F = ma), , M F1 - F2, y, L, M, , (using eq. (1)), , yö, æ, æ yö, Þ T = F1 ç 1 - ÷ + F2 ç ÷, è Lø, è Lø, (b) The net acceleration of the system is given by, 1 1, 1, =, +, a a1 a 2, 1 1, 1, 1, =, +, + ......, = 1 + 2 + 3 + ......... + n, a a1 a 2, an, , =, , n, n, 2, [2 + (n – 1) 1] = [n + 1] =, 2, 2, n(n + 1), , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 10, , 29, , (10) (a) As the mass of the system is 6 + 4 + 2 = 12 kg and, applied force is 60 N, the acceleration of the system, F 60, =, = 5 m/s2, m 12, Now at point A as tension in pulling the rope of mass, 2kg and block Q of mass 4kg., TA = (2 + 4) × 5 = 30N, Similarly for B and C,, TB= (1 + 4) × 5 = 25N, and TC = (0 + 4) × 5 = 20N, (11) (a) In case (a), the pulling force = 2mg – mg = mg, and the mass is 2m + m = 3m, so acceleration a = mg/3m = g/3, While in case (b), the pulling force = 2mg – mg = mg, but, the mass in motion = m + 0 = m, Acceleration, a = mg/m = g, (12) (c) It this problem as the pulling force is 2mg while, opposing force is mg, so net force, F= 2mg – mg = mg,, and as the mass in motion = m + m + m = 3m, a=, , (15) (b) The string is massless and inextensible the tension T is, same. Let mass B move down the inclined plane., For B the equation of motion m1g sin q – T = m1a, 30 × 9.8 × sin 53º – T = 30a, Þ 235.2 – T = 30 a, ...(1), and for A the equation of motion, T – 20 × 9.8 × sin 37º = 20a, T – 117.6 = 20a, ...(2), From (1) & (2) T = 164.64 N, (16) (c), , T, , 4, mg, 3, Now to calculate tension in the string BC we consider, the downward motion of C,, i.e. T2 = m (g – a) = m (g – g/3) = (2/3) mg, (13) (a) As pulley Q is not fixed so if it moves a distance d the, length of string between P and Q will changes by, 2d (d from above and d from below) i.e. M will move 2d., This in turn implies that if a (®2d) is the acceleration of, M, the acceleration of Q and so 2M will be of (a/2), Now if we consider the motion of mass M, it is, accelerated down so T = M(g – a), ...(1), And for the motion of Q,, 2T – T' = 0 × (a/2) = 0 Þ T' = 2T, ...(2), And for the motion of mass 2M,, T' = 2M (a/2) Þ T' = Ma, ...(3), , From equation (2) and (3) T =, , 1, Ma, so eq. (1) reduces, 2, , æ 1ö, 2, çè 2 ÷ø Ma = M (g – a) Þ a = g, 3, (14) (a) The tension is same in two segments, For B the equation is, (40 × 9.8 – T) = 40a, ...(1), For C the equation is, , 1, ) = 50a, ...(2), 2, From equation (1) and (2) a = 1.63 m/s2, distance of fall, , (T – 50 × 9.8 ×, , S=, , 1 2 1, at = × 1.63 × 42 = 13.04 m, 2, 2, , a, mg, , (Force diagram in the frame of the car), Applying Newton’s law perpendicular to string, , force mg g, =, =, So the acceleration =, mass 3m 3, Now as A is accelerated up while B and C down. so, tension T1, is such that mg, < T1 < 2mg, Actually for the motion of A,, , T1 = m (g + a) = m(g + g/3) =, , a, , ma m, , mg sin q = ma cos q, , a, g, Applying Newton’s law along string, tan q =, , Þ, , T - m g 2 + a 2 = ma, , or, , T = m g 2 + a 2 + ma, , (17) (a) As A moves up and B moves down with acceleration a, for the motion of A ,, T – 11 g = 11 a ... (i), for the motion of B,, 11.5 g – T = 11.5 a ...(ii), From (i) & (ii) ,, m1 - m 2, a= m +m, 1, 2, , (11.5 - 11)9.8, = 0.218 m/sec2, 11.5 + 11, Assuming that the particles are initially at rest, their, velocity at the end of 4 sec will be, v = u + at = 0 + 0.218 × 4 = 0.872 m/s, (18) (a) The height ascended by A in 4 sec, , g=, , 1 2, 1, at = 0 + (0.218) 42 = 1.744 m, 2, 2, This is also the height descended by B in that time., (19) (c) At the end of 4 sec the string is cut. Now A and B are no, longer connected bodies but become free ones, falling, under gravity., Velocity of A, when the string was cut, = 0.872 m/s upwards., Acceleration a = – g (acting downwards),, displacement from this position in the subsequent 2 sec, , h = ut +
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t.me/Magazines4all, DPP/ P 10, , 30, 1 2, 1, at = (0.872) × 2 + (–9.8) 22, 2, 2, = 1.744 – 4.9 × 4 = –17.856m, A descends down by a distance of 17.856 m from the, position it occupied at the end of 4 sec from its start. B, has a free fall. Its position is given by, So the acceleration of mass M is (2/3)g, while tension in the string PQ will be, T = M(g – (2/3)g) = (1/3)Mg, The force exerted by clamp on the pulley, , h = ut +, , 2, Mg, = T +T =, 3, (20) (a) Here the system behaves as a rigid system, therefore, every part of the system will move with same acceleration. Thus applying newton’s law, mg – T = ma, ......... (i), 2T – mg = ma, ......... (ii), Doubling the first equation and adding, 2, , gö, æ, T = m (g – a) = m çè g - ÷ø, 3, 2, mg, 3, , (22) (a), (1) Inertia µ mass, (2) 1 Newton = 105 dyne, , 3R, , N, F, , 60° mg, , 30°, , same force will be, , 1 1, 1, 1, = +, + ......, a a1 a 2, an, , (2) Newton's Ist and IIIrd law can be derived from second, law therefore IInd law is the most fundamental law out, of the three law., (24) (a), (1) For equilibrium of a body under the action of concurrent, ®, , ®, , Extension in the spring is, \, , x = AB – R = 2R cos 30° – R = ( 3 - 1) R, Spring force, , ( 3 - 1) mg, ( 3 - 1) R = 2mg, R, FBD of bead is, F = kx =, , 3 3 3, mg, =, 2, 2, Tangential force F1 = F sin 30° – mg sin 30°, , N = F (mg cos 30°) = (2mg + mg), , = (2mg – mg) sin 30° =, , r, DM r, r, v - Mg, (3) Thrust on rocket F =, Dt, (4) Apparent weight of a body in the lift accelerated up is, W = m (g + a)., (23) (b), (1) If a1, a2, ... an be the accelerations produced in n, different bodies on applying the same force, the, acceleration produced in their combination due to the, , ®, , Initial elongation = 2R cos 30° =, , 2, , 1, mg = 3ma or acceleration a = g, 3, (21) (c) Tension in the string, , T=, , (2) If the downward acceleration of the lift is a = g, then the, body will enjoy weightlessness., (3) If the downward acceleration of the body is a > g, then, the body will rise up to the ceiling of lift, (25) (d), (26) (d), (27) (a)., , ®, , forces F1 + F2 + F3 + .....Fn = 0, , mg, 2, , \ tangential acceleration = g/2, 28. (d) Here the acceleration of both will be same, but their, masses are different. Hence, the net force acting on, each of them will not be same., 29. (c) The FBD of block A in Figure is, N, mg, The force exerted by B on A is N (normal reaction). The, force acting on A are N (horizontal) and mg (weight, downwards). Hence statement I is false., 30. (d) T – m1g = m1a – .... (1), m2g – T = m2a – .... (2), æ m 2 - m1 ö, æ 2m1m2 ö, Solving (1) and (2), T1 = çè m + m ÷ø g a = çè m + m ÷ø g, 1, 2, 1, 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 11, , 31, , DAILY PRACTICE, PROBLEMS, (1), , (a) Let the contact force on the block by the surface be F, which makes an angle q with the vertical. The component, of F perpendicular to the contact surface is the normal, force N and the component F parallel to the surface is the, friction f. As the surface is horizontal, N is vertically upward., For vertical equilibrium, , minimum possible force F that can be applied, the friction, is limiting and hence, f = mN, where N is normal force., in the vertical direction, there is no acceleration, \ N = mg, in the horizontal direction,, , N, , F, , 11, , PHYSICS, SOLUTIONS, , let the acceleration be a, then, mN = ma, , q, , mmg = ma, a = mg, f, , N = Mg = (0.400) (10) = 4.0 N, The frictional force is f = 3.0 N, , (2), , f, 3, =, Þ q = tan–1 (3/4) = 37º, N, 4, (c) The magnitude of the contact force is, , (3), , 2, 2, N 2 + f 2 = (4) + (3) = 5.0 N, (c) The forces on the block are, , f = mN, , Next consider the motion of M, The equation of motion is, F = mN = Ma, F – mmg = Mmg, F = mg (M + m), , tan q =, , N, , F=, , f = mN, , f, , N, , Mg, , (5), q, , (4), , F, , mg, , (i) the weight mg downward by the earth, (ii) the normal contact force N by the incline, and, (iii) the friction f parallel to the incline up the plane, by the, incline., As the block is at rest, these forces should add up to zero., Also since q is the maximum angle to prevent slipping, this, is a case of limiting equilibrium and so, f = mSN, Taking component perpendicular to the lncline,, N – mg cos q = 0 Þ N = mg cos q ....... (1), Taking component parallel to the incline, f – mg sin q = 0 Þ f = mg sin q ........ (2), \ mSN = mg sin q, Dividing (2) by (1) ms = tanq, q = tan–1mS = tan–1 (0.3), (a) When the maximum force F is applied, both the blocks, move together towards right. The only horizontal force on, the upper block of mass m is that due to the friction by the, lower block of mass M. Hence this force on m should be, towards right. The force of friction on M by m should be, towards left by Newton's third law. As we are talking of the, , (b) When A moves with B the force opposing the motion is, the only force of friction between B and S the horizontal, and velocity of the system is constant, R2, A, R1, , F, , B, f1, , (6), , F = f1 = mR1 = 0.25 (4 + 8) = 3N, (d) When A is held stationary the friction opposing the, motion is between A and B and B and S. So, , R2 A, , F, , B, , f2, , R1, S, , f1, , F = mR1 + mR2 = 3 + 0.25 (4), F= 3+1= 4N
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t.me/Magazines4all, DPP/ P 11, , 32, (7), , (d) In this situation for dynamic equilibrium of B, , (10), , R2 A, f2, F, , f2, , R1, , L–y, R, , f1, , (8), , f1, , F = mR1 + mR2 + T, ..... (1), While for the uniform motion of A, T = mR2, ....... (2), Substituting T from Equation (2) in (1) we get, F = mR1 + 2mR2 = 3 + 2 x 1 = 5N, (a) Figure shows the forces acting on the two blocks. As, we are looking for the maximum value of M/m, the, equilibrium is limiting. Hence the frictional forces are equal, to m times corresponding normal force., Equilibrium of the block m gives, , W’, , y, , W, , W = fL, , T, , m, , mN1, , M, m, =, m sin q - m cos q, (a) The situation is shown in figure in the limiting, equilibriums the frictional force f will be equal to mN., , N, , M, M, yg = m, (L – y) g, L, L, , (11), , (d) The insect will crawl up the bowl till the component of, its weight along the bowl is balanced by limiting friction, so, resolving weight perpendicular to the bowl and along, the bowl we get, R, , y, fL, , q, , R = mg sin, , mg sinq mg cos, q, , F, , fS = m N, , q, fL = mg cos q, , Mg, For horizontal equilibrium, F sin q = mN, For vertical equilibrium, F cos q + mg = N, Eliminating N from these equations, F sin q = mF cos q + mmg, F=, , M, (L – y)g, L, , mM, (L – y) g, L, Substituting these values of W and fL in equation (1) we, get, , q, , T = mN1 and N1 = mg Þ T = mmg, .... (1), Next consider the equilibrium of the block M. Taking, components parallel to the incline, T + mN2 = Mg sin q, Taking components normal to the Incline, N2 = Mg cos q, These give T = Mg (sin q – m cos q), ...... (2), From (1) and (2) mmg = Mg (sin q – m cos q), , q, , M, y g and, L, , So that fL = mR =, , M, , mg, , F, , R = W' =, , N2, , mg, , (9), , ....... (1), , But from figure W =, , N1, mN1, , (c) If y is the maximum length of chain which can be hang, out side the table without sliding, then for equilibrium of, the chain, the weight of hanging part must be balanced by, force of friction from the portion on the table, , m, (sin q - m cos q), , tanq =, , R, R, 1, = 1 =, ;, f L mR1 m, , y, 2, , R -y, , 2, , =, , 1, m, , R1, , m2y2 = R2 – y2 ;, , y=, , So, h = R – y = R –, , é, ù, 1, ú, = R ê1 2, ê, ú, m2 + 1, (, m, +, 1, ), ë, û, R, , m2 + 1, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., DPP/ P 11, (12), , 33, (18), , (a), , m, L, , h, , Loss in P.E. in reaching the bottom = mgh and gain in K.E., 1, mv2, 2, where v is velocity gained by the body in reaching the, bottom, , reaching the bottom =, , 1, mv2, 2, work done against friction = FL, , \ Net loss in energy = mgh –, , \ mgh –, (13), , (14), , 1, mv2 = FL ; v =, 2, , 2, (mgh - FL), m, , (a) Let R be the normal reaction on the block exerted by the, floor. The limiting (maximum) force of static friction is, fs = msR = msmg, = 0.4 × 2kg × 9.8 ms–2 = 7.84 N, The applied force F is 2.5 N, that is less than the limiting, frictional force. Hence under the force F, the block does, not move. So long the block does not move, the (adjustable), frictional force is always equal to the applied force. Thus, the frictional force is 2.5 N., (b) When the block does not slip on the table surface, it, performs simple harmonic motion along with the table., x = a sin wt, The instantaneous acceleration of the block is, , d2 x, dt 2, , = –w2a sin wt, , The maximum acceleration is, , d2 x, dt 2 max, , = w2a, , The maximum force on the block is fmax = mw2a, where m is its mass. The frictional force on the block is, mmg. since the block is at rest with respect to the table, we, have mw2a = mmg, (2pf)2a = mg, Þa=, (15), (16), , (17), , mg, 4p 2f 2, , t.me/Magazines4all, , =, , 0.72 ´ 10, 4 ´ (3.14) 2 ´ 32, , = 0.02 m, , (c) Stopping distance is independent on mass., (a) (i) coefficient of static friction is always greater than the, coefficient of kinetic friction, (ii) limiting friction is always greater than the kinetic friction, (iii) limiting friction is never less than static friction, (d) The system can not remain in equilibrium, , (a) (i) In the force applied v/s friction graph : The graph is, a straight line of slope 45º for small F and a straight line, parallel to the F-axis for large F., (ii) There is small kink on the graph, (19) (a) (i) force of friction between two bodies may be equal to, zero, (ii) bodies may be rough, (20) (b) It is easier to pull a body than to push, because the, friction force is more in pushing than that in pulling, (21) (a)ma = µmg, a = µg, (22) (a), (1) Kinetic friction is lesser than limiting friction., (2) In rolling the surfaces at contact do not rub each other., (3) If a body is at rest and no pulling force is acting on it,, force of friction on it is zero., (23) (a), (1) Force of friction is partically independent of microscopic, area of surface in contact and relative velocity between, them. (if it is not high), (2) Normally with increase in smoothness friction, decreases. But if the surface area are made too smooth by, polishing and cleaning the bonding force of adhesion will, increase and so the friction will increase resulting in 'Cold, welding', (3) Friction is a non conservative force, i.e. work done, against friction is path dependent., (24) (c), (2) Friction may opposes the motion, (4) If the applied force is increased the force of static friction, also increases upto limiting friction., (25) (a), (26) (a), (27) (a)., Fmax = kx + µ mg, Fmin = kx – µ mg, \ Fmax + Fmin = 2µ mg, or 2 = 2 µ 10, \ µ = 0.1, Fmax + Fmin = 2kx, ......... (1), From graph, Fmax + Fmin = 5 and x = 0.1, Putting in eq. (1), t = 2k (0.1) ; k = 25 N/m, When x = 0.03, kx = 25 × 0.03 = 0.75 N, which is less than µ mg, = 0.1 × 10 = 1N, \ The block will be at rest, without applying force F., (28) (b) It is easier to pull a heavy object than to push it on a, level ground. Statement-1 is true. This is because the, normal reaction in the case of pulling is less as compared, by pushing. (f = m N). Therefore the functional force is, small in case of pulling., Statement-2 is true but is not the correct explanation of, statement-1., (29) (c) W = (force) × (displacement of point of application), (30) (d) Statement – 2 is false because friction force may be, more than applied force when body is retarding and, external force is acting on body.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 12, , 35, mg, l, mgl, ×, =, 5, 10, 10, \ Work done = U = mg l/50, (a) At maximum speed all the power is used to overcome, the resistance to motion. Hence if the maximum speed is v,, then 50000 = 1000 × v or v = 50 m/s, At 25 m/s, let the pull of the engine be P, then the power, , U=, , (12), 1, × 5 × (10)2, 2, or 50 – F = 12.5, or F = 50 – 12.5, \ F = 37.5 N, Work done by the force = – 37.5 × 20 = – 750 joule, (The negative sign is used because the push of the air is, upwards while the displacement is downwards.), (a), , or (50 – F) 20 =, , (10), , 50, 000, = 2000 N, 25, Now resultant force = 2000 – 1000 = 1000 N, Applying Newton's law ; F = ma, we have, 1000 = 1000 a or a = 1.0 m/s2, (a) 1 mole i.e.235 gm of uranium contains 6 × 1023 atoms, so, 2 kg i.e. 2 × 103 gm of uranium will contain, , or P =, , (13), , R, F sin 45°, , 2 ´ 103 ´ 6 ´ 1023, atoms = 5.106 × 1024 atoms, 235, Now as in each fission only one uranium atom is consumed i.e. Energy yield per uranium atom, = 185 MeV = 185 × 1.6 × 10 –13 J = 2.96 × 10–11 J, So Energy produced by 2 kg uranium, = (No. of atoms ) × (energy /atom), = 5.106 × 1024 × 2.96 × 10–11 = 1.514 × 10–14 J, As 2 kg uranium is consumed in 30 days i.e. 1.51 × 10–14 J, of energy is produced in the reactor in 30 days i.e., 2.592 × 106 sec, So, power output of reactor, , =, , F cos 45°, , mR, , mg, , The different forces acting on the block are shown in fig., Now we have, R + F sin 45° = m g, ............(1), F cos 45° = m R, ............(2), From equation (1) and (2), \ F=, , =, , m mg, cos 45º +m sin 45º, , (14), , Substituting the given values, we have, , (c) When the vehicle of mass m is moving with velocity v,, 1, mv2 and if S is the, 2, stopping distance, work done by the friction, W = FS cos q = m MgS cos 180º = – m MgS, So by Work-Energy theorem,, , the kinetic energy of the where K =, , 0.20 ´ (5 ´ 9.78), = 11.55 N, (0.707) + (0.20 ´ 0.707), The block is pulled through a horizontal distance, r = 20 metre, Hence, the work done, W = F cos 45° × r = (11. 55 × 0.707) × 20 = 163. 32 Joule, (c), F=, , (11), , W = D K = Kf – ki, v2, 1, 2, Þ – m MgS = 0 –, Mv Þ S =, 2mg, 2, , (15), , l/5, , (a) As T = (2p/w),, so w = 2p/(3.15 × 107) = 1.99 × 10–7 rad/s, Now v = rw = 1.5 × 1011 × 1.99 × 10–7 » 3 × 104 m/s, Now by work - energy theorem ,, W = Kf – Ki = 0 –, , Mass of the hanging part of the chain = (m/5) The weight, mg/5 acts at the centre of gravity of the hanging chain, i.e.,, at a distance = l/10 below the surface of a table., The gain in potential energy in pulling the hanging part on, the table., , 1.514 ´ 1014 J, E, =, = 58.4 MW, t, 2.592 ´ 106 S, , 1, mv2, 2, , 1, × 6 × 1024 (3 × 104)2 = – 2.7 × 1033 J, 2, Negative sign means force is opposite to the motion., (b) As the particle is moving in a circle, so, , =–, , (16), , mv2, k, 1, k, = 2 Now K.E =, mv2 =, r, 2, 2r, r
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t.me/Magazines4all, DPP/ P 12, , 36, Now as, , F=–, , (22), (23), , dU, dr, r, , r, , Þ P.E, U = – ò Fdr =, ¥, , æ kö, , ò + çè r 2 ÷ø dr = –, , ¥, , k, r, , k k, k, +, =r 2r, 2r, Negative energy means that particle is in bound state ., (c) Let the mass of the person is m, Work done, W = P.E at height h above the earth surface, = (M + m) gh, or 4900 = (M + 10) 9.8 × 10 or M = 40 kg, (b) As the rod is kept in vertical position the shift in the, centre of gravity is equal to the half the length = l/2, , So total energy = U + K.E = –, , (17), , (18), , l, 4, = 20 × 9.8 × = 392 J, 2, 2, (a) We know that the increase in the potential energy, , Work done W = mgh = mg, , (19), , é1 1 ù, DU = GmM ê - ú, ë R R 'û, According to question R' = R + R = 2R, 1 ù GMm, é1, DU = GMm ê ú =, 2R, ë R 2R û, , (20), , 1, (c) In first case, W1 =, m(v1)2 + mgh, 2, 1, m(12)2 + m × 10 × 12, 2, = 72 m + 120 m = 192 m, and in second case,W2 = mgh = 120 m, The percentage of energy saved, , =, , =, (21), , (c) Given that, U (x) =, , We know F = –, or, , 6b, x, , 7, , =, , 192m - 120m, × 100 = 38%, 192m, , a, 12, , x, , -, , b, x6, , x=, , 12a, x13, 1/6, , mg, = 0.2m, 5k, , \ Extension of vertical spring = 0.2m, Extension of horizontal spring = 2x = 0.4m, From conservation of energy, mgx =, , 1 2 1, 1, 1, kx + k(2x)2 + mv2 + m(2v)2, 2, 2, 2, 2, , mgx =, , 3 2 3, kx + mv 2, 2, 2, , 7, 3, mgx = mv 2, 10, 2, v=, , du, = (–12) a x–13 – (– 6b) x–7 = 0, dx, , æ 2a ö, or x6 = 12a/6b = 2a/b or x = ç ÷, è bø, , (a) W = 0, (b), (1) There will be an increase in potential energy of the, system if work is done upon the system by a conservative, force., (2) The work done by the external forces on a system equals, the change in total energy, (24) (a), (1) The work done by all forces equal to change in kinetic, energy, (2) The work done by conservative forces equal to change, in potential energy, (3) The work done by external and nonconservative forces, equal to change in total energy, (25) (b), (26) (b), 27. (c), For vertical block, mg = kx + 2T, ....... (1), For horizontal block, T = k (2x), ....... (2), From eq. (1) and eq. (2), , 7, gx, 15, , Required speed = 2v = 1.9 m/s, (28) (d) Statement – 1 is true but statement – 2 is false., (29) (a) Work done by action reaction force may be zero only if, displacement of both bodies are same., (30) (b) Both statements are true and independent., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 13, , 37, , (1), , (a) Let m1 and m2 be the masses of bullet and the rifleman, and v1 and v2 their respective velocities after the first shot., Initially the rifleman and bullet are at rest, therefore initial, momentum of system = 0., As external force is zero, momentum of system is constant, i.e. initial momentum = final momentum, = m1v1 + m2v2, , (10 ´ 10-3 kg)(800m / s), m1 v1, =–, = – 0.08 m/s, 100 kg, m2, Velocity acquired after 10 shots, = 10 v2 = 10 × (–0.08) = – 0.8 m/s, i.e, the velocity of rifle man is 0.8 m/s in a direction opposite, to that of bullet., (c) Let the mass of block and bullet be M and m respectively, If v is the velocity of bullet and V is the velocity of block, with bullet embedded in it,, Now according to conservation of momentum ,, mv = (M + m) V, (10×10 –3)(300) = (290×10–3 + 10 ×10–3) V or V = 10 m/s, , m1. Applying conservation of linear momentum for the, collision of bullet with plate m2., i.e. mv1 = (m2 + m) v, 0.02 v1 = (2.98 + 0.02) v, 3, v = 150 v, .........(2), .02, Required percentage loss in initial velocity of bullet, , i.e. v1 =, , or v2 =, , (2), , (4), , 2, , 2, , 2, 1, 1 V, 1, 10 2, ×, =, =, 3, 2 gd, 2, (10)(15), (a) Let the in itial velocity of the bullet of mass, m = 20 g = 0.020 kg be u and v the velocity with which each, plate moves., The initial momentum of system (bullet + plate) = mu, , m=, , cos2 30o, æ 20 ´ 10-3 ö, ÷ .(200)2 ., = 0.15 m, = çç, ÷, (2) (10), 2, ø, è, (5), , (c) Initial velocity of bullet, u1 = 500 m/s, Let v1 and v2 be the speeds of bullet and block after collision, , m, , 0.1m, m4, , m2, , Final momentum of system = m1v + (m2 + m) v, (Since bullet remains in 2nd plate), \ According to principle of conservation of momentum, i.e. mu = m1v + (m2 + m) v,, i.e. 0.02u = 4v, 4, = 200 v, ..........(1), .02, Let v1 be the velocity of the bullet as it comes out of plate, , or, , u=, , ..........(2), , æ m ö u 2 cos 2 a, ÷ ., h= ç, èM+mø, 2g, , 1, (M + m) V2 = m (M + m) gd, 2, , (3), , u - v1, 200v - 150v, × 100% =, × 100 = 25%, u, 200 v, (a) Part (I) - The horizontal component of the momentum of, the bullet is equal to the momentum of the block with the, bullet, mu cos a = (M + m) V, ..........(1), Where V is the velocity of the block plus bullet embedded, in it., Part (II) - As the block can move as a pendulum, the block, rises till its kinetic energy is converted into potential energy., So, if the block rises upto a height h,, , 1, (M + m) V2 = (M + m) gh, 2, From (1) & (2), , 1, (M + m) V2, which, 2, is lost due to work done on it by the force of friction F., Since force of friction F = m (M+m)g and the work done is, given by Fd, we have, , The kinetic energy just after impact is, , or, , 13, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , u1, m1 m, 2, , respectively then,, Þ v2 =, , v1, v2, , 1, m v 22 = mgh, 2, , 2gh = 2 ´ 9.8 ´ 0.1 = 1.4 m/s, , According to principle of conservation of linear momentum,, We have, m1u1 + 0 = m1v1 + m2 v2, or 0.01 × 500 = 0.01 v1 + 2 × 1.4 Þ v1 = 220 m/s
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t.me/Magazines4all, DPP/ P 13, , 38, (6), , \ (v1 – v2) = 0, Solving equations (3) and (4), we have, , (a) The rate of change of momentum is equal to force, F=, , dp, dm, =v, (Here v is constant), dt, dt, , ......(4), , v1 = 3 3 m/s and v2 = 3 3 m/s, , dm, = 50 × 10–3 kg/s, dt, \ F = 4 × 103 × 50 × 10 –3 = 200 N, , Here v = 4 × 103 m/s &, , (7), , v1, , (a) Given that, Initial velocity = u, , 30°, 30°, , u, 4, So by conservation of momentum, we have, , Final velocity =, , u, 3u, + m × v2 Þ mv2 =, 4, 4, and by conservation of energy, we have, , 1×u+0=1×, , v2, ....... (1), , According to law of conservation of energy, Energy before collision = Energy after collision, 1, 1, 1, 1, m u12 + m u 22 = m v12 + m v 22, 2, 2, 2, 2, , 2, , 1, 1, 1, æ uö, × 1 × u2 + 0 = × 1 ç ÷ + m v 22, è 4ø, 2, 2, 2, , 15 2, u, or, =, 16, From equation (1) and (3),, , v 22, , (mv2 )2, mv 22, (8), , =, , (9 / 16)u 2, (15 / 16)u 2, , 1, 1, 1, m (9)2 + 0 = m (3 3 )2 + m (3 3 )2, 2, 2, 2, , ....... (2), , or m = 0.6 kg, , (a) Initial momentum of the balls, =m×9+m ×0= 9m, ........(1), where m is the mass of each ball., Let after collision their velocities are v1 and v2 respectively., Final momentum of the balls after collision along the same, line = mv1 cos 30° + mv2 cos 30°, , (9), , 81 m, 54 m, =, 2, 2, L.H.S. # R.H.S., i.e., energy is not conserved in this collision or this is a, case of inelastic collision., (a) The situation is shown in fig., Let v1 and v2 be the velocities of two pieces after explosion., Applying the law of conservation of energy, we have, , m1 = 4kg, m = 8kg, , q, 30°, , mv1 3, mv2 3, +, ........(2), 2, 2, According to law of conservation of momentum, =, , 9m=, 9´ 2, 3, , u = 50m/s, , v2, , mv1 3, mv2 3, +, 2, 2, = v1 + v2, , v1, , 1, 1, 1, (8) (50)2 + 15000 = (D) v12 + (D) v 22, 2, 2, 2, , .......(3), , ......(1), or 25000 = 2 ( v12 + v22 ), Applying the law of conservation of momentum along, x-axis and y-axis respectively, we get, 8 (50) = 4 v1 cos q + v2 cos 30°, ......(2), and 0 = 4 v1 sin q, = 4 v2 sin 30° = 2 v2, ......(3), , Stationary ball, (a) Before collision, The initial momentum of the balls along perpendicular, direction = 0 ., Final momentum of balls along the perpendicular direction, m, (v – v ), 2 1 2, Again by the law of conservation of momentum, (m/2) (v1 – v2) = 0, , = mv1 sin 30° – mv2 sin 30° =, , or sin q =, , (10), , v2, 2v1, , ......(4), , From eq. (2), 100 = v1 cos q + v2 cos 30°, (a) Let m be the mass of the rocket and vr the relative, velocity of the gas ejecting from the rocket. Suppose the, fuel is burnt at a rate (dm/dt) to provide the rocket an, acceleration a., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 13, , 40, (18), , y, (M+m)v¢, , (19), (20), , A, , q, , mv1, , x, , (21), (22), , Mv2, , (23), , The final momentum, P = (M + m) v' = [(mv1 ) 2 + (Mv 2 ) 2 ], , (24), (25), , [form eqn. (3)], Dividing eqn. (2) by eqn. (1), we have, tan q =, (16), , (17), , Mv2, or, mv1, , æ Mv2 ö, q = tan–1 ç mv ÷, è, 1ø, , (a) Let the angle of reflection be q' and the magnitude of, velocity after collision be v'. As there is no force parallel to, the wall, the component of velocity parallel to the surface, remains unchanged., Therefore, v' sin q ' = v sin q, ......(1), As the coefficient of restitution is e, for perpendicular, component of velocity, Velocity of separation = e x velocity of approach, –(v' cos q ' – 0) = –e (v cos q – 0), ......(2), From (1) and (2), v' = v sin 2 q + e2 cos 2 q, and tan q ' = tan q/e, (a) The fraction of energy lost is given by,, , mg(h - h '), DE, h - h', =, =, mgh, E, h, given that, h = 2 meter and h' = 1.5 meter, \, , DE, 2 - 1.5 1, =, =, E, 2, 4, , (a) A bullet is fired from the gun. The gun recoils, the kinetic, energy of the recoil shall be less than the kinetic energy of, the bullet., (a) Conservation of linear momentum is equivalent to, Newton's second law of motion, (a) In an inelastic collision momentum is conserved but, kinetic energy is not., (a) Inelastic collision is the collision of electron and, positron to an inhilate each other., (a) Total kinetic energy is not conserved in inelastic, collisions but momentum is conserved, (a) (1) when m1 = m2 and m2 is stationary, there is maximum, transfer of kinetic energy in head an collision, (2) when m1 = m2 and m2 is stationary, there is maximum, transfer of momentum in head on collision, (3) when m1 >> m2 and m2 is stationary, after head on, collision m2 moves with twice the velocity of m1., (a) Momentum remains conserved, (a) Speed of particle after the collision, 2, , æ 15, ö, = ç ´ 3 ÷ + 25 = 5.036 m /s, è 43, ø, , (26), (27), (28), (29), (30), , 30, m/s, 43, (a) Angular speed of sphere is zero as impulse due to, collision passes through centre of sphere., (c) When e = 0, velocity of separation along common normal, zero, but there may be relative velocity along common, tangent., (c) Statement – 1 is false but statement – 2 is true., (d) Momentum remains constant before, during and after, the collision but KE does not remain constant during, the collision as the energy gets converted into elastic, potential energy due to deformation., , (b) Speed of the sphere just after collision =, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 14, , 42, 11., , (a) For translatory motion the force should be applied on, the centre of mass of the body, so we have to calculate the, location of centre of mass of 'T' shaped object., Let mass of rod AB is m so the mass of rod CD will be 2m., Let y1 is the centre of mass of rod AB and y2 is the centre of, mass of rod CD. We can consider that whole mass of the, rod is placed at their respective centre of mass i.e., mass m, is placed at y1 and mass 2 m is placed at y2., , 0+2´, =, , 80, 80, + 4´, +0, 30, 2, 2, =, 16, 2, , Similarly y =, , 14., , 30, , 2, 2, so, r = x + y = 30 cm, , 2, (b) Linear density of the rod varies with distance, dm, = l (Given ) \ dm = ldx, dx, , y, D, , A, , y1, , l, , B, dx, , y2, l, x, , x, C, Taking point 'C' at the origin, position vector of point y1, ur, ur, and y2 can be written as r1 = 2l ˆj, r2 = l ˆj , and m1 = m and, m2 = 2m, Position vector of centre of mass of the system, r, r, m r + m2 r2 m2l ˆj + 2ml ˆj, r, =, rcm = 1 1, m1 + m2, m + 2m, , Position of centre of mass, xcm =, , 3, , 4ml ˆj 4 ˆ, = lj, =, 3m, 3, , 12., , 13., , ò (l dx) ´ x, , =, , 3, 0, , 3, , 3, , ò (2 + x) ´ xdx, , =0, , (b) According to figure let A is the origin and co-ordinates, of centre of mass be (x, y) then,, , 3, , ò (2 + x)dx, 0, , y, 2 kg, , 4 kg, , C, , r, , 8 kg, , A, , 15., 2 kg, , B, , x, , m1 x1 + m 2 x 2 + m3 x 3 + m 4 x 4, m1 + m 2 + m3 + m 4, , =, , é 2 x3 ù, êx + ú, 3 úû, ëê, 0, 3, , é, x3 ù, ê2x + ú, 2 úû, êë, 0, , 9 + 9 36 12, =, = m., 9 21 7, 6+, 2, (c) Centre of mass lies always on the line that joins the, two particles., For the combination cd and ab this line does not pass, through the origin., For combination bd, initially it pass through the origin, but later on it moves toward negative x-axis., But for combination ac it will always pass through, origin. So we can say that centre of mass of this, combination will remain at origin., , =, , (x,y), , x=, , 0, , ò l dx, , 4, Hence the distance of centre of mass from C = l, 3, (a) Initial acceleration is zero of the system. So it will, always remain zero because there is no external force on, the system., , D, , ò dm ´ x, ò dm, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 15, , 45, , 1, ´ (15) ´ (10)2, 2, = 750 rad, 18. (d) As the block remains stationary therefore, For translatory equilibrium, = 0+, , å Fx = 0 \ F = N, , 22. (a) As mechanical contact is not made, total angular, momentum remains constant., \ Iw0= constant, Differentiating both sides,, D (Iw0) = 0, Þ I Dw0 + w0 DI = 0, Þ, , and å Fy = 0 \ f = mg, , Dw 0, DI, Dw DI, =+, =0 Þ, w0, I, w, I, Dw 0, DI, =w0, I, , Also,, , f, , =-, , 2 DR æ DI 2DR ö, çQ =, ÷, R è I, R ø, , 1, L2, = K (given) \ K µ (If L = constant), I, 2I, When child stretches his arms the moment of inertia of, system get doubled so kinetic energy will becomes half i.e., K/2., 24. (c). Angular impulse = change in angular momentum : Frt =, L Þ L1 < L2, , 23. (a), F, , O, N, , mg, , = -2 a DT, , E=, , L2, Þ K1 = K2, 2I, 25. (b); 26. (a); 27. (c), Drawing the F.B. D of the plank and the cylinder., , K=, , For rotational equilibrium, , åt = 0, , By taking the torque of different forces about point 0, uuur uur uuur uuuur, t F + t f + t N + t mg = 0, , F sin q, , N1, , As F and mg passing through point O, uur uuur, \ t f + tN = 0, As t f ¹ 0 \t N ¹ 0, and torque by friction and normal reaction will be in, opposite direction., 19. (c) The velocity of the top point of the wheel is twice that, of centre of mass and the speed of centre of mass is, same for both the wheels (Angular speeds are different)., 2p (n2 - n1 ), =, 20. (d) a =, t, , F cos q, f1, mg, , æ 4500 - 1200 ö, 2p ç, ÷ø, è, 60, rad/s2, 10, , f1, N1, Mg, , 3300, 2 p, 3 6 0 d e g ree, 60, =, ´, 10, 2 p, s2, , a = 1980 degree/s 2, , 1, 21. (b) q = w 0 t + a t 2, 2, Þ q = 100 rad, , 100, \ Number of revolution =, = 16 (approx.), 2p, , N2, Equations of motion are, F cos q – f1 = ma, F sinq + N1 = mg, f1+ f2 = MA, f1R – f2R = Ia, A = Ra, , f2, ....(1), ....(2), .....(3), .....(4), .....(5), , 1, 4 ´ 55 ´, 4 F cos q, 2, a=, =, = 10 m/s 2, 3M + 8m [ ( 3 ´ 1) + (8 ´ 1)]
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DPP/ P 15, , 46, 1, 3´ 1 ´ 55 ´, 3MF cos q, 2 = 7.5 N, f1 =, =, 3M + 8 m, 3 ´ 1+ 8 ´1, , 28., , t.me/Magazines4all, , 1, 1 ´ 55 ´, 2 = 2.5 N, and f = MF cos q =, 2, 3M + 8 m 3 ´ 1 + 8 ´ 1, ur, r dL, (b) t =, and L = I w, dt, , 29., 30., , (b) t = rF sin q. If q = 90° then t max = rF, Unit of torque is N-m., (d) Torque = Force × perpendicular distance of the line of, action of force from the axis of rotation (d)., Hence for a given applied force, torque or true tendency, of rotation will be high for large value of d. If distance, d is smaller, then greater force is required to cause the, same torque, hence it is harder to open or shut down, the door by applying a force near the hinge., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 16, , 48, (13) (a) M.I. of rod (1) about Z – axis I1 =, , Ml 2, 3, , Moment of inertia of loop about given axis =, , 3, MR 2, 2, , 2, , 3 æ Lö, 3rL3, = rL ç ÷ =, 2 è 2p ø, 8p 2, , 3, , (19) (b) M.I. of disc =, , 2, , 1, , æ, M, M ö, Therefore R 2 =, çè As r =, 2, p, t r ÷ø, pR t, , Ml 2, M.I. of rod (2) about Z-axis, I 2 =, 3, M.I. of rod (3) about Z – axis, I3 = 0, Because this rod lies on Z-axis, , If mass and thickness are same then, I µ, \, , 2Ml 2, 3, (14) (c) Distribution of mass about BC axis is more than that, about AB axis i.e. radius of gyration about BC axis is, more than that about AB axis., i.e. KBC > KAB \ IBC > IAB > ICA, , \, , (15) (a), , Isystem = I1 + I2 + I3 =, , I=, , 1, 1 æ M ö 1 M2, =, MR 2 = M ç, 2, 2 è pt r ÷ø 2 pt r, , Ml 2 0.12 ´ 12, =, = 0.01 kg - m 2, 12, 12, , I1 r2 3, =, = ., I 2 r1 1, , (20) (c) According to problem disc is melted and recasted into, a solid sphere so their volume will be same., 2, VDisc = VSphere Þ pRDisc, t=, , 4 3, pRSphere, 3, , RDisc, é, ù, æ RDisc ö 4 3, 3, Þ pRDisc, çè 6 ÷ø = 3 pRSphere êt = 6 , given ú, ë, û, 3, 3, Þ RDisc, = 8RSphere, Þ RSphere =, , (16) (c), , 1, r, , RDisc, 2, , Moment of inertia of disc, 1, , IDisc =, , 2, , 1, 2, MR Disc, = I (given), 2, , \ M ( R Disc ) = 2I, 2, , x, , Moment of inertia of sphere Isphere =, 1, 2, I1 = M.I. of ring about its diameter = mR, 2, I2 = M.I. of ring about the axis normal to plane and, passing through centre = mR2, Two rings are placed according to figure. Then, 1, 3, I xx ¢ = I1 + I 2 = mR 2 + mR 2 = mR 2, 2, 2, (17) (a) Mass of the centre disc would be 4M and its moment, 1, 2, of inertia about the given axis would be (4 M ) R ., 2, For the given section the moment of inertia about the, , same axis would be one quarter of this i.e., , 1, MR 2 ., 2, , (18) (d) Mass per unit length of the wire = r, Mass of L length, M = rL, and since the wire of length L is bent in a or of circular, loop therefore 2pR = L Þ R =, , L, 2p, , 2, 2, MRSphere, 5, , 2, , =, , 2 æ RDisc ö, 2I 1, M, = ( RDiscs )2 =, =, M, 5 çè 2 ÷ø, 10, 10 5, , (21) (d) Moment of inertia of system about YY', I = I 1 + I2 + I3, =, , 1, 3, 3, MR 2 + MR 2 + MR 2, 2, 2, 2, , =, , 7, MR 2, 2, , Y, 1, 2, , 3, , (22) (d) As C is the centre of mass, so, IC will be minimum., Also more mass is towards B so IA > IB., (23) (a) Applying the theorem of perpendicular axis,, I = I1 + I2 = I3 + I4, Because of symmetry, we have I1 = I2 and I3 = I4 Hence, I = 2I1 = 2I2 = 2I3 = 2I4 or I1 = I2 = I3 = I4, i.e. sum of two moment of inertia of square plate about, any axis in a plane (Passing through centre) should be, equal to moment of inertia about the axis passing, through the centre and perpendicular to the plane of, the plate., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 16, , 49, , (24) (d) Moment of inertia depends on all the three factors given, in (1), (2) & (3)., , (28) (c) Radius of gyration of a body is not a constant quantity., Its value changes with the change in location of the, axis of rotation. Radius of gyration of a body about a, , é2, ù, (25) (d) I = 4 ê MR2 + M (R 2)2 ú, ë5, û, , r12 + r22 + ..... + rn2, n, (29) (c) The moment of inertia of a particle about an axis of, rotation is given by the product of the mass of the, particle and the square of the perpendicular distance, of the particle from the axis of rotation. For different, axis, distance would be different, therefore moment of, inertia of a particle changes with the change in axis of, rotation., (30) (a) When earth shrinks, it angular momentum remains, , given axis is given as K =, , é2, ù, = 4MR 2 ê + 2ú, 5, ë, û, =, , 4 MR 2 ´ 12 48 MR 2, =, ., 5, 5, , (26) (b) Let a be the acceleration of centre of mass, Mg – T = 0, ... (i), F.x = T.2x, ... (ii), , constant. i.e. L = I w =, , 2, 2p, = constant., mR 2 ´, T, 5, , \ T µ I µ R 2 . It means if size of the earth changes, then its moment of inertia changes., In the problem radius becomes half so time period, , F, x, M, , (Length of the day) will becomes, , (27) (c) remain the same, i.e., , 24, = 6 hr., 4, , 1, of the present value, 4
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t.me/Magazines4all, DPP/ P 17, , 50, , 1., , (a) Since disc is rolling (without slipping) about point O., Hence, w, , 6., , (d), , P, , 7., Q v = rw, , \w =, , (d) Applying the theorem of perpendicular axis,, I = I1 + I2 = I3 + I4, Because of symmetry, we have, I1 = I2 and I3 = I4, Hence I = 2I1 = 2I2 = 2I3 = 2I4, , 3., , K, , 2, , =, , R2, , g sin q g / 2 5 g, =, =, 2, 7 / 5 14, 1+, 5, K2, , or I1 = I2 = I3 = I4, i.e. sum of two moment of inertia of square plate about, any axis in a plane (Passing through centre) should be, equal to moment of inertia about the axis passing, through the centre and perpendicular to the plane of, the plate., (a) By the conservation of energy, , Þw=, 8., 9., , 2, 5, , k2, 1+ 2, r, , 2 gh, 2, , r + k2, 2mgh, 2, , mr + mk, , 2, , =, , 2mgh, 2, , mr + I, , 2mgh, , =, , K2, R2, , I + mr 2, ) is minimum for, , sphere., (b) As body is moving on a frictionless surface. Its, mechanical energy is conserved. When body climbes, up the inclined plane it keeps on rotating with same, angular speed, as no friction force is present to provide, retarding torque so, 1 2 1, 1, Iw + mv 2 ³ Iw2 + mgh Þ v ³ 2gh, 2, 2, 2, , a, , P.E. of rod = Rotational K.E., l, 1, mg sin a = Iw 2, 2, 2, , =, , (a) Because its M.I. (or value of, , 10. (a), , l/2, , v, =, r, , R, , 2, , 2 gh, , (b) We know v =, , \ vQ > vC > vP, 2., , g sin q, , As q = 30o and, , C, , O, , a=, , 1+, , Q, , OQ > OC > OP, , 17, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 11., , 1, MR 2 = I Þ MR 2 = 2I, 2, Moment of inertia of disc about a tangent in a plane, , 5, 5, 5, = MR2 = (2I ) = I, 4, 4, 2, (d) Moment of inertia of system about YY’, I = I1 + I2 + I3, , 3 g sin a, l, 1 ml2 2, Þ mg sin a =, w Þw=, l, 2, 2 3, But in the problem length of the rod 2L is given, , 1, 3, 3, 7, = MR 2 + MR 2 + MR 2 = MR 2, 2, 2, 2, 2, Y, , 4., , 3g sin a, 2L, (c) Graph should be parabola symmetric to I- axis, but it, should not pass from origin because there is a constant, value Icm is present for x = 0 ., , 1, , 5., , (b), , \w =, , v=, , 2 gh, 1+, , K2, R2, , =, , 2 gh, 4, =, gh, 1, 3, 1+, 2, , 2, , 3, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 17, 12. (b), , 51, 19., , MR 2, =, = 2 :1, I Disc 1, MR 2, 2, , I Ring, , (a) M.I. of complete disc about ‘O’ point, 1, ITotal - (9M ) R 2, 2, , 13. (a), 14. (b) It follows from the theorem of parallel axes., 15. (a), , l, , A, , B, , 2R/3, O, , P, , R, O, , Moment of inertia of Rod AB about point P and, MI2, 12, M.I. of rod AB about point ‘O’, , perpendicular to the plane =, , R/3, , 2, , =, , MI 2, MI2, æ Iö, + Mç ÷ =, è 2ø, 12, 3, , O, R, , (By using parallel axis theorem), but the system consists of four rods of similar type so, by the symmetry, æ Ml2 ö, Isystem = 4 ç, ÷, è 3 ø, , Radius of removed disc = R, 3, , 16. (a), 17. (d), , l, , \, , b, , M.I. of plate about O and parallel to length =, , Iz = Ix + I y, , 9M, =M, 9, , [As M = pR 2 t \ M ¥R 2 ], M.I. of removed disc about its own axis, , O, , 18. (d), , Mass of removed disc =, , Iz, , 2, , Mb, 12, , =, , 2, , 1 æRö, MR 2, Mç ÷ =, 2 è3ø, 18, , Moment of inertia of removed disc about ‘O’, , Iremoved disc = Icm + mx2 =, Ix, , 2, , MR2, MR2, æ 2R ö, +M ç ÷ =, 18, 2, è 3 ø, , M. I. of complete disc can also be written as, , ITotal = I Re moved disc + I Re maining disc, MR 2, + I Re maining disc, 2, Equating (i) and (ii) we get, ITotal =, , Iy, , 200 = I D + I D = 2 I d, \ I D = 100 gm ´ cm, , 2, , ..... (ii), , MR 2, 9MR 2, + I Re maining disc =, 2, 2, \ I Re maining disc =, , 9MR 2 MR 2 8MR 2, =, = 4MR 2, 2, 2, 2
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t.me/Magazines4all, DPP/ P 17, , 52, 20., , (d), , I = Icm + Mx 2 =, , ML2, æLö, +M ç ÷, 12, è4ø, , I cm, , I, , L/4, =, , 21., , (c), , 2, , 24. (a), , 2, , æ 2mR 2 ö æ a ö, (T – f) R = Ia = ç 2 ÷ çè R ÷ø Þ T – f = ma, è, ø, For linear motion of block, , 2, , ML, ML, 7ML, +, =, 12, 16, 48, , w 2 = w 20 - 2aq Þ 0 = 4p 2 n2 - 2aq, , æ 1200 ö, 4p2 ç, ÷, è 60 ø = 200p2 rad, q=, 2´ 4, , T, 2m, , \ 2pn - 200p 2 Þ n = 100 p = 314 revolution, , (b) Rotational K.E. =, T.E. =, , 1 2, 1, w (MR 2 + MR 2 ) = w2 ´ 2MR 2, 2, 2, , (ii), , \a =, , 1 2, w ´ 2MR 2, 2, , =, , plane is given by a =, , 1, 2, , 2, For a solid sphere I = MR 2, 5, , 1 æ2, ö 1, 7, \ T.E. = w2 ç MR 2 + MR 2 ÷ = w2 MR 2 ´, 2 è5, 5, ø 2, Rotational K.E. = 1 ´ 2 MR 2 w2, 2 5, 1 2, ´ MR 2 w2, 2, 2, 5, b=, =, 1 2, 7, 7, w MR 2 ´, 2, 5, , 23., , (a) Time of descent µ, , K2, , . Time of descent depends upon, R2, the value of radius of gyration (K) or moment of inertia, (I). Actually radius of gyration is a measure of moment, of inertia of the body., , æ2, ö æ 3 + 4sin q ö mg, T = mg - 2m ç g (1 - sin q)÷ = ç, ÷ø, 7, è7, ø è, , æ 1 + 6sin q ö, (iii) F = T – ma = ç, ÷ø mg, è, 7, 28. (c) The acceleration of a body rolling down an inclined, , 1, Rotational K.E. = MR 2 w2, 2, , 1, MR 2 w2, 2, , f, , m 2a, 2mgsinq, mg, q, ///////////////////////////////////////////, , 1 2 1, Iw + MV 2, 2, 2, , \ T.E. =, , T, R, , 1 2, Iw &, 2, , 1, 1, 1, = Iw2 + MR 2 w2 = w2 (I + MR 2 ), 2, 2, 2, For ring I = MR2, , 2, (1 - sin q) g, 7, , mg – T = m (2a) Þ a =, , 2, , 22., , æ K2 ö, 1, 1, K2, mv2 = mv2 ç, Þ\, =1, ÷, 2, 2, R2, è R2 ø, , This value of K 2 / R 2 match with hollow cylinder.., 25. (b) 26. (c) 27. (d), (i) Let acceleration of centre of mass of cylinder be a then, acceleration of block will be 2a., For linear motion of cylinder T + f – 2mgsinq = 2m(a), For rolling motion of cylinder, , L/4, , 2, , KT = KR Þ, , For hollow cylinder, , g sin q, I, 1+, MR 2, , I, MR 2, , =, , MR 2, MR 2, , =1, , 1, MR 2, 1, 2, For solid cylinder, =, =, 2, 2, 2, MR, MR, Þ Acceleration of solid cylinder is more than hollow, cylinder and therefore solid cylinder will reach the, bottom of the inclined plane first., \ Statement -1 is false, • Statement - 2, In the case of rolling there will be no heat losses., Therefore total mechanical energy remains conserved., The potential energy therefore gets converted into, kinetic energy. In both the cases since the initial, potential energy is same, the final kinetic energy will, also be same. Therefore statement -2 is correct., 29. (b) Frictional force on an inclined plane, , I, , 30., , 1, = g sin a ( for a disc ) ., 3, (c) The moment of inertia about both the given axes shall, be same if they are parallel. Hence statement–1 is false., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 18, , 53, , (1), , (c) If r is the distance between m and (M – m), the gravitational force will be -, , m(M - m), , (7), , r2, , (2), , The gain in P E is U2 – U1 =, , m 1, = (as M and r are constants), M 2, , (c) mg =, , (3), , (8), , c, 3 kg , v = 2, m0, , m=, , 1 - (v 2 / c 2 ), , 3, 1-, , c, , 2, , =, , 3, 3, 2, , kg, , 4xc 2, (at equator l = 0), , (a) g' = g – Rew2, If a body is weightless,, g' = 0 , g – Rew2 = 0, , (9), , 10, g, =, = 1.25 × 10-3 rad/sec., 6400 ´ 103, R, (b) The apparent weight of person on the equator, (latitude l = 0) is given by, W' = W – m Rew2,, , Þ w=, , (4), , W' =, , 3, 3, W = mg, 5, 5, , 2, 9.8, 2g, w=, rad/ sec, = 5´, 6400 ´ 103, 5R, = 7.826 × 10–4 rad/sec, (c) According to question,, , g' =, , G ´ 4M p, R 2p, , on the planet and g =, , Q Rp = Re and Mp = Me, , (d), , Vg = Vg1 + Vg2 = -, , or Fr =, , Gm2, , Gm2, mv2, +, 2, 2, 2 =, r, 2r, 4r, , or v =, , Gm æ 2 2 + 1ö, r çè, 4 ÷ø, , (b) The resultant gravitational force on each particle provides it the necessary centripetal force, \, , mv2, = F2 + F2 + 2F2 cos 60o = 3F ,, r, , But r =, , 3, 2, l, l´ =, 2, 3, 3, , GM, l, (10) (b) The acceleration due to gravity on the surface of the, earth, in terms of mass Me and radius Re of earth, is, , G Me, R 2e, , ëê 0.5, , +, , GM e, R e2, , if Mm be the mass of the moon, Rm its radius, then the, acceleration due to gravity on the surface of the moon, on the earth, , Gm1 Gm 2, r1, r2, , é102, , = – 6.67 × 10–11 ê, , 2 F + F', , given by g =, , g', Now,, = 4 Þ g' = 4g = 40 m/sec2, g, Energy needed to lift 2 kg mass through 2m distance, = mg'h = 2 × 40 × 2 = 160 J, (6), , GMm 1, = mgR, 2R, 2, , \ v=, , 3, 3, mg = mg – mRw2 or mRw2 = mg – mg, 5, 5, , (5), , GMm, (R + R), , GM, é, ù, êQ g = 2 on surface of earth ú, ë, R, û, (c) Resultant force on particle '1', , Fr =, , =, , GMm, R, , The P.E. of object at a height R, U2 = –, , d éG, ù, (mM - m 2 ) ú = 0, i.e,, dm êë r 2, û, , or, , (c) The P.E. of the object on the surface of earth is, U1 = –, , G, , = 2 (mM – m2), r, dF, The force will be maximum if,, =0, dm, F=G, , 18, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 103 ù, –7, ú, 0.5 ûú = – 1.47 × 10 Joule/kg, , will be given by g' =, , GM m, R m2, , Dividing eq. (ii) by eq. (i), we get, 2, , 2, Mm æ Re ö, g', 1 æ 4ö, 1, ´, =, = M ç, =, ç, ÷, ÷, g, 80 è 1 ø, 5, e è Rm ø, , \ g' = g/5., (11) (b) The value of g at the height h from the surface of earth, æ 2h ö, g ' = g ç1 - ÷, è, Rø, The value of g at depth x below the surface of earth, xö, æ, g ' = g ç1 - ÷, è Rø
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t.me/Magazines4all, DPP/ P 18, , 54, xö, æ 2h ö æ, These two are given equal, hence çè1 - ÷ø = çè 1 - ÷ø, R, R, , On solving, we get x = 2h, (12) (a) If g be the acceleration due to gravity at the surface of, the earth, then its value at a height h above the earth's, surface will be g¢ =, , \, , g, , æ, h ö, çè1 + R ÷ø, e, , Here, , 2, , g' 1, =, g 9, , h, 1, 1, or 1 +, =3, =, 2, R, 9 æ, e, h ö, çè 1 + R ÷ø, e, , or h = 2 Re = 2 × 6400 = 12800 km., (13) (c) Consider the case of a body of mass m placed on the, earth's surface (mass of the earth M and radius R). If g, is acceleration due to gravity, then, mg = G, , Me m, , or g =, , GM e, , R, R2, where G is universal constant of gravitation., Now when the radius is reduced by 1%, i.e., radius, becomes 0.99 R, let acceleration due to gravity be g',, then g' =, , 2, , (16) (d) In the position of solar eclipse, net force on earth, FE = FM + FS, In the position of lunar eclipse, net force on earth, F'E = FS – FM, , GM e, , (0.99R) 2, , From equation (A) and (B), we get, , g', 1, R2, =, =, g, (0.99) 2, (0.99R) 2, , \ Change in acceleration of earth,, Df =, , 2GM, , é 1, 1 ù, W = GMe m ê R - R + h ú, ë e, û, e, , =, =, , gR e2 mh, GM e mh, =, [\ GMe = gRe2], R e (R e + h), R e (R e + h), mgh, æ, h ö, çè1 + R ÷ø, e, , (18) (a) The P.E of the mass at d/2 due to the earth and moon is, Earth, , 2, , æ F1 ö, H = R ç F - 1÷ = 350 km where (F2 = .9F1), è 2 ø, (15) (a) The extension in the length of spring is, x=, , GMm, mg, =, ,, k, r2 k, , 2 ´ 6.67 ´ 10 -11 ´ 7.36 ´ 10 22, , 3.822 ´ 1016, R2, = 6.73 × 10-5 m/s2, (17) (c) Let Me be the mass of the earth. The work required, , æ 1 ö, \ g' = g × ç, or g' > g, è 0.99 ÷ø, Thus , the value of g is increased., (14) (a) Force of gravity at surface of earth,, F1 = Gm M/R2, .......... (1), Force of gravity at height H is, F2 = Gm M (R + H)2, .......... (2), Dividing (A) by (B) and Rearranging, , =, , O1, M, , U=– 2, , R1, , m, 1, , 2, , æ 6400 ö, or x2 = 1 × ç, = 0.79 cm ., è 7200 ÷ø, , Moon, O2, R2, M 2, , d, , GM1m, GM 2 m, –2, d, d, , or U = –, , 2Gm, (M1 + M2), d, , (Numerically), , 1, G, (M1 + M 2 ), m Ve2 = U ÞVe = 2, 2, d, (19) (d) Let m be the mass of the body. The gravitational potential energy of the body at the surface of the earth is, GM e m, Re, The potential energy at a height 10 Re above the surface of the earth will be, , U=–, , x2, R2, =, \ xµ 1 , \, x1 (R + h) 2, r2, , P, , U' = –, , GM e m, (R e +10R e ), , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 18, , 55, , \ Increase in potential energy, GM e m æ GM e m ö 10 GMe m, + ç R ÷ = 11 R, è, 11R e, e ø, e, This increase will be obtained from the initial kinetic, energy given to the body. Hence if the body be thrown, with a v velocity then, , (23) (a) All statements except (4) are wrong., (24) (a) Value of g decreases when we go from poles to equator., , U' – U = –, , 20Gme, 10 GMe m, 1, mv2 = 11 R, Þ v = 11R, e, e, 2, Substituting the given values, we get, v=, , æ 20 ´ (6.67 ´ 10 -11 ) ´ (6 ´ 10 24 ) ö, ç, ÷, 11 ´ (6.4 ´ 106 ), è, ø, , = 1.07 × 104 m/s., (20) (b), , F<, , Gm ∋ M , m(, r2, , For maximum force, , Þ, , dF, <0, dm, , d æç GmM Gm2 ö÷÷, ç, ,, ÷< 0, dm èçç r 2, r 2 ø÷÷, , m 1, Þ M , 2m < 0 Þ <, M 2, , (21) (b), , 2, , 2, , g ' = g - w R cos l, , For weightlessness at equator l = 0 and g ' = 0, g, 1 rad, =, R 800 s, (22) (a) k represents gravitational constant which depends only, on the system of units., , \ 0 = g - w2R Þ w =, , r2, , (25) (b), , r1, , Gravitational PE at perihelion < ,GMm / r1 as r1 is, minimum Therefore, PE is minimum., (26) (c) Total energy = constant., (27) (c) As Pluto moves away, displacement has component, opposite to air force, hence work done is –negative., (28) (b) For two electron, , Fg, Fe, , = 10 -43 i.e. gravitational force is, , negligible in comparison to electrostatic force of, attraction., (29) (c) The universal gravitational constant G is totally, different from g., FR 2, Mm, The constant G is scalar and posses the dimensions, G=, , é M -1L3T -2 ù ., ë, û, g=, , GM, R2, , 0, -2, g is a vector and has got the dimensions éë M LT ùû ., It is not a universal constant., (30) (a) As the rotation of earth takes place about polar axis, therefore body placed at poles will not feel any centrifugal force and its weight or acceleration due to gravity remains unaffected.
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t.me/Magazines4all, DPP/ P 19, , 58, (20) (d) If the period of revolution of a satellite about the earth, be T, then, T2 =, , 4p 2 (R e + h)3, GMe, , (24) (a), , (25) (d) T 2 µ R3, 3, Te2 = KRe3 ; Tm2 = kRm, ; T 2 = kR3, , where h is the height of the satellite from earth's surface., \ Me =, , R + Rm, R= e, 2, , 4p 2 (R e + h)3, , GT2, The satellite is revolving just above the earth, hence h, is negligible compared to Re., \ Me =, , éT 2 / 3 T 2 / 3 1 ù, Þ T = k ê e1/ 3 + m1/ 3 ´ ú, 2 úû, k, êë k, , GT2, , é T 2 / 3 + Tm2 / 3 ù, ÞT =ê e, ú, 2, ëê, ûú, , 4, p Re3 r where r is the density of the earth., 3, , (26) (a), , 4p 2 R e3, 4, Thus p Re3 r =, 3, GT2, , (21) (a), , E KQ, E KP, , =, , 2, vQ, v2P, , Orbital velocity, V0 = R e g, , = 7.9 × 103 m/s, , According to question,, VP = V0 + Ve = 7900 – 463 = 7437 m/s, VQ = V0 + Ve = 7900 + 463 = 8363 m/s, \, , (22) (b), , (23) (a), , E KP, , æ 8363 ö, =ç, è 7437 ÷ø, , 2, , 2GM, i.e. escape velocity depends upon the, R, mass and radius of the planet., , ve =, , ve =, , 2 Re Ee, M, ´, Me, æ Re + Rm ö, 2ç, ÷ø, è, 2, , 2M æ Re ö, Ee, M e çè Re + Rm ÷ø, , (27) (c) Areal velocity of the artificial planet around the sun, will be more than that of earth., (28) (a), , 2pR e 6.28 ´ 6.4 ´ 10 6, = 463 m/s, =, 24 ´ 3600, Te, , E KQ, , =, , 2GM, 8, = R pGr, R, 3, , If mean density is constant then ve µ R, vp, ve R e 1, =, = Þ ve =, vp R p 2, 2, , 3/ 2, , GM s M e, GM s M, =–, 2 Re, 2R, , ., , Linear velocity of earth,, Ve=, , Ee = =, , 3p, \ r T2 =, ., G, which is universal constant. To determine its value,, 3 ´ 3.14, 3p, =, r T2 =, G, 6.67 ´ 10 –11 m3 / kg-s2, , 3, , 2, , 4p 2 R e3, , But Me =, , GM, r, , v0 =, , v0 = R e, , g, Re + h, , For satellite revolving very near to earth R e + h = R e, As ( h << R ), v 0 = R e g ; 64 ´ 105 ´ 10 = 8 ´ 103 m / s = 8 kms -1, , Which is independent of height of a satellite., (29) (d) Due to resistance force of atmosphere, the satellite, revolving around the earth losses kinetic energy., Therefore in a particular orbit the gravitational, attraction of earth on satellite becomes greater than, that required for circular orbit there. Therefore satellite, moves down to a lower orbit. In the lower orbit as the, potential energy ( U = -GMm / r ) becomes more, negative, Hence kinetic energy ( Ek = GMm / 2r ), increases, and hence speed of satellite increases., (30) (a) Because gravitational force is always attractive in nature, and every body is bound by this gravitational force of, attraction of earth., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 20, , 59, , 1., , 2., , 3., , 8., , MgL, , (a). Y =, , W=, , 1, 1, F × l = × stress x volume x strain, 2, 2, , W=, , 1, × Y × strain2 × volume, 2, , æ 4Aö, \ D = çè, ÷, p ø, , 9., , (c). Stress =, , 10., , (c). Strain =, , 11., , 2 ´ 1011 ´ 10-6 ´ 10-6, = 0.1 J, 2 ´1, , 12., , 5., , 1, × 5.4 × 106 × 3, 2, W = 8.1 × 106 ergs, (d). By Hook's law, , Y=, 6., , 16 ´ 1, (4 ´ 10 ) (0.2 ´10 -2 ), , 13., , = 2 × 1011 N/m2, , DPV, DV, Given, DP = hdg = 200 x 103 x 10, DP = 2 × 105 N/m2, DV 0.1, =, = 10–3, V 100, , 7., , 2 ´ 106, 10-3, , (b). Y =, \l=, , 1, DV, == 5 × 10–10, K, VDp, , DV, = c D p = 5 × 10–10 × 15 × 106, V, = 7.5 × 10–3, (c). Increase in length on heating Dl = a L DT, To annul this increase if pressure applied is p then, Dl, = Ya DT, L, = 2 × 1011 × 1.1 × 10–5 × 100 = 2.2 × 108 N/m2, (c). y = 2h (1 + s), y = 2.4 × h, 2.4 h = 2h (1 + s), (1 + s) = 1.2, s = 0.2, (c). Stress = F/A = 10/(2 × 10-6) = 5 × 106 N/m2, , p=Y, , 14., , (a). B = –, , \B=, , F, 4.8 ´ 103 N, =, = 4.0 × 107 N/m2, A 1.2 ´ 10-4 m 2, , =–, , F / A FL, =, l / L Al, -8, , 1/2, , \ Fractional decrease in volume, , W=, , Y=, , æ 4 ´ 10-6 ö, =ç, ÷, p ø, è, , Dl 1 ´10-3, =, = 5 × 10–4, longitudinal, l, 2, (c). F = Y A a Dt, = 2 × 1011 × 3 × 10–6 × 10-5 × 20, F = 120 N., (c). Compressibility, , c=, , 1, × load × elongation, 2, , (d). W =, , 1/2, , = 1.13 × 10–3 m = 1.13 mm, , Dl 2, 1, YADL2, W = × Y × 2 × AL =, 2, 2L, L, , 4., , (a). Limiting stress = 4.0 × 108 N/m2, F 400, =, = 4.0 × 108, A, A, or A = 10–6 m2, , pr 2 Dl, but Mg/pr2 = 20 × 108 & Dl = L then, Y = 20 × 108 N/m2, (b). F = Y a DtA, A = 2 × 10–6 m2, Y = 2 × 11 N/m2, a = 1.1 × 10–5, t = 50 – 30 = 20°C, F = 2 × 1011 × 1.1 × 10–5 × 20 × 2 × 10–6 = 88N., (d). Work done on the wire, , W=, , 20, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 15., , Strain =, , = 2 × 109 N/m2, , stress F / A F L, =, =, strain l / L Al, , FL, 20 ´ 9.8 ´ 4, =, A Y p ´ (10 -3 )2 ´ 1.96 ´ 1011, , = 1.27 × 10–3 m = 1.27 mm, , Stress 5 ´ 106, =, = 2.5 × 10–5, Y, 2 ´ 1011, , l = L × strain = 1 × 2.5 × 10–5, l = 2.5 × 10–5 m, 16., , Dl Mg 1000 ´ 980 ´ 100, (b). L = AY =, 1012 ´ 0.01, , 17., , Dl = 0.0098 cm., (a). Volume = Mass/density, Area of cross-section = volume/length
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t.me/Magazines4all, DPP/ P 20, , 60, 15.6 ´ 10-3, mass, =, =, = 8 × 10–7 m2, 7800 ´ 2.5, density ´ length, , Y=, 18., , Fl, 8 ´ 9.8 ´ 2.5, =, ADL (8 ´ 10 -7 ) ´ 1.25 ´ 10-3, , Y = 1.96 × 1011 N/m2, (c). Potential energy per unit volume = u, 1, × stress × strain, 2, , =, , stress, strain, \ stress = Y x strain = Y x S, \ Potential energy per unit volume = u, , But Y =, , =, , 19., , 1, 1, × (YS)S = YS2, 2, 2, , 21., , (c) F = YAaDq \ F µ A, , 22., , (a) For twisting, angle of shear f µ, , 23., , (a) Y = 2h (1 + s ), , 24., , (a) Y = 2h (1 + s ) Þ s =, , 25., , (a) Tensile stress =, , 26., , (a) Tensile stress is maximum when cos 2 q is maximum,, i.e., q = 0°, , 27., , (b) Shearing stress =, , 28., , L1 = L, L2 = 2L, r1 = 2R., r2 = R, , 20., , l1, L R2, 1, =, . 2 =, l 2 2L 4R, 8, , (c). stress =, , 29., , Force, F, = 2, Area pr, , 1, \ stress S µ 2, r, æ S1 ö æ r2 ö, \ ç ÷ =ç ÷, è S2 ø è r1 ø, , 0.5Y - h, h, , Fcos q Fcos2 q, =, a / cos q, a, , Fsin q, Fsin q cos q, =, a / cos q, a, , Fsin 2q, 2a, (a) Elasticity is a measure of tendency of the body to regain, its original configuration. As steel is deformed less than, rubber therefore steel is more elastic than rubber., (a) Bulk modulus of elasticity measures how good the body, is to regain its original volume on being compressed., Therefore, it represents incompressibility of the material., =, , l1 L1 r22, (d). l =, L 2 r12, 2, , \, , 1, L, i.e. if L is more then f will be small., , 30., 2, , r1 2, Given r = 1 \, 2, , S1 1, =, S2 4, , - PV, where P is increase in pressure, DV is change, DV, in volume., (a) A bridge during its use undergoes alternating strains, for a large number of times each day, depending upon the, movement of vehicles on it when a bridge is used for long, time, it losses its elastic strength. Due to which the amount, of strain in the bridge for a given stress will become large, and ultimately, the bridge may collapse. This may not, happen, if the bridges are declared unsafe after long use., K=, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 21, , 61, , 1. (a) Force required to separate the plates, F=, , 2. (c) T, , Let the width of each plate is b and due to surface, tension liquid will rise upto height h then upward force, due to Surface tension, .......(i), = 2Tb cos q, Weight of the liquid rises in between the plates, , 2TA 2 ´ 70 ´ 10-3 ´ 10 -2, =, = 28 N, t, 0.05 ´ 10 -3, , q, , 21, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , q, , T, , = Vdg = (bxh)dg, , ........(ii), , Equating (i) and (ii) we get, 2T cos q = bxhdg, \h =, , 10. (b), , æ D2 d 2 ö, 3. (d) W = T ´ 8p(r22 - r12 ) = T ´ 8p ç, ÷, ç 4, 4 ÷ø, è, , 11., 12., , = 2p( D 2 - d 2 )T, , 4. (b) Increment in area of soap film = A2 - A1, = 2 ´ [(10 ´ 0.6) - (10 ´ 0.5)] ´10 -4 = 2 ´10 -4 m 2, Work done = T ´ DA, = 7.2 ´ 10-2 ´ 2 ´ 10-4 = 1.44 ´10 -5 J, , 5. (a), 6. (c) Excess pressure inside soap bubble is inversely proportional, 1, r, This means that bubbles A and C posses greater, pressure inside it than B. So the air will move from A, and C towards B., , to the radius of bubble i.e. DP µ, , 13., 14., , x, b, , h, , (b) Tc = T0 (1- µ t ), i.e. surface tension decreases with increase in, temperature., (a), (d) Tension in spring T = upthrust - weight of sphere, = (h - 1)V rg = (h - 1) mg., , 15., 16., 17., 18., , (b), (c), (c) A stream lined body has less resistance due to air., (a) Weight of cylinder = upthrust due to both liquids, , æA 3 ö, æ A Lö, V ´ D ´ g = ç ´ L÷ ´ d ´ g + ç ´ ÷ ´ 2d ´ g, è5 4 ø, è 5 4ø, A´ L´d ´ g, æA, ö, Þ ç ´ L÷ ´ D ´ g =, è5, ø, 4, Þ, , 8. (c) The radius of resultant bubble is given by R 2 = r12 + r2 2 ., (b), , \ M µ R . If radius becomes double then mass will, becomes twice., (d) In the satellite, the weight of the liquid column is zero., So the liquid will rise up to the top of the tube., , = V sg - V rg = V hrg - V rg (As s = hr), , rr, 5´ 4, = 20 cm, 7. (b) r = 1 2 =, r1 - r2 5 - 4, , 9., , Mass of liquid in capillary tube, , æ 1ö, M = pR 2 H ´ r\ M µ R 2 ´ ç ÷ (As H µ1/ R), è Rø, , Weight of metal disc = total upward force, = upthrust force + force due to surface tension, = weight of displaced water + T cos q (2pr ), = W + 2prT cos q, , 2T cos q, xdg, , 19., , D d, 5, = \D = d, 5 4, 4, , (b) Velocity of efflux when the hole is at depth h, v = 2 gh, Rate of flow of water from square hole, Q1 = a1v1 = L2 2 gy, Rate of flow of water from circular hole, Q2 = a2 v2 = pR 2 2 g (4 y ), , According to problem Q1 = Q2, , Þ L2 2 gy = pR 2 2 g (4 y) Þ R =, , L, 2p
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t.me/Magazines4all, DPP/ P 21, , 62, 20., , (a) Let A = cross-section of tank, , (, , ), , 23., , o, (d) At critical temperature Tc = 370 C = 643 K , the, , 24., 25., , surface tension of water is zero., (d), (b), , A, , 3m, , B, , v, , 52.5 cm, , PBottom > PSurface . So bubble rises upward., , a = cross-section hole, V = velocity with which level decreases, v = velocity of efflux, av, A, By using Bernoulli’s theorem for energy per unit volume, Energy per unit volume at point A, = Energy per unit volume at point B, , From equation of continuity av = AV Þ V =, , 1, 1, P + rgh + rV 2 = P + 0 + rv 2, 2, 2, Þ v2 =, , 21., , 2 gh, æaö, 1- ç ÷, è Aø, , 2, , =, , 2 ´ 10 ´ (3 - 0.525), 1 - (0.1), , 2, , = 50( m / sec) 2, , (c) If the liquid is incompressible then mass of liquid, entering through left end, should be equal to mass of liquid, coming out from the right end., \ M = m1 + m 2 Þ Av1 = Av2 + 1.5A.v, Þ A ´ 3 = A ´ 1.5 + 1.5A.v Þ v = 1 m / s, , 22., , (b), , T=, , F, 6.28 ´ 10 -4, =, = 5 ´ 10 -2 N / m, 2 pr 2 ´ 3.14 ´ 2 ´ 10 -3, , 26., 27., , At constant temperature V µ 1 ( Boyle 's law ), P, Since as the bubble rises upward, pressure decreases, then, from above law volume of bubble will increase i.e. its size, increases., (a), (d)., F = Patm × Area = 105 × 1 × 10–6 = 0.1 N, 2T ö, æ, F = ç Patm +, ´ A = 0.10023 N, è, r ÷ø, , 28. (a) Since the excess pressure due to surface tension is, inversely proportional to its radius, it follows that, smaller the bubble, greater is the excess pressure. Thus, when the larger and the smaller bubbles are put in, communication, air starts passing from the smaller into, the large bubble because excess pressure inside the, former is greater than inside the latter. As a result, the, smaller bubble shrinks and the larger one swells., 29. (b) Statement-1 is True, Statement-2 is True; Statement-2, is NOT a correct explanation for Statement-1., 30. (a) In a stream line flow of a liquid, according to equation, of continuity,, av = constant, Where a is the area of cross-section and v is the velocity, of liquid flow. When water flowing in a broader pipe, enters a narrow pipe, the area of cross-section of water, decreases therefore the velocity of water increases., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 22, , 63, , 1., , 2., , (c), , (a), Þ, , 3., , Þ, Þ, Þ, (c), , Due to volume expansion of both mercury and flask,, the change in volume of mercury relative to flask is, given by DV = V0 [ g L - g g ]Dq = V [ g m - 3a g ]Dq, = 50[180 × 10–6 – 3 × 9 × 10–6] (38 – 18), = 0.153 cc, greal = gapp. + gvessel, So (gapp. + gvessel)glass= (gapp. + gvessel)steel, 153 × 10–6 + (gvessel)glass= (144 × 10–6 + gvessel)steel, Further, (gvessel)steel = 3a = 3 × (12 × 10–6) = 36 × 10–6/°C, 153 × 10–6 + (gvessel)glass = 144 × 10–6 + 36 × 10–6, (gvessel)glass = 3a = 27 × 10–6/°C, a = 9 × 10–6/°C, Initial diameter of tyre = (100 – 6) mm = 994 mm,, , 5., , 45, é 75, ù, -6, -6, = x ê ´ 10 ´ 100 + (20 - x) ´ ´ 10 ´ 100 ú, 2, ë2, û, On solving we get x = 10 cm., (b) Due to volume expansion of both liquid and vessel,, the change in volume of liquid relative to container is, given by DV = V0 [ g L - g g ]Dq, Given V0 = 1000 cc, ag = 0.1× 10–4/°C, \, , DV = 1000[1.82 ´ 10-4 - 0.3 ´ 10-4 ] ´ 100 = 15.2cc, (b) g r = g a + g v ; where g r = coefficient of real expansion,, g a = coefficient of apparent expansion and, g v = coefficient of expansion of vessel., , 6., , 994, = 497mm, 2, and chan ge in diameter DD = 6 mm, so, , For copper g r = C + 3a Cu = C + 3A, , 6, DR = = 3 mm, 2, After increasing temperature by Dq, tyre will fit onto, wheel, Increment in the length (circumference) of the iron tyre, , Þ C + 3A = S + 3a Ag Þ a Ag =, , g, DL = L ´ a ´ Dq = L ´ ´ Dq, 3, , Dq Þ, , For silver g r = S + 3a Ag, , 7., , g, [As a ´ ], 3, , ægö, 2pDR = 2pR ç ÷ Dq, è3ø, , 8., , 3 DR, 3´ 3, =, g R 3.6 ´ 10-5 ´ 497, , V = V0 (1 + gDq) Þ Change in volume, V – V0 = DV = A.Dl = V0gDq, , Þ, , Dl =, , = 45 × 10–3m = 4.5 cm, (b) Loss of weight at 27°C is, = 46 – 30 = 16 = V1 × 1.24 r1 × g, Loss of weight at 42°C is, = 46 – 30.5 = 15.5 = V2 × 1.2 r1 × g, Now dividing (i) by (ii), we get, , DL = L0 aDq, Rod A : 0.075 = 20 × aA × 100, , But, , 45, ´ 10-6 / °C, 2, For composite rod: x cm of A and (20 – x) cm of B we, have, , aA, , (20 – x), , A, , B, , ...(i), ...(ii), , V2, 15.5 ´ 1.24, = 1 + 3a(t2 - t1 ) =, = 1.001042, V1, 16 ´ 1.2, , Þ 3a (42° – 27°) = 0.001042, Þ a = 2.316 × 10–5/°C, , aB =, , x, , V0 .Dq 10-6 ´ 18 ´ 10-5 ´ (100 - 0), =, A, 0.004 ´ 10-4, , V 1.24, 16, = 1´, 15.5 V2 1.2, , 75, Þ aA =, ´ 10-6 / °C, 2, Rod B : 0.045 = 20 × aB × 100, Þ, , C - S + 3A, 3, , (d), , Þ Dq = 500°C, , (b), , g g = 3a g = 3 ´ 0.1 ´ 10-4 / °C = 0.3 ´ 10-4 / °C, , \, , So, initial radius of tyre R =, , 4., , 22, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , aB, , 20 cm, 0.060 = x aA × 100 + (20 – x) aB × 100, , 9., , mL, . This, t, must be the heat supplied for keeping the substance in, molten state per sec., , (b) Heat lost in t sec = mL or heat lost per sec =, , mL, Pt, = P or L =, t, m, 10. (b) Initially ice will absorb heat to raise it's temperature to, 0°C then it's melting takes place, If m1 = Initial mass of ice, m1' = Mass of ice that melts, and mW = Initial mass of water, , \
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t.me/Magazines4all, DPP/ P 22, , 64, , 11., , By Law of mixture, Heat gained by ice = Heat lost by water Þ m1 × (20) +, m1' × L = mWcW [20], Þ 2 × 0.5 (20) + m1' × 80 = 5 × 1 × 20, Þ m1' = 1 kg, So final mass of water = Initial mass of water + Mass, of ice that melts = 5 + 1 = 6 kg., (a) If mass of the bullet is m gm,, then total heat required for bullet to just melt down, Q1 = mcDq + mL = m × 0.03 (327 – 27) + m ×6, = 15m cal = (15m × 4.2)J, Now when bullet is stopped by the obstacle, the loss in, 1, -3 2, its mechanical energy = (m ´ 10 )v J, 2, (As m gm = m × 10–3kg), As 25% of this energy is absorbed by the obstacle,, , Q2 =, , 17., , W, , ( Joules ), , 18., , CA 3, =, CB 4, , and CB (23 - 19) = CC (28 - 23) Þ, , CB 5, =, CC 4, , Þ, , C A 15, =, CC 16, , 19., , 13., , mc cc (Dq)c 50 ´ 10-3 ´ 420 ´ 10, =, = 5°C, mW cW, 10 ´ 10 -3 ´ 4200, (b) Heat lost by hot water = Heat gained by cold water in, beaker + Heat absorbed by beaker, Þ 440 (92 – q) = 200 × (q – 20) + 20 × (q – 20), Þ q = 68°C, (a), (b) Firstly the temperature of bullet rises up to melting, point, then it melts. Hence according to W = JQ., , 14., , 15., 16., , g, = 10-4 / °C, 3, (b) As we know, g real = g app. + g vessel, Þ g app. = g glycerine – g, = 0.000597 – 0.000027, = 0.00057/°C, , (a), , t=, , ( Pt - P0 ), ´ 100°C, ( P100 - P0 ), , (60 - 50), ´ 100 = 25°C, (90 - 50), (c) Since specific heat = 0.6 kcal/gm × °C, = 0.6 cal/gm × °C, From graph it is clear that in a minute, the temperature, is raised from 0°C to 50°C., Þ Heat required for a minute, = 50 × 0.6 × 50 = 1500 cal., Also from graph, boiling point of wax is 200°C., (b) The horizontal parts of the curve, where the system, absorbs heat at constant temperature must depict, changes of state. Here the latent heats are proportional, to lengths of the horizontal parts. In the sloping parts,, specific heat capacity is inversely proportional to the, slopes., (d) Let L0 be the initial length of each strip before heating., Length after heating will be, , 22., , 23., , DqW =, , Þ, , 1 2, mv = J .[m.c.Dq + mL] = J [mS (475 - 25) + mL], 2, , Þ, , mS (475 - 25) + mL =, , mv 2, 2J, , glass, , =, , C A 28 - q, =, ...(ii), CC q - 12, , On solving equation (i) and (ii) q = 20.2°C, (a) Same amount of heat is supplied to copper and water, so mcccDqc = mWcWDqW, Þ, , 20., , ....(i), , C A (q - 12) = CC (28 - q) Þ, , Þ Mgh = mL, Þ 3.5 × 10 × 2000 = m × 3.5 × 105, Þ m = 0.2 kg = 200 gm, (d) Coefficient of volume expansion, , a=, , 21., , If q is the temperature when A and C are mixed then,, , H, , ( Joules ), , (r - r ), Dr, (10 - 9.7), = 1 2 =, = 3 ´ 10-4, r.DT, r.( Dq), 10 ´ (100 - 0), Hence, coefficient of linear expansion, , Now the bullet will melt if Q2 ³ Q1, , C A (16 - 12) = CB (19 - 16) Þ, , =, , g=, , 75 1 2, 3, ´ mv ´ 10-3 = mv 2 ´ 10-3 J, 100 2, 8, , 3 2, mv ´ 10-3 ³ 15m ´ 4.2 Þ vmin = 410 m/s, i.e., 8, 12. (c) Heat gain = heat lost, , (b) Suppose m kg of ice melts then by using, , Brass Strip, , d, , Copper Strip, , R, , q, , LB = L0 (1 + α BΔT ) = ( R + d ) θ, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 22, , 65, , LC = L0 (1 + αC ΔT ) = Rθ, Þ, , r2 =, , R + d 1 + αB ΔT, =, R, 1 + αC ΔT, , Þ 1+, , r 2 v 2 A2 = r1v1 A1, , d, = 1 + ( α B - α C ) ΔT, R, , Þ R=, , Þ 1350 × v2 = 1500 × 0.1, v2 = 1/9 m/s, \ Volume rate of flow at the end of tube, , d, , ( α B - αC ) ΔT, , = A2v2 = 4 × 10–4 ´, , 1, 1, ÞRµ, and R µ, ΔT, α, ( B αC ), , 24. (a) A bimetallic strip on being heated bends in the form of, an arc with more expandable metal (A) outside (as, shown) correct., , B, , A, , 25. (a), , aA > aB, B, , A, , aA, , aB, , aB, , aA, , 26. (c), 27. (c), r1v1A1 = r2v2A2, m = 1500 kg/m3 × 0.1 m/s × 4 (cm)2, msDT = 10000, 1500 × 0.1 × 4 × 10–4 × 1500 × DT = 10000, DT =, , 10000 1000, =, °C, 90, 9, , r1, 1500, =, = 1350 kg/m3, (1 + gDT ) æ, -3 1000 ö, çè1 + 1 ´ 10 ´, ÷, 9 ø, , 1, 9, , 4, 40, ´ 10-4 m3 =, ´ 10-5 m3, 9, 9, Volume rate of flow at the entrance = A1v1, = 0.1 × 4 × 10–4 = 4 × 10–5m3, Hence, difference of volume rate of flow at the two, ends, , =, , 4, æ 40 ö, -5, = ç - 4÷ ´ 10, = ´ 10-5 m3, è 9, ø, 9, , 28. (d) Celsius scale was the first temperature scale and, Fahrenheit is the smallest unit measuring temperature., 29. (a) Linear expansion for brass (19 × 10–4) > linear expansion, for steel (11 × 10–4). On cooling the disk shrinks to a, greater extent than the hole and hence it will get loose., 30. (b) The latent heat of fusion of ice is amount of heat, required to convert unit mass of ice at 0°C into water, at 0°C. For fusion of ice, L = 80 cal/gm = 80000 cal/gm = 8000 × 4.2 j/kg, = 336000 J/kg.
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t.me/Magazines4all, DPP/ P 23, , 68, =, , rrc 4.2, ´, ´ 10-6, s, 48, , 50 - q, æ 50 + q, ö, =Kç, - 25 ÷, è 2, ø, 10, On dividing, we get, , 7 r rc, 7 rrc, µs ;, µs, [As J = 4.2], 80 s, 72 s, (c) Suppose temperature difference between A and B is, =, , 12., , Þ, , 100° C and q A > qB, 15., , C, , (d), , H/2, , H/2, , A, , H, H/2, , H/2, , Heat current will flow from A to B via path ACB and, ADB. Since all the rod are identical so, , (Dq) AC = (Dq) AD, , d q (T 4 - T04 ), µ, dt, mc, , Dq, ; here R = same for all), R, , Þ q A - qC = q A - q D, , Þ, , Þ qC = q D, , C, R, , R, , B, , A, , 16., T, , R, , R, , 17., D, , 13., , (a) Initially at t = 0, , Rateof cooling (R) µ Fall in temperature of body (q – q0), Þ, , R1 q1 - q0, =, R2 q 2 - q0, , =, , 14., , (c), , 100 - 40 3, =, 80 - 40 2, , 60 - 50, æ 60 + 50, ö, = Kç, - 25 ÷, è 2, ø, 10, , ........... (i), , d q (T 4 - T04 ), µ, dt, V rc, , Now if we consider that equal volume of liquid (V) are, taken at the same temperature then, , i.e. temperature difference between C and D is zero., , 2T, , d q sA 4, (T - T04 ). If the liquids put in exactly similar, =, dt mc, calorimeters and identical surrounding then we can, consider T0 and A constant then, , dq 1, µ, dt c, So for same rate of cooling c should be equal which is, not possible because liquids are of different nature., Again from equation (i),, , D, , (Because heat current H =, , 10, 60, =, 50 - q q, q = 42.85° C, , d q (T 4 - T04 ), µ, dt, mc, If we consider that equal masses of liquid (m) are taken, at the same temperature then, , B, , H, , .......... (ii), , 1, dq, µ ., dt rc, So for same rate of cooling multiplication of r × c for, two liquids of different nature can be possible. So,, option (d) may be correct., (d) For cooking utensils, low specific heat is preferred for, it's material as it should need less heat to raise it's, temperature and it should have high conductivity,, because, it should transfer heat quickly., (b), , K1 A1 (q1 - q2 ), æQö, çè ÷ø =, l, t 1, K A (q - q ), Q, and æç ö÷ = 2 2 1 2, è t ø2, l, , æQö, æQö, Given, ç ÷ = ç ÷, è t ø1 è t ø 2, , Þ K1 A1 = K 2 A2, 18. (a) Convection may be stopped, 19. (d) Heated fluid becomes less dense than the cold fluid, above it, 20. (c) According to Kirchoff's law, the ratio of emissive power, to absorptive power is same for all bodies is equal to, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 24, , 71, , 19. (c) From graph it is clear that P3 > P1 ., , C, , P3, P1, , C, , P3, P1, , A, B, , P2, , B, , D (V2 ) V, Since area under adiabatic process (BCED) is greater, than that of isothermal process (ABDE).Therefore net, work done, W = Wi + (-WA ) \WA > Wi Þ W < 0, , E(V1), , PV 2 = constant represents adiabatic equation. So, during the expansion of ideal gas internal energy of, gas decreases and temperature falls., (a) For adiabatic process, , = 1×, , V, , 3, 9, R × (4T0 – T0) = RT, T0, 2, 2, , Heat absorbed =, , 9, 11, RT, T0 + RT0 =, RT, T0, 2, 2, , 25. (b) AB is an isothermal process then, , 20. (b), , T1Vbg-1, , D(V2), , 24. (d) Work done = Area of ABC with V-axis, = P0(2V0 –V0) + 0 = P0V0 = nRT0 = RT0, Change in internal energy = nCVDT, , E (V1 ), , 21., , A, , P2, , P, 3, , P × 2V = PB × 6V Þ PB =, , P, , = Constant, , A, P, , For bc curve, , C, , T1Vbg-1 = T2Vcg-1, , or, , T2 æ Vb ö, =ç ÷, T1 è Vc ø, , g-1, , ....(i), , B, , For ad curve, , V, 6V, 2V, Now BC is an isochoric process then, , g -1, , T1Vag -1 = T2Vd, or, , T2 æ Va ö, =, T1 çè Vd ÷ø, , g -1, , .....(ii), , From equation (i) and (ii), Vb Va, =, Vc Vd, , 22. (d) There is a decrease in volume during melting on an ice, slab at 273K. Therefore, negative work is done by, ice-water system on the atmosphere or positive work, is done on the ice-water system by the atmosphere., Hence option (b) is correct. Secondly heat is absorbed, during melting (i.e. DQ is positive) and as we have, seen, work done by ice-water system is negative (DW, is negative). Therefore, from first law of, thermodynamics ΔU = ΔQ - ΔW, Change in internal energy of ice-water system, DU will, be positive or internal energy will increase., 23. (a) From graph it is clear that P3 > P1., Since area under adiabatic process (BCED) is greater, than that of isothermal process (ABDE). Therefore net, work done, , W = Wi + ( - WA ) Q WA > Wi Þ W < 0, , PB PC, P, P, =, =, ;, TB TC, 3T0 TC ; TC = 3T0, 26. (a) Heat absorbed during BC is given by, , Q = nCv DT = n ´, , 3R, (TC - TB ), 2, , 3R, (2T0 ) = 3nRT0., 2, 27. (b) Heat capacity is given by, , = n´, , 1 dQ, 1 Q, ; C=, n dT, n 2T0, 28. (c) As isothermal processes are very slow and so the, different isothermal curves have different slopes so, they cannot intersect each other., 29. (d) Adiabatic compression is a rapid action and both the, internal energy and the temperature increases., , C=, , 30. (a), , Q, ; a gas may be heated by putting pressure,, m.Dq, so it can have values for 0 to ¥ ., CP and CV are it’s two principle specific heats, out of, infinite possible values., In adiabatic process C = 0 and in isothermal process, C= ¥., c=
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 26, , 75, , the translational kinetic energy i.e.,, ETrans, , 20. (b), , 3, 3, = RT = ´ 8.31´ (273 + 0) = 3.4 ´103 J, 2, 2, , VA μ A, =, ¹ 1 i.e. VA ¹ VB, VB μ B, , Similarly if VA = VB then, , ( D Q ) P = m CP D T, Þ 2 ´ CP ´ (35 - 30) Þ C P = 7, , cal, mole - K, , 25. (b) n1Cv (T – T0) + n2 Cv (T – 2T0) = 0, 3, T = T0, 2, , Q CP - CV = R, Þ CV = C P - R = 7 - 2 = 5, , cal, mole - kelvin, , Pf =, , \ (DQ)V = mCV DT, = 2 ´ 5 ´ (35 - 30) = 50 cal, 21. (a) Average kinetic energy per molecule per degree of, freedom = 1 kT . Since both the gases are diatomic, 2, and at same temperature (300 K), both will have the, same number of rotational degree of freedom i.e. two., Therefore, both the gases will have the same average, rotational kinetic energy per molecule, , 1, æ, ö, ç = 2 ´ kT = kT ÷ ., 2, è, ø, Thus, , 22. (a), Coefficient of volume expansion at constant pressure is, 1, for all gases. The average transnational K.E. is same, 273, , for molecule of all gases and for each molecules it is, , 23. (b), v=, , v rms =, , kT, 2πd 2 P, , 3RT, , vP =, M, , 26. (c), , PT, i f, Ti, , =, , 3, P0, 2, , DQ = n1Cv (T f – T0 ), =, , P0V0 3, æ3, ö 3, ´ R ´ ç T0 – T0 ÷ = P0V0, 2 RT0 2, 2, è, ø 8, , 27. (c) Let DV is change in volume in any compartment then, æV, ö, Pf ç 0 – DV ÷, P0V0, 2, è, ø and, n1 =, =, 2 RT0, RT f, æV, ö, Pf ç 0 + DV ÷, 2 P0V0, 2, è, ø, n2 =, =, Þ DV = 0, 2 RT0, RT f, , E1 1, =, E2 1, , Mean free path λ =, , PA μ A, =, ¹ 1 i.e. PA ¹ PB ., PB μ B, , 3, kT, 2, , (as P decreases, l increases), 2RT, = 0.816 vrms, M, , 8RT, = 0.92 v rms Þ v P < v < v rms, πM, , 1, 1 3, 3, mv 2rms = m v 2P = mv2P, 2, 2 2, 4, 24. (d) According to problem mass of gases are equal so, , Further E av =, , 28. (b) Internal energy of an ideal gas does not depend upon, volume of the gas, because there are no forces of, attration/repulsion amongest the molecular of an ideal, gas., Also internal energy of an ideal gas depends on, temperature., 29. (b) Helium is a monoatomic gas, while oxygen is diatomic., Therefore, the heat given to helium will be totally used, up in increasing the translational kinetic energy of its, molecules; whereas the heat given to oxygen will be, used up in increasing the translational kinetic energy, of the molecule and also in increasing the kinetic energy, of rotation and vibration. Hence there will be a greater, rise in the temperature of helium., 30. (d) Maxwell speed distribution graph is asymmetric graph,, because it has a long “tail” that extends to infinity., Also vrms depends upon nature of the gas and it’s, temperature., dN, dv, , number of moles will not be equal i.e. μ A ¹ μ B, From ideal gas equation PV = μRT Þ, , PA VA PB VB, =, μA, μB, , [As temperature of the container are equal], From this relation it is clear that if PA = PB then, , vmp vav vrms, , v
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 27, , 77, , This excess pressure is responsible for providing the, restoring force (F) to the piston of mass M., Hence F = DP. A =, , pö, æ, and y3 = a sin ç wt + ÷, è, 4ø, On superimposing, resultant SHM will be, , PAx, h, , é æ, pö, æ, y = a êsin ç wt - ÷ + sin wt + sin ç wt +, è, ø, è, 4, ë, , PA, Comparing it with F = kx Þ k = M w =, h, 2, , p, é, ù, = a ê2 sin wt cos + sin wt ú = a [ 2 sin wt + sin wt], 4, ë, û, , PA, Mh, Þ T = 2p, Mh, PA, 20. (b) Time taken by particle to move from x = 0 (mean, position) to x = 4 (extreme position), Þw=, , = a (1 + 2) sin wt, Resultant amplitude = (1 + 2)a, Energy in SHM µ (Amplitude)2, , T 1.2, =, = 0.3 s, 4, 4, Let t be the time taken by the particle to move from, x = 0 to x = 2 cm, =, , y = a sin wt Þ 2 = 4sin, , 2, , E Resultant æ A ö, = ç ÷ = ( 2 + 1)2 = (3 + 2 2), \ E, è aø, Single, , Þ Eresultant = (3 + 2 2) Esingle, , 2p, 1, 2p, t Þ = sin, t, T, 2, 1.2, , a, , p 2p, Þ =, t Þ t = 0.1s ., 6 1.2, , 21. (b), , 45°, , OR, , Hence time to move from x = 2 to x = 4 will be equal, to 0.3 - 0.1 = 0.2 s, Hence total time to move from x = 2 to x = 4 and, back again = 2 ´ 0.2 = 0.4sec, , a, , a, , a+a 2, , º ¾¾¾¾, ®, , 45°, , 25. (a) Acceleration µ - displacement, and direction of, acceleration is always directed towards the equilibrium, position., 26. (d), 27. (b), Compare given equation with, , Force constant (k) µ, , 22. (c), , y = Kt 2 Þ, , d2 y, dt, , 2, , Dividing,, , T1, =, T2, , l, and T2 = 2p, g, g + ay, g, , Þ, , dt, , 2, , + w2 x = 0 ; w2 =, , b, a, , x = A sin (wt + f) = A cos, , b, t, a, , 28. (b), x = a sin ωt and v =, , 29. (b), , = a y = 2K = 2 ´ 1 = 2 m/s 2 (Q K = 1m/s 2 ), , Now, T1 = 2p, , d2 x, , a max w 2 A, b, =, =w =, wA, vmax, a, At t = 0, f = p/2, , 1, Length of spring, , 2, l, 3, K l1 3, Þ, = =, Þ K1 = K ., K1 l, l, 2, , pöù, 4 ÷ø úû, , l, +, g, ( ay ), , T2 6, 6, Þ 12 =, 5, T2 5, , dx, = aω cos ωt, dt, , p, ., 2, 30. (a) The total energy of S.H.M. = Kinetic energy of particle, + potential energy of particle., The variation of total energy of the particle in SHM, with time is shown in a graph., , It is clear that phase difference between ‘x’ and ‘a’ is, , Energy, , Zero slope, , A, , Total energy, , 23. (a) At x = 0, v = 0 and potential energy is minimum so, particle will remain at rest., 24. (d) Let simple harmonic motions be represented by, pö, æ, y1 = a sin ç wt - ÷ , y2 = a sin wt, è, 4ø, , Kinetic energy, , Potential energy, , T/4, , 2T/4, , 3T/4
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t.me/Magazines4all, DPP/ P 28, , 78, , 1., , 28, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , (b) When the particle of mass m at O is pushed by y in, the direction of A . The spring A will compressed by, y while spring B and C will be stretched by, , Putting v = 2 gh and w =, , 2 p 2p, =, =p, T, 2, , we get T = m( g + p 2 gh), , y ' = y cos 45o . So that the total restoring force on the, mass m along OA., , k, , FB, , FC, , C, , O, , B, , k, , 5., , m, , Force constant (k ) µ, , FA, , Fnet = FA + FB cos 45o + FC cos 45o, , 6., , (b) Initially time period was, T = 2p, , = ky + 2ky ' 45o = ky + 2k ( y cos 45o ) cos 45o = 2ky, Also, Fnet = k ' y Þ k ' y = 2ky Þ k ' = 2k, , l, g, , When train accelerates, the effective value of g becomes, , ( g 2 + a 2 ) which is greater than g., Hence, new time period, becomes less than the initial time, period., , m, m, T = 2p, = 2p, k', 2k, (b) When mass 700 gm is removed, the left out mass (500, + 400) gm oscillates with a period of 3 sec, , \ 3 = t = 2p, , 1, Length of spring, , 2, l, l1 3, 3, k, Þ, = =, Þ k1 = k, k1, l, l, 2, , A, , 2., , (b), , (500 + 400, k, , ......(i), When 500 gm mass is also removed, the left out mass, is 400 gm., \ t ' = 2p, , 400, k, , ...(ii), a, , 3., , 3, 900, =, Þ t ' = 2sec, t', 400, (a) Slope is irrelevant hence, , 4., , æM ö, T = 2p ç ÷, è 2k ø, (a) Tension in the string when bob passes through lowest, , Þ, , geff, , 1/ 2, , point T = mg +, , mv 2, = mg + mvw (\ v = r w), r, , 7., , g, , (b) In accelerated frame of reference, a fictitious force, (pseudo force) ma acts on the bob of pendulum as, shown in figure., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 28, , 80, 21., , (c) Energy of particle is maximum at resonant frequency, i.e., w2 = w0 . For amplitude resonance (amplitude, maximum) frequency of driver force, w = w02 - b 2 2 m2 Þ w1 ¹ w0, , 22., 23., , Þ T - Mg cosθ =, , 4p 2, , A = mg , T = 0.2 sec,, T2, At t = 0.05 sec., , v, 2, , 1, 1, 1, 1, mv 2 = mV 2 + mV 2 + kx 2, 2, 2, 2, 2, Where x is the maximum compression of the spring., , Also, , æ mö, On solving the above equations, we get x = v ç ÷, è 2k ø, At maximum compression, kinetic energy of the, A – B system =, 24., , y = A sin wt = 1 sn, , 28., 29., , 1, = 0.1 Joule, 100, (c) Statement -1 is False, Statement-2 is True., (a) The time period of a oscillating spring is given by,, , 30., , m, 1, ÞT µ, k, k, Since the spring constant is large for hard spring,, therefore hard spring has a less periodic time as, compared to soft spring., (d) Time period of simple pendulum of length l is,, , 1, 1, mv2, mV 2 + mV 2 = mV 2 =, 2, 2, 4, , –f +f, q, , L, , T = 2p, , T = 2p, , mg sinq, , l, ÞT µ l, g, , Þ, , DT 1 Dl, =, T, 2 l, , \, , DT 1, = ´ 3 = 1.5%, T, 2, , mg mg cosq, From following figure it is clear that, , 2p, × 0.05cm. = 1cm., 0.2, , PE = mgy = 1 × 10 ×, , 1/ 2, , (d), , L, , Mv 2, L, , Also tangential acceleration a T = g sin θ., 25. (a) Except (4) all statements are wrong., 26. (b), 27. (b)., For minimum time period w2A = mg, , c, ; when b = 0, a = c, amplitude, (b) A =, a+b-c, A ® ¥. This corresponds to resonance., (b) Let the velocity acquired by A and B be V, then, mv = mV + mV Þ V =, , T - Mg cos θ = Centripetal force, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 29, , 82, 16., , (b) With path difference, , l, , waves are out of phase at the, 2, , For the given wave ω = 2πn = 15π, k =, , point of observation., 2, , 2, , 2, , 2, , 17., , (b), , 18., , (c) For interference, two waves must have a constant phase, relationship. Equation ‘1’ and ‘3’ and ‘2’ and ‘4’ have, p, a constant phase relationship of out of two choices., 2, , 19., , Now v =, , 1, 2p, A = a = a + a + 2a cos q Þ cos q = - Þ q =, 2, 3, 2, , Only one S2 emitting ‘2’ and S4 emitting ‘4’ is given, so only (c) option is correct., (a) The resultant amplitude is given by, , 20., , (d), , y=, , 1, a, , sin wt ±, , p2 v, wBA, 1, , I = wp m A Þ I = m, v, 2, 2B, 25. (d), 26. (c), 27. (a)., , 24., , Iinitial 9, I, = Þ Ifinal = initial, Ifinal, 1, 9, (ii) During the first half time, wavelength first increases as, the component of velocity of source increases till it becomes equal to the velocity of source itself, then it decreases till it becomes zeros., , pö, æ, sin ç wt + ÷, è, 2ø, b, , 1, , (iii) t1 =, , p, 2, \ The resultant amplitude, , Here phase difference =, , 2, , 30°, , (a) In a wave equation, x and t must be related in the form, , We rewrite the given equations y =, , t1, , 1 + ( x - vt ) 2, , For t = 2 , this becomes, [1 + ( x - 2v ) 2 ], , =, , 28., , 1, , [1 + ( x - 1) 2 ], , (c), , y, l, , 29., , R, , P, , x, l/2, 23., , Q, , (b), Standard wave equation which travel in negative x-direction is y = A sin ( ωt + kx + f0 ), , p, rad / s , t1 = 1, 7, 13, ....... t2 = 5, 11, 17,, 3, (c) The velocity of every oscillating particle of the medium, is different of its different positions in one oscillation, but the velocity of wave motion is always constant i.e.,, particle velocity vary with respect to time, while the, wave velocity is independent of time., Also for wave propagation medium must have the, properties of elasticity and inertia., , when w =, , 2v = 1or v = 0 m/s, 22., , 60°, , 1, , 1, For t = 0 , this becomes y =, , as given, (1 + x 2 ), , 1, , 60°, , t=0, , 30°, , ( x - vt ), , y=, , 5p / 3 2p + 5p / 3, p / 3 2p + p / 3, ,, ....... , t 2 =, ,, w, w, w, w, , t2, , a+b, æ 1 ö æ 1 ö, 1 1, = ç, + =, ÷ +ç, ÷ =, a b, ab, è aø è bø, , 21., , pm =, , (a), , 1, (i) I µ, , dinitial = R, dfinal = 3R,, d2, , (\1 + cos q = 2 cos 2 q / 2), , 2, , Coefficient of t ω 15π, = =, = 1.5m / sec, Coefficient of x k 10π, , 2π 2π, =, = 0.2 m., k 10π, , and λ =, , AR = A2 + A2 + 2 AA cos q = 2 A2 (1 + cos q ), = 2 A cos q / 2, , 2π, = 10π, λ, , 30., , Distance travelled by wave (l), Time period (T ), Wavelength is also defined as the distance between two, nearest points in phase., (b) Transverse waves travel in the form of crests and, through involving change in shape of the medium. As, liquids and gases do not possess the elasticity of shape,, therefore, transverse waves cannot be produced in, liquid and gases. Also light wave is one example of, transverse wave., (b) Velocity of wave =, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 30, , 84, Hence number of beats heard per second, , OA i.e y = 0, 0 £ x £ L, , æ v ö, æ v ö, n-ç, n, =ç, è v - vs ÷ø, è v + vs ÷ø, , AB i.e. y = L, 0 £ y £ L, BC i.e. y = L, 0 £ x £ L, , =, 16., 17., , 2nuus, , =, , u2 - u2s, , OC i.e. x = 0, 0 £ y £ L, The above conditions are satisfied only in alternatives, (b) and (c)., , 2 ´ 256 ´ 330 ´ 5, = 7.8 Hz, 335 ´ 325, , (a) In closed pipe only odd harmonics are present., (a) Maximum pressure at closed end will be atmospheric, pressure adding with acoustic wave pressure, , Note that u ( x, y) = 0 for all four values e.g. in, alternative (d ), u ( x, y) = 0 for y = 0, y = L But it is not, , So Pmax = PA + P0 and Pmin = PA - P0, , zero for x = 0 or x = L . Similarly in option (a), , Pmax PA + P0, =, Thus P, PA - P0, min, , x = 0 or y = 0 , while in options (b) and (c),, , u ( x, y) = 0 at x = L, y = L but it is not zero for, , 1, 1, (nopen ) = ´ 320 = 160 Hz, 2, 2, , 18., , (b) nclosed =, , 19., , (a) l = 1.21Å, , A, , N, , N, , (a), , 21., , (c) Loudness depends upon intensity while pitch depends, upon frequency., , 22., , (d) Using l = 2 ( l2 - l1 ) Þ v = 2n ( l2 - l1 ), , 7, 2, , 300 301, , 25., , 303, , 2, 307, , 28., , 1, 308, , 8, 8, (a) Doppler shift doesn't depend upon the distance of listner, from the source., (b) Since the edges are clamped, displacement of the edges, , 29., , u ( x, y) = 0 for line -, , y, (0,L), , C, , B, , (L,L), , 3 ´ 2.87 ´ 10 4, s, , = 1.2 × 106 Þ s = 0.07175 cm., , 3, , 7, 5, , 5, 308 300 305, , Y 1 2.87 ´ 10 4, ´, =, r 2s, s, , For 3rd harmonic, f = 3f0 = 1.2 × 106 Hz Þ, , Þ 2 ´ 215(63.2 - 30.7) = 33280cm / s, Actual speed of sound v0 = 332m / s = 33200cm / s, Hence error = 33280 – 33200 = 80cm / s, 3, , 24., , v 2.87 ´ 104, From f0 = =, Þ, s, l, Þ Y = 8.76 × 1012 N/m2, , æn ö, n, 1 l, 270, n µ Þ 2 = 1 Þ l2 = l1 ç 1 ÷ = 50 ´, = 13.5cm, l, l1 n2, 1000, è n2 ø, , 1, , Y, r where Y is Young's modu-, , lus of quartz and r is its density., , 20., , (c), , l, = s Þ l = 2s, 2, , Velocity of waves is, v =, , 1.21 Å, , 23., , u ( x, y) = 0 for x = 0, y = 0, x = L and y = L, 27. (d)., , For fundamental force,, A, , N, , 26. (b),, , 30., , (d) As emission of light from atom is a random and rapid, phenomenon. The phase at a point due to two, independent light source will change rapidly and, randomly. Therefore, instead of beats, we shall get, uniform intensity. However, if light sources are LASER, beams of nearly equal frequencies, it may possible to, observe the phenomenon of beats in light., (d) The person will hear the loud sound at nodes than at, antinodes. We know that at anti-nodes the displacement, is maximum and pressure change is minimum while at, nodes the displacement is zero and pressure change is, maximum. The sound is heared due to variation of, pressure., Also in stationary waves particles in two different, segment vibrates in opposite phase., (a) Stationary wave, , A, N, O, , A, (0, 0), , x, , N, , N, , A, A node is a place of zero amplitude and an antinode is, a place of maximum amplitude., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 31, , 85, , 1., , 31, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , A, , (b), 20 division of vernier scale = 8 div. of main scale, , æ 8ö, æ 2ö, Þ 1 V.S.D. = çè ÷ø M.S.D. = çè ÷ø M.S.D., 20, 5, , c=, , b, , a+ b+ c, 3, , B, , Least count, , (Q 1 M.S.D. =, , 1, cm. = 0.1 cm.), 10, , Directly we can use, , æ b -aö, L.C. = M - V = ç, M, è b ÷ø, 3, æ 20 - 8 ö æ 1 ö, cm. = 0.06 cm., = çè, ÷ø çè ÷ø cm. =, 20, 10, 50, 2. (c) Within elastic limit it obeys Hooke's Law i.e., stress µ, strain., 3., , (c) Least count = 1 ´ 1 cm = 1, N 10, 10N, , 4., , (b) 5th division of vernier scale coincides with a main scale, , 5., , 1, = 0.1mm, division. L.C. =, 10, \ Zero error = – 5 × 0.1 = – 0.5 mm, This error is to be subtracted from the reading taken for, measurement. Also, zero correction = + 0.5 mm., (b) If Y = Young's modulus of wire, M = mass of wire,, g = acceleration due to gravity, x = extension in the wire,, A= Area of cross-section of the wire,, l = length of the wire., , Y=, , Þ, , 6., 7., , curvature is given by r =, , C, , c2 h, +, 6h 2, , Pitch, No. of circular divisions, 9., (b) The specific heat of a solid is determined by the method, of mixture., 10. (a), , 8., , (b) L.C. =, , 11., , (d) Least count of screw gauge =, , 0.5, mm = 0.01mm, 50, , \ Reading = [Main scale reading + circular scale, reading × L.C] – (zero error), , = [3 + 35 × 0.01] – (–0.03) = 3.38 mm, 12. (d) 30 Divisions of vernier scale coincide with 29 divisions, of main scales, 29, MSD, 30, Least count = 1 MSD – 1VSD, , Therefore 1 V.S.D =, , = 1 MSD -, , 29, MSD, 30, , 1, , MSD, 30, 1, ´ 0.5° = 1 minute., =, 30, , =, , 0.5, = 0.01mm, 50, Zero error = 5 × L.C, = 5 × 0.01 mm, = 0.05 mm, Diameter of ball = [Reading on main scale] + [Reading, on circular scale × L . C] – Zero error, = 0.5 × 2 + 25 × 0.01 – 0.05, = 1.20 mm, , 13. (c) Least count =, , Mgx, DY DM Dx DA Dl, Þ, =, +, +, +, Al, Y, M, x, A, l, , DY 0.01 0.01 2 ´ 0.001 0.001, =, +, +, +, = 0.065, Y, 3.00 0.87, 0.041, 2.820, , DY, ´ 100 = ± 6.5%, Y, (b) The instrument has negative zero error., (c) If A, B and C be the points corresponding to the, impressions made by the legs of a spherometer then, , or, , a, , If h is the depression or elevation then the radius of, , æ 2ö, = 1 M.S.D – 1 V.S.D. = 1 M.S.D. – çè ÷ø M.S.D, 5, 3, æ 2ö, æ 3ö, = çè1 - ÷ø M.S.D. = çè ÷ø M.S.D. = ´ 0.1 cm. = 0.06 cm., 5, 5, 5, , d, , 14. (d), , Dg Dl, DT, =, +2, g, l, T, , Dl and DT are least and number of readings are, maximum in option (d), therefore the measurement of, g is most accurate with data used in this option.
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t.me/Magazines4all, DPP/ P 31, , 86, 15., , mg, , (b) We know that Y =, , p, Þ, , Y=, , 4mgL, 2, , pD l, , =, , ´, , 2, , D, 4, , L, l, , 18., 4 ´ 1 ´ 9.8 ´ 2, , (, , p 0.4 ´ 10 -3, , ) ´ (0.8 ´ 10 ), 2, , 2, , 2, , dy, æ dy ö, + xç ÷ - y, =0, 2, è, ø, dx, dx, dx, (a) The condition for terminal speed (vt) is, Weight = Buoyant force + Viscous force, , Þ xy, , -3, , d y, , B=Vr2 g, , Fv, , = 2.0 ´ 1011 N/m 2, , Now DY = 2DD + Dl, l, Y, D, [Q the value of m, g and L are exact], 0.01 0.05, +, = 2 × 0.025 + 0.0625, 0.4 0.8, = 0.05 + 0.0625 = 0.1125, Þ DY = 2 × 1011 × 0.1125 = 0.225 × 1011, , W=V r 1g, , = 2´, , 16., , = 0.2 ´ 1011 N/m 2, (b) The time period of a simple pendulum is given by, , T = 2p, , 19., , l, l, l, \ T 2 = 4p 2 Þ g = 4p 2 2, g, g, T, , Dg, Dl, DT, ´ 100 =, ´ 100 + 2, ´ 100, l, g, T, , Þ, , Case (i), Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 128s, , \, , Vg (r1 - r1 ), k, (d) From the figure it is clear that liquid 1 floats on liquid, 2. The lighter liquid floats over heavier liquid., , \ V r1 g = V r2 g + kvt2, , 20., , Dg, ´ 100 = 0.3125, g, , Therefore we can conclude that r1 < r2, Also r3 < r2 otherwise the ball would have sink to the, bottom of the jar., Also r3 > r1 otherwise the ball would have floated in, liquid 1. From the above discussion we conclude that, r1 < r3 < r2., (c) In case of water, the meniscus shape is concave, upwards. Also according to ascent formula, , h=, , Case (ii), Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 64s, , \, , Dg, \, ´ 100 = 1.055, g, Clearly, the value of, , vT, , 2, , 0.2, , =, , l, , (c), , A, , 2, , 17., , 21., , 2r ( d1 - d 2 ) g, 9h, , (10.5 - 1.5), 9, Þ vT = 0.2 ´, 2, (19.5 - 1.5), 18, , \ vT = 0.1 m / s, 2, , 2, ìï, d2y, dy, æ dy ö üï, x í - By 2 - B ç ÷ ý + By, =0, è, ø, dx ïþ, dx, dx, ïî, , Y, Wire (1), , Dg, ´ 100 will be least in case (i), g, , (c) Terminal velocity, vT =, , 2T cos q, rrg, , The surface tension (T) of soap solution is less than, water. Therefore rise of soap solution in the capillary, tube is less as compared to water. As in the case of, water, the meniscus shape of soap solution is also, concave upwards., , Dg, ´ 100 = 0.46875, g, , Case (iii), Dl = 0.1 cm, l = 20cm, DT = 0.1s, T = 36s, , \ vt =, , 3A, , Y, l/3, Wire (2), , As shown in the figure, the wires will have the same, Young’s modulus (same material) and the length of the, wire of area of cross-section 3A will be l/3 (same, volume as wire 1)., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 31, , 87, , For wire 1,, Y=, , For wire 2 ,, , F/A, D x/l, , Y=, , 25. (a) Screw gauge is used to measure the diameter (d) of the, wire so that the area of cross-section is calculated by the, formula, , ...(i), , F '/ 3 A, Dx /( l / 3), , ...(ii), , F l, F', l, ´, =, ´, Þ F ' = 9F, A Dx 3 A 3Dx, 22. (b) Lower the vernier constant, more accurate measurement, is possible by it., , From (i) and (ii) ,, , 23. (b) Effective length = MC = MN + NC = l +, , M, l, , N, d/2, C, 24. (c) Here, original length (L) = y,, Extension (l) = x, Force applied (F) = p, Area of cross-section (A) = q, Now, Young's modulus (Y) =, ÞY=, , yp, xq, , pd 2, 4, 26. (b) Both the statements (1) & (2)are precautions to be taken, during the experiment., 27. (a) The liquid cools faster first and slowly later on when its, temperature gets close to surrounding temperature., 28. (a) Maximum percentage error in measurement of e, as given, by Reyleigh’s formula., (Given error is measurement of radius is 0.1 cm), De = 0.6 DR = 0.6 × 0.1 = 0.06 cm., Percentage error is, A=, , FL, AL, , d, 2, , De, 0.06, ´ 100 =, ´ 100 = 3.33%, e, 0.6 ´ 3, 29. (b) Speed of sound at the room temperature., l1 = 4.6 cm, l2 = 14.0 cm.,, l = 2 (l2 – l1) = 2 (14.0 – 4.6) = 18.8 cm., 18.8, = 376 m / s, 100, 30. (c) End correction obtained in the experiment., , v = f l = 2000 ´, , e=, , l 2 - 3l1 14.0 - 3 ´ 4.6, =, = 0.1 cm., 2, 2
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 33, , 91, , 1., , 2., , (c) ABCDE is an equipotential surface, on equipotential, surface no work is done in shifting a charge from one, place to another., (c) Potential at centre O of the square, , Q, , Q, a, , a, , 2, , 9., , (b) Potential at A= Potential due to (+q) charge, + Potential due to (– q) charge, , =, , Q, , W = -Q(V¥ - V0 ) = QV0, , x, , V=, , 11., , 4., , ur, æ ¶V $ ¶V $ ¶V ö, Now E = - ç $i, +j, +k, ÷, ¶y, ¶z ø, è ¶x, , 8., , =–, , KE = q(V1 - V2 ) = 2 ´ (70 - 50) W = 40 eV, , (a) The electric potential V ( x, y , z ) = 4 x 2 volt, , 7., , 1 (-6 ´ 10-6 ), 1 (12 ´ 10-6 ), +, = 0 Þ x = 0.2 m, 4pe0, x, 4pe0 (0.2 + x), , -6, -6, é, and V B = 1 0 1 0 ê ( 2 ´ 1 0 ) - 5 ´ 1 0, -2, 5 ´ 10 -2, êë 1 5 ´ 1 0, , (a) Work done in moving a charge from P to L , P to M, and P to N is zero while it is q (VP – Vk) > 0 for motion, from P to k., , 6., , ¶V, ¶V, ¶V, =0, Now ¶x = 8 x, ¶y = 0 and, ¶z, ur, Hence E = -8$i, so at point (1m, 0, 2m), ur, E = -8 xi$ volt/meter or 8 along negative X - axis., (b) Electric fields due to electrons on same line passing, through centre cancel each other while electric potential, due to each electron is negative at centre C. Therefore,, ur, at centre E = 0,V ± 0, ur r, (a) By using W = Q( E.Dr ), Þ W = Q[(e1$i + e2 $j + e3 k$ ).(ai$ + b $j )] = Q( e1a + e2b ), , 20 cm, , é (-5 ´10-6 ) 2 ´10-6 ù 1, 6, VA = 1010 ê, +, ú = ´10 volt, -2, -2, 5 ´ 10 úû 15, êë 15 ´ 10, , v, QA, 2QV, 1, q, Þvµ Q Þ A =, =, =, M, vB, QB, 4q 2, , (a), , 12mC, , q = 3 ´10-6 coulomb where, , (b) Using, , 5., , =0, , (a) Work done W = q -6 (V A - VB ); where, , 4 2Q, 2Q 2, = Q., =, 4pe0a pe 0a, , v=, , a 2 + b2, , -6mC, , P, , Work done in shifting (– Q ) charge from centre to, infinity, , (-q), , 10. (c) Point P will lie near the charge which is smaller in, magnitude i.e. -6mC . Hence potential at P, , æ, ö, Q, V0 = ç, ç 4pe (a / 2) ÷÷, 0, è, ø, , 3., , q, 1, 1, ., +, 4pe0 a2 + b 2 4pe0, , O, , Q, , 33, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , ù, ú, úû, , 13, ´ 106 volt, 15, , é1, æ 13, öù, \ W = 3 ´ 10-6 ê ´ 106 - ç ´ 106 ÷ ú, è 15, øû, ë15, , = 2.8 J, 12. (c), , A, , p, , +q, l, , B, +q, , Pnet =, , p net, , l, , l, , C, – 2q, , 60°, p, , p 2 + p 2 + 2 pp cos 60o = 3 p = 3 ql (\ p = ql )
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 33, , 93, y, , B, q/3, 30°, , C, –2q/3, , F–, x, F+, , O, , A, q/3, The potential energy of the system is non zero, Force between B & C, =, , F, , 60°, , 120°, 60° 60°, 30°, 60°, , Net force F :, , 1 (q / 3) ( -2q / 3), q2, =, 4 pe0, ( 3R) 2, 54pe 0 R 2, , Potential at O =, , 1 æ q q 2q ö, + =0, 4pe 0 çè 3 3 3 ÷ø, , 24. (d) The given graph is of charged conducting sphere of, radius R0. The whole charge q distributes on the surface of sphere, 25 (b), 26 (b), 27 c, +, , F+, , –, F–, , (F+ > F– as E+ > E–), , Net torque immediately after it is released Þ clockwise, A body cannot exert force on itself., 28. (d) When the bird perches on a single high power line, no, current passes through its body because its body is at, equipotential surface i.e., there is no potential, difference. While when man touches the same line,, standing bare foot on ground the electrical circuit is, completed through the ground. The hands of man are, at high potential and his feet’s are at low potential., Hence large amount of current flows through the body, of the man and person therefore gets a fatal shock., 29. (a) Electron has negative charge, in electric field negative, charge moves from lower potential to higher potential., 30. (b) Potential is constant on the surface of a sphere so it, behaves as an equipotential surface. Free charges, (electrons) are available in conductor. The two, statements are independent.
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t.me/Magazines4all, DPP/ P 35, , 96, , 1., , (a) By using, V = V0 e -t / CR Þ 40 = 50e -1/ CR Þ e -1/ CR = 4 / 5, , across A is V and potential difference across B is, , Potential difference after 2 sec, , hence energy of the system now is, 2, , 2., , 35, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 2, , æ4ö, V ' = V0 e-2 / CR = 50(e-1/ CR )2 = 50 ç ÷ = 32V, è5ø, Fraction of energy after 1 sec, , 1, 1, 10, æV ö, U 2 = (3C )V 2 + (3C ) ç ÷ = CV 2 ......(ii), 2, 2, 6, è3ø, , 1, C (V f ) 2, 2, 16, æ 40 ö, = 2, =ç ÷ =, 1, è, ø, 50, 25, C (Vi ) 2, 2, , So,, 4., , U1 3, =, U2 5, , (c) Plane conducting surfaces facing each other must have, equal and opposite charge densities. Here as the plate, areas are equal, Q2 = -Q3 ., , (c) The given circuit can be redrawn as follows. All, capacitors are identical and each having capacitance, , The charge on a capacitor means the charge on the, inner surface of the positive plate (here it is Q2 ), , e A, C= 0, d, , Potential difference between the plates, , =, , 1 2, , 4, 4, 5, , + –, V, , charge, Q, 2Q, = 2 = 2, capacitance, C, 2C, , Q - ( -Q2 ) Q2 - Q3, = 2, =, 2C, 2C, , 3 2, , 5., , 6., , (b) While drawing the dielectric plate outside, the, capacitance decreases till the entire plate comes out, and then becomes constant. So, V increases and then, becomes constant., (b) Given circuit can be reduced as follows, , 3C, , 3C, B, , A, | Charge on each capacitor | = | Charge on each plate |, , ( C = capacitance of each capacitor), , e A, = 0 V, d, Plate 1 is connected with positive terminal of battery, , e0 A, .V, d, Plate 4 comes twice and it is connected with negative, terminal of battery, so charge on plate 4 will be, , so charge on it will be +, , 2e 0 A, V, d, (c) Initially potential difference across both the capacitor, is same hence energy of the system is, , 7., , 1, 1, CV 2 + CV 2 = CV 2, .....(i), 2, 2, In the second case when key K is opened and dielectric, medium is filled between the plates, capacitance of both, the capacitors becomes 3C, while potential difference, , The capacitor 3C ,3C shown in figure can with stand, maximum 200 V ., \ So maximum voltage that can be applied across, A and B equally shared. Hence maximum voltage applied, cross A and B be equally shared. Hence Maz. voltage, applied across A and B will be (200 + 200) = 400 volt., (c) Common potential, V'=, , -, , 3., , V, 3, , C1V + C2 ´ 0, C1, =, .V, C1 + C2, C1 + C2, C1V1 - C2V2 6 ´ 12 - 3 ´ 12, =, = 4 volt, C1 + C2, 3+ 6, , 8., , (b) V =, , 9., , A, 2 = K1e0 A, (d) C1 =, d, æd ö, ç ÷, è2ø, , U1 =, , K1e0, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 35, , 99, , 28. (b) The electric field due to one charged plate at the, location of the other is E =, , 30. (b) By the formula capacitance of a capacitor, , s, and the force per unit, 2e 0, , s2, ., area is F = sE =, 2e0, , 29. (d) A charged capacitor, after removing the battery, does, not discharge itself. If this capacitor is touched by, someone, he may feel shock due to large charge still, present on the capacitor. Hence it should be handled, cautiously otherwise this may cause a severe shock., , C1 = e0 ´, , Hence,, or, , KA K, µ, d, d, , C1 K1 d 2, K, d /2 1, =, ´, = 1´, =, C2 d1 K 2 K 2 3K, 6, , C2 = 6C1, , Q, V, Therefore, capacity of a capacitor does not depend, upon the nature of the material of the capacitor., , Again for capacity of a capacitor C =
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 36, , 101, , 3W, 12., , 2W, A, , X, , (c)., , 4W, B, , C, 4W, , Y, , 8W, , 13., , 4 8, + = 4ohm., 3 3, Corresponding to points X and Y, the resistances 3 ohm, 4, ohm and 6 ohm are in parallel, hence effective resistance, RXY is, , Now, RACB = RAC + RCB =, , 1, 1 1 1 4+3+ 2 9, = + + =, =, R XY 3 4 6, 12, 12, 12 4, = ohm., 9 3, , 2, =1A, 2, Power dissipated in the circuit = i2 R = 1 × 2 = 2 watts, Potential difference between X and, 4 4, Y = i × RXY = 1 × = V.., 3 3, 4, V.., 3, , 4/3 4, = = 0.44 amp., 3, 9, , ( 30 + 30 ) 30 60 ´ 30, (c). Requivalent = 30 + 30 + 30 =, = 20 W, (, ), 90, i=, , 2, 1, V, =, = ampere, 20, 10, R, , P1, P2, R/4 1, =, =, =, P2, P1, R, 4, , 60 ´ 8 ´ 301, = 14.4, 1000, Hence cost = 14.4 ´ 1.25 = ` 18, , (c). Total kWh consumed =, , 15., , (d). Since all bulbs are identical they have the same, resistances. The current I flowing through 1 branches at, A. So current in 2 and 3, as well as in 4 will be less than I., The current through 5 is also I. Thus 1 and 5 glow equally, brightly and more than 2, 3 or 4., (b). Let R1 and R2 be the resistances of the coils, V the, supply voltage, Q the heat required to boil the water., Heat produced by first coil of resistance R1 in time t1, V 2 t1, V 2 ´ 6 ´ 60, (= 6 min) = Q =, =, cal, JR1, 4.2R1, , Current in the circuit =, , R, 4, , 14., , 16., , 4 2, Total resistance R of the circuit = + = 2W., 3 3, , \, , V2, . If resistance of heater coil is R, then resistance, R, , So, , 1 1 3, 8, RCB = + =, \RCB = ohm., 4 8 8, 3, , 11., , (c). P =, , of parallel combination of two halves will be, , 1, 1 1 3, 3, = + =, R AC 2 4 4 \RAC = 4 ohm, The effective resistance RCB between C and B, , Current in 3 ohm resistor =, , 4 ± 16 + 32 2 ± 2 3, =, 2, , R cannot be negative, hence R = 2 ± 2 3 = 5.46 W, , + –, 2V, 3W, The effective resistance RAC between A and C, , \ Potential difference across 3 ohm resistor =, , 2´ R, Þ 2R + R2 = 8 + 4R + 2R, 2+R, , R2 – 4R – 8 = 0 Þ R =, , Þ, , 6W, , \ RXY =, , R= 2+2 +, , ......(a), , Heat produced in second coil of resistance R2 in time, t2 (= 8 min), V 2 t1 V 2 ´ 6 ´ 60, =, =Q=, JR 2, 4.2R 2, , .....(b), , Equating (a) and (b), we get, 6, 8, R, =, i.e. 2 = 8 = 4, R2 R2, R1 6 3, 4, R, .....(c), 3 1, (i) When the two heating coils are in series, the effective, resistance is, , or R2 =, , 4, 7, .R = R ., 3 1 3 1, with two coils in series, let the kettle take t' time to boil., , R' = R1 + R2 = R1 +, , Q=, , V2 t ', =, JR ', , V2 t ', æ7 ö, 4.2 ´ ç R1 ÷, è3 ø, , .....(d)
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 37, , 105, , (1), , (d). Let R is resistance of the voltmeter. The effective, resistance across points A, B is, , (7), , V, 25, =, A, R 1000, Let R' be the required resistance to be connected in, series with voltmeter., , (a). \ i =, , 60 ´ R, ..............(1), 60 + R, The current in the circuit is I = 12/(50 + r), The p.d. across AB points is V = Ir, , r=, , 12, ×r, 50 + r, or r = 50 W, using it in (1),, , or 6 =, , or, , (8), , G, G, G, =, =, W, n - 1 100 - 1 99, , (b). S =, , (4), , 80 ´ 80, (c). R = 20 +, = 60W, 80 + 80, , (5), , 4V, , B, , VAB, , R2, 4kW, , 4kW, RV, , V, , Current flowing in the circuit, I=, , I=, , 1A, B, , 2A, , (6), , (), , 10, × 99 = 111W, 100 - 10, , 1, \ V = iR' =, × 40 = 1.33 volt., 30, (a). According to Kirchhoff's first law, At junection A, iAB = 2 + 2 = 4A, At junection B, iAB = iBC – 1 = 3A, , A, , K, , R1, 4kW, , A, , V 2, 1, =, =, amp., R 60 30, , 2A, , 25, 250, =, Þ R' = 9000 W., 1000 1000 + R ', (d). The potential difference between A and B in the absence, of voltmeter = 2 volt., , + –, , (3), , i=, , 25, A, 1000, , \, , (c). S =, , i - ig, , V', R+R', , Here V' = 250, R = 1000 W and i =, , ..............(2), , (2), , G=, , So i =, , 50 + r = 2r, , 60, we get 50 =, 60 + R, 300 + 5R = 6R, or R = 300 W, , ig, , 37, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 1.3 A, C, , i, At junection C, i = iBC – 1.3 = 3 – 1.3 = 17 amp, (b). The current required for a full-scale deflection of the, galvanometer is, i = 4.0 x 10-4 x 25 = 10-2 A, Let a resistance R W is to be connected in series, V, G+R, Here G = 50 W, V = 2.5 V and i = 10-2 A, , Then by the ohm's law, we have i =, , V, 2.5, = -2 = 250, \ G+R=, i 10, Þ R = 250 – G = 250 – 50 = 200W., , E, E, =, RR, R2 + R ', R2 + 1 V, R1 + R V, , 4, 2, = ampere, 4´4 3, 4+, 4+4, , Potential difference measured by voltmeter, 2, 4, ×2=, 3, 3, Error in the reading of voltmeter, , V'AB = IR' =, , = VAB – V'AB = 2 –, , 4, 2, = volt, 3, 3, , The error in voltmeter reading for 2 volt p.d. =, , 2, volt, 3, , The error in voltmeter reading for 1 volt p.d., 2 1 1, ´ = volt, 3 2 3, the error in voltmeter reading for 100 volt p.d., , =, , =, , 100, = 33.3% volt, 3
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t.me/Magazines4all, DPP/ P 37, , 106, (9), , (d). E = V + I r = I R + I r, Þ E = 0.25 x 10 + 0.25 × r, In second stage, Þ E = 0.5 × 4 + 0.5 r, Subtracting eq. (b) from eq. (a), 2.5 + 0.25 r – 2.0 – 0.5 r = 0, 0.5 = 0.25 r, , 13., , (b) Cells area joined in parallel when internal resistance is, higher then a external resistance. (R << r), E, , i=, , R+, , 0.5, = 2W., 0.25, (10) (d) Suppose current through different paths of the circuit, is as follows., , 14., , r, n, , (b). Current in the ammeter I =, , r=, , E, é R 'ù, R '+ r ê1 + ú, ë Rû, , On increasing the value of R, the denominator will, decrease and consequently the value of I will increase., 15., , (a), , I/6, , 54 W, , 28 W, , I, 6V, I3, , 1, , 8V, , I/3, , I/6, I/3, , 2, , I/3, 12 V, , I/6, I/3, , I/6, , I/3, , After applying KVL for loop (1) and loop (2), , and, , I/6, , 1, i1 = - A, 2, , We get 28i1 = – 6 – 8 Þ, , Let ABCDEFGH be skeleton cube formed of twelve, equal wires each of resistance R. Let a battery of e.m.f., E be connected across A and G. Let the total current, entering at the corner A and leaving the diagonally, opposite corner G be I. By symmetry the distribution, of currents in wires of cube, according to Kirchoff's I st, law is shown in fig. ApplyingKirchoff's IInd law to mesh, ADCGEA, we get, , 1, 54i2 = – 6 – 12 Þ i2 = - A, 3, , 5X + 2 × 10, Þ = 20 W, X + 10, (b) The circuit can be simplified as follows, , (11) (d) VAB = 4 =, , B, , i1, A, , i3, i2, , F, , i3, 40 W, 40 W, , –, , C, , 30 W, , I, , E, , 5, Hence i3 = i1 + i2 = - A, 6, , 12., , I/3, , I/3, , 1, 1, 1, R– R– R+E=0, 3, 6, 3, , 5, IR, ......(a), 6, If RAB is equivalent resistance between comers A and B,, then from Ohm’s law comparing (a) and (b), we get, , or E =, , D, , 40 V, , E, 80 V, , Applying KCL at junction A, i3 = i1 + i2, ...(i), Applying Kirchoff's voltage law for the loop ABCDA, – 30i1 – 40i3 + 40 = 0, Þ, – 30i1 – 40(i1 + i2) + 40 = 0, Þ, 7i + 4i2 = 0, ...(ii), Applying Kirchoff's voltage law for the loop ADEFA., – 40i2 – 40i3 + 80 + 4 = 0, Þ, – 40i2 – 40(i1 + i2) = – 120, Þ, i2 + 2i2 = 3, ...(iii), On solving equation (ii) and (iii) i1 = – 0.4 A., , IRAB =, 16., , 5, IR, 6, , (d)., , (y – z), , H, I = x + 2y, , y, , (y – z), , z, , A, , z, , x, (y – z), , y E, x + 2y B, , 2 (y–z), , D, , y, E, , G, , F, (y – z), , y, , C, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 37, , 107, , Let I = x + 2y current enter at point A, when a battery of, e.m.f. E and no internal resistance is connected across edge, AB. The edges AD and AH are symmetrically connected, to A, therefore they will carry equal currents. The, distribution of currents according to Kirchoff's Ist law is, shown in fig., If RAB is equivalent resistance, then from Ohm's law,, E = RAB I = RAB (x + 2y), ....(a), and from Kirchoff's law applied to mesh containing AB, and cell E is, Rx=E, .....(b), (since R is resistance of each wire), Applying Kirchoff's II law to mesh AHEB, yR + zR + yR – xR = 0 or x – 2y – z = 0, .....(c), Applying Kirchoff's II law to mesh DGFC, (y – z) R + 2 (y – z) R – zR = 0, or 4(y – z) –z = 0 or 4y = 5 z, .....(d), i.e. z = (4/5) y, .....(E), Substituting this value in (c), we get, , E = RAB, , 24, x= R. x, 14, , I, , I, , ·, , B, , 2, IR = E', 5, , [Using (b)], ......(c), , R ö, æ 2r + 2R ö, æ, I' = I çè, ÷ø = I çè 1 +, ÷, 2r + R, 2r + R ø, , G, I/2-I1, D, , y, E, , 1, 1, R + I1 R + R = E, 2, 2, , 7, 7, 7, IR i.e. RAB = R = × 2 = 2.8 W, 5, 5, 5, 18. (c). In the first case I = E/(r + R) and in the second case, I' = E/(r + R/2) = 2E/(2r + R), Using E = I(r + R), we get, , (I/2-I1), I1, , E, , 2, I, ...... (b), 5, nd, Applying Kirchoff's II law to external circuit AHEBE',, we get, , or 2I – 5I1 = 0 or I1 =, , RAB I =, , H, I/2, A, , æ1, ö, or 2 çè - I1÷ø + (I – 2I1) – I1 = 0, 2, , 7, IR = E, 5, Comparing (a) and (c), wet get, , 24, 7, RAB =, R \ RAB =, R., 14, 12, , 17. (a)., , æ1, ö, R + çè - I1÷ø R – I1 R = 0, 2, , or, , 14, 5, or, y = x i.e. y =, x, 5, 14, Substituting value of y in (a), we get, , E = RAB, , æ1, ö, çè - I1÷ø R + (I – 2I1), 2, , IR +, , 4, x – 2y – y = 0, 5, , 10 ö, æ, çè x + x÷ø, 14, , The distribution of currents, keeping in mind symmetry, condition, is shown in fig., Let R (= 2W) be the resistance of each wire., Applying Kirchoff's II law to mesh DGFC, we get, , I-2I1, I1, , I/2-I1, , F, , I/2, , I/2-I1, I/2, , C, , Let a battery of e.m.f. E is applied between points A and B., Let a current I, enter through point A., If RAB is equivalent resistance between points A and B,, then from Ohm's law, RAB I = E, , Thus the term in bracket is greater than 1 but less than, 2. Thus 2I > I' > I, 19. (b). Let R be the combined resistance of galvanometer and, an unknown resistance and r the internal resistance of, each battery. When the batteries, each of e.m.f. E are, connected in series, the net e.m.f. = 2E and net internal, resistance = 2r, 2 ´ 15, R + 2r, \ R + 2r = 3.0., ...(i), When the batteries are connected in parallel, the e.m.f., remains E and net internal resistance becomes r/2. therefore, , \ Current i1 =, , Current i2 =, , \ 2R + r =, , 2E, R + 2r, , E, R+, , r, 2, , =, , or 1.0 =, , 2E, 2R + r, , 2E 2 ´ 15, =, = 5.0, i2, 0.6, , Solving (i) and (ii), we get r = 1/3 W., , ...(i)
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t.me/Magazines4all, DPP/ P 37, , 108, 20., , (a). The circuit with current distribution is shown in fig., , 21., , C, , G, H, , 100W, E, i1, i, , D, , I=, , i, , F, , D, , (a). After full charging, the steady current in the condenser, is zero, hence no current will flow in 4W resistance., , 200W, i - i1, i, , B, , 24., , (d)., , Ig, , 110V, Applying Kirchoff's second law to the loop DEFGHID,, , I–Ig, , ......(1), , Now applying Kirchoff's second law to loop ADIHGCBA,, we have. (i – i1) 200 + i × 300 = 110, 500i – 200 i1 = 110, , .....(2), 25., , Solving eqs. (1) and (2), we get, i=, , 3, 1, amp and i1 = amp., 10, 5, , Current in 100 ohm resistance i1 =, , 1, amp., 5, , Current in 200 ohm resistance i – i1 =, Current in 300 ohm resistance i =, , 1, 10, , 26., , 27., 28., , 3, amp., 10, , resistance, \ VA – VC = current × resistance, , 29., , 1, × 100 = 20 volt., 5, , Potential difference between C and B is given by, 3, VC – VB = i × 300 =, × 300 = 90 volt., 10, , (d), , VAD VAC AD 4 2, =, =, = = ;, VAB VAB AB 6 3, , 200, cm., 3, (a) D is balance point, hence no current, (a) Voltameter measures current indirectly in terms of mass, of ions deposited and electrochemical equivalent of, , mö, æ, the substance çè I = ÷ø . Since value of m and Z are, Zt, , = Potential difference across 100 ohm, or potential difference across 200 ohm resistance, , For Ammeter IgG = (I – Ig) R, 50 × 10–6 × 100 = 5 × 10–3 × (R) Þ R » 1W, For voltmeter Ig (R + G) = V, 50 mA (R + G) = 10V Þ R + G = 200 kW Þ R » 200kW, (a) Potential at A = 6V, VA – VC = 4, Þ VC = 2V, , AD =, , Potential difference between A and C, , = i1 × 100 =, , G, R, , we have i1 × 100 – (i – i1) × 200 = 0, 300 i1 – 200 i = 0, , 6, 6, =, =1.5 A, æ 2 ´ 3 ö 28 + 12, 28 + ç, è 2 + 3 ÷ø, , Let current flowing in 2W resistance is I1, \ 2W and 3W resistance are connected in parallel, \ 2I1 = (1.5 – I1) x 3, 5I1 = 4.5, I1 = 0.9 amp., , 300W, , A, , E, =, R+R', , 30., , measured to 3rd decimal place and 5th decimal place, respectively. The relative error in the emasurement of, current by voltmeter will be very small as compared to, that when measured by ammeter directly., (c) The e.m.f. of a dry cell is dependent upon the electrode, potential of cathode and anode which in turn is, dependent upon the reaction involved as well as, concentration of the electrolyte. It has nothing to do, with size of the cell., So, statement-1 is false and statement-2 is true., (d) V = E – ir = 4 – 2 × 2 = 0, During charging V > E., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ P 38, , 109, , 17. (c) Potential gradient, x=, , V, e, R, =, L ( R + Rh + r ) L, , Þ 2.2 ´ 10-3 =, , 18. (a), , If P is slightly icnreased, potential of C will decrease., , 2.2, ´1 Þ R ' = 990W, (10 + Rh ), , E = xl = irl Þ i =, , Hence current will from A to C., If Q is slightly increased, potential of C will increase., Hence current will flow from C to A., 25-27, We have, , E 2.4 ´ 10-3, =, = 4 ´ 10-4 A, rl, 1.2 ´ 5, , Rs =, , 19. ( b) Give circuit is a balanced Wheaststone bridge circuit,, hence it can be redrawn as follows, , 12 ´ 6, RAB = (12 + 6 ) = 4 W., , 25., 28., , 20. (a) Balancing length is independent of the cross sectional, area of the wire., 21. (a) In meter bridge experiment, it is assumed that the, resistance of the L shaped plate is negligible, but, actually it is not so. The error created due to this is, called, end error. To remove this the resistance box and, the unknown resisance must be interchanged and then, the mean reading must be taken., 22. (a) Ammeter is always connected in series with circuit., 23. (a) In balanced Wheastone bridge, the arms of, galvanometer and cell can be interchanged without, affecting the balance of the bridge., 24. (d), W, , W, , W, , W, , V, 10.0V, - Rc =, - 20.0W = 9980W, Ifs, 0.00100A, , At full-scale deflection, Vab = 10.0V, voltage across the, meter is 0.0200 V, voltage across Rs is 9.98 V, and current, through the voltmeter is 0.00100 A. In this case most of the, voltage appears across the series resistor., The equivalent meter resistance is Req = 20.0 W + 9980 W, = 10,000W . Such a meter is described as a "1,000 ohmsper-volt meter" referring to the ratio of resistance to fullscale deflection. In normal operation the current through, the circuit element being measured is much greater than, 0.00100 A, and the resistance between points a and b in, the circuit is much less than 10,000W. So the voltmeter, draws off only a small fraction of the current and disturbs,, only slightly the circuit being measured., (d), 26., (c), 27. (b), (d) The resistance of the galvanometer is fixed. In meter, bridge experiments, to protect the galvanometer from, a high current, high resistance is connected to the, galvanometer in order to protect it from damage., , 29. (a) Sensitivity µ, , 1, µ (Length of wire), Potential gradiant, , 30. (a) If either the e.m.f. of the driver cell or potential, difference across the whole potentiometer wire is lesser, than the e.m.f. of the experimental cell, then balance, point will not obtained., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ P 40, , 113, , DAILY PRACTICE, PROBLEMS, , 40, , PHYSICS, SOLUTIONS, , (1), , (b) Ekp = eV, \ Ek = qV,, \ Ek µ q, \ V = constant, Ekp : Ekd : Eka : : 1 : 1 : 2., , (2), , (c) EK =, , q 2 r 2 B2, 2m, , Now r =, , \, , mp, q 2 qa2, =, ´, 2, m mp E kp, , or, , Ek µ, , =, , region between PQ and RS. The particle moves in a, circular path of radius r in the magnetic field. It can just, enter the region x > b for r ³ (b – a), , (3), , (a) B =, , (4), , (a) F = qvB sinq, , r2, , =, , 10-7 ´ 3 ´1.66 ´10-19, (2 ´10-10 )2, , (6), , (7), , = 1.2 Tesla., , = 1.28 × 10–14 N, [Q charge on a particle = 2e], r r, r, (b) The direction of F is along (V ´ B) which is towards, the right. Thus the beam deflects to your right side., (b) The particle is moving clockwise which shows that, force on the particle is opposite to given by right hand, palm rule of fleming left hand rule. These two laws are, used for positive charge. Here since laws are disobeyed,, we can say that charge is negative., (b) The point lies at the circumference hence it will come, back after a time period T, 2p m, qB, (a) The magnetic force on a current carrying wire of length, L, placed in a magnetic field B at an angleq with the, field is given by, F = i l Bsinq., Here B = 5.0 × 10–4 N/A.m. i = 2.0 A,, l = 50 cm = 0.50 m,, q = 60º, F = 2.0 × 0.50 × (5.0 × 10–4) × sin 60º, = 4.33 × 10–4 N, According to the flemings left - hand rule, this force, will act perpendicular to both the wire and the magnetic, field., , T=, , (8), , m, 1 4 16, :, :, Þ rH : rHe : ro =, =1:2:2, q, 1 1, 2, Radius is smallest for H+, so it is deflected most., (10) (b) In the figure, the z-axis points out of the paper, and the, magnetic field is directed into the paper, existing in the, (9), , (a, c), , rµ, , Þ vmin =, , r, , q, , q (b - a) B, m, , b, q, , q, , æ 1ö, = 2 × 1.6 × 10–19 × 105 × 0.8 × çè ÷ø, 2, (5), , q (b - a) B, , v³, , m, (11) (a) From figure it is clear that, , 4 1, ´ = E Ka = 8eV.., 1 4, , KVe, , mv, ³ (b - a ), qB, , d, p, d, also r = qB, r, , sin q =, , \ sin q =, , Bqd, p, , (12) (a) For on wire Q due to wire P is, 2 ´ 30 ´ 10, ´ 0.1 = 6 ´ 10– 5 N (Towards left), 0.1, Force on wire Q due to wire R is, , FP = 10, , -7, , ´, , 2 ´ 20 ´ 10, ´ 0.1 = 20 ´ 10–5 (Towards right), 0.02, Hence Fnet = FR – FP = 14 ´ 10–5N = 14 ´ 10–4 N, (Towards right), r, r r, (13) (d) F = q (v ´ B), , FR = 10, , -7, , r r, v´B =, , ´, , ˆi, , ˆj kˆ, , 3, 5 ´10, , 5, , 2 0 = k̂ (–10 × 105) = (– k̂ 106), 0 0, , q = 2e = 2 × 1.6 x 10-19 = 3.2 × 10-19 Coulomb, r, F = 3.2 × 10–19 (– k̂ × 106), r, Þ F = – 3.2 × 10–13 k̂ ., \ |F| = 3.2 × 10–13 Coulomb., (14) (b), r, \ F = q ( vr × B ) = 2evB sin 90º, or F = 2evB, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, DPP/ P 40, , 114, (15) (a) Force on side BC AND AD are equal but opposite so, their net will be zero., 10 cm, , B, 2A, , C, 15 cm, , FAB, , FCD, D, , A, 2 cm, , But FAB = 10, , -7, , ´, , and FCD = 10-7 ´, , 2 ´ 2 ´1, 2 ´10, , -2, , 2 ´ 2 ´1, , ( 2 ´10 ), -2, , 1.6 ´ 10-19 ´ 6.37 ´ 106, , (17) (c) We have F = qvB =, , =, , 6.8 ´ 10-27, , mv, mv2, = evB or r =, ., eB, r, Here m = 9.1 × 10–31 kg, v = 5 × 106 m/s., e = 1.6 × 10–19 coul and B = 2 × 10–3 weber/m2, , ´ 15 ´10 -2 = 0.5 ´ 10–6 N, , \, , = 2.6 × 107 m/s., , The frequency of rotation n =, , v, 2 pr, , 2.6 ´ 107, = 9.2 × 106 sec–1., 2 ´ 3.14 ´ 0.45, Kinetic energy of a-particle,, , 1, × 6.8 × 10–27 × (2.6 × 107)2, 2, = 2.3 × 10–12 joule., , EK =, , 2.3 ´ 10-12, , eVolt = 14 × 106 eV = 14 MeVolt., 1.6 ´ 10-19, If V is accelerating potential of a-particle, then Kinetic, energy = qV, 14 × 106 eVolt = 2eV (since charge on a-particle = 2e), , (9.1 ´ 10 -31 ) (5 ´ 106 ), (1.6 ´ 10-19 ) (2 ´10 -3 ), , = 1.43 × 10-2 m = 1.43 cm., (19) (a) For L length or wire, to balance,, Fmagnetic = mg Þ ILB = mg,, Therefore B = mg/IL = (m/L)g/I, 45 ´10-3 ´ 9.8, = 1.47 × 10-2 tesla., 30, = 147 Gauss., (20) (b) According to Fleming's left hand rule, magnetic force, on electrons will be downward., , =, , e– e – e – e – e– e – e – ×, e – e – e – e – e– e– e – ×, (21) (b) \ F = mg = Bil, or 1 × 9.8 = 0.98 × i × 1, Þ i = 10A., (22) (b) When currents flow in two long, parallel wires in the, same direction, the wires exert a force of attraction on, each other. The magnitude of this force acting per meter, length of the wires is given by, µ0 i1i 2, i1i 2, = 2 × 10–7, N/m., 2p R, R, Here i1 = 10 A, i2 = 15 A, R = 30 cm = 0.3 m, , F=, , =, , =, , r=, , = 1.67 × 10–8 weber/m2., , qBr, mv2, or v =, m, r, , 3.2 ´ 10-19 ´ 1.2 ´ 0.45, , E 1 ´ 104, =, = 5 × 106 m/s., B 2 ´ 10-3, If electric field is removed, the electron traverses a, circular path of radius r given by, , mv, mv2, i.e. qv B =, or B =, qr, r, –27, Here m = 1.7 × 10 kg, v = 1.0 × 107 m/s, q = e = 1.6 × 10-19 coulomb, r = 6.37 × 106 m, , 1.7 ´ 10-27 ´ 1.0 ´ 107, , V=, , v=, , ´ 15 ´10-2 = 3 ´ 10–6 N, , Þ Fnet = FAB – FCD = 2.5 ´ 10–6 N, = 25 ´ 10–7 N, towards the wire., (16) (b) In order to make a proton circulate the earth along the, r, equator, the minimum magnetic field induction B, should be horizontal nad perpendicular to equator., The magnetic force provides the necessary centripetal, force., , B=, , 14 ´ 106, = 7 × 106 Volt., 2, (18) (a) If electron beam passes undeflected in simultaneous, r, r, electric and magnetic fields E and B velocity of beam, r, v much be mutually perpendicular and the required, speed v is given by-, , \, , 10 ´ 15, = 1 × 10–4 N/m., 0.3, \ Force on 5m length of the wire, = 5 × (1 × 10–4) = (5 × 10-4) = 5 × 10-4 N (attraction)., (23) (d) The electron will pass undeviated if the electric force, and magnetic force are equal and opposite. Thus, E.e. = Bev or B = E/v but E = V/d, , \, , F = 2 × 10–7, , \, , 600, Therefore, B = V =, v.d. 3 ´ 10-3 ´ 2 ´ 106, B = 0.1 Wb/m2., The direction of field is perpendicular to the plane of, paper vertically downward.
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t.me/Magazines4all, , DPP/ P 40, (24) (b) The component of velocity of the beam of protons,, parallel to the field direction, = v cosq = 4 × 105 × cos 60º = 2 × 105 m/sec., and the component of velocity of the proton beam at, right angle to the direction of field, = v sinq = 4 × 105 × sin 60º = 2 3 × 105 m/sec., therefore,the radius of circular path = (mv sinq /Be), , 1.7 ´ 10-27 ´ 2 3 ´ 105, , = 12.26 × 10–3 metre, 0.3 ´ 1.6 ´ 10-19, or r = 1.226 × 10–2 metre., Pitch of the Helix = v cosq x (2pm/Be), or r =, , \, , Pitch =, , 2 ´ 105 ´ 2 ´ 3.14 ´ 1.7 ´ 10-27, , 0.3 ´ 1.6 ´ 10-19, = 44.5 × 10–3 m = 4.45 × 10–2 m., , 115, (25) (a), (26) (a), (27) (c)., r, r r, F = q (v ´ B) = q (x 2 - y2 ) kˆ, (28) (c) When two long parallel wires, are connected to a battery in series. They carry currents in opposite directions, hence they repel each other., (29) (c) No net force will act on charged particle if, r, r r r, F = q [ E + v ´ B] = 0, r, r r, Þ E = -v ´ B Þ v need not to be perpendicular to B, (30) (c) In this case we can not be sure about the absence of the, magnetic field because if the electron moving parallel, to the direction of magnetic field, the angle between, velocity and applied magnetic field is zero, (F = 0). Then also electron passes without deflection., Also F = evBsin q Þ F µ B ., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 41, , 116, Þ, 20., 21., 22., , 23., , r, æ1, ö, A3 = ç ´ 10 ´ 10 ´ 10-4 ÷ iˆ m 2, è2, ø, , μ1, μ1, ´ 2 =, m2 = M 2 =, 2, 2, , (a), (d), (a) Couple of force on loop S will be maximum because, for same perimeter the area of loop will be maximum, and magnetic moment of loop = i ×A. So, it will also, be maximum for loop S., , Magnetic moment vector,, r, r, 2, ˆ, m = iA = 10(0.01iˆ + 0.005jˆ + 0.005k)Am, 2, ˆ, = (0.1iˆ + 0.05ˆj + 0.05k)Am, c. Torque,, r, ˆ ´ (2iˆ - 3jˆ + k), ˆ, t = (0.1iˆ + 0.05jˆ + 0.05k), , q nBA, =, i, C, t = mBsin q is zero for q = 0°, 180°., , (b) Sensitivity S =, , 24. (b), 25-27, , ˆi, , =, 28., , 29., , 25., , (a) 26. (b), 27. (b), (a) The net force on a current carrying loop of any, arbitrary shape in a uniform magnetic field is zero., r, Fnet = 0, (b) The given loop can be con sidered to be a, superposition of three loops as shown in figure. The, area vector of the three loops (1), (2) and (3) are, , r, æ1, ö, A1 = ç ´ 10 ´ 10 ´ 10-4 ÷ ˆjm 2, è2, ø, r, æ1, ö, A 2 = ç ´ 10 ´ 10 ´ 10-4 ÷ kˆ m 2, è2, ø, , 30., , ˆj, , kˆ, , = 0.1 0.05 0.05 = -0.1iˆ - 0.4kˆ Nm, 2, , -3, , 1, , (a) Due to metallic frame the deflection is only due to, current in a coil and magnetic field, not due to vibration, in the strings. If string start oscillating, presence of, metallic frame in the field make these oscillations, damped., (b) The torque on the coil in a magnetic field is given by t, = nIBA cos q, For radial field, the coil is set with its plane parallel to, the direction of the magnetic field B, then q = 0° and, cos q = 1 Þ Torque = nIBA (1) = nIBA (maximum)., (c) Loop will not oscillate if in unstable equilibrium, position., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 42, , 117, , 1., , (a). Force between magnetic poles in air is given by, m 0 m1 m 2, ´, 4p, r2, , F=, , 8., 9., , Given that m1 = 50 Am, m2 = 100Am,, r = 10 cm = 0.1 m and, µ0 = permeability of air = 4 p × 10–7 Hm–1., 4 p ´ 10-7 50 ´ 100, ., = 50 × 10–3 N, 4p, 0.1 ´ 0.1, (a) Strength of a magnetic field due to a pole of strength m, is given by, 1 m, ., H=, 4 p r2, , \ 10-7 ×, , 3., , H=, , = 5 × 0.2 × 15 ×, 4., , (a) F = mB =, , =, 5., , 1, = 7.5 Nm., 2, , 7., , = 0.4 × 10-4, , =, , 11., , l ´ mBsin 30, r, , 25 ´ 10-2 ´ 24 ´ 0.25, , (b). B =, , -2, , 12 ´ 10, , 1, = 6.25 N, 2, , m 0 2M, ´, = constant, 4p x3, , æ M2 ö, \ x2 = x1 ç, è M ÷ø, , 1/3, , 1, , 12. (d). B =, , ´, , æ 1 ö, = 20 × ç, è 2 ´ 4 ÷ø, , 1/3, , = 10 cm, , m0, m 2M, 2M, =, = 0, 4 p ( l2 + x 2 )3/2 4p x 3, 3, , m 0 2m ' l, m, 4p x 3, , 10 -7 ´ 2 ´ 200 ´ 0.05 ´ 100, 8 ´ 10 -3, , 13. (c). According to tangent law, BA = BB tan q, or, , = 2.5 × 10-2 N, (d). W = MB (cosq1 - cosq2), , W2 = MB (cos 30º – cos 90º) =, , (0.1)3, , B1 æ x 2 ö, =, \, = 8 : 1 approximately.., B2 çè x1 ÷ø, , m 0 2M m0 M, tanq, =, 4p d13, 4p d13, , d1, \ d = (2 cot q)1/3, 2, , MB, \ W1 = MB (cos 0º – cos 60º) =, 2, , 6., , F=, , 1, 40, ´, = 79.57 Am–1, 4 p (20 ´ 10 -2 ) 2, , Now, magnetic induction at the same point :, B = µ0 H = 4p × 10–7 × 79.57 = 10–4 wb/m2, (c) Couple acting on a bar magnet of dipole moment M, when placed in a magnetic field, is given by, t = MB sin q, where q is the angle made by the axis of magnet with, the direction of field., Given that m = 5 Am, 2l = 0.2 m, q = 30°, and B = 15 Wbm–2, \ t = MB sin q = (m × 2l) B sin q, , 2M, , M = 0.2 A m2, 10. (d). t = F × r = MB sin 30, , Given that m = 40 Am, r = 20 cm = 20 × 10–2 m., \, , (c), (d) Here, d = 10 cm = 0.1 m ,, H = 0.4 gauss = 0.4 x 10-4 T, M = ?, Neutral points in this case, lie on axial line of magnet,, such that, , m 0 2M, =H, 4p d3, , \F=, , 2., , 42, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 3 MB, 2, , \ W2 = 3 W1., (b). Loss in P.E. = gain in K.E., \ Ek = Ui – Uj = – MB cos 90º – (– MB cos 0º), = 4 × 25 × 10-6 = 10-4 J, (a). t = MB sin q = ml B sin q, = 10-3 × 0.1 × 4p × 10-3 × 0.5, = 2p × 10-7 N-m, , 14. (a). T = 2p, , T' = 2p, , I, = 2p, MB, , m l2, = 4 sec, 12 ´ m p lB, , m 2, l, 2, = 4 sec, mp, 12 ´, lB, 2
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t.me/Magazines4all, DPP/ P 42, , 118, 15., , I, MB, , (c). T = 2p, , = 4p × 10–1 T mA–1,, , Given that µ0, M = 13.4 Am2,, r = 15 cm = 0.15 m and l = 5.0 cm = 0.05 m, , 1, , or T µ, , M, , T1, 3M - 2M, =, or T2 = 5 5 s, T2, 3M + 2M, 16., , (b). T = 2p, , I, MB, , \ BH = 10–7 ×, = 10–7 ×, 20., , 2, , m l / 12, or T µ ml, m p lB, , = 2p, æ m lö, T' ç, ÷, = n n, T ç ml ÷, ç, ÷, è, ø, , 17., , 1/2, , or T' =, , T, n, , 6.6 ´ 10 -3 kg, 7.9 ´ 103 kg / m3, , = 3.0 × 106 A/m, (d). The compass box will be on the axial line of the magnet,, , 1 2M 1 2 ´ 2lm, =, ., ., = H tan q, 4p r 3, 4p, r3, , Given that H = horizontal component of the earth’s, magnetic field = 30 Am–1, q = 45°,, r = 20 cm = 0.02 m,, M = 2 l m = 4 × 10–2 m, Hence,, , \m=, , 19., , 2 ´ 4 ´ 10, , -2, , 4 p (0.2), , m, , 3, , 30 ´ 4p ´ (0.2), , M 8.0, =, = 26.7 Am, 2l 0.30, (a). The situation is shown in figure. The horizontal, component of earth’s magnetic field at the location of, the cable (angle of dip q = 0 is), , m=, , 21., , = 30 × tan 45°=30 × 1;, 3, , = 37.7 Am, 2 ´ 4 ´ 10-2, (c). As the magnet is placed with its south pole pointing, south, hence the neutral point lies on the equatorial, line. At the neutral point, the magnetic field B due to, the magnet becomes equal and opposite to horizontal, component of earth’s magnetic field i.e., BH., Hence, if M be magnetic dipole moment of the magnet, of length 2l and r the distance of the neutral point from, its centre, then, B=, , µ0, 2Mr, ., = BH, 4p (r 2 - l 2 ) 2, , (0.34 ´ 10-4 ) - [(0.40)2 - (0.15)2 ]2, 2 ´ 0.40, = 8.0 Am2, The pole strength of the magnet is,, , 2.5 A - m, M, =, V, 8.3 ´ 10-7 m 2, , Hence,, , = 0.34 × 10–4 T, , = 10–7×, , 2, , 18., , 0.025 0.025, , 4p BH (r 2 - l 2 ) 2, ., \M=, µ0, 2r, , = 8.3 × 10-7 m3., , The intensity of magnetization is, I=, , 1.34, , Given that µ0 = 4p × 10–7 TmA–1 ,, r = 40 cm = 0.40 m, l = 15 cm = 0.15 m and, HH = 0.34 Gauss = 0.34 × 10–4 T, , mass, V =, density, =, , [(0.15) + (0.5) 2 ]3/2, , (a). As the magnet is placed with its north pole pointing, south, the neutral points are obtained on the axial line., At the neutral points the magnetic field B due to the, magnet becomes equal and opposite to the horizontal, component of earth’s magnetic field i.e., BH., Hence, if M be the magnetic dipole moment of the, magnet of length 2l and r the distance of neutral point, from the centre of the magnet, then we have, B=, , (a). The volume of the bar magnet is, , 1.34, 2, , µ0, M, = BH, 4p (r 2 + l 2 )3/2, , Geographic, West, , Magnetic North, North, 10° Geographic, Meridian, Magnetic, Meridian, I, 10°, Current, , Cable, Geographic, East, , South Magnetic, South, BH = B cos q = B cos 0 = B = 0.33 Gauss, = 0.33 × 10–4 Tesla, BH is directed horizontally in the magnetic meridian., The magnetic field produced by the cable at a distance, of R meter is given by, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 42, , 119, , µ0 I, 2 pR, , Figure shows a point Q on the axis of a short bar magnet, lying at a distance r from the centre of the magnet. It is, given that resultant magnetic field at Q is inclined at, an angle of 45° w.r.t. earth’s field B., From the figure, it is clear that, Ba = B tan 45° = B, Now for a short bar magnet in this position,, , B=, , 2.5 5.0 ´ 10 -7, =, T, R, R, According to right-hand-palm rule no, 1, the field B is, directed horizontally along BH at a point below the, cable, and opposite to BH at a point above the cable., Therefore, neutral points will be obtained above the, cable. At these points, will be equal and opposite to, BH. Thus, , = 2 × 10–7 ×, , 5.0 ´ 10-7, = BH = 0.33 × 10–4 ., R, , 5.0 ´ 10-7, , = 15 × 10–3 m = 1.5 cm, 0.33 ´ 10-4, Thus, the line of neutral points lies above and parallel, to the cable at a distance of 1.5 cm from it., 22. (a) The magnetic lines of force are in the form of closed, curves whereas electric lines of force are open curves., 23. (a) Inside a magnet, magnetic lines of force move from, south pole to north pole., 24. (b) Given that pole strength, m = 5.25 × 10–2 JT–1, q =, 45°, and B = 0.42 G = 0.42 × 10–4 T., or R =, , Beq., , B, 45°, P, r, , N, , S, , Figure shows a point P on the normal bisector of a, short bar magnet lying at a distance r from the, centre O of the magnet. It is given that resultant, magnetic field at P makes an angle of 45° w.r.t. earth’s, field B., From the figure, it is clear that, Beq = B tan 45° = B., Now, for a short bar magnet in this position, Beq = B =, , µ0 M, ., 4p r3, , 5.25 ´ 10-2, , 0.42 ´ 10-4, –2, \ r = 5 × 10 m = 5 cm., , B, 45°, N, , O, , m 0 2M, ., = 2 × 125 × 10–6, 4p B, \ r = (2 × 125 × 10–6)1/3 = 6.3 cm, 25-27, Given that earth’s magnetic field, B = 0.39 G and angle, of dip, q = 35°, Horizontal and vertical components of earth’s magnetic, field B at the location of the cable are, BH = B cos q = 0.39 cos 35°, = 0.39 × 0.82 = 0.32 Gauss, and, BV = B sin q = 0.39 sin 35°, = 0.39 × 0.57 = 0.22 Gauss, The magnetic field produced by four current carrying, straight cable wires at a distance R is, µ0 I, 1.0, × 4 = 2× 10–7 ×, ×4, 2pR, 0.04, = 0.2 × 10–4 T = 0.2 Gauss, Resultant magnetic field below the cable, According to right - hand, palm rule no. 1, the direction, of B¢ below the cable will be opposite to that of, BH.Therefore, at a point 4 cm below the cable, resultant, horizontal component of earth’s magnetic field, RH = BH – B¢ = 0.32 – 0.2 = 0.12 Gauss., Resultant vertical component of earth’s magnetic field, RV = BV = 0.22 Gauss (unchanged), \ Resultant magnetic field below the cable is, , B¢ =, , R=, , [R 2 H + R 2 V ], , [(0.12)2 + (0.22)2 ], = 0.25 Gauss, The angle that R makes with the horizontal is given by, =, , Q, , S, r, , Ba, , RV, RH, , 0.22, = tan–1(1.8) @ 62°, 0.12, Resultant magnetic field above the cable, Again, according to right - hand - palm no. 1, the, direction of B¢ at a point above the cable is the same as, that of BH., Therefore, at a point 4 cm below the cable, the, horizontal component of resultant magnetic field will, be, RH = BH + B¢ = 0.32 + 0.20 = 0.52 Gauss, , = tan–1, , = 125 × 10–7., , m 0 2M, ., 4p r 3, , or r3 =, , q = tan–1, , µ0 M, ., or r3 =, 4p B, , = 10–7 ×, , Ba = B =
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DPP/ P 42, , 120, Vertical component of resultant magnetic field will be, RV = BV – 0.22 Gauss ( unchanged), Hence, magnitude of resultant magnetic field below, the cable, R =, =, , [(0.52)2 + (0.52)2 ], , æ RV ö, q = tan–1 ç R ÷ = tan–1, è Hø, , (a), , 28., 29., , [R 2 H + R 2 V ], , = 0.56 Gauss, The angle that R makes with the horizontal is given by, , 25., , t.me/Magazines4all, , æ 0.22 ö, çè, ÷, 0.52 ø, , = tan–1 (0.43) @ 23°, 26. (a), 27. (b), , 30., , (a), (d) The earth has only vertical component of its magnetic, field at the magnetic poles. Since compass needle is, only free to rotate in horizontal plane. At north pole, the vertical component of earth's field will exert torque, on the magnetic needle so as to aligh it along its direction. As the compass needle can not rotate in vertical, plane, it will rest horizontally, when placed ont he magnetic pole of the earth., (c) It is quite clear that magnetic poles always exists in, pairs. Since, one can imagine magnetic field configuration with three poles. When north poles or south poles, of two magnets are glued together. They provide a three, pole field configuration. It is also known that a bar, magnet does not exert a torque on itself due to own its, field., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 43, , 121, , DAILY PRACTICE, PROBLEMS, 1., 2., 3., , (a) Susceptibility of ferromagnetic substance is greater, than 1., (c) Susceptibility of diamagnetic substance is negative and, it does not change with temperature., (c), , m d = 0 and m p ¹ 0, , 4., , (d) m r =, , 5., 6., 7., 8., 9., , (b), (b), (d), (a), (a), , 10. (c), 11. (b), 12. (b), 13. (d), 14. (b), 15. (a), 16. (b), , 17., 18., 19., 20., 21., , (b), (d), (c), (d), (d), , B, =4, B0, , Dimagnetic, Paramagnetic, A super conductor exhibits perfect diamagnetism., Soft iron is highly ferromagnetic., Diamagnetism is the universal property of all, substances., Diamagnetic substances are repelled by magnetic field., Coercive force, Because, diamagnetic substance, moves from stronger, magnetic field to weaker field., m r > 1, c > 0, Hysteresis curve for a given material estimates, hysteresis loss., Neon atom is diamagnetic, hence it’s net magnetic, moment is zero., On heating, different domains have net magnetisation, in them which are randomly distributes. Thus the net, magnetisation of the substance due to various domains, decreases to minimum., Repelled due to induction of similar poles., From the characteristic of B - H curve., Diamagnetic substances are feebly repelled by magnets., Net magnetic induction B = B0 + Bm = µ0H + µ0M, c m = (mr - 1) Þ c m = (5500 - 1) = 5499, , 22. (c) Q mr =, , 8p, , = 104, 2 ´ 10 ´ 4 p ´ 10-7, m, (a) WH = VAft =, Aft, d, 0.6, or WH =, × 0.722 × 50, 7.8 ´ 103, = 277.7 × 10-5 Joule, or mr =, , 23., , m, B, =, m 0 Hm 0, 3, , 43, , PHYSICS, SOLUTIONS, , 24. (c), 25. (b) Given that : H = 1600 Am–1 , f =2.4 ×10–5 Wb,, A = 0.2 cm2 = 0.2 × 10–4 m2., B = magnetic flux per unit cross - sectional are, f 2.4 ´ 10 -5, =, = 1.2 Wbm–2, A 0.2 ´ 10 -4, Magnetic permeability :, , =, , m=, , B 1.2 Wbm -2 or T, =, = 7.5 × 10–4 T A–1 m, H, 1600 Am -1, , As µ = µ0 (1 + xm), \ xm =, , µ, 7.5 ´ 10-4, -1 =, µ0, 4 ´ 3.14 ´ 10 -7, , = 597.1 – 1 = 596.1., 26. (a) The energy lost per unit volume of a substance in a, complete cycle of magnetisation is equal to the area of, the hysteresis loop., 27. (a) Statement (4) is the only true statement among the given, choices., 28. (b) The susceptibility of ferromagnetic substance decreases, with the rise of temperature in a complicated manner., After Curie's point the susceptibility of ferromagnetic, substance varies inversely with its absolute, temperature. Ferromagnetic substance obey’s Curies, law only above its Curie point., 29. (d) A paramagnetic sample display greater magnetisation, when cooled, this is because at lower temperature, the, tendency to disrupt the alignment of dipoles (due to, magnetising field) decreases on account of reduced, random thermal motion., 30. (d) The permeability of a ferromagnetic material dependent, ur, ur, on magnetic field, B = K m B0 , where B0 is applied, ur, field. The total magnetic field B inside a ferromagnet, may be 103 or 104 times the applied field B0 The, permeability Km of a ferromagnetic material is not, uur, ur, constant, neither the field B nor the magnetization M, ur, increases linearly with B even at small value of B0., From the hysteresis curve, magnetic permeability is, greater for lower field.
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t.me/Magazines4all, DPP/ P 4 4, , 122, , DAILY PRACTICE, PROBLEMS, r r, f = B.A, , (1), , (a), , (2), , = ( 0.02iˆ ) . ( 30iˆ + 16ˆj + 23kˆ ) × 10–4, = 0.6 × 10-4 Wb = 60m Wb, (c) The induced emf, d, (3t2 + 2t + 3) × 10-3, dt, (because given flux is in mWb)., Thus E = ( – 6t – 2) × 10–3, At t = 2 sec,, E = ( – 6 × 2 – 2) × 10–3 = –14 mV., (a) The direction of current in the solenoid is clockwise., On displacing it towards the loop a current in the loop, will be induced in opposite sense so as to oppose its, approach. Therefore the direction of induced current, as observed by the observer will be anticlockwise., (b) When north pole of the magnet is moved away, then, south pole is induced on the face of the loop in front of, the magnet i.e. as seen from the magnet side, a clockwise, induced current flows in the loop. This makes free, electrons to move in opposite direction, to plate a. Thus, excess positive charge appear on plate b., (d) If electron is moving from left to right, the flux linked, with the loop (which is into the page) will first increase, and then decrease as the eletron passes by. So the, induced current in the loop will be first anticlockwise, and will change direction as the electron passes by., , (9), , e = 4 × 10–5 × 3.14 × (.5)2 × 2 = 6.28 × 10–5 volt, (d) Rate of decrease of area of the semicircular ring, dA, = (2R) V, dt, According to Faraday’s law of induction, induced emf, –, , E = – df/dt = –, , (3), , (4), , (5), , (6), , (7), , (8), , 44, , PHYSICS, SOLUTIONS, , df, dt, or df = – Edt = (0 - f), or f = 4 × 10–3 × 0.1, = 4 × 10–4 weber, , (c) E = –, , df, d, =–, [10t2 + 5t + 1] × 10–3, dt, dt, = – [10 × 10–3 (2t) + 5 × 10–3], , (d) e =, , e= -, , ×, ×, ×, ×, M, , ×, ×, ×, ×, , × ×, N, ×, × ×, × ×, , 2R, , ×, ×, ×, ×, Q vt, , The induced current in the ring must generate magnetic, field in the upward direction. Thus Q is at higher, potential., (10) (b) E = Blv, 360 ´ 1000, =2V, 60 ´ 60, (11) (c) The flux through the area is, f = BA cos 57º = 42 × 10-6 × 2.5 × 0.545, = 57 × 10-6 Wb., (12) (a) The magnetic flux linked with the loop at any instant of, time t is given by, f = BAN cos wt, or f = 10Ba2 cos wt, Here N = 10, A = a2, (13) (b) According to Lenz’s Law, , = 4 × 10–4 × 50 ×, , 300 ´ 103, ´ 3 = 10–2 V, 60 ´ 60, (15) (b) Magnetic flux passing through the disc is f = BA, , (14) (c) e = Bvl = 0.4 × 10–4 ×, , weber, , × 3.14 × (15 × 10–2 meter)2, meter 2, = 7.065 × 10–4 weber., The line joining the centre and the circumference of the, disc cuts 7.065 x 10-4 weber flux in one round. So, the, rate of cutting flux (i.e. induced emf), = flux × number of revolutions per second, = 0.01, , at t = 5 second, e = –[10 × 10–2 + 5 × 10–3] = [0.1 + 0.005], |e| = 0.105V, (a) For each spoke, the induced emf between the centre O, and the rim will be the same, 1, BwL2 = BpL2 f (Q w = 2pf), 2, Further for all spokes, centre O will be positive while, rim will be negative. Thus all emf's are in parallel giving, total emf e = BpL2 f, independent of the number of the spokes., Substituting the values, , df, dA, = – B(2RV), V), = –B, dt, dt, , 100, = 3.9 × 10–4 volt., 60 ´ 3, The magnetic flux passing through each turn of a coil of area, A, perpendicular to a magnetic field B is given by, f1 = BA., The magnetic flux through it on rotating it through 180º, will be, f2 = – BA.(- sign is put because now the flux lines, enters the coils through the outer face), , = 7.065 × 10–4 ×, , e=, , (16) (c), , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 4 4, , 123, , \ change in magnetic flux, D f = f1 – f2 = – BA – (BA) = – 2BA., Suppose this change takes in time Dt. According to, Faraday's law, the emf induced in the coil is given by, Df, 2NBA, =, ,, Dt, Dt, where N is number of turns in the coil. The current in, the coil will be, , e=–N, , e 1 2NBA, =, R R Dt, where R is the resistance of the circuit. The current, persists only during the change of flux i.e. for the time, interval Dt second. So, the charge passed through the, circuit is, , i=, , 2NBA, ., R, Here N = 500, B = 0.2 weber/meter2,, A = 4.0 cm2 = 4.0 × 10–4 meter2 and R = 50 ohm., , q = i × Dt =, , 2 ´ 500 ´ 0.2 ´ 4.0 ´ 10-4, 50, = 1.6 × 10–3 coulomb., , \ q=, , (17) (a) T, (18) (c) The induced emf between centre and rim of the rotating, disc is, 1, 1, BwR2 = × 0.1 × 2p × 10 × (0.1)2, 2, 2, = 10p × 10–3 volt,, (19) (a) The induced emf, E = Blv = 0.2 × 10–4 × 1 × 180 × 1000/3600, = 0.2 × 18/3600 = 1 × 10–3. V = 1mV, (20) (d) The induced emf is obtained by considering a strip on, the disc fig. Then, the linear speed of a small element dr, at a distance r from the centre is = wr. The induced emf, across the ends of the small element isde = B(dr)v = B wr dr, Thus the induced emf across the inner and outer sides, of the disc is, , E=, , 1, Bw (b2 – a2), 2, r r r, (21) (c) The induced emf e = – (v ´ B).l, r r, r r, For the part PX, v ^ B , and the angle between (v ´ B), r, direction (the dotted line in figure and l is (90 – q)., Thus, eP –ex = vBl cos (90 – q/2) = vBl sin (q/2), Similarly ey – ep = vBl sin (q/2), Therefore induced emf, between X and Y is eyx = 2 B v l sin (q/2), (22) (a) Given, area = 10 × 20 cm2 = 200 × 10-4 m2, B = 0.5 T, N = 60, w = 2p × 1800/60, , e=, , b, , ò a Bwr dr, , =, , Qe=–, , d ( Nf ), dt, , d, (BA cos wt), dt, = NBAw sin wt, \ emax = NABw, = 60 × 2 × 10–2 × 0.5 × 2p × 1800/60 = 113 volt., Hence, induced emf can be 111 V, 112 V and 113.04 V, (23) (b) The change is flux linked with the coil on rotating it, through 180º is, = nAB – (–nAB) = 2nAB, , = –N, , \ induced e.m.f. = –, , df, dt, , 2 ´ 1 ´ 0.1, = 20 V, 0.01, The coil is closed and has a resistance of 2.0 W., Therefore i = 20/2 = 10A., (24) (d) Initial magnetic flux f1 = 5.5 × 10–4 weber., Final magnetic flux f2 = 5 × 10–5 weber., \ change in flux, Df = f2 - f1 = (5 × 10-5) – (5.5 × 10-4) = – 50 × 10–5, weber., Time interval for this change, Dt = 0.1 sec., \ induced emf in the coil, , = 2nAB/dt (numerically) =, , Df, (-50 ´ 10-5 ), = 5 volt., = – 1000 ×, Dt, 0.1, Resistance of the coil, R = 10 ohm. Hence induced, current in the coil is, , e=–N, , i=, , e, 5 volt, =, = 0.5 ampere., R 10ohm, , (25) (a), (26) (c), (27) (b), (a) The current of the battery at any instant, I = E/R., The magnetic force due to this current, , EBl, R, This magnetic force will accelerate the rod from its, position of rest. The motional e.m.f. developed in the, rod id B/v., The induced current,, FB = IBL =, , Blv, R, The magnetic force due to the induced current,, Iinduced =, , B2l 2 v, R, From Fleming’s left hand rule, foce FB is to the right, and Finduced is to the left., Net force on the rod = FB – Finduced., Finduced = Iinduced =
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t.me/Magazines4all, DPP/ P 4 4, , 124, From Newton’s law,, FB - Finduced = m, , dv, dt, , EBl B2l 2 v, dv, =m, R, R, dt, On separating variables and integrating speed from v0, to v and time from 0 to t, we have, dv, Bl, dt, =, E - Bvl mR, Bl, , dv, , v, , t, , ò0 E - Bvl = mR ò0 dt, 2 2, æ E - Bvl ö B l, t, - ln ç, ÷=, è E ø mR, 2 2, , l, , B, t, E - Bvl, = e mR, E, , v=, where, , t=, , E, (1 - e- t / r ), Bl, mR, (Bl) 2, , (b) The rod will attain a terminal velocity at t ® ¥, i.e.,, when e–t/t = 0, the velocity is independent of time., E, Bl, (d) The induced current Iinduced = Blv/R. When the rod, has attained terminal speed., vT =, , Bl æ E ö, ´, = E/ R, R çè Bl ÷ø, The current of battery and the induced current are of, same magnitude, hence net current through the circuit, is zero., (28) (c) Since both the loops are identical (same area and number, of turns) and moving with a same speed in same, magnetic field. Therefore same emf is induced in both, the coils. But the induced current will be more in the, copper loop as its resistance will be lesser as compared, to that of the aluminium loop., (29) (c) As the aircraft flies, magnetic flux changes through its, wings due to the vertical component of the earth’s, magnetic field. Due to this, induced emf is produced, across the wings of the aircraft. Therefore, the wings, of the aircraft will not be at the same potential., (30) (c) Lenz’s Law is based on conservation of energy and, induced emf opposes the cause of it i.e., change in, magnetic flux., , Iinduced =, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 45, , 125, , 1., , 2., , 3., , 4., 5., , dI, 1.0, = – 10 × 10–3, = –1Volt., dt, 0.01, \ | e | = 1 volt., , (a). e = – L, , dI, d, =–L, [3t2 + 2t], dt, dt, = – L [6t + 2] = –10 × 10–3 [6t + 2], (e)at t = 2 = – 10 × 10–3 (6 × 2 + 2), = – 10 × 10–3 (14) = – 0.14 Volt, |e| = 0.14 volt., (a). The mutual inductance between two coils depends on, their degree of flux linkage, i.e., the fraction of flux linked, with one coil which is also linked to the other coil. Here,, the two coils in arrangement (a) are placed with their planes, parallel. This will allow maximum flux linkage., , 6., , 0 - 10, dI1, == 3 × 104 volt = 30 kV.., dt, 3 ´10-3, (d). L2 and L3 are in parallel. Thus their combination gives, , (b). e2 = – M, , L 2 L3, = 0.25 H, L 2 + L3, , 7, , 2 ´ 10-3, (ii) DI = 5 – 7 = – 2A,, dt = (5 – 2) ms = 3 ms, Thus e = – 4.6 ×, , 3 ´ 10-3, (b). I = I0 (1 – e– t/t), Where I =, Thus, or, , = 3.07 × 103 V, , 1, I and t = L/R, 2 0, , 1, I = I (1 – e – t/t ), 2 0 0, , 1, = e – t/t or 2 = e+ t/t or log 2 = t/t, 2, , 50 ´10-3, × 0.693 = 1.385, 0.025, (a). VS = IS ZS Þ 22 = IS × 220, \ IS = 0.1A, , Thus t = t loge 2 =, 8., , (-2), , VS IP, =, VP IS, , Pp, 2, , Q VsIs =, , 10., , , Is = ?, , Vp Ip, , 2, After putting the given value you will find, Is = 0.25 A., (d). During decary of current, -, , 11., , 12., , = – 16.1 × 103 volt, , I, 22, = P Þ IP = 0.01 A., 220 0.1, (c). Vp = 220V, Ip = 5S, Vs = 2200V, , Ps =, , The L' and L1 are in series, thus the equivalent inductance, is L = L1 + L' = 0.75 + 0.25 = 1H, (a). We use e = – L DI/Dt to determine the value of induced, emf. (i) DI = (7 – 0) = 7A,, Dt = (2 – 0) ms = 2ms, Thus e = – 4.6 ×, , 7., , 9., , (a). e = – L, , L' =, , 45, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , Rt, , 100´10-3, , 100 100´10-3 1, E, e, i = i0 e, = A., = eL =, 100, R, e, (c). In a generator e.m.f. is induced according as Lenz’s, rule., The minus sign indicates that the direction of the induced, e.m.f. is such as to oppose the change in current., (a). 'Immediately' after pressing the switch S, the current in, the coil L, due to its self-induction will be zero, that is, i2 = 0., The current will only be found in the resistance R1 and this, will be the total current in the circuit., , \ i = i1 =, 13., , Rt, L, , E, 10 volt, =, = 2.0 ampere., R 2 5.0 volt, , (a). (i) In series the same current i will be induced in both, the inductors and the total magnetic-flux linked with them, will be equal to the sum of the fluxes linked with them, individualy, that is,, F = L1i + L2i., If the equivalent inductance be L, then F = Li., \ Li = L1i + L2i, or L = L1 + L2., (ii) In parallel, let the induced currents in the two coils be i1, and i2. Then the total induced current is, di di1 di 2, =, +, dt dt, dt, In parallel, the induced e.m.f. across each coil will be the, same., , i = i1 + i2 \, , Hence e = – L1, , di1, di 2, = –L2, ., dt, dt, , If the equivalent inductance be L, then e = – L, \, , æ di1 di 2 ö, e, di, e, e, +, =+, = – çè, =, dt, dt ÷ø, L, dt, L1 L2, , di, ., dt
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 46, , 127, , 1., , (b) The coil has inductance L besides the resistance R ., Hence for ac it’s impedance resistance, , R 2 + X L2 will, , be larger than it’s impedance R for dc., 2., , i, 4, (b) ir.m.s. = 0 =, = 2 2 amperee, 2, 2, , 3., , T, (d) The current takes sec to reach the peak value., 4, , In the given question, , 4., 5., , (b) E = E0 cos wt = E0 cos, , 6., 7., , (c), , irms =, , 2pt, T, , 2 p ´ 50 ´ 1, p, = 10 cos = 5 3 volt., 600, 6, i12, , + i22, 2, , =, , 1, , 9., , 120, =, = 84.8 V, (b) Vrms =, 1.414, 2, (d) Peak value to r. m. s. value means, current becomes, 1, , times., 2, If peak is at t = 0, current is of the form, , i = i0 cos100pt Þ, Þ cos, , 1, 2, , 2 2 2, , \ Z = R + 4p f L, 11., , (b) The applied voltage is given by V = VR2 + VL2, V = (200)2 + (150)2 = 250 volt, , 2 2, , =, , VL = V 2 - VR2 = 400 - 144 = 256 = 16 volt., , 14. (b) Reading of ammeter,, i rms =, , =, , Vrms Vo wC, =, Xc, 2, , 200 2 ´ 100 ´ (1 ´ 10-6 ), , 2, = 2 × 10–2 A = 20 m A, R, , C, , 15. (a), VR, , VR, ~, , Let the applied voltage be V, volt., Here, VR = 12 V, VC = 5V, V=, 16. (c), , VR2 + VC2 =, , (12 )2 + ( 5)2, , = 13 V., , Z = X L = 2p´ 60 ´ 0.7, , 120, 120, =, = 0.455 amperee, Z, 2p ´ 60 ´ 0.7, 17. (d) Current will be max at first time when, 100 pt + p/3 = p/2 Þ 100 pt = p/6 Þ t = 1/600 s., \i =, , 18. (d), , Z = R 2 + X 2 = R 2 + (2p fL)2, , 0.4 ö, æ, = (30) 2 + ç 2 p ´ 50 ´, ÷, è, p ø, , p, 1, sec = 2.5 ´10 -3 sec., = cos100 pt Þ t =, 4, 400, , 2, , 2, , V 2 = VR2 + VL2, , ´ i0 = i0 cos100pt, , 10. (b) Z = R 2 + X L2 , X L = wL and w = 2pf, , 120, , = 0.016 A, R +w L, 100 + 4p2 ´ 602 ´ 202, 13. (a) The voltage across a L-R combination is given by, , 2, (c) Hot wire ammeter reads rms value of current. Hence, , V0, , V, , i=, , (i12 + i22 )1/ 2, , its peak value = irms ´ 2 = 14.14 amp, 8., , 12. (b), , 2p, 1, = 200p Þ T =, sec, T, 100, , 1, \ Time to reach the peak value = 400 sec, (b) 50 c/s or Hz, , = 10 cos, , 46, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , i=, , 19., , (d), , 20. (a), , 2, , = 900 + 1600 = 50W, , V 200, =, = 4 ampere, Z, 50, , In purely inductive circuit voltage leads the current by 90o ., Current through the bulb i =, 60 W, 10 W, i, 10 V, , L, VL, , i, 100 V, 50 Hz, , P 60, =, = 6A, V 10
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t.me/Magazines4all, DPP/ P 46, , 128, V=, , V = V0 sin ( wt - f), , VR2 + VL2, , (100)2 = (10)2 + VL2 Þ VL = 99.5 Volt, Also VL = iXL = i ´ (2pvL), Þ 99.5 = 6 ´ 2 ´ 3.14 ´ 50 ´ L = Þ L = 0.052 H., 21., , (a) Phase angle tan f =, , 4, 3, (a) The root mean square voltage is effective voltage., (c) E = 141 sin (628t),, , Erms =, , E0, 2, , =, , 141, = 100V and 2pf = 628, 1.41, , Þ f = 100 Hz, 24., , Vrms =, 26., , 27., 28., , T 1, 1, =, =, = 5 ´ 10-3 sec, 4 4v 4 ´ 50, , and i0 = irms 2 = 10 2 = 14.14 amp, (a) As in case of ac,, , 29., 30., , V0, 2, , ;Vrms = 220 V, , (b) In case of ac,, Vav =, , (d) Time taken by the current to reach the maximum value, t=, , 25., , and as in case of ac,, , wL 2p ´ 200 1 4, =, ´ =, R, 300, p 3, , \ f = tan -1, , 22., 23., , The peak value V0 = 220 2 = 311V, , 2, 2, 622, V0 = ´ 311 =, V, p, p, p, , 314, = 50Hz, 2´ p, (d) When ac flows through an inductor current lags behind, the emf., by phase of p/2, inductive reactance,, , (a) As w = 2pf ,2pf = 314i.e., f =, , X L = wL = p.2 f .L, so when frequency increases, correspondingly inductive reactance also increases., (c) Like direct current, an alternating current also produces, magnetic field. But the magnitude and direction of the, field goes on changing continuously with time., (b) We can use a capacitor of suitable capacitance as a, choke coil, because average power consumed per cycle, in an ideal capacitor is zero. Therefore, like a choke, coil, a condenser can reduce ac without power, dissipation., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 47, , 129, , 1., , 2., 3., 4., , 13. (b) Resonance frequency, , (a) For resonant frequency to remain same, LC should be const. LC = const, L, Þ LC = L' × 2C Þ L ' =, 2, (b) At resonance, LCR circuit behaves as purely resistive, circuit, for purely resistive circuit power factor = 1, (a) If the current is wattless than power is zero. Hence, phase difference f = 90°, , (c), , VL = 46 volts, VC = 40 volts, VR = 8 volts, E.M.F. of source V = 82 + (46 - 40)2 = 10 volts, , 1, , 5., , (c) Resonant frequency =, , 6., , resistance., (a) At resonance LCR series circuit behaves as pure, , 2p LC, , does not depend on, , resistive circuit. For resistive circuit f = 0o, 7., , R, (b) cos f = Z =, =, , f =, , R, R2 + w2 L2, 12, , 2, , (12) + 4 ´ p ´ (60) 2 ´ (0.1) 2, , 1, , 2, , Þ f µ, , Þ cos f = 0.30, , (a), , 9., , (b) In non resonant circuits, , 1 ö, æ, + wC ÷, 2 ç, wL ø, è, R, , 2, , , with rise in, , frequency Z decreases i.e. current increases so circuit, behaves as capacitive circuit., , 11., , R 10 1, cos f = =, = Þ f = 60o, Z 20 2, , (d), , 12. (a), , 1, - 2pfL, 2pfC, Þ tan 45o =, R, , tan f =, , XC - X L, R, , ÞC=, , 1, 2pf (2pfL + R), , =, , 1, 8 ´10, , -3, , Resonance current =, , ´ 20 ´10-6, , = 2500 rad/sec, , V 220, =, = 5A, R 44, , P = Vr.m.s. ´ ir.m.s. ´ cos f =, , 14. (c), , 100 10 -3, p, ´, ´ cos, 3, 2, 2, , 102 ´ 10-3 1 10-1, ´ =, = 0.025 watt, 2, 2, 4, 15. (b) R = XL = 2XC, =, , Z = R 2 + (X L - X C ) 2, , =, , (2X C )2 + (2X C - X C ) 2, , =, , 4X 2C + X 2C, , XL–XC, , 5R, = 5X C =, 2, X L - XC 2XC - XC, =, tan f =, R, 2XC, , 17. (b) P =, , 1, , 10. (c), , LC, , Z, , f, , R, , 1, -1 æ 1 ö, ; f = tan çè ÷ø, 2, 2, , 16. (d) Phase angle f = 90o , so power P = Vrms I rms cos f = 0, , C, , 1, , impedance Z =, , 1, , w=, , tan f =, , 1, , 8., , 2p LC, , 47, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 2, Vrms, (30)2, =, = 90 W, R, 10, , 18. (c) At A : X C > X L, At B : X C = X L, At C : X C < X L, 19. (d) The instantaneous values of emf and current in, inductive circuit are given by E = E0 sin at and, , pö, æ, i = i0 sin ç w - ÷ respectively.., 2ø, è, pö, æ, So, Pinst = Ei = E0 sin wt ´ i0 sin ç wt - ÷, 2ø, è, = E0i0 sin wt cos wt, 1, = E0i0 sin 2wt (sin 2wt = 2sin wt cos wt ), 2, Hence, angular frequency of instantaneous power is, 2w .
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t.me/Magazines4all, DPP/ P 47, , 130, 20., , (d) The voltage VL and VC are equal and opposite so, voltmeter reading will be zero., Also R = 30W, X L = X C = 25W, So i =, , V, , V 240, = =, = 8A, R 2 + ( X L - X C )2 R 30, , 21. (d) Since quality factor, Q =, 22., 23., , (d), (a), , 24., , (c) Reactance Z =, =, , 1, R, , L, C, , (DVL + DVC )max = DVC - DVL = 7.4 - 2.6 = 4.8 volt, (c), , 27., , (8.8)2 + (4.8) 2 = 10 volt, (a) If f then (DVL)max , , 28., , (a) Capacitive reactance XC =, , 29., , æ, ö, E, ç I=, ÷, increase ç, 2, 2 ÷ and hence brightness of, R, X, +, C ø, è, source (or electric lamp) will also increase., (b) The phase angle for the LCR circuit is given by, , =, , ( X L - X C )2 + R 2, , (80 - 50 )2 + 402, , = 50 W, 40, R, Power factor = cos f =, =, = 0.8, 50, Z, 200V, Vrms, =, =4A, 50W, Z, Power current = Irms.cos f = 4 × 0.8, = 3.2 A, Wattless current = Irms.sin f = 4 × 0.6, = 2.4 A, 25-27, , Irms =, , Sol., , DVL, , 25., , DVR, , (d), , DVC, , 2, 2, E m = (DVR )max, + ( DVC - DVL )max, , 26., , 30., , 1, . When capacitance, wC, (C) increase, the capacitive reactance decreases. Due, to decrease in its values, the current in the circuit will, , 1, X L - X C wL - wC, tan f =, =, R, R, where XL, XC are inductive reactance and capacitive, reactance respectively when XL > XC then tan f is, positive i.e. f is positive (between 0 and p/2). Hence, emf leads the current., (a) If resistor is used in controlling ac supply, electrical, energy will be wasted in the form of heat energy across, the resistance wire. However, ac supply can be, controlled with choke without any wastage of energy., This is because, power factor (cos f) for resistance is, unity and is zero for an inductance. [P = EI cos f]., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, DPP/ P 48, , 132, , æ 2px, ö, + 2pnt ÷, = r sin ç, è l, ø, 11, 2pn = 1.5 × 10, 1.5 ´ 1011, = 23.9 × 109 Hz = 23.9 Hz, 2p, (c) The given equation, Ey = 0.5 cos[2p × 108 (t – x/c)] ..... (1), indicates that the electromagnetic waves are, propagating along the positive direction of X-axis., The standard equation of electromagnetic wave is given, by Ey = E0 cosw(t – x/c), ..... (2), Comparing the given eq. (1) with the standard eq. (2),, we get w = 2p × 108, or 2pn = 2p × 108, \ n = 108 per second, , n=, , 14., , 15., , 3 ´ 108, c, =3m, =, n, 108, (a) The maximum value of magnetic field (B0) is given by, , 16., 17., , E0, E0, =, = 10–6 tesla, c, c, The magnetic field will be along Z-axis, The maximum magnetic force on the electron is, Fb = |q (v × B)| = q n B0, = (1.6 × 10–19) × (2.0 × 107) × (10–6), = 3.2 × 10–18 N, (c) b-rays are beams of fast electrons., (b), , 18., , (d) v =, , 19., , (c) lm > lv > lx, , \ Speed and wavelength of wave becomes half and, frequency remain unchanged., For 25 - 27, h = 150 km = 150 × 103 m, Nm = 9 × 1010 per m3, D = 250 km = 250 × 103 m, 25. (a) Critical frequency of layer, fc = 9 N m = 9 ´ 9 ´ 1010 = 2.7 ´ 106 Hz., , 26., , f = fc 1 +, , 27., , Now, l =, , B0 =, , 20., , c, , mr er =, , 3 ´ 108, , 1.3 ´ 2.14, , 21 . (c) Wave impedance Z =, , =, , 50, ×376.6=1883W, 2, , 22., , (a), , 23., , (a) b- rays are beams of fast electrons., , 24., , (a) Refractive index =, , me, m 0e 0, , Þ Then refractive index =, , e, =2, e0, , 4h 2, , æ 250 ´ 103 ö, = 2.7 ´ 10 ´ 1 + ç, 3÷, è 4 ´ 150 ´ 10 ø, , 2, , 6, , f, 3.17 ´ 106, =, = 1.174, fc, 2.7 ´ 106, , fi = sec -1 (1.174) = 31.6°, , 28., , 29., , 30., , m, mr, ´ 0, er, e0, , D2, , = 3.17 × 106 Hz., (c) If angle of incidence at this layer, is fi , from second law of f = f c sec, f i., sec fi =, , = 1.8 ×108m/sec, , r r, (d) Direction of wave propagation is given by E×B ., , (b) Maximum usuable frequency, , (d) The electromagnetic waves of shorter wavelength do, not suffer much diffraction from the obstacles of earth's, atmosphere so they can travel long distance., (b) The wavelength of these waves ranges between 300 Å, and 4000 Å that is smaller wavelength and higher, frequency. They are absorbed by atmosphere and, convert oxygen into ozone. They cause skin diseases, and they are harmful to eye and cause permanent, blindness., (b) Radio waves can be polarised because they are, transverse in nature. Sound waves in air are longitudinal, in nature., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, DPP/ P 49, , 134, , (1), , 49, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , (5), , 360 360, =, =4, q, 90, \ n is an even number., , (a) Here, n =, , (6), , M2, I2, , O, , c•, , (c) Small and erect image is formed only by convex mirror., Plane mirror from images equal to object and concave, mirror form images bigger than object., (b) The image will be formed by the plane mirror at a 30, cm behind it, while the image by convex mirror will be, formed at 10 cm behind the convex mirror., Since for convex mirror u = – 50 cm, v = 10 cm, 1, 1, 1 -1 + 5 4, =, + =, =, f -50 10, 50, 50, , M1, , 50, = 12.5 cm, 4, Therefore the radius of curvature of convex mirror is, r = 2 f = 25 cm, (a) The image of object O from mirrror M1 is I1 and the, image of I1 (the vitual object) from mirror M2 is I3., The image of object O from mirror M2 is I2 and the, image of I2 (the virtual object) from mirror M1 is I4., Notice that this interpretation, according to ray diagram, rules, is valid only for Fig. (A). All others are, inconsistent., (a) Angle betwen incident ray and mirror = 90º - 30º = 60º, , f=, , I3, , (2), , I1, , Thus, number of images formed = n – 1 = 3. All these, three images lie on a circle with centre at C (The point, of intersection of mirrors M1 and M2 ) and whose radius, is equal to the distance between C and object., (c) Here, [n] = 5 Þ n – 1 £ 5 £ n, \, , 360, 360, –1£5£, q, q, , (7), , (8), , (3), , 360, 360, or q £, 6, 5, \ 60º £ q £ 72º, (d) For the given q = 50º,, , (4), , 360, 360, =, = 7.2, q, 50, The integer value of (7.2) is 7., Thus number of images formed is 7., (a) The situation is illustrated in figure., , or, q ³, , 30º, , 60º 30° 30°, , X, , 60º, , n=, , B, , q, O, , (9), , M2, q, a, a, b, , C, , N2, N1, i, i, 90º-i, A, , X, , By law of reflection Ð i = Ð r, So angle of reflection Ð r = 30º., Hence angle between mirror and reflected ray = 60º, (b) As shown in figure, ray AB goes to mirror M1, gets, reflected and travels along BC and then gets reflected, by M2 and goes in CD direction. If the angle between, M1 and M2 be a, then, M1, , M1, , XA is the incident ray. BC is the final reflected ray. It, is given that BC is parallel to mirror M1. Look at the, assignment of the angles carefully. Now N2 is normal, to mirror M2. Therefore b = q, Then from D OAB, q + b + 90º – i = 180º, or q + q + 90º – i = 180º or i = 2q – 90º, Thus if the angle of incidence is i = 2q - 90º, then the, final reflected ray will be parallel to the first mirror., , B, , E, , a, D, , a, a a, , O, , a, C, , A, M2, , In D OBC , ÐOBC and ÐOCB are equal to a, \ 3a = 180º, a = 60º
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t.me/Magazines4all, , DPP/ P 49, , 135, (12) (d) u = –50 cm , f = 25cm, , (10) (a) Here, Object distance , u = – 15cm, focal length , f = – 5 cm, Object height, h0 = 0.2 cm, , 1, 1 1, =- + ;, 25, 50 v, , O, , 1 1, 1 2 +1 3, =, +, =, =, v 25 50, 50, 50, , I, , v=, , We know, mirror formula,, , uR, 2u - R, dv (2u - R).R - uR.2 du, ., =, dt, dt, (2u - R)2, , Þv=, , uf, u-f, , (-15)( -5), = – 7.5 cm, -15 + 5, Again, magnification,, , 2, , æ R ö du, ., = –ç, è 2u - R ÷ø dt, , =, , m =-, , 1 1 2, + =, v u R, , (13) (a) We know,, , 1 1 1, + =, v u f, , ÞV =, , 50, = 16.6 cm., 3, , 2, dv, æ R ö, =ç, \ Speed of image, ., è 2u - R ÷ø, dt, , v, 1, -7.5, ==u, 2, -15, , Now, | m | =, , hi, ho, , 0.2, cm = 0.1 cm, 2, Thus, the image is formed at 7.5 cm from the pole of, the mirror and its size is 0.1 cm., , du æ R ö2, V, =, dt çè 2u – R ÷ø 0, (14) (a) For shrot linear object du and dv represent size of, object and image respectively., \, \\, \, \, \ \, \ \, \ \ \ \ \, , Þ hi = | m | h o =, , (11) (a), , u, , P, , O, , F I, , C, , We know,, Þ dv = –, Here, Object distance, u = – 30 cm, Focal length, f = + 20 cm, 1 1 1, We know, mirror formula, + =, v u f, uf, -30 ´ 20, =, = + 12 cm, u - f -30 - 20, Again, magnification ,, , Þ |db| =, , -12cm, -v, 2, =, =, 30cm, u, 5, Now, Image height = m × object height, , 2, ´ 0.5 cm = 0.2 cm, 5, Thus the image is formed behind the mirror at a distance, of 12 cm from the pole. Image height is 0.2 cm., , =, , u+ du, , b, , 1 1 1, + =, v u f, v2, u2, v2, u2, , .du, , |du|, 2, , æ f ö, =ç, ÷ .b, èf–uø, , Þv=, , m=, , 2, , 1, , v, 1 1 1, u u, 15. (b) \ m = - u .Also f = v + u Þ f = v + 1, Þ-, , u, u, -v, f, = 1- Þ, =, v, f, u, f -u, , so m =, , f, f -u, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, DPP/ P 49, , 136, 16., , (b) Given u = – 15 cm, f = – 10 cm, O = 1 cm, –n =, , 1 1 1 1 1 1, 1, 1, + = , = - =, v u f v f u, -10 -15, \ v = – 30 cm, , 17., , I, v, -30, =- ==–2, O, u, -15, I = – 2 × 1 = – 2 cm, Image is inverted and on the same side (real) of size 2, cm., (b) As shown in the figure, when the object (O) is placed, between F and C, the image (I) is formed beyond C. It, is in this condition that when the student shifts his eyes, towards left, the image appears to the right of the object, pin., , -f, Þ nf + nu = –f, -f - u, , -(n + 1), f, n, (d) Here image can be real or virtual. If the image is real, f = –30, u = ?, m = –3, , nu = –f –nf, , 22., , Þu=, , f, -30, Þ –3 =, ; u = – 40 cm., f -u, -30 - u, If the image is virtual, , m=, , f, -30, Þ 3=, Þ u = –20 cm., f -u, -30 - u, (b) By keeping the incident ray is fixed, if plane mirror, rotates through an angle q reflected ray rotates through, an angle 2q, , m=, , 23., , Movement towards left, F, I, , C, , q, , O, , q, q, , 18., , A, , (a), I, 30, º, , N, , 30º, , O, , 19., 20., , P, , 30º, , 60º, , Q, , B, , (d) Real image, , 24., , Ð i = Ð r = 30º, \ Ð OIQ = 60º, \ IOQ = 90º – 60º = 30º, (d) The image formed by a convex mirror is always virtual., (d) From the ray diagram., M, 2L X N, d/2, , A, D L, , B, , I, , O, Virtual object, , 25., , (b,c) Convex mirror and concave lens form, virtual image, for all positions of object., (26) (a); (27) (d), , d/2, X, , In DANM and DADB, ÐADB = ÐANM = 90°, ÐMAN = ÐBAN, (laws of reflection), Also ÐBAN = ÐABD, Þ ÐMAN = ÐABD, \ DANM is similar to DADB, x, d /2, =, or x = d, 2L, L, So, required distance = d + d + d = 3d., , \, , 21., , (a) m = –n ; m =, , f, f -u, , O, I, , 70 cm, , 100 cm, 700 cm, 700 cm, µ2 µ1 µ2 – µ1, , when x = 70 cm, –, =, v, u, R, 1.5 1.2 1.5 – 1.2, –, =, v –70, 20, Þv=, , 20 ´ 70 ´ 1.5, –1.2 ´ 20 + 0.3 ´ 70
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t.me/Magazines4all, , DPP/ P 49, Þ v = – 700 cm, µ2 µ1 µ2 – µ1, –, =, v, u, R, 1.2 1.5 1.2 –1.5, –, =, –20, v 900, 900 ´ 200 ´ 1.2, 1.5 ´ 200 – 900 ´ 3, Þ v = – 90 cm, Þv=, , 137, Similarly, for x = 80 cm, v = 80 cm, and for x = 90 cm, v = 70 cm, 28. (d), 29. (d) When an object is placed between two plane parallel, mirrors, then infinite number of images are formed., Images are formed due to multiple reflections. At each, reflection, a part of light energy is absorbed. Therefore,, distant images get fainter., 30. (d) The size of the mirror does not affect the nature of the, image except that a bigger mirror forms a brighter, image., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, DPP/ P 50, , 138, , (1), , 50, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, (a) We know that d = A (m – 1), , (, , ), , 2, 2, Þ sin2 a < m1 - m 2 sin a <, , d, or m = 1 +, A, Here A = 6 °, d = 3° , therefore, , amax = sin–1, , 3, = 1.5, 6, (c) According to given problem, A = 30°, i1 = 60° and d = 30° and as in a prism, d = (i1 + i2) – A, 30° = (60 + i2) – 30 i.e., i2 = 0, So the emergent ray is perpendicular to the face from, which it emerges., Now as i2 = 0, r2 = 0, But as r1 + r2 = A , r1 = A = 30°, So at first face, , m12 - m22, , B, , m=1+, , (2), , 1× sin 60° = m sin 30° i.e., m =, (3), , (4), , (5), , i, a O, (6), , 3, , c 3, c, Þ v = = × 3 × 108, m 4, v, = 2.25 × 108 m/s, As, c = n l0 and v = n l, , (7), , R, 4, =, = 1.55, A 2.58, (d) The apparent shift of the bottom point upwards will be, x = x1 + x2, , æ, 1 ö, çè 1 - m ÷ø, 2, , æ, 1 ö, = 4 çè 1 - (4 / 3) ÷ø + 2, , 3, i.e., l = l0/m = × 6000Å = 4500Å, 4, (d) Herer + 90º + r’ = 180º, Þ r’ = 90º – r, or, r’ = (90º – i) as Ð i = Ðr, Now, according to Snell’s law :, sin i = m sin r’ = m sin (90º – i), or, tan i = m, or, i = tan–1 m = tan–1 (1.5), (b) Here the requirement is that i > c, , æ, 1 ö, çè 1 - (3 / 2) ÷ø, , æ 3ö, æ 2ö, = 4 çè 1 - ÷ø + 2 çè 1 - ÷ø = 1.67 cm., 4, 3, (8), , C, n, The time taken are, , (d) Since v =, , 20 (1.63), 20 (1.47), , t1 =, C, C, Therefore , the difference is, , t2 =, , m2, Þ sin i > sin c Þ sin i > m, 1, , 20(1.63 - 1.47) 20 ´ 0.16, =, = 1.07 × 10–8 sec., C, 3 ´ 108, (b) As the beam just suffers TIR at interface of region III, and IV, t2 - t1 =, , sin a, sin r, , (9), , Also in DOBA, r + i = 90° Þ r = (90 – i), Hence from equation (ii), sin a = m1 sin(90 – i), sin a, Þ cos i = m, 1, æ sin a ö, sin i = 1 - cos 2 i = 1 - ç, ÷, è m1 ø, From equation (i) and (ii), , A, , (b) From the information given, it is clear that the apparent, depth is 2.58 mm and the real depth is 4mm. Therefore,, the refractive index will be, , æ, 1ö, = t1 ç 1 - m ÷ + t2, è, 1ø, , v 1, =, c m, , From Snell's law m1 =, , r, , m=, , (a) As are know, m =, , Þ l / l0 =, , m12 - m22, , Region I Region II, q, n0, , n0, 2, q, , 2, , ...(iii), , nosin q, , Region III, n0, 6, , 0.2 m, , n0, 8, 0.6 m, , n, n, n0, sin q1 = 0 sin q2 = 0 sin 90°,, 6, 8, 2
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t.me/Magazines4all, , DPP/ P 50, , 139, , 1, -1 1, Þ q = sin, 8, 8, (10) (b) The ray of light returns back from the polished face, AC., sin q, , A, , i, , E 30º, , 60º, D, 30º, F, , B, , C, , \ ÐADE = 90º. From the figure it is clear that the angle, of refraction at face AB is 30º. Hence from Snell’s law, m=, , sin i, sin r, , As m =, , or sin i =, , 2 and r = 30º \, , sin i, 2 = sin 30º, , 2, 1, =, = sin 45º, 2, 2, , .......(1), .......(2), , sin, , A + dm, 60 + d m, sin, 2, 2, or 1.532 =, A, °, sin, 30, sin, 2, , 60 + d m 1.532, =, = 0.766, 2, 2, \ dm = 40º, (12) (b) d1 = d2, (m1 – 1) A1 = (m2 – 1) A2, (1.5 – 1) 6 = (1.6 – 1) A2, A2 = 5º, (13) (a) The deviation produced by the crown prism is, d = (m – 1) A, and by the flint prism is, d ' = (m' – 1) A', The prisms are placed with their angles inverted with, respect to each other . The deviations are also in, opposite directions. Thus, the net deviation is, D = d – d ' = (m – 1) A – (m' – 1) A' ...........(1), sin, , m v + m r 1.5230 + 1.5145, =, 2, 2, m = 1.5187, , w=, , m v - mr, m -1, , 1.5230 - 1.5145 0.0085, =, = 0.0163, 1.5187 - 1, 0.5187, (15) (c) Here, w = 0.021; m = 1.53; w ' = 0.045;, m' = 1.65; A' = 4.2°, For no dispersion,, wd+w'd'=0, or w A (m – 1) + w ' A ' (m – 1) = 0, , =, , sin i sin 50º, =, = 1.532, sin r sin 30º, , n=, , 1.517 - 1, (m - 1), ´ 5° = 4.2°., A=, 1.620 - 1, (m '- 1), The angular dispersion produced by the crown prism, is, dv – dr = (mv – mr) A, and that by the flint prism is, d 'v – d 'r = (m'v – m'r) A', The net angular dispersion is ,, d = (mv – mr) A – (m'v – m'r) A', = (1.523 –1.514) × 5° – (1.632–1.613) × 4.2°, = – 0.0348°., The angular dispersion has magnitude 0.0348°., (14) (a) mv = 1.5230, mr = 1.5145, w = ?, Mean refractive index,, , or, A' =, , m=, , \ i = 45º, (11) (a) A = r1 + r2 = 60, In minimum deviation position, r1 = r2, From eqs. (1) and (2), A = 2r1 = 60º, \ r1 = 30º, \n=, , If the net deviation for the mean ray is zero,, (m – 1) A = (m' – 1) A', , or A = –, , w ' A ' ( m '- 1), w ( m - 1), , 0.045 ´ 4.2 ´ (1.65 - 1), = – 11.04°, 0.021 ´ (1.53 - 1), Net deviation,, d + d ' = A (m – 1) + A' (m ' – 1), = – 11.04 (1.53 –1) + 4.2 (1.65 –1), = – 11.04 × 0.53 + 4.2 × 0.65, = – 5.85 + 2.73 = 3.12°, (16) (d) At first face of the prism as i1 = 0,, sin 0 = 1.5 sin r 1 i.e., , r1 = 0, And as for a prism, r1 + r2 = A, so r2 = A, ......(1), But at second face, as the ray just fails to emerge, i.e., r2 = qC, ......(2), So from Eqn,.(1) and (2), A = r2 = q C, , =–, , é1ù, é2ù, But as qC = sin –1 ê ú = sin –1 ê ú = 42°, m, ë3û, ë û, So A = 42°, (17) (a) Here, mb= 1.532 and mr = 1.514 A = 8° ., Angular dispersion, = (mb – mr) A = (1.532 – 1.514) × 8, = 0.018 × 8 = 0.144° ., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, DPP/ P 50, , 140, (18) (c) Velocity of the ball when it reaches at the height of 12.8, m. above the surface is v = 2 ´10 ´ 7.2 = 12 m/s, Height of the ball from surface as seen by fish, h¢ = mh Þ, , dh ¢, dh, =m, dt, dt, , (21) (c), , dPrism = (µ - 1)A = (1.5 - 1)4° = 2°, , \ dTotal = dPrism + dMirror, = (µ – 1)A + (180 – 2i) = 2° + (180 – 2 × 2) = 178°, (22) (a) From the following figure, , u=0, u = 12 m/s, , 20, , a, , 12.8 m, , n, , 4, ´ 12 = 16 m/s., 3, (19) (a) Suppose, the angle of the crown prism needed is A and, that of the flint prism is A’. We have, m - mr, w= v, or,, mv – mr = (m – 1) w, m -1, The angular dispersion produced by the crown prism, is, (mv – mr) A = (m –1) w A, Similarly , the angular dispersion produced by the flint, prism is (m´ – 1) w´ A´, For achromatic combination , the net dispersion should, be zero. Thus ,, (m – 1) w A = (m´ – 1) w´ A´, , Þ v¢ = mv =, , or ,, , A ' (m - 1) w, 0.517 ´ 0.03, =, = 0.50 ......(1), =, A (m '- 1) w ' 0.621 ´ 0.05, , The deviation in the yellow ray produced by the crown, prism is d = (m – 1) A and by the flint prism is d´ = (m´, – 1) A´. The net deviation produced by the combination, is, d – d´ = (m – 1) A – (m´ – 1) A´, or 1° = 0.517 A – 0.621 A´, ......(2), Solving (1) and (2 ), A = 4.8° and A´ = 2.4°. Thus, the, crown prism should have its refracting angle 4.8° and, that of the flint prism should be 2.4°., (20) (a) For TIR at AC, q>C, A, B, q, Þ sin q ³ sin C, Þ sin q ³, , µw, µg, , Þ sin q ³, , 8, 9, , r + i = 90° Þ i = 90° – r, For ray not to emerge from curved surface i > C, Þ sin i > sin C Þ sin (90° – r) > sin C Þ cos r > sin C, 1ü, 1, ì, 1 - sin 2 r >, í\ sin C = ý, Þ, nþ, n, î, Þ 1-, , sin 2 a, 2, , >, , 1, 2, , Þ 1>, , 1, 2, , n, n, n, Þ n2 > 1 + sin2 a Þ n > 2, Þ Least value =, , (1 + sin 2 a ), {sin i ® 1}, , 2, , (23) (a) w depends only on nature of material., (24) (a), (25) (c);, 26. (b), 27 (a), The normal shift produced by a glass slab is,, æ, 1ö, æ 2ö, Dx = ç1 - ÷ t = ç1 - ÷ (6) = 2cm, è 3ø, è mø, , i.e., for the mirror, the object is placed at a distance, (32 – Dx) = 30 cm from it., Applying mirror formula,, , 1 1 1, + =, v u f, 1 1, 1, +, =v 30, 10, or, v = – 15 cm, (a) When x = 5 cm: The light falls on the slab on its return, journey as shown. But the slab will again shift it by a, distance., , 6 cm, , q, , 1, w µg, , Þ sin q ³, , i, , r, , I, C, , Dx, , 15 cm, Dx = 2 cm. Hence, the final real image is formed at a, distance (15 + 2) = 17 cm from the mirror.
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t.me/Magazines4all, , DPP/ P 50, , 141, , (b) When x = 20 cm: This time also the final image is at a, distance of 17 cm from the mirror but it is virtual as, shown., , (29) (b) The velocity of light in a material medium depends upon, its colour (wavelength). If a ray of white light is incident, on a prism, then on emerging the different colours are, deviated through different angles., Also dispersive power w =, , ( mV - m R ), ( mY - 1), , i.e. w depends upon only m, (30) (c) Apparent shift for different coloured letter is, , I, , Dx, , (28) (b) As rays of all colours emerge in the same direction (of, incidence of white light), hence there is no dispersion,, but only lateral displacement in a glass slab., , æ 1ö, d = h ç1 - ÷, è mø, Þ lR > lV so mR < mV, Hence dR < dV i. e. red coloured letter raised least., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 51, , 141, , 1., , 51, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , (a). According to sign convention, it is given that, u = – 0.6 m, R = 0.25 m, µ1 = 1 (air), µ2 = 1.5, , v=, , 2 - 60, 40, ´, == – 3.63 cm, 3 11, 11, , \ AI = 3.63 cm, 3., , (c)., , u, Therefore, using, -, , •, O, , f, , f, , x1, , µ1 µ2 µ2 - µ1 , we get, +, =, u, v, R, , v, , •, F1, , •, , x2, •, F2, , •, I, , 1.5, 1, 1.5 - 1, 1, 0.5, +, =–, =, +, v, (- 0.6) 0.25, 0.6 0.25, 5, 1, +2=, 3, 3, Þ v = 4.5 m, The image is formed on the other side of the object (i.e., inside the refracting surface)., (a). On viewing from the closer surface A (near to object), :, The final image is formed at I., , =–, , 2., , As in case of thin lens the distance of either foci from the, optical centre is f ,, | u | = ( f + x1 ) and, | v | = ( f + x2 ), Substituting thses values of u and v in lens formula with, proper sign, 1, 1, 1, =, ( f + x 2 ) - ( f + x1 ) f, x1 + x 2 + 2f, 1, =, ( f + x1 )( f + x 2 ) f, , or, , A, , I, , O, , O C, A, ·, 1cm, , C, , Nearing, surface, , B, Far, surface, , 4., , 1, = (m – 1), f, , 2/3 1 2/3 -1, m 1 µ -1, - =, Þ v +4 = -5, v u, R, , Þ, , 2, 1 1 4 - 15 - 11, = - =, =, 3v 15 4, 60, 60, , é 1, 1 ù, mL, ê ú with m =, R, R, m, ë 1, 2û, M, , Here f = 30 cm and R1 =10 cm and R2 = ¥, , From sign convention, u = OA = – 4 cm, v = ?, R = AC = – 5cm, µ2 2, µ= µ = 3, 1, , i.e., fx1 + fx2 + 2f 2 = f 2 + fx1+ fx2 + x1 x2, or, x1 x2 = f 2, (b). According to Lens-maker’s formula :, , So,, , 5., , é1 1ù, 1, =(m–1) ê - ú, 30, ë10 ¥ û, , 3m – 3 = 1, or, m = ( 4 / 3 ), (a). For combination of lenses, 1 1 1, 1 1, 25 1, = + = + =, =, f f1 f 2 10 15 150 6, , Therefore, f = 6 cm.
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t.me/Magazines4all, DPP/ P 51, , 142, 6., , (c). Let f2 is the focal length of the diverging lens. Then ,, , 9., , 1 1 1, = +, f f1 f2, , (b). The focal length of lens in water is given by, fl =, , It is given that f1 = +20 cm, f = 30 cm, 1, 1, 1, =, +, 30 20 f 2, , Thus f2 = – 60 cm, (a). As radius of curvature of silvered surface is 22 cm, so,, , 10., , 1 2 (m - 1), =, f, R, , 11., R -22, =, = –11 cm = – 0.11 m, 2, 2, , and hence, PM = –, , 1, 1, 1, D, =, =, f M -0.11 0.11, , Further as the focal length of lens is 20 cm, i.e. 0.20 m, its, power will be given by :, , It is given that, , 1, = + 5 and m = 1.5, f, , Therefore, 5 =, , 2 (1.5 - 1), R, , Now as in image formation, light after passing through the, lens will be reflected back by the curved mirror the lens, again P = PL + PM + PL = 2PL + PM, 2, 1, 210, D, +, =, 0.20 0.11 11, So the focal length of equivalent mirror, , =, 12., , i.e. P =, , 8., , 1, 11, 110, m=cm, =P, 210, 21, , 1, metre = 20 cm, 5, , (b). The question is based on the conventional method of, measurement of focal length by displacement method., According to this method where D is the distance between, object and the image, and x is the displacement given to, the object., From the data x = 25 cm and D = 75 cm ., Thus, f=, , 1, 1, D, =, PL =, f L 0.20, , F=–, , 1.2 - 1, fa, fa = 1.2, -1, -1, 1.33, a ml, , 0.2 ´ 1.33, fa, 0.13, Hence f is negative and as such it behaves as a divergent, lens., (a). The focal length of an equiconvex lens is given by, , or R =, , fM =, , -1, , a mg, , fl = -, , 1, 1, 1 2-3, 1, or f = 30 - 20 = 60 = - 60, 2, , 7., , a mg, , (75)2 - (25) 2, 4 ´ 75, , 13., 14., , 1, 1, 21, +, =i.e. v = –11 cm, v -10, 110, i.e. image will be 11 cm in front of the silvered lens, (b). On covering the lens half by a black paper will reduce, the intensity of image and not the part of image. So full, image is formed., , 15., , (75 - 25) (75 + 25), 4 ´ 75, , 50 ´ 100 50, =, = 16.7 cm, 4 ´ 75, 3, , A1, A2, and m2, O, O, A1A 2, Þ m1m2 =, O2, Also it can be proved that m1m2 = 1, , (c). m1 =, , So O =, , i.e. the silvered lens behaves as a concave mirror of focal, length (110/21) cm. So for object at a distance 10 cm in, front of it,, , =, , A1A 2, , (c). Effective power P = P1 + P2 = 4 – 3 = 1D, (d). One part of combination will behave as converging, lens and the other as diverging lens of same focal length., As such total power will be zero., (c). Let the image of an object at O is formed at the same, point as shown in figure. The distance of O from the plane, surface is x. The rays suffer refraction at first surface, (curved) as they reach lens. After wards become parallel, and gets reflected form plane surface and so retrace the, path and image is formed at O itself., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 51, , 143, So –, , i.e., u = – 7 cm, i.e., film is at a distance of 7 cm in front of projection lens., , x, , ·, O, , 40 ´ 100, 20 ´ 100 cm, =, u, 3.5 cm, , And from lens formula 1 - 1 = 1 , here we have, v u f, 1, 1 1, = or f @ 7 cm = 70 mm, 4000 -7 f, , m 1 m -1, - =, v u, R, , u = – x, v = ¥, 20., , m 1 m -1, + =, ¥ x, R, , x=, , 25, 1, 1 1, = i.e., ue = –, 6, -25 u e 5, , R, m -1, , And so, me =, , As such O behaves as equivalent to centre of curvature of, equivalent concave mirror., , (a). As here f = 10 cm and v = 5m = 500 cm, So from lens formula, 1, 1 1, - =, 500 u 10, , So that, , 17., , m=, , 1, =, diopter = 20 D, P=, -2, 0.05, 5 ´ 10 m, , (b). For relaxed eye, MP is minimum and will be, MP =, , D 25, =, = 5×, f, 5, , While for strained eye, MP is maximum and will be, MP = 1 +, , 19., , Also, m =, , D, = 1 + 5 = 6×, f, , (a). As in case of projector, m =, , v, (95 / 6), é 94 ù, =, = -ê ú, u (-95 / 94), ë6û, , ...(2), , So total magnification,, , v, 500, =, = – 49, u -500 / 49, , 1, , 18., , i.e. object is at a distance (95/94) cm in front of field lens., , é 500 ù, cm, u=– ê, ë 49 úû, , Here negative sign means the image is inverted with respect, to object. Now as here object is (2 cm × 2cm ) so the size, of picture on the screen, Ai = ( 2 × 49 cm) × ( 2 × 49cm) = ( 98 × 98) cm2, (a). As power of a lens is reciprocal of focal length in m,, , ...(1), , 6 1, 1, 95, - =, i.e., u = –, cm, 95 u 0.95, 94, , 1 1 1, - = , we have, v u f, , i.e.,, , ve, -25, =, =6, u e (-25 / 6), , Now for objective,, v = L – ue = 20 – (25/6) = (95/6), So if object is at a distance u from the objective,, , R, \ Radius = x =, m -1, 16., , [as ( 1/4000) << (1/7) ], i.e., focal length of projection lens is 70 mm., (a). As final image is at 25 cm in front of eye piece, , é 94 ù, M = m xme = – ê ú × (6) = – 94, ë6û, , 21., , i.e., final image is inverted, virtual and 94 times that of, object., (a). As object is distant, i.e., u = – ¥ , so, 1 1, 1, =, i.e. v = f0 = 12 cm, v -¥ f 0, , i.e. objective will form the image IM at its focus which is, at a distance of 12 cm from O. Now as eye- piece of focal, length – 4 cm forms image I at a distance of 24 cm from it,, 24, 1, 1, 1, =, Þ ue =, = 4.8 cm., 5, -24 u e -4, , i.e, the distance of IM from eye lens EA is 4.8 cm. So the, length of tube L = OA – EA = 12 – 4.8 = 7.2 cm., I v, =, O u
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t.me/Magazines4all, DPP/ P 51, , 144, 26., f0=12 cm, , I, , >, >>, , f0, , µ2, µ, + w =0, – R2 x ', , fe, 4.8 cm, , (, , q0, , >, , O, , q0, , ), , (, , >>, , ue, , q, , E, , L= 7.2 cm, , A, )q, , IM, , >> >, , B, , 24 cm, , 22., , (d) In case of astronomical telescope if object and final, image both are at infinity., MP = – (f0/fe), and L = f0 + fe, So here –(f0/fe) = – 5 and f0 + fe = 36, Solving these for f0 and fe, we get, f0 = 30 cm and fe = 6 cm, , 1, l, , 23., , (a) Resolving power of microscope µ, , 24., , (a) Light gathering power µ Area of lens aperture or d2, (b) For lens L1, ray must move parallel to the axis after, refraction, , 25., , µ1 µw µ1 – µw, +, =, Þ x = 10cm, ¥, x, R1, , (a) For lens L2, image must form at centre of curvature of, the curved surface after refraction through plane part, , Þ x¢ = 8 cm, 27. (b) Length of tube = x + x¢ = 18 cm, 28. (a) Focal length of lens immersed in water is four times, the focal length of lens in air. It means, fw = 4fa = 4×10 =40 cm, 29. (d) Focal length of the lens depends upon its refractive, index as, , 30., , 1, µ ( m - 1) ., f, , Since mb > mr so fb < fr, Therefore, the focal length of a lens decreases when red, light is replaced by blue light., (b) The light gathering power (or brightness) of a telescope, µ (diameter)2. So by increasing the objective diameter even, far off stars may produce images of optimum brightness., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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DPP/ P 52, , 148, , 28., , 29., , t.me/Magazines4all, , Here n' = n0 + kt, n = refractive index of plate., d sin f, n = n0 + kt +, b, (d) When d is negligibly small, fringe width b which is, proportional to 1/d may become too large. Even a single, fringe may occupy the whole screen. Hence the pattern, cannot be detected., (d) In Young's experiments fringe width for dark and white, fringes are same while in Young's double slit experiment, , 30., , when a white light as a source is used, the central fringe, is white around which few coloured fringes are, observed on either side., (d) When one of slits is covered with cellophane paper,, the intensity of light emerging from the slit is decreased, (because this medium is translucent). Now the two, interfering beam have different intensities or, amplitudes. Hence intensity at minima will not be zero, and fringes will become indistinct., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, DPP/ P 53, , 148, , 1., , (b). The distance of first diffraction minimum from the, central principal maximum x = lD/d, \ sin q =, , For first maxima, sin q =, , 3l1D, 3l 2 D, and x2 =, 2a, 2a, \ spacing between the first maxima of two sodium lines, , 3., , 4., , 3 ´ 2(5896 - 5890) ´ 10-10, 2 ´ 2 ´ 10-6, , =9×, , According to question, l = 2 × 10–3 m, a = 4 × 10–3 m, From equation (1) and (2), , b 2l, =, D, a, , 2 ´ 6328, = 0.36°, 0.2, (a). Here distance of the screen from the slit,, D = 2m,, a = ?,, x = 5 mm, = 5 × 10–3 m,l = 5000Å, = 5000 × 10–10 m, For the first minima, sin q = l/a = x/D,, , q=, , 7., , Dl 2 ´ 5000 ´ 10-10, =, = 2 × 10–4 m, x, 5 ´ 10-3, (d). Here, l = 6500Å = 6.5 × 10–7 m, a = 0.5 mm =, 5 × 10–4 m,, D = 1.8 m, Angular separation of two dark bands on each side of central, bright band 2q = 2l/a, Actual distance between them,, 2x = 2l/a x D, a=, , 10–4 m., 8., , 2x =, , 2 ´ 6.5 ´ 10-7 ´ 18, , 5 ´ 10-4, 2x = 4.68 × 10–3 m, , .....(1), 9., , (c). Width of central maxima =, , .....(2), =, , æ 1ö, q = sin–1 çè ÷ø, 2, 5., , 2ld, a, , Angular width Wq =, , ax, 0.3 ´ 10-3 ´ 5 ´ 10-3, =, n.f, 3 ´1, –7, l = 5 × 10 m, l = 5000Å, , æ lö, (b). q = sin–1 çè ÷ø, a, , 12.5 ´ 10-3, 12.5 ´ 10-3 ´ 10-3, ×a=, 2, 2, l = 6.25 × 10–6 m = 6250 mm, (a). Slit width = a = 0.2mm,, , b=, , 3D, (l – l1), 2a 2, , ax, (d)., = nl, f, , l=, , 6., , 3l1 x1, =, 2a, D, , Þ x1 =, , = x2 – x1 =, , 12.5 ´ 10-3, 2, , l=, , 5000 ´ 10-8, = 2 × 5 × 10–5, sin 30º, Þ d = 1.0 × 10–4 cm., (b). Here, l1 = 5890Å = 5890 × 10–10 m, l2 = 5896 Å = 5896 × 10–10 m, a =2mm = 2 × 10–6 m, D = 2m, , =, , q = l/a =, , x l, l, = Þd=, D d, sin q, , Þd=, , 2., , 53, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , q = 30°, (a). Here the width of principal maxima is 2.5 mm, therefore, its half width is, b 2.5, =, = 1.25 × 10–3m, 2, 2, , b / 2 12.5 ´ 10 -3, Diffraction angle q =, =, D, 2, \aq=l, , 10., , (a). q =, , l, a, , x, f, From eqs. (a) and (b), , q=, , 2fl, a, , 2 ´ 2 ´ 6000 ´ 10-10, 0.2 ´ 10-3, , = 12 mm, , .....(a), .....(b), , l x, =, a f, x=, , fl, a, , .....(c)
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t.me/Ebooks_Encyclopedia27., , DPP/ P 53, , 150, 24., , (a) It magnitude of light vector varies periodically during, it's rotation, the tip of vector traces an ellipse and light, is said to be elliptically polarised. This is not in nicol, prism., , 25., , (a) Multiple focii of zone plate given by f p =, , 28., , rn2, ( 2 p - 1) l ,, , where p = 1, 2, 3 ........, 26., , (a) Angular width is the angle subtended by the distance, between first minima on either side at the centre of the, slit. It is given by f = 2 q , where q is the angle of, diffraction., For first diffraction minimum, a sin q = 1 l, l, a, , or, , sin q = l/a, , \, , Angular width f = 2 q =, , or, , q=, , 29., , 2l, i.e. f µl, a, , f1 l1, f, 70, ; \ l 2 = l1 2 = 6000 ´, =, = 4200 Å, f2 l 2, f1, 100, , 27., , t.me/Magazines4all, , (b) On immersing in liquid, a wavelength l = 6000 Å must, be behaving as l' = 4200 Å to get the same decrease in, angular width. Therefore, refractive index of medium, m=, , l 6000, =, = 1.43., l¢ 4200, , 30., , (a) When a polaroid is rotated in the path of unpolarised, light, the intensity of light transmitted from polaroid, remains undiminished (because unpolarised light, contains waves vibrating in all possible planes with, rotated in path of plane polarised light, its intensity, will vary from maximum (when the vibrations of the, plane polarised light are parallel to the axis of the, polaroid) to minimum (when the direction of the, vibrations becomes perpendicular to the axis of the, crystal). Thus using polaroid we can easily verify that, whether the light is polarised or not., (d) The nicol prism is made of calcite crystal. When light, is passed through calcite crystal, it breaks up into two, rays, (i) the ordinary ray which has its electric vector, perpendicular to the principal section of the crystal and, (ii) the extra ordinary ray which has its electric vector, parallel to the principal section. The nicol prism is made, in such a way that it eliminates one of the two rays by, total internal reflection, thus produces plane polarised, light. It is generally found that the ordinary ray is, eliminated and only the extra ordinary ray is transmitted, through the prism. The nicol prism consists of two, calcite crystal cut at - 68° with its principal axis joined, by a glue called Canada balsam., (d) The clouds consists of dust particles and water droplets., Their size is very large as compared to the wavelength, of the incident light from the sun. So there is very little, scattering of light. Hence the light which we receive, through the clouds has all the colours of light. As a, result of this, we receive almost white light. Therefore,, the cloud are generally white.
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t.me/Magazines4all, , DPP/ P 54, , 151, , 1., , 0.101, , (a). la =, , V, , Å,, , V =, , 8., , 0.101, 0.004, , (d). l =, , 9., , c 3 ´ 108, =, = 1.5 × 107 m/sec, 20, 20, h = 6.626 × 10–34 J-s, m = 1.67 × 10–27 kg, , Q v=, , 6.626 ´10-34, , me, , 1, m, , (a). lp =, , \, , ld, =, lp, , 12.27, 20, , 1, mp, , 11., , = 0.061 Å, 40 ´ 103, \ Resolving limit of electron microscope = 0.061 Å, (d). The linear momentum of the photon, =, , = 7.2 × 105 m/s, , ,, , h 6.63 ´ 10-34, kg - m, =, = 5.43 × 10–27, 9, l 122 ´ 10, s, , Q p = mv Þ v =, Þv=, , mp, , me, , 1836, 1, , =, , 1.67 ´ 10-27, , 150, le2, , V=, , h, , volt, to determine the p.d. through which it was, , 150, volt, V = 3765 Volt = 3.76 kV, 0.2 ´ 0.2, , 2md ed V, , ld, =, ed = ep Þ, lp, , = 3.25 m/s, , accelerated to achieve the given de-broglie wavelength., Then the same p.d. will retard it to rest. Thus,, , 2m p ep V, , m d ed, , p, m, , 5.43 ´ 10-27, , 12. (b). V =, , h, , m p ep, , Å = 2.75 Å, , 150, , l=, , 9.1 ´ 10-31 ´ 10 ´ 10-10, , Þ ld =, , 7., , , lp µ, , 6.6 ´ 10-34, , (b). l µ, , Å, V, Þ V = 40–20 = 20 Volt, , 150, Å, V, Now, electrons have energy of 40 KeV, therefore they, are accelerated through a potential difference of 40 ×, 103 volt., , me, , le, =, lp, , 6., , 1, , h, h, (c). l =, Þv=, ,, mv, ml, , v=, , 5., , m, , Þ le µ, , 12.27, , 10. (d). Wavelength of electrons is l =, , mp, , le, \ l =, p, , 4., , 1, , (a). Q l µ, , (c). l =, , Þ l=, , 1.67 ´10-27 ´ 1.5 ´ 107, Þ l = 2.64 × 10–14 m, , 3., , 6.62 ´ 10-34, , 2 ´ 9.1 ´ 10-31 ´ 13.6 ´ 1.6 ´ 10-19, Þ l = 3.3 × 10–10 m = 3.3 Å, , h, mv, , \ l=, , (b)., l=, , V = 25.25 V , V = 637.5 V, Ea =qa × Va » 1275 eV, , 2., , 54, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 13. (b). lphoton =, Q md = 2mp,, Þ, mp ep, 2m p e p, , =, , 1, 2, , (d). Kmax of photoelectrons doesn't depends upon intensity, of incident light., , hc, and lproton =, E, , λ photon, λ electron, , =c, , h, 2me, , λ photon, 1, 2m, µ, Þ, λ electron, E, E, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ P 54, 26. (a) DE = W0 + E ; (Ek) = DE – W0, For maximum value of (Ek), W0 should be minimum, W0 for lithium = 2.3 eV, \ (Ek) = 2.75 – 2.3 = 0.45 eV, 27. (c) The maximum magnitude of stopping potential will be, for metal of least work function., \ required stopping potential is, Vs =, , hv – f0, = 0.45 volt., e, , 153, , hv h, 28. (c) Mass of moving photon m = 2 =, and E = mc2, cl, c, 29. (c) Less work function means less energy is required for, ejecting out the electrons., 30. (a) de-Broglie wavelength associated with gas molecules, varies as l µ, , 1, T, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 56, , 155, , 1., , 106, , eV 6.25 ×, 1.6 ´ 10-19, The number of fissions should be, thus, = 1000 ×, , number =, 2., , 9., , (c). The energy produced per second is, 103, , J=, , 1024, , 92, , U235, , 2, =, × NA, 235, , 5.12 ´ 10, = 1.975 × 1018 per sec, 30 ´ 24 ´ 60 ´ 60, Usable energy per fission = 185 MeV, \ Power output, = (185 × 106)(1.975 × 1018)(1.6 × 10–19) watt, = 58.4 × 106 watt = 58.46 MW, , Fission rate =, , 6 ´1023, and each, 2, atom of 6C12 contains electron, protons and neutrons, = 6, 6, 6, \ No. of electron, protons and neutron in 6 gm of, 12, 23, 23, 23, 6C = 18 × 10 , 18 × 10 , 18 × 10, (c). Use r = Mass/volume, , (d). \ 6 gm of 6C12 contains atoms =, , =, 5., 6., , 7., 8., , 1.66 ´ 10-27 ´ 16, (4 / 3)p(3 ´ 10 -15 ), , = 2.35 × 1017 kg m-3, , (a). Mass defect Dm = M (Ra 226) – M(Rn 222) – M (a), = 226.0256 – 222.0175 – 4.00026 = 0.0053 u., (a). E = mc2 = (1.66 × 10–27) (3 × 108)2 J, = 1.49 × 10–10 J, =, , or log 20 =, , = 3.125 × 1016, 200 ´ 106, , 24, , 4., , N, 1, 0.6931, and l =, Þ 20 = e, Given N =, 20, 3.8, 0, Taking log of both sides, , eV, , 2, =, × (6.02 × 1026) = 5.12 × 1024, 235, , 3., , (b). By the forumula N = N0e–lt, , 6.25 ´ 1024, , (b). No. of atoms in 2kg, , 1.49 ´ 10-10, 1.6 ´ 10-13, , 56, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , MeV = 931.49 MeV, , (b). E = mc2, = (9.1 × 10–31) (3 × 108)2 J = 0.51 MeV, (c). DE = D mc2, 0.5, kg = 0.005 kg, 100, c = 3 × 108 m/s, DE = 0.005 × (3 × 108)2, DE = 4.5 × 1014 J or watt-sec, , 0.6931 ×t, log10 e, 3.8, , 0.6931 ×t ´ 0.4343, Þ t = 16.5 days, 3.8, (b). A = 238 - 4 = 234, Z = 92 - 2 = 90, (a). Dm = 0.03 a.m.u., A = 4, , or 1.3010 =, , 10., 11., , Þ DE =, , Dm ´ 931, A, , 0.03 ´ 931, = 7 MeV, 4, 12. (a). Q DE = Dm × 931 MeV, , Þ DE =, , Þ Dm =, , DE 2.23, =, = 0.0024 a.m.u., 931 931, , R Al, (27)1/3 3 6, =, = =, 13. (a). R, (125)1/3 5 10, Te, , 14. (d). R = R0 A1/3 = 1.2 × 10-15 × (64)1/3, = 1.2 × 10–15 × 4 = 4.8 fm, 15. (b). Number of protrons in nucleus = atomic number = 11, Number of electrons = number of protons = 11., Number of neutrons = mass number A – atomic number Z, N = 24 – 11 = 13, 16. (d). Q equivalent mass of each photon = 1/2000 amu, Q 1 amu = 931 MeV, 931, = 0.465 MeV, 2000, 17. (c). Deuterium, the isotope of hydrogen consits of one proton, and neutron. Therefore mass of nuclear constituents of, deuterium = mass of proton + mass of neutron, = 1.00759 + 1.00898 = 2.01657 amu., mass of nucleus of deuterium = 2.01470 amu., Mass defect = 2.01657 – 2.01470 = 0.00187 amu., Binding energy = DE = 0.00187 × 931 MeV = 1.741 MeV., , \ Energy of each photon =, , Dm =, , 4.5 ´ 1014, = 1.25 × 10111 watt hour, 60 ´ 60, DE = 1.25 × 108 kWH, , DE =, , 0.631×t, 3.8, , DE Dm ´ 931, =, MeV, A, A, Dm = (3mp + 4mn) – mass of Li7, Dm = (3 × 1.00759 + 4 × 1.00898) – 7.01653, Dm = 0.04216 a.m.u., , 18. (a). E =, , DE =, , 0.04216 ´ 931 39.25, =, = 5.6 MeV, 7, 7
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t.me/Magazines4all, DPP/ P 56, , 156, 19., , (d). The sun radiates energy in all directions in a sphere. At, a distance R, the energy received per unit area per, second is 1.4 KJ (given). Therefore the energy released, in area 4pR2 per sec is = 1400 × 4pR2 Joule the energy, released per day = 1400 × 4pR2 × 86400J, where R = 1.5 × 1011 m, Thus, DE = 1400×4 × 3.14 × (1.5 × 1011)2 × 86400, The equivalent mass is, Dm = DE/c2, , 25., , NA, 6.022 ´ 1023, ´mÞ N =, ´1, 12, 12, = 5.02 × 1022 atoms., The ratio of 14C/12C atoms = 1.3 × 10–12 (Given), \, Number of 14C atoms = 5.02 × 1022 × 1.3 × 10–, , 12, , = 6.5 × 1010, , 1400 ´ 4 ´ 3.14 ´ (1.5 ´ 10 ) ´ 86400, , 0.693, N0, \ Rate of decay R0 = lN0 = T, 1/ 2, , 9 ´ 1016, , Dm = 3.8 × 1014 kg, 20., , (b), , \ R0 =, , B.E., A, +, , Fusion, , 21., 22., , æ1ö, (c) Nt = N0 ç ÷, è2ø, , 26., , = 12500, , (d). Power received from the reactor,, P = 1000 KW = 1000 × 1000 W = 106 J/s, , 6.25 ´ 1018, = 3.125 × 1016, 200, Energy released per hour = 106 × 60 × 60 Joule, , =, , Mass decay per hour = Dm =, , Þ Dm =, , 27., , 28., 29., , DE, c2, , 106 ´ 60 ´ 60, (3 ´ 108 )2, , 10–8 kg, , 23., 24., , Þt=, , 106, , eV/sec., 1.6 ´ 10-19, P = 6.25 × 1018 MeV/sec, \ number of atoms disintegrated per sec, P=, , Þ Dm = 4 ×, (a), (a) In fusion two lighter nuclei combines, it is not the, radioactive decay., , (c) For 10g sample, number of decays = 0.5 per second., i.e. R = 0.05 and R0 = 0.25 for each gram of 14C, ln ( R0 / R ), R, 1 ln ( R0 / R ), = e -lt Þ t =, =, l, 1, (0.693 / T1/ 2 ), R0, , 10 / 5, , æ1ö, = 50000 ç ÷, è2ø, , 0.693 ´ 6.5 ´ 1010, 5730 ´ 365 ´ 24 ´ 3600, , = 0.25 Bq = 0.25 (decays/s), , +, , Fission, A, , t /T, , 1g of carbon,, , N=, , 11 2, , Dm =, , (c) The number, , of 12C atoms in, , 30., , 5730 years, æ 0.25 ö, ´ ln ç, = 13310 years, è 0.05 ÷ø, 0.693, , (d) If there are no other radioactive ingredients, the sample, is very recent. But the error of measurement must be, high unless the statistical error itself is large. In any, case, for an old sample, the activity will not be higher, than that of a recent one., (d) The penetrating power is maximum in case of gamma, rays because gamma rays are an electromagnetic, radiation of very small wavelength., (b) b-particles, being emitted with very high velocity (up, to 0.99 c). So, according to Einstein's theory of, relatively, the mass of a b-particle is much higher, compared to is its rest mass (m0). The velocity of, electrons obtained by other means is very small, compared to c (Velocity of light). So its mass remains, nearly m0. But b-particle and electron both are similar, particles., (b) Electron capture occurs more often than positron, emission in heavy elements. This is because if positron, emission is energetically allowed, electron capture is, necessarily allowed, but the reverse is not true i.e. when, electron caputre is energetically allowed, positron, emission is not necessarily allowed., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 57, , 157, , 1., 2., 3., 4., , 5., 6., 7., 8., 9., 10., 11., 12., 13., 14., , (a) With temperature rise conductivity of semiconductors, increases., (b), (b) In insulators, the forbidden energy gap is very large, in, case of semiconductor it is moderate and in conductors, the energy gap is zero., (c) In intrinsic semiconductors, the creation or liberation, of one free electron by the thermal energy creates one, hole. Thus in intrinsic semiconductors ne = nh, (b) Both P-type and N-type semiconductors are neutral, because neutral atoms are added during doping., (d) Conductor has positive temperature coefficient of, resistance but semiconductor has negative temperature, coefficient of resistance., (d), (c) At zero Kelvin, there is no thermal agitation and, therefore no electrons from valence band are able to, shift to conduction band., (c) Antimony is a fifth group impurity and is therefore a, donor of electrons., (d) At 0K temperature semiconductor behaves as an, insulator, because at very low temperature electrons, cannot jump from the valence band to conduction band., (b) Formation of energy bands in solids are due to Pauli’s, exclusion principle., (), (a) In conductors valence band and conduction band may, overlaps., (b) With rise in temperature, conductivity of semiconductor, increases while resistance decreases., , 15. (a) Because vd =, , i, (ne )eA, , 16. (b) In reverse biasing, width of depletion layer increases., 17. (b) Because in case (1) N is connected with N. This is not, a series combination of transistor., 18. (d) Resistance in forward biasing R fr » 10W and, resistance in reverse biasing, , RRw » 105 W Þ, , R fr, RRw, , =, , 57, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 1, 104, , 19. (b) In forward biasing the diffusion current increases and, drift current remains constant so not current is due to, the diffusion., In reverse biasing diffusion becomes more difficult so, net current (very small) is due to the drift., 20. (a) In a triclinic crystal a ¹ b ¹ c and a ¹ b ¹ g ¹ 90º, 21. (a), 22. (a) In figure (1), (2) and (3). P-crystals are more positive, as compared to N-crystals., 23. (a) Wood is non-crystalline and others are crystalleine., 24. (a) Resistance of conductors (Cu) decreases with decrease, in temperature while that of semi-conductors (Ge), increases with decrease in temperature., DV p, , = 1.2 ´ 104 ohm, , rp =, , 26. (a), , gm =, , 27. (a), , m = rp ´ g m = (1.2 ´ 10 4 ) ´ ( 5.33 ´ 10 -3 ) = 64, , DI p, DI p, DVg, , =, , (180 - 120), , 25. (b), , (15 - 10) ´10, , =, , -3, , (15 - 7 ) ´ 10 -3, = 5.33 ´ 10 -3 ohm -1, ( - 2 .0 ) - ( - 3 .5 ), , 28. (a) According to law of mass action, ni2 = ne nh . In, intrinsic semiconductors ni = ne = nh and for P-type, semiconductor ne would be less than ni, since nh is, necessarily more than ni., 29. (d) Resistivity of semiconductors decreases with, temperature. The atoms of a semiconductor vibrate with, larger amplitudes at higher temperatures there by, increasing its conductivity not resistivity., 30. (c) We cannot measure the potential barrier of a PNjunction by connecting a sensitive voltmeter across its, terminals because in the depletion region, there are no, free electrons and holes and in the absence of forward, biasing, PN- junction offers infinite resistance.
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t.me/Magazines4all, DPP/ P 58, , 158, , 1., , 12., , (b), , N, , (a), , N, , Forward, biased, , 2., , reverse, biased, , P, , (d) a is the ratio of collector current and emitter current, , C = A.B = A + B = A + B (De morgan’s theorem), , while b is the ratio of collector current and base current., 10, , 3., , (a) Emitter current Ie =, Base current, , Ne 10 ´1.6 ´10, =, t, 10-6, , Ib =, , Hence output C is equivalent to OR gate., , -19, , = 1.6 mA, , 2, × 1.6 = 0.032 mA, 100, , But, Ie = Ic + Ib, \ Ic = Ie – Ib = 1.6 – 0.032 = 1.568 mA, \ a=, , C = AB. AB = A.B + A.B = AB + AB = AB, In this case output C is equivalent to AND gate., 13. (c), , (d), , 5., , 90, ´ iE Þ 10 = 0.9 ´ iE Þ I E = 11mA, (d) iC =, 100, , 0.1 + 0.1 = 1.0 + 0.1 = 0 + 0 = 0, 1.0 + 1.0 = 0.1 + 1.0 = 0 + 0 = 0, , 14. (d), , Also i E = i B + i C Þ i B = 11 - 10 = 1mA, , 7., , If A = B = C = 0 then, , D = (0 + 0) + 0 = 0 + 0 = 1 + 1 = 1, , V, i, 0.01, = 10-5 A, b = c and ib = i =, Ri 103, ib, , If A = B = 1, C = 0 then, , Hence ic = 50 ´ 10-5 A = 500 mA, , D = (1 + 1) + 0 = 1 + 0 = 0 + 1 = 1, , a = 0.8 Þ b =, , 15. (a), , 0.8, =4, (1 - 0.8), , 8., , (b) ie = ib + ic Þ ic = ie - ib, , 9., , (b), , 10., , (d) For CE configuration voltage gain = b´ RL / Ri, Power gain = b2 ´ RL / Ri Þ, , The Boolean expression for ‘NOR’ gate is Y = A + B, i.e. if A = B = 0 (Low), Y = 0 + 0 = 0 = 1 (High), , Di, Also b = c Þ Dic = b ´ Dib = 4 ´ 6 = 24mA, Dib, , 11., , 1.1 + 1.1 = 0.0 + 1.1 = 0 + 1 = 1, The output D for the given combination, , D = ( A + B).C = ( A + B) + C, , (d) b = 50, R = 1000W, Vi = 0.01V, , (c), , For ‘XNOR’ gate Y = A B + AB, i.e. 0.0 + 0.0 = 1.1 + 0.0 = 1 + 0 = 1, , Ic, Ic, 1.568, 1.568, =, = 0.98 and b = I =, = 49, Ie, 1.6, 0.032, b, , 4., , 6., , 58, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, , 16., , (d) The Boolean expression for ‘AND’ gate is R = P.Q, , 17., , Þ 1.1 = 1,1.0 = 0, 0.1 = 0, 0.0 = 0, (a) The given Boolean expression can be written as, , Y = ( A + B).( A.B) = ( A.B).( A + B) = ( A A).B + A( B.B), , Power gain, =b, Voltage gain, , (b) For ‘OR’ gate X = A + B, i.e. 0 + 0 = 0, 0 + 1 = 1,1 + 0 = 1, 1 + 1 = 1, , = A.B + A B = A B, A, 0, 1, 0, 1, , B, 0, 0, 1, 1, , Y, 1, 0, 0, 0, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 58, , 159, , 18. (c) The Boolean expression for the given combination is, output Y = ( A + B ).C, The truth table is, A, B, C, Y = (A + B).C, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, Hence A = 1, B = 0, C = 1, 19. (d) For ‘NAND’ gate C = A.B, i.e. 0.0 = 0 = 1, 0.1 = 0 = 1, 1.0 = 0 = 1,1.1 = 1 = 0, , 20. (d), , But I e = I c + I b = 0.96 I e + I b, , Þ I b = 0.04 I e, \ Current gain, b =, 23., 24., 25., 26., , I c 0.96 I e, =, = 24, I b 0.04I e, , (a), (b), (a), (b) The probability that a state with energy E is occupied, is given by, 1, P(E) = (E -E ) / kT, , where EF is the Fermi energy,,, F, e, +1, T is the temperature on the Kelvin scale, and k is the, Boltzmann constant. If energies are measured from the, top of the valence band, then the energy associated with, a state at the bottom of the conduction band is E = 1.11, eV. Furthermore, kT = (8.62 × 10–5 eV/K) (300K) =, 0.02586 eV. For pure silicon, EF = 0.555 eV and, (E – EF)/kT = (0.555eV) / (0.02586eV) = 21.46. Thus,, , P(E) =, , 1, 21.46, , = 4.79 ´ 10 -10, , +1, e, For the doped semi-conductor,, (E – EF) / kT = (0.11 eV) / (0.02586 eV) = 4.254, Hence option (d) is true., 21. (c), , 1, = 1.40 ´ 10 -2 ., and P(E) = 4.254, +1, e, 27. (a) The energy of the donor state, relative to the top of the, valence bond, is 1.11 eV – 0.15 eV = 0.96 eV. The, Fermi energy is 1.11 eV – 0.11 eV = 1.00 eV. Hence,, (E - E F ) / kT = (0.96eV - 1.00eV), /(0.02586eV) = -1.547, and P(E) =, , True Table, X, 0, 1, 1, 0, , Y, , X Y, P = X + Y Q = X.Y R = P+Q, 1 1, 0, 1, 1, 0, 1 0, 0, 1, 1, 0, 0 0, 1, 0, 0, 1, 0 1, 1, 1, 1, 0, Hence X = 1, Y = 0 gives output R = 1, , 1, , = 0.824, +1, e, 28. (d) In diode the output is in same phase with the input, therefore it cannot be used to built NOT gate., 29. (a) This is Boolean expression for ‘OR’ gate., -1.547, , 30. (d) Statement -1 is true but statement -2 is false., , Ic, 22. (c) I = 0.96, e, , Þ, , I c = 0.96 I e, , If A = 1, B = 0, C = 1 then Y = 0
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t.me/Magazines4all, DPP/ P 59, , 160, , DAILY PRACTICE, PROBLEMS, 1., , (a) By using fc ; 9 ( N max )1/ 2 Þ f c = 2.84MHz, , 2., 3., , 4., 5., , (d) Carrier frequency > audio frequency, (a) A maximum frequency deviation of 75 kHz is permitted, for commercial FM broadcast stations in the 88 to 108, MHz VHF band., (c) Carrier + signal ® modulation., (c) 270W, , 6., , (a), , m1 > m2, , 7., , (c), , MUF =, , 8., 9., , (d) coordinated waves of a particular wavelength, (d) Surgery needs sharply focused beam of light and laser, can be sharply focused., (d) Laser beams are perfectly parallel. So that they are very, narrow and can travel a long distance without, spreading. This is the feature of laser while they are, monochromatic and coherent, these are characteristics, only., (b) The formula for modulating index is given by, , 10., , 11., , mf =, , fc, 60, =, = 175MHz, cos q cos 70o, , fa < f f, , (b), , 13., 14., 15., , (c) An antenna is a metallic structure used to radiate or, receive EM waves., (a) Varying the local oscillator frequency, (b) In the channel or in the transmission line, , 16., , (a) Carrier swing =, , 17., , (c) In optical fibre, light travels inside it, due to total, internal reflection., (b) The process of changing the frequency of a carrier wave, (modulated wave) in accordance with the audio, frequency signal (modulating wave) is known as, frequency modulation (FM)., (d) Following are the problems which are faced while, transmitting audio signals directly., (i) These signals are relatively of short range., (ii) If every body started transmitting these low frequency, signals directly, mutual interference will render all of, them ineffective., (iii) Size of antenna required for their efficient radiation, would be larger i.e. about 75 km., , 19., , 20., , (a) The critical frequency of a sky wave for relection from, a layer of atmosphere is given by fc = 9( N max )11/ 2, Þ 10 ´106 = 9( N max )11/ 2, 2, , æ 10 ´106 ö, 12 -3, Þ N max = ç, ÷ ; 1.2 ´10 m, ç 9 ÷, è, ø, , 21., , (b) Core of acceptance angle q = sin -1 n12 - n22, , 22, , (a) Remote sensing is the technique to collect information, about an object in respect of its size, colour, nature,, location, temperature etc without physically touching, it. There are some areas or location which are, inaccessible. So to explore sensing is used. Remote, sensing is done through a satellite., (b), (a) Laser beams are perfectly parallel. They are, monochromatic and coherent. These are characteristics, only., , 23., 24., , 25., , d, Frequency var iation, 10 ´ 103, =, =, =5, vm Modulating frequency 2 ´ 103, , 12., , 18., , 59, , PHYSICS, SOLUTIONS, , (d) f c = 9 N m = 9 ´ 9 ´ 1010, = 2.7 ´ 106 Hz = 2.7 MHz, , 26., , (b) f = f c 1 +, , D2, 4h 2, , = 2.7 ´ 10 6 ´ 1 +, , ( 250 ´ 103 )2, 2, 4 ´ (150 ´ 103 ), , = 3.17 × 106Hz, 27., , Frequency deviation, 50, =, = 7.143, Modulating frequency 7, , (c), , f = fc = sec fi, , Secfi =, , f 3.17×106, =, = 1.174, f c 2.7 ´106, , fi = sec –1 (1.174) = 31.6° ., , 28., , 29., , 30., , (d) TV signals (frequency greater than 30 MHz) cannot, be propagated through sky wave propagation., Above critical frequency, an electromagnetic wave, penetrates the ionosphere and is not reflected by it., (d) The electromagnetic waves of shorter wavelength do, not suffer much diffraction from the obstacles of earth’s, atmosphere so they can travel long distance., Also, shorter the wavelength, shorter is the velocity of, wave propagation., (b) A dish antenna is a directional antenna because it can, transmit or signals in all direction., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ P 60, , 161, , DAILY PRACTICE, PROBLEMS, 1., , (b) Voltmeter measures voltage across two points so it is, connected in parallel and ammeter measures current, so it is connected in series., , 2., , (a), , 3., , P R, =, so by changing the gap resistance of copper, Q S, strip gets cancelled., (b) Potential gradient = Potential drop/unit length, V = IR = I, , rl, A, , (d), , 5., , (a), , 6., , (a), , 7., , (c), , 8., , (b), , 9., , (c), , V Ir, = ], l A, Refractive index is the property of the material, hence, it does not depend on angle of the prism., A meter-bridge is a device which is based on the, principle of Wheatstone bridge., A potentiometer is device which is used to compare, e.m.f.'s of two cells as well as to determine the internal, resistance of a cell. It is based on the principle that when, a current flows through a wire of uniform thickness and, material, potential difference between its two points is, directly proportional to the length of the wire between, the two points., When a p-n junction diode is connected in reverse, biased mode, a reverse current flows., A Zener diode is a heavily doped p-n junction diode, which operates on reverse bias beyond breakdown, voltage., In a transfer characteristics Vi is plotted along x-axis, and V0 along y-axis., , 10. (a) Current gain in CE configuration is b =, , Χ IC, ., ΧI B, , Ic >> IB, hence it is maximum., , 11., , rl V, Ir, \ <k< ,, A l, A, increase in I, will increase k, so it will decrease, sensitivity., , (c) V = IR = I, , 12. (c), 13. (c) According to the figure the voltmeter and the resistor, are connected in parallel., 14. (b) Here I = 4A, c, , pc, æ 30 ´ p ö, q = 30° = ç, ÷ =, 6, è 180 ø, Now, k =, , 15. (c) In reverse-bias mode, a reverse current flows., Therefore, (c) represents the form., 16. (c) Larger the length, lesser will be the potential gradient,, so more balancing length will be required., 17. (a) LED is a p-n junction diode which always operates on, forward bias., 18. (b) Emitter current is the sum of base and collector current, by Kirchhoff's 1st law., Irl, \ potential gradient = k, A, when A is decreased, k will increase., 20. (d) Positive terminal is at lower potential, (0V) and negative terminal is at higher potential + 5V., 21. (d), , 19. (b) V = IR =, , \, , 4., , 60, , PHYSICS, SOLUTIONS, , 22. (d) V = 30.0, I = 0.020 A, R =, Error : As R =, , V, DR DV DI, \, =, +, I, R, V, I, , æ DV DI ö, + ÷, Þ DR = R çè, V, Iø, æ 0.1 0.001ö, +, = 1.50 × 103 çè, ÷ = 0.080 kW, 30.0 0.020 ø, 23. (a) u = – 0.30 m, v = – 0.60m, By mirror formula,, 1 1 1, + =, v u f, , Þ, , -3.0, 1, -1, 1, Þ f = 0.20m, =, Þ f=, f 0.30 0.60, 0.60, , Þ, , 1 1 1, -df -dv -du, = + Þ 2 = 2 - 2, f v u, f, v, u, , é 0.01, 0.01 ù, +, ú, Þ df = (0.20)2 ê, 2, (0.30)2 ûú, ëê (0.60), Þ df = 0.0055 » 0.01m, Þ Focal length f = (0.20 ± 0.01) m, 24. (d) As shown in the figure., , P, l, l, =, , Pµ, Q 100 - l, 100 - l, , P(unknown)Q, , I, 4, 4´6´7 2´6´7, =, =, =, q æpö, 22, 11, ç ÷, è6ø, , 84, =, = 7.6 A / rad, 11, , V 30.0, =, = 1.50 kW, I 0.020, , G, B, l, , 100l
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t.me/Magazines4all, DPP/ P 60, , 162, 27., , DP Dl D (100 - l), =, +, l, P, 100 - l, , 28., , Dl, Dl, =, +, l 100 - l, , 29., , (c) because balance point depends upon the value of, unknown and known resistance only., (c) Potentiometer is used to measure the potential, difference between the two points of a wire., (a) Let j be the current density., Then j ´ 2pr 2 = I Þ j =, , 0.1, 0.1, DP, +, Þ, ´ 100 = 0.42%, 40.0 60.0, P, (b) As shown in the figure, when the object (O) is placed, between F and C, the image (I) is formed beyond C. It, is in this condition that movement towards left, =, , 25., , \ E = rj =, , ////, , |||, , ||||, , //, , 26., , when the student shifts his eye towards left, the image, appears to the right of the object pin., (c) For a spherical mirror, the graph plotted between (1/u), and (1/v) will be a straight line with a negative slope, of (– 1) and position intercept (1/f) on the (1/v) axis, 1, 1 1, =- +, v, u f, 1/ f, , tan q = -1, , q, O, , 1/ f, , (1/u), , 2pr 2, a, , ò, , a+b, , a, r uur, E. dr = - ò, , rI, , a + b 2 pr, , 2, , dr, , a, , =-, , rI é 1 ù, rI, rI, =, 2p ëê r ûú a + b 2pa 2p (a + b), , On applying superposition as mentioned we get, , ///, , |||, , | |||, , 2 pr 2, , rI, , ', Now, DVBC, =-, , //////////////, ////, ///, / //, , Movement towards left, F, C, O, I, , I, , ', DVBC = 2 ´ DVBC, =, , 30., , (c), , E=, , rI, 2 pr 2, , rI, rI, pa p(a + b), , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , INDEX/CHAPTERS, Page No., , DPP-1, , PHYSICAL WORLD, UNI TS & MEASUREMENTS, , P-1 – P-4, , DPP-2, , MOTION IN A STRAIGHT LINE, , P-5 – P-8, , DPP-3, , MOTION IN A PLANE, , DPP-4, , LAWS OF MOTI ON, , DPP-5, , WORK, ENERGY AND POWER, , P-17 – P-20, , DPP-6, , SYSTEM OF PARTICLES AND ROTATIONAL MOTION, , P-21 – P-24, , DPP-7, , GR AVIT AT IO N, , P-25 – P-28, , DPP-8, , MECHANICAL PROPERTIES OF SOLIDS, , P-29 – P-32, , DPP-9, , MECHANICAL PROPERTIES OF FLUIDS, , P-33 – P-36, , DPP-10, , TH ERMAL PROPERTI ES OF MATT ER, , P-37 – P-40, , DPP-11, , TH ER MODY NA MI CS, , P-41 – P-44, , DPP-12, , KINETIC THEORY, , P-45 – P-48, , OSCI LLATI ONS, , P-49 – P-52, , DPP-14, , WAVES, , P-53 – P-56, , DPP-15, , ELECTRIC CHARGES AND FIELDS, , P-57 – P-60, , DPP-16, , ELECTROSTATIC POTENTIAL & CAPACITANCE, , P-61 – P-64, , DPP-17, , CURRENT ELECTRICITY, , P-65 – P-68, , DPP-18, , MOVING CHARGES AND MAGN ETISM, , P-69 – P-72, , DPP-19, , MAGNETI SM AND MATTER, , P-73 – P-76, , DPP-20, , ELECTROMAGNETIC IN DUCTION, , P-77 – P-80, , DPP-21, , ALTERNATING CURRENT, , P-81 – P-84, , DPP-13, , ., , P-9 – P-12, P-13, , [b], , – P-16
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t.me/Magazines4all, , EBD_7156, , t.me/Ebooks_Encyclopedia27., , DPP-22, , ELEC TROMAGN ETI C W AVES, , P-85 – P-88, , DPP-23, , RAY OPTICS AND OPTICAL INSTRUMENTS, , P-89 – P-92, , DPP-24, , WAVE OPTICS, , P-93 – P-96, , DPP-25, , DUAL NATURE OF RADIATION AND MATTER, , DPP-26, , ATOMS, , P-101 – P- 104, , DPP-27, , NU CL EI, , P-105 – P- 108, , DPP-28, , SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS, , P-109 – P- 112, , Solutions To Chapter-wise DPP Sheets (1-28), , [c], , P-97 – P-100, , S -1 – S -115
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP01, , SYLLABUS : Physical World, Units & Measurements, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , The density of material in CGS system of units is 4g/cm3. In, a system of units in which unit of length is 10 cm and unit of, mass is 100 g, the value of density of material will be, (a) 0.4 unit, (b) 40 unit, (c) 400 unit, (d) 0.04 unit, The time period of a body under S.H.M. is represented by:, T = Pa Db Sc where P is pressure, D is density and S is, surface tension, then values of a, b and c are, 3 1, (a) - , , 1, (b) -1, - 2, 3, 2 2, 1, 3, 1, 1, ,- ,(c), (d) 1, 2,, 3, 2, 2, 2, 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, , 4., , RESPONSE, GRID, , 3., , 1., 6., , 2., , 5., , 6., , Young’s modulus of a material has the same unit as that of, (a) pressure, (b) strain, (c) compressibility, (d) force, Of the following quantities, which one has dimensions, different from the remaining three?, (a) Energy per unit volume, (b) Force per unit area, (c) Product of voltage and charge per unit volume, (d) Angular momentum, The pressure on a square plate is measured by measuring, the force on the plate and length of the sides of the plate by, using the formula P =, , F, , . If the maximum errors in the, l2, measurement of force and length are 4% and 2%, respectively, then the maximum error in the measurement of, pressure is, (a) 1%, (b) 2%, (c) 8%, (d) 10%, , Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP01, , P-2, , 7., , The siemen is the SI unit of, (a) resistivity, (b) resistance, (c) conductivity, (d) conductance, 8., An object is moving through the liquid. The viscous, damping force acting on it is proportional to the velocity., Then dimensions of constant of proportionality are, (a) [ML–1T–1], (b) [MLT–1], (c) [M0LT–1], (d) [ML0T–1], 9., The least count of a stop watch is 0.2 second. The time of 20, oscillations of a pendulum is measured to be 25 second., The percentage error in the measurement of time will be, (a) 8%, (b) 1.8%, (c) 0.8%, (d) 0.1%, 10. Weber is the unit of, (a) magnetic susceptibility, (b) intensity of magnetisation, (c) magnetic flux, (d) magnetic permeability, 11. The physical quantity which has the dimensional formula, [M1T–3] is, (a) surface tension, (b) solar constant, (c) density, (d) compressibility, 12. The dimensions of Wien’s constant are, (a) [ML0 T K], (b) [M0 LT0 K], (c) [M0 L0 T K], (d) [MLTK], 13. If the capacitance of a nanocapacitor is measured in terms, of a unit ‘u’ made by combining the electric charge ‘e’,, Bohr radius ‘a0’, Planck’s constant ‘h’ and speed of light ‘c’, then, hc, e2 h, (a) u =, (b) u = 2, e a0, a0, (c), , u=, , e2 c, ha 0, , (d), , u=, , e2 a 0, hc, , 15., , 17., , 18., , 19., , 20., 21., 22., , 23., , 1 e2, 14. The dimensions of, are, Îo hc, (a) M–1 L–3 T4 A2, , 16., , m=, , RESPONSE, GRID, , 8., 13., 18., 23., , real depth, apparent depth, , is found to have values of 1.34, 1.38,, , 1.32 and 1.36; the mean value of refractive index with, percentage error is, (a) 1.35 ± 1.48 %, (b) 1.35 ± 0 %, (c) 1.36 ± 6 %, (d) 1.36 ± 0 %, , (b) ML3 T–4 A–2, , (c) M0 L0 T0 A0, (d) M–1 L–3 T2 A, The density of a cube is measured by measuring its mass, and length of its sides. If the maximum error in the, , 7., 12., 17., 22., , measurement of mass and length are 4% and 3%, respectively, the maximum error in the measurement of, density will be, (a) 7%, (b) 9%, (c) 12%, (d) 13%, Which is different from others by units ?, (a) Phase difference, (b) Mechanical equivalent, (c) Loudness of sound, (d) Poisson’s ratio, DV, A quantity X is given by e 0 L, where Î0 is the, Dt, permittivity of the free space, L is a length, DV is a potential, difference and Dt is a time interval. The dimensional formula, for X is the same as that of, (a) resistance, (b) charge, (c) voltage, (d) current, If the error in the measurement of the volume of sphere is, 6%, then the error in the measurement of its surface area will, be, (a) 2%, (b) 3%, (c) 4%, (d) 7.5%, If velocity (V), force (F) and energy (E) are taken as fundamental, units, then dimensional formula for mass will be, –2 0, 0 2, –2 0, –2 0, (a) V F E (b) V FE, (c) VF E, (d) V F E, Multiply 107.88 by 0.610 and express the result with correct, number of significant figures., (a) 65.8068 (b) 65.807, (c) 65.81, (d) 65.8, Which of the following is a dimensional constant?, (a) Refractive index, (b) Poissons ratio, (c) Strain, (d) Gravitational constant, If E, m, J and G represent energy, mass, angular momentum, and gravitational constant respectively, then the, dimensional formula of EJ2/m5G2 is same as that of the, (a) angle, (b) length, (c) mass, (d) time, The refractive index of water measured by the relation, , 9., 14., 19., Space for Rough Work, , 10., 15., 20., , 11., 16., 21., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP01, , P-3, , 24. If e is the charge, V the potential difference, T the temperature,, eV, are the same as that of, then the units of, T, (a) Planck’s constant, (b) Stefan’s constant, (c) Boltzmann's constant (d) gravitational constant, 25. The dimensions of mobility are, (a) M–2T 2A, (b) M–1T 2A, (c) M–2T 3A, (d) M–1T 3A, 26. Two quantities A and B have different dimensions which, mathematical operation given below is physically, meaningful?, (a) A/B, (b) A + B, (c) A – B, (d) A = B, 27. The velocity of water waves (v) may depend on their, wavelength l, the density of water r and the acceleration, due to gravity, g. The method of dimensions gives the, relation between these quantities is, (a) v, (b) v2 µ gl, 2, 2, (c) v µ gl, (d) v2 µ g–1l2, 28. The physical quantities not having same dimensions are, (a) torque and work, (b) momentum and Planck’s constant, (c) stress and Young’s modulus, (d) speed and (m0e0)–1/2, 29. A physical quantity of the dimensions of length that can be, e2, is [c is velocity of light, G is, 4pe0, universal constant of gravitation and e is charge], , formed out of c, G and, 1/ 2, , (a), , é, e2 ù, c êG, ú, ëê 4pe 0 ûú, 2, , 32., , 33., , 34., , p(r 2 - x 2 ), . Here v is velocity of oil at a distance x from, 4vl, the axis of the tube. From this relation, the dimensional, formula of h is, h=, , 35., , 1/ 2, , (b), , 1 é e2 ù, ê, ú, c 2 ëê G4 pe0 ûú, , 1/ 2, , 30., 31., , 1 é, e2 ù, 1, e2, G, êG, ú, (c), (d), 2, c 4pe0, c ëê 4pe 0 ûú, The unit of impulse is the same as that of, (a) energy, (b) power, (c) momentum, (d) velocity, If Q denote the charge on the plate of a capacitor of, , capacitance C then the dimensional formula for, (a) [L2M2T], (c) [L2MT–2], , RESPONSE, GRID, , (b) [LMT2], (d) [L2M2T2], , 24., 29., 34., , 25., 30., 35., , Q2, is, C, , The mass of the liquid flowing per second per unit area of, cross-section of the tube is proportional to (pressure, difference across the ends)n and (average velocity of the, liquid)m. Which of the following relations between m and n, is correct?, (a) m = n, (b) m = – n, (c) m2 = n, (d) m = – n2, The Richardson equation is given by I = AT2e–B/kT. The, dimensional formula for AB2 is same as that for, (a) I T2, (b) kT, (c) I k2, (d) I k2/T, Turpentine oil is flowing through a capillary tube of length, l and radius r. The pressure difference between the two, ends of the tube is p. The viscosity of oil is given by :, , 36., 37., , (a), , [ML-1T -1 ], , (b) [MLT -1 ], , (c), , [ML2 T -2 ], , (d) [M 0 L0 T 0 ], , éæ 2 p, öù, Given that y = A sin êç (ct - x) ÷ú , where y and x are, øû, ëè l, measured in metre. Which of the following statements is, true?, (a) The unit of l is same as that of x and A, (b) The unit of l is same as that of x but not of A, 2p, (c) The unit of c is same as that of, l, 2p, (d) The unit of (ct – x) is same as that of, l, If L = 2.331 cm, B = 2.1 cm, then L + B =, (a) 4.431 cm (b) 4.43 cm, (c) 4.4 cm, (d) 4 cm, In the relation x = cos (wt + kx), the dimension(s) of w is/are, (a) [M0 LT], (b) [M0L–1T0], 0, 0, –1, (c) [M L T ], (d) [M0LT–1], , 26., 31., 36., Space for Rough Work, , 27., 32., 37., , 28., 33.
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t.me/Magazines4all, , DPP/ CP01, , P-4, , 38., , 39., , In a vernier callipers, ten smallest divisions of the vernier, scale are equal to nine smallest division on the main scale. If, the smallest division on the main scale is half millimeter,, then the vernier constant is, (a) 0.5 mm (b) 0.1 mm, (c) 0.05 mm (d) 0.005 mm, Which two of the following five physical parameters have, the same dimensions?, (A) Energy density, (B) Refractive index, (C) Dielectric constant, (D) Young’s modulus, (E) Magnetic field, (a) (B) and (D), (b) (C) and (E), (c) (A) and (D), (d) (A) and (E), , 40., , æ, a, In the eqn. ç P + 2, V, è, , 41., , (a) dyne cm5, (b) dyne cm4, 3, (c) dyne/cm, (d) dyne cm2, The dimensions of Reynold’s constant are, (a) [M0L0T0], (b) [ML–1T–1], –1, –2, (c) [ML T ], (d) [ML–2T–2], , 43., , 44., , 45., , 39., 44., , Which of the following do not have the same dimensional, formula as the velocity?, Given that m0 = permeability of free space, e0 = permittivity, of free space, n = frequency, l = wavelength, P = pressure, r, = density, w = angular frequency, k = wave number,, (a) 1, , ö, ÷ (V - b) = constant, the unit of a is, ø, , 38., 43., , RESPONSE, GRID, , 42., , m 0 eo, , (b) n l, , (c), , (d) wk, , P/r, , Unit of magnetic moment is, (a) ampere–metre2, (b) ampere–metre, (c) weber–metre2, (d) weber/metre, An experiment is performed to obtain the value of, acceleration due to gravity g by using a simple pendulum of, length L. In this experiment time for 100 oscillations is, measured by using a watch of 1 second least count and the, value is 90.0 seconds. The length L is measured by using a, meter scale of least count 1 mm and the value is 20.0 cm. The, error in the determination of g would be:, (a) 1.7%, (b) 2.7%, (c) 4.4%, (d) 2.27%, The dimensional formula for magnetic flux is, (a) [ML2T–2A–1], (b) [ML3T–2A–2], 0, –2, 2, –2, (c) [M L T A ], (d) [ML2T–1A2], , 40., 45., , 41., , 42., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP01 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP02, , SYLLABUS : Motion in a Straight Line, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , A particle starts moving rectilinearly at time t = 0 such that, its velocity v changes with time t according to the equation, v = t2 – t where t is in seconds and v is in m/s. Find the time, interval for which the particle retards., (a), , 1, <t <1, 2, , (b), , 1, >t >1, 2, , 1, 1, 3, <t <1, (d), <t<, 4, 2, 4, The co-ordinates of a moving particle at any time ‘t’are given, by x = at3 and y = bt3. The speed of the particle at time ‘t’, is given by, , (c), , 2., , (a), , 2, , 3t a + b, , 2, , (b), , 3t, , 2, , 2, , a +b, , 5., , 6., , 2, , (d), a 2 + b2, t 2 a 2 + b2, If a car covers 2/5th of the total distance with v1 speed and, 3/5th distance with v2 then average speed is, , 3., , 2v1v2, 5v1v2, v +v, 1, v1v2 (b) 1 2 (c), (d), v1 + v2, 3v1 + 2v2, 2, 2, Choose the correct statements from the following., (a) The magnitude of instantaneous velocity of a particle, is equal to its instantaneous speed, , (a), 4., , RESPONSE, GRID, , 1., 6., , 2., 7., , v (in ms -1 ), , (c), , (b) The magnitude of the average velocity in an interval is, equal to its average speed in that interval., (c) It is possible to have a situation in which the speed of, the particle is never zero but the average speed in an, interval is zero., (d) It is possible to have a situation in which the speed of, particle is zero but the average speed is not zero., A particle located at x = 0 at time t = 0, starts moving along, with the positive x-direction with a velocity 'v' that varies as, v = a x . The displacement of the particle varies with time as, (a) t 2, (b) t, (c) t 1/2, (d) t 3, Figure here gives the speed-time graph for a body. The, displacement travelled between t = 1.0 second and t = 7.0, second is nearest to, 4, (a) 1.5 m, 8, 6, (b) 2 m, 0, t, 2, 4, (in sec .), (c) 3 m, -4, (d) 4 m, A particle is moving in a straight line with initial velocity, and uniform acceleration a. If the sum of the distance, travelled in tth and (t + 1)th seconds is 100 cm, then its, velocity after t seconds, in cm/s, is, (a) 80, (b) 50, (c) 20, (d) 30, , 7., , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP02, , P-6, , 8., , 9., , A thief is running away on a straight road on a jeep moving, with a speed of 9 m/s. A police man chases him on a motor, cycle moving at a speed of 10 m/s. If the instantaneous, separation of jeep from the motor cycle is 100 m, how long, will it take for the police man to catch the thief?, (a) 1 second, (b) 19 second, (c) 90 second, (d) 100 second, The displacement x of a particle varies with time according, a, to the relation x = (1 - e - bt ). Then select the false, b, alternative., , 1, (a) At t = , the displacement of the particle is nearly 2 æç a ö÷, b, 3èbø, (b) The velocity and acceleration of the particle at t = 0 are, a and –ab respectively, a, (c) The particle cannot go beyond x =, b, (d) The particle will not come back to its starting point at, t®¥, 10. A metro train starts from rest and in five seconds achieves a, speed 108 km/h. After that it moves with constant velocity, and comes to rest after travelling 45m with uniform, retardation. If total distance travelled is 395 m, find total, time of travelling., (a) 12.2 s (b) 15.3 s (c) 9 s, (d) 17.2 s, 11. The deceleration experienced by a moving motor boat after, its engine is cut off, is given by dv/dt = – kv3 where k is a, constant. If v0 is the magnitude of the velocity at cut-off,, the magnitude of the velocity at a time t after the cut-off is, v0, (a), (b) v0 e –kt, 2, (2 v 0 kt + 1), , (c) v0 / 2, (d) v 0, 12. The velocity of a particle is v = v0 + gt + ft2. If its position is, x = 0 at t = 0, then its displacement after unit time (t = 1) is, (a) v0 + g /2 + f, (b) v0 + 2g + 3f, (c) v0 + g /2 + f/3, (d) v0 + g + f, 13. A man is 45 m behind the bus when the bus starts accelerating, from rest with acceleration 2.5 m/s2. With what minimum, velocity should the man start running to catch the bus?, (a) 12 m/s (b) 14 m/s (c) 15 m/s (d) 16 m/s, 14. 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 x2(t) after the same time interval. Which of, the following graphs correctly describes (x1 – x 2) as a, function of time ‘t’?, , RESPONSE, GRID, , 8., 13., 18., , 9., 14., 19., , (x1 – x 2), , (x1 – x2), , (a), , t, , O, , (b), , (x1 – x2), , (c), , t, , O, (x1 – x 2), , t (d), , O, , t, , O, , 15., , From the top of a building 40 m tall, a boy projects a stone, vertically upwards with an initial velocity 10 m/s such that it, eventually falls to the ground. After how long will the stone, strike the ground ? Take g = 10 m/s2., (a) 1 s, (b) 2 s, (c) 3 s, (d) 4 s, 16. Two bodies begin to fall freely from the same height but the, second falls T second after the first. The time (after which, the first body begins to fall) when the distance between the, bodies equals L is, (a), 17., , 1, T, 2, , (b), , T, L, (c), +, 2 gT, , L, gT, , (d) T + 2L, gT, , Let A, B, C, D be points on a vertical line such that, AB = BC = CD. If a body is released from position A, the, times of descent through AB, BC and CD are in the ratio., (a) 1 : 3 - 2 : 3 + 2 (b) 1 : 2 - 1 : 3 - 2, , 18., , 19., , 20., , 21., , (c) 1 : 2 - 1 : 3, (d) 1 : 2 : 3 - 1, The water drops fall at regular intervals from a tap 5 m above, the ground. The third drop is leaving the tap at an instant, when the first drop touches the ground. How far above the, ground is the second drop at that instant? (Take g = 10 m/s2), (a) 1.25 m (b) 2.50 m (c) 3.75 m (d) 5.00 m, The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in, kg) moving in one dimension under the action of a force, is, related to time ‘t’ (in sec) by t = x + 3 . The displacement, of the particle when its velocity is zero, will be, (a) 2 m, (b) 4 m, (c) zero, (d) 6 m, A body moving with a uniform acceleration crosses a, distance of 65 m in the 5 th second and 105 m in 9th second., How far will it go in 20 s?, (a) 2040 m (b) 240 m (c) 2400 m (d) 2004 m, An automobile travelling with a speed of 60 km/h, can brake, to stop within a distance of 20m. If the car is going twice as, fast i.e., 120 km/h, the stopping distance will be, (a) 60 m, (b) 40 m, (c) 20 m, (d) 80 m, , 10., 15., 20., Space for Rough Work, , 11., 16., 21., , 12., 17., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP02, , P-7, , 22. A particle accelerates from rest at a constant rate for some, time and attains a velocity of 8 m/sec. Afterwards it, decelerates with the constant rate and comes to rest. If the, total time taken is 4 sec, the distance travelled is, (a) 32 m, (b) 16 m, (c) 4 m, (d) None of the above, 23. The equation represented by the graph below is :, (a) y =, , 1, gt, 2, , (b) y =, , -1, gt, 2, , 30., , -1 2, gt, 2, 24. A particle moves a distance x in time t according to equation, x = (t + 5)–1. The acceleration of particle is proportional to:, (a) (velocity) 3/2, (b) (distance)2, –2, (c) (distance), (d) (velocity)2/3, 25. A particle when thrown, moves such that it passes from, same height at 2 and 10 seconds, then this height h is :, (a) 5g, (b) g, (c) 8g, (d) 10g, 26. The distance through which a body falls in the nth second, is h. The distance through which it falls in the next second is, , 31., , g, (c) h – g, (d) h + g, 2, 27. A stone thrown upward with a speed u from the top of the, tower reaches the ground with a velocity 3u. The height of, the tower is, (a) 3u2/g (b) 4u2/g (c) 6u2/g (d) 9u2/g, 28. A particle moves along a straight line OX. At a time t (in, seconds) the distance x (in metres) of the particle from O is, given by x = 40 + 12t – t3. How long would the particle travel, before coming to rest?, (a) 40 m, (b) 56 m, (c) 16 m, (d) 24 m, 29. The graph shown in figure shows, the velocity v versus time t for a, body., Which of the graphs represents the, corresponding acceleration versus, time graphs?, , 33., , (b), , h+, , a, , (a), , RESPONSE, GRID, , 32., , 34., , 35., , a, t, , 22., 27., 32., , A particle moving along x-axis has acceleration f, at time t,, , 1, 1, f0T2 (b) f0T2, (c), f T (d) f0T, 2, 2 0, A body is thrown vertically up with a velocity u. It passes, three points A, B and C in its upward journey with velocities, u u, u, , and respectively. The ratio of AB and BC is, 2 3, 4, (a) 20 : 7, (b) 2, (c) 10 : 7, (d) 1, A boat takes 2 hours to travel 8 km and back in still water, lake. With water velocity of 4 km h–1, the time taken for, going upstream of 8 km and coming back is, (a) 160 minutes, (b) 80 minutes, (c) 100 minutes, (d) 120 minutes, A body starts from rest and travels a distance x with uniform, acceleration, then it travels a distance 2x with uniform speed,, finally it travels a distance 3x with uniform retardation and, comes to rest. If the complete motion of the particle is along, a straight line, then the ratio of its average velocity to, maximum velocity is, (a) 2/5, (b) 3/5, (c) 4/5, (d) 6/7, A man of 50 kg mass is standing in a gravity free space at a, height of 10 m above the floor. He throws a stone of 0.5 kg, mass downwards with a speed 2 m/s. When the stone, reaches the floor, the distance of the man above the floor, will be:, (a) 9.9 m, (b) 10.1 m (c) 10 m, (d) 20 m, A boy moving with a velocity of 20 km h–1 along a straight, line joining two stationary objects. According to him both, objects, (a) move in the same direction with the same speed of, 20 km h–1, (b) move in different direction with the same speed of, 20 km h–1, (c) move towards him, (d) remain stationary, , (a), , (d) y =, , (a) h, , t, , (d), , t, given by f = f0 æç 1 - ö÷ , where f0 and T are constants. The, è Tø, particle at t = 0 has zero velocity. In the time interval between, t = 0 and the instant when f = 0, the particle’s velocity (vx) is, , y, (m), , 1 2, (c) y = gt, 2, , t, , (c), , t(s), , O, , a, , a, , t, , (b), , 23., 28., 33., , 24., 29., 34., Space for Rough Work, , 25., 30., 35., , 26., 31.
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t.me/Magazines4all, , DPP/ CP02, , P-8, , A rubber ball is dropped from a height of 5 metre on a plane, where the acceleration due to gravity is same as that onto, the surface of the earth. On bouncing, it rises to a height of, 1.8 m. On bouncing, the ball loses its velocity by a factor of, 16, 2, 3, (a), (b) 9, (c), (d), 25, 5, 5, 25, 37. A stone falls freely from rest from a height h and it travels a, 9h, in the last second. The value of h is, 25, (a) 145 m (b) 100 m (c) 122.5 m (d) 200 m, Which one of the following equations represents the motion, of a body with finite constant acceleration ? In these, equations, y denotes the displacement of the body at time t, and a, b and c are constants of motion., , distance, 38., , y = at + bt 2, a, (b) y = at + bt 2 + ct 3, (d) y = + bt, t, 39. A particle travels half the distance with a velocity of 6 ms–1., The remaining half distance is covered with a velocity of, 4 ms–1 for half the time and with a velocity of 8 ms–1 for the, rest of the half time. What is the velocity of the particle, averaged over the whole time of motion ?, (a) 9 ms–1 (b) 6 ms–1 (c) 5.35 ms–1, (d) 5 ms–1, 40. A bullet is fired with a speed of 1000 m/sec in order to, penetrate a target situated at 100 m away. If g = 10 m/s2, the, gun should be aimed, (a) directly towards the target, (b) 5 cm above the target, (c) 10 cm above the target, (d) 15 cm above the target, 41. A body covers 26, 28, 30, 32 meters in 10th, 11th, 12th and, 13th seconds respectively. The body starts, (a) from rest and moves with uniform velocity, (b) from rest and moves with uniform acceleration, (c) with an initial velocity and moves with uniform, acceleration, (d) with an initial velocity and moves with uniform velocity, , (a) y = at, , 42., , (a) 9.6 ms–1 (b) 12.87 ms–1, 64 ms–1, , 36., 41., , 37., 42., , O, , P, , 327 ms–1, , S, , (a) 60 m, (b) 55 m, (c) 25 m, (d) 30 m, 44., , 30, 20, 10, 0, , A stone falls freely under gravity. It covers distances h 1, h2, and h3 in the first 5 seconds, the next 5 seconds and the, next 5 seconds respectively. The relation between h 1, h2, and h3 is, (a) h1 =, , h2, h, = 3, 3, 5, , (c) h1 = h2 = h3, 45., , 1 2 3 4, Time in second, , (b) h2 = 3h 1 and h3 = 3h2, (d) h1 = 2h 2 = 3h 3, , A car, starting from rest, accelerates at the rate f through a, distance S, then continues at constant speed for time t and, f, to come to rest. If the total, 2, distance traversed is 15 S , then, 1, (a) S = ft 2, (b) S = f t, 6, 1 2, 1 2, ft, (c) S = ft, (d) S =, 4, 72, , then decelerates at the rate, , 38., 43., , 39., 44., , 40., 45., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP02 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , Q, , (c), (d), 43. The variation of velocity of a particle with time moving along, a straight line is illustrated in the figure. The distance travelled, by the particle in four seconds is, , (b), , RESPONSE, GRID, , A particle is moving with uniform acceleration along a, straight line. The average velocity of the particle from P to Q, is 8ms–1 and that Q to S is 12ms–1. If QS = PQ, then the, average velocity from P to S is, , Velocity (m/s), , 36., , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP03, , SYLLABUS : Motion in a Plane, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , A projectile is given an initial velocity of (iˆ + 2 ˆj ) m/s, where, iˆ is along the ground and ĵ is along the vertical. If g = 10, m/s2 , the equation of its trajectory is :, (a) y = x – 5x2, (b) y = 2x – 5x2, 2, (c) 4y = 2x – 5x, (d) 4y = 2x – 25x2, An aircraft moving with a speed of 250 m/s is at a height of, 6000 m, just overhead of an anti aircraft–gun. If the muzzle, velocity is 500 m/s, the firing angle q should be:, , 5., , 6., , (a) 30°, (b) 45°, (c) 60°, (d) 75°, 3., , 4., , Two racing cars of masses m1 and m2 are moving in circles, of radii r 1 and r2 respectively. Their speeds are such that, each makes a complete circle in the same duration of time t., The ratio of the angular speed of the first to the second car, is, (a) m1 : m2, (b) r1 : r2, (c) 1 : 1, (d) m1r1 : m2r2, A boy playing on the roof of a 10 m high building throws a, ball with a speed of 10m/s at an angle of 30º with the, horizontal. How far from the throwing point will the ball be, at the height of 10 m from the ground ?, , RESPONSE, GRID, , 1., 6., , 2., 7., , 7., , 1, 3, ], [ g = 10m/s2 , sin 30o = , cos 30o =, 2, 2, (a) 5.20m (b) 4.33m (c) 2.60m (d) 8.66m, A bomber plane moves horizontally with a speed of 500 m/s, and a bomb released from it, strikes the ground in 10 sec., Angle at which it strikes the ground wil be (g = 10 m/s2), æ 1ö, -1 æ 1 ö, (a) tan ç ÷, (b) tan çè ÷ø, è 5ø, 5, (c) tan–1 (1), (d) tan–1 (5), Two particles start simultaneously from the same point and, move along two straight lines, one with uniform velocity v, and other with a uniform acceleration a. If a is the angle, between the lines of motion of two particles then the least, value of relative velocity will be at time given by, v, v, v, v, sin a (b), cos a (c), tan a (d), cot a, (a), a, a, a, a, Initial velocity with which a body is projected is 10 m/sec, and angle of projection is 60°. Find the range R, y, 15 3m, (a), 2, 40, m, (b), 3, (c) 5 3m, R, 20, 30°, m, (d), x, 3, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , P-10, , 8., , The position vectors of points A, B, C and D are, Ù, , Ù, , Ù, , Ù, , Ù, , Ù, , Ù, , Ù, , Ù, , 16., , A = 3 i + 4 j + 5 k] B = 4 i + 5 j + 6 k] C = 7 i + 9 j + 3 k, , 9., , 10., , 11., , 12., , 13., , 14., , 15., , uuur, Ù, Ù, and D = 4 i + 6 j then the displacement vectors AB and, uuur, CD are, (a) perpendicular, (b) parallel, (c) antiparallel, (d) inclined at an angle of 60°, A person swims in a river aiming to reach exactly on the, opposite point on the bank of a river. His speed of swimming, is 0.5 m/s at an angle of 120º with the direction of flow of, water. The speed of water is, (a) 1.0 m/s (b) 0.5 m/s (c) 0.25 m/s (d) 0.43 m/s, A projectile thrown with velocity v making angle q with, vertical gains maximum height H in the time for which the, projectile remains in air, the time period is, (a), (b), H cos q / g, 2H cos q / g, (d), (c), 4H / g, 8H / g, A ball is thrown from a point with a speed ' v0 ' at an, elevation angle of q. From the same point and at the same, ' v0 ', instant, a person starts running with a constant speed, 2, to catch the ball. Will the person be able to catch the ball? If, yes, what should be the angle of projection q ?, (a) No, 0° (b) Yes, 30° (c) Yes, 60° (d) Yes, 45°, r, r, wt, wt, If vectors A = cos wtiˆ + sinwtjˆ and B = cos ˆi + sin ˆj, 2, 2, are functions of time, then the value of t at which they are, orthogonal to each other is :, p, p, p, (b) t =, (c) t = 0, (d) t =, (a) t =, 2w, w, 4w, A bus is moving on a straight road towards north with a, uniform speed of 50 km/hour turns through 90°. If the speed, remains unchanged after turning, the increase in the velocity, of bus in the turning process is, (a) 70.7 km/hour along south-west direction, (b) 70.7 km/hour along north-west direction., (c) 50 km/hour along west, (d) zero, The velocity of projection of oblique projectile is, (6î + 8ˆj) m s -1 . The horizontal range of the projectile is, (a) 4.9 m, (b) 9.6 m, (c) 19.6 m (d) 14 m, A point P moves in counter-clockwise direction on a circular, y, path as shown in the figure. The, movement of 'P' is such that it, B, sweeps out a length s = t3 + 5,, P(x,y), where s is in metres and t is in, seconds. The radius of the path, m, 20, is 20 m. The acceleration of 'P', when t = 2 s is nearly, x, O, A, (a) 13 m/s2 (b) 12 m/s2 (c) 7.2 ms2 (d) 14 m/s2, , RESPONSE, GRID, , 8., 13., 18., , 9., 14., 19., , ®, , ®, , DPP/ CP03, , The resultant of two vectors A and B is perpendicular to, ®, the vector A and its magnitude is equal to half the, ®, , ®, , 17., , ®, , magnitude of vector B . The angle between A and B is, (a) 120°, (b) 150°, (c) 135°, (d) 180°, A man running along a straight road with uniform velocity, r, u = u ˆi feels that the rain is falling vertically down along – ĵ ., If he doubles his speed, he finds that the rain is coming at, an angle q with the vertical. The velocity of the rain with, respect to the ground is, (a) ui – uj, , 18., , 19., , 20., , (b), , (c) 2uiˆ + u cot qˆj, (d), Two projectiles A and B thrown with speeds in the ratio, 1 : 2 acquired the same heights. If A is thrown at an angle, of 45° with the horizontal, the angle of projection of B will be, (a) 0°, (b) 60°, (c) 30°, (d) 45°, A projectile can have the same range ‘R’ for two angles of, projection. If ‘T1’ and ‘T2’ be time of flights in the two, cases, then the product of the two time of flights is directly, proportional to, 1, 1, (a) R, (b), (c), (d) R2, R, R2, A man standing on the roof of a house of height h throws, one particle vertically downwards and another particle, horizontally with the same velocity u. The ratio of their, velocities when they reach the earth's surface will be, (a), , 2gh + u 2 : u, , (c) 1 : 1, 21., , 22., , u ˆ, j, tan q, ui + u sin qˆj, ui -, , (b) 1 : 2, (d), , 2gh + u 2 : 2gh, , If a unit vector is represented by 0.5iˆ + 0.8 ˆj + ckˆ , the value, of c is, (a) 1, (b), 0.11 (c), 0.01 (d) 0.39, An aeroplane is flying at a constant horizontal velocity of, 600 km/hr at an elevation of 6 km towards a point directly, above the target on the earth's surface. At an appropriate, time, the pilot releases a ball so that it strikes the target at, the earth. The ball will appear to be falling, (a) on a parabolic path as seen by pilot in the plane, (b) vertically along a straight path as seen by an observer, on the ground near the target, (c) on a parabolic path as seen by an observer on the, ground near the target, (d) on a zig-zag path as seen by pilot in the plane, , 10., 15., 20., Space for Rough Work, , 11., 16., 21., , 12., 17., 22., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP03, , P-11, , 23. A particle is projected with a velocity v such that its range, on the horizontal plane is twice the greatest height attained, by it. The range of the projectile is (where g is acceleration, due to gravity), 4g, 4v 2, v2, 4v 2, (b), (c), (d), g, 5g, 5g, 5v 2, 24. Two stones are projected from the same point with same, speed making angles (45° + q) and (45° – q) with the horizontal, respectively. If q £ 45° , then the horizontal ranges of the, two stones are in the ratio of, (a) 1 : 1, (b) 1 : 2, (c) 1 : 3, (d) 1 : 4, 25. Three forces acting on a body are shown in the figure. To, have the resultant force only along the y-direction, the, magnitude of the minimum additional force needed is:, y, (a) 0.5 N, 4N, 1N, (b) 1.5 N, , 30., , (a), , (c), (d), , 3, N, 4, , 31., , 30°, 60°, , 3N, , x, , 32., , 2N, , 26. A particle moves in x-y plane under the action of force F, and p at a given time t px = 2 cos q, py = 2sinq. Then the, angle q between F and p at a given time t is :, (a) q = 30° (b) q = 180° (c) q = 0° (d) q = 90°, 27. A person sitting in the rear end of the compartment throws, a ball towards the front end. The ball follows a parabolic, path. The train is moving with velocity of 20 m/s. A person, standing outside on the ground also observes the ball. How, will the maximum heights (ym) attained and the ranges (R), seen by the thrower and the outside observer compare with, each other?, (a) Same ym different R (b) Same ym and R, (c) Different ym same R (d) Different ym and R, 28. A car moves on a circular road. It describes equal angles, about the centre in equal intervals of time. Which of the, following statement about the velocity of the car is true ?, (a) Magnitude of velocity is not constant, (b) Both magnitude and direction of velocity change, (c) Velocity is directed towards the centre of the circle, (d) Magnitude of velocity is constant but direction, changes, 29. Three particles A, B and C are thrown from the top of a, tower with the same speed. A is thrown up, B is thrown, down and C is horizontally. They hit the ground with speeds, vA, vB and vC respectively then,, (a) vA = vB = vC, (b) vA = vB > vC, (c) vB > vC > vA, (d) vA > vB = vC, , RESPONSE, GRID, , 23., 28., 33., , 24., 29., 34., , A particle is moving such that its position coordinate (x, y), are, (2m, 3m) at time t = 0, (6m, 7m) at time t = 2 s and, (13m, 14m) at time t = 5s., r, Average velocity vector (Vav ) from t = 0 to t = 5s is :, 7 ˆ ˆ, 1, ˆ, (i + j), (13iˆ + 14j), (a), (b), 3, 5, 11 ˆ ˆ, (i + j), (c) 2(iˆ + ˆj), (d), 5, A particle moves so that its position vector is given by, r, r = cos wtxˆ + sin wtyˆ . Where w is a constant. Which of the, following is true ?, r, (a) Velocity and acceleration both are perpendicular to r, r, (b) Velocity and acceleration both are, parallel to r, r, (c) Velocity is perpendicular to r and acceleration is, directed towards the origin r, (d) Velocity is perpendicular to r and acceleration is, directed away from the origin, Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a direction, perpendicular to AB with velocity v1. The boy at A starts, running simultaneously with velocity v and catches the, other boy in a time t, where t is, (a), , 33., , 34., , a / v2 + v12, , (b) a /(v + v1), , (c) a /(v–v1), (d), a 2 /(v 2 - v12 ), A projectile is fired at an angle of 45° with the horizontal., Elevation angle of the projectile at its highest point as seen, from the point of projection is, æ 3ö, -1 æ 1 ö, (a) 60° (b) tan çè ÷ø (c) tan -1 ç, (d) 45°, è 2 ÷ø, 2, r, The position vector of a particle R as a function of time is, r, given by R = 4 sin(2pt)iˆ + 4 cos(2 pt) ˆj, where R is in meter, t in seconds and î and ĵ denote unit, vectors along x-and y-directions, respectively., Which one of the following statements is wrong for the, motion of particle?, (a), , Magnitude of acceleration vector is, , the velocity of particle, Magnitude of the velocity of particle is 8 meter/second, Path of the particle is a circle of radius 4 meter., r, (d) Acceleration vector is along - R, uur, uur, uur uur uur uur, The vectors A and B are such that | A + B |=| A - B |, The angle between the two vectors is, (a) 60°, (b) 75°, (c) 45°, (d) 90°, (b), (c), , 35., , v2, , where v is, R, , 25., 30., 35., Space for Rough Work, , 26., 31., , 27., 32.
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t.me/Magazines4all, , DPP/ CP03, , P-12, , 36., , 37., , 38., , 39., , 40., , The velocity of projection of oblique projectile is, (6î + 8ˆj) m s -1 . The horizontal range of the projectile is, (a) 4.9 m, (b) 9.6 m, (c) 19.6 m (d) 14 m, An artillary piece which consistently shoots its shells with, the same muzzle speed has a maximum range R. To hit a, target which is R/2 from the gun and on the same level, the, elevation angle of the gun should be, (a) 15°, (b) 45°, (c) 30°, (d) 60°, A car runs at a constant speed on a circular track of radius, 100 m, taking 62.8 seconds in every circular loop. The average, velocity and average speed for each circular loop, respectively, is, (a) 0, 10 m/s, (b) 10 m/s, 10 m/s, (c) 10 m/s, 0, (d) 0, 0, A vector of magnitude b is rotated through angle q. What is, the change in magnitude of the vector?, q, q, (a) 2b sin (b) 2b cos (c) 2b sin q (d) 2b cos q, 2, 2, A stone projected with a velocity u at an angle q with the, horizontal reaches maximum height H1. When it is projected, æp, ö, with velocity u at an angle çè - q÷ø with the horizontal, it, 2, reaches maximum height H2. The relation between the, horizontal range R of the projectile, heights H1 and H2 is, (a), , R = 4 H1H 2, , (d), , R=, , 36., 41., , 37., 42., , b, a, a, 2b, (d), (c), (b), a, b, 2b, a, ur, A vector A is rotated by a small angle Dq radian (Dq << 1), ur ur, ur, to get a new vector B In that case B - A is :, ur, ur, ur, (a) A Dq, (b) B Dq - A, ur æ Dq2 ö, (c) A çç 1 (d) 0, ÷, 2 ÷ø, è, If a body moving in circular path maintains constant speed, of 10 ms–1, then which of the following correctly describes, relation between acceleration and radius ?, , 43., , 44., , a, , a, , (a), , (b), r, , r, , a, , a, , (c), , H12, H 22, , The vector sum of two forces is perpendicular to their vector, differences. In that case, the forces, (a) cannot be predicted, (b) are equal to each other, (c) are equal to each other in magnitude, (d) are not equal to each other in magnitude, , RESPONSE, GRID, , A particle crossing the origin of co-ordinates at time t = 0,, moves in the xy-plane with a constant acceleration a in the, y-direction. If its equation of motion is y = bx2 (b is a, constant), its velocity component in the x-direction is, (a), , (b) R = 4(H1 – H2), , (c) R = 4 (H1 + H2), 41., , 42., , 45., , (d), r, , r, , The position of a projectile launched from the origin at t = 0, is given by rr = 40iˆ + 50 ˆj m at t = 2s. If the projectile was, launched at an angle q from the horizontal, then q is, (take g = 10 ms–2), 3, -1 2, -1 4, -1 7, (a) tan, (d) tan, (b) tan -1, (c) tan, 3, 5, 2, 4, , 38., 43., , (, , 39., 44., , ), , 40., 45., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP03 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP04, , P-13, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP04, , SYLLABUS : Laws of Motion, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , A player stops a football weighing 0.5 kg which comes flying, towards him with a velocity of 10m/s. If the impact lasts for, 1/50th sec. and the ball bounces back with a velocity of 15, m/s, then the average force involved is, (a) 250 N (b) 1250 N, (c) 500 N, (d) 625 N, For the given situation as shown in the figure, the value of, q to keep the system in equilibrium will be, T1, , 4., , q, , 5., T2, W = 60N, , 3., , (a) 30°, (b) 45°, (c) 0°, (d) 90°, A 5000 kg rocket is set for vertical firing. The exhaust speed, is 800 m/s. To give an initial upward acceleration of 20 m/s2,, the amount of gas ejected per second to supply the needed, thrust will be (Take g = 10 m/s2), (a) 127.5 kg/s, (b) 137.5 kg/s, (c) 155.5 kg/s, (d) 187.5 kg/s, , RESPONSE, GRID, , 1., 6., , 2., , 6., , Which one of the following statements is correct?, (a) If there were no friction, work need to be done to move, a body up an inclined plane is zero., (b) If there were no friction, moving vehicles could not be, stopped even by locking the brakes., (c) As the angle of inclination is increased, the normal, reaction on the body placed on it increases., (d) A duster weighing 0.5 kg is pressed against a vertical, board with force of 11 N. If the coefficient of friction is, 0.5, the work done in rubbing it upward through a, distance of 10 cm is 0.55 J., A stone is dropped from a height h. It hits the ground with, a certain momentum P. If the same stone is dropped from a, height 100% more than the previous height, the momentum, when it hits the ground will change by :, (a) 68% (b) 41%, (c) 200%, (d) 100%, A 3 kg ball strikes a heavyrigid wall with a speed, 60º, of 10 m/s at an angle of 60º. It gets reflected with, the same speed and angle as shown here. If the, 60º, ball is in contact with the wall for 0.20s, what is, the average force exerted on the ball bythe wall?, (a) 150N, (b) zero, (c) 150 3N, , 3., Space for Rough Work, , (d) 300N, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP04, , P-14, , 7., , 8., , The upper half of an inclined plane of inclination q is perfectly smooth while lower half is rough. A block starting, from rest at the top of the plane will again come to rest at the, bottom, if the coefficient of friction between the block and, lower half of the plane is given by, 2, (a) m =, (b) m = 2 tan q, tan q, 1, (c) m = tan q, (d) m =, tan q, A block of mass m is in contact with the cart C as shown in, the figure., , C, , g, g, g, mg, (b) a >, (c) a ³, (d) a <, mm, m, m, m, 9., A bridge is in the from of a semi-circle of radius 40m. The, greatest speed with which a motor cycle can cross the bridge, without leaving the ground at the highest point is, (g = 10 m s–2) (frictional force is negligibly small), (a) 40 m s–1, (b) 20 m s–1, (c) 30 m s–1, (d) 15 m s–1, 10. An explosion blows a rock into three parts. Two parts go off, at right angles to each other. These two are, 1 kg first part, moving with a velocity of 12 ms–1 and 2 kg second part, moving with a velocity of 8 ms–1. If the third part flies off, with a velocity of 4 ms–1, its mass would be, (a) 5 kg (b) 7 kg, (c) 17 kg, (d) 3 kg, 11. A monkey is decending from the branch of a tree with, constant acceleration. If the breaking strength is 75% of the, weight of the monkey, the minimum acceleration with which, monkey can slide down without breaking the branch is, , a>, , 3g, g, g, (c), (d), 4, 4, 2, A car having a mass of 1000 kg is moving at a speed of 30, metres/sec. Brakes are applied to bring the car to rest. If the, frictional force between the tyres and the road surface is, 5000 newtons, the car will come to rest in, (a) 5 seconds, (b) 10 seconds, (c) 12 seconds, (d) 6 seconds, A spring is compressed between two toy carts of mass m1 and, m2. When the toy carts are released, the springs exert equal, and opposite average forces for the same time on each toy, , (a) g, , 12., , 13., , RESPONSE, GRID, , m, , mm, 2mm, g (b) mm g (c) 2mm g (d), g, M, (, M, + m), (M + m), M, 15. The rate of mass of the gas emitted from rear of a rocket is, initially 0.1 kg/sec. If the speed of the gas relative to the, rocket is 50 m/sec and mass of the rocket is 2 kg, then the, acceleration of the rocket in m/sec2 is, (a) 5, (b) 5.2, (c) 2.5, (d) 25, 16. A plank with a box on it at one end, is gradually raised about the other, end. As the angle of inclination with, mg, the horizontal reaches 30º the box, q, starts to slip and slides 4.0 m down, the plank in 4.0s. The coefficients of static and kinetic friction, between the box and the plank will be, respectively :, (a) 0.6 and 0.5, (b) 0.5 and 0.6, (c) 0.4 and 0.3, (d) 0.6 and 0.6, 17. Four blocks of same mass connected by cords are pulled by, a force F on a smooth horizontal surface, as shown in fig., The tensions T1, T2 and T3 will be, T1, T2, T3, F, M, M, M, M, , (b), , 7., 12., 17., , 8., 13., 18., , 2m mg, , M, , (a), , m, , The coefficient of static friction between the block and the, cart is m. The acceleration a of the cart that will prevent the, block from falling satisfies:, (a), , 14., , cart. If v1 and v2 are the velocities of the toy carts and there is, no friction between the toy carts and the ground, then :, (a) v1/v2 = m1/m2, (b) v1/v2 = m2/m1, (c) v1/v2 = –m2/m1, (d) v1/v2 = –m1/m2, A plate of mass M is placed on a horizontal frictionless, surface (see figure), and a body of mass m is placed on this, plate. The coefficient of dynamic friction between this body, and the plate is m. If a force 2m mg is applied to the body of, mass m along the horizontal, the acceleration of the plate, will be, , 18., , 1, 3, 1, F , T2 = F , T3 = F, 4, 2, 4, 1, 1, 1, (b) T1 = F , T2 = F , T3 = F, 4, 2, 2, 3, 1, 1, (c) T1 = F , T2 = F , T3 = F, 4, 2, 4, 3, 1, 1, (d) T1 = F , T2 = F , T3 = F, 4, 2, 2, A body of mass M is kept on a rough horizontal surface, (friction coefficient µ). A person is trying to pull the body, by applying a horizontal force but the body is not moving., The force by the surface on the body is F, then, (a) F = Mg, (b) F = mMg, , (a), , T1 =, , (c), , Mg £ F £ Mg 1 + µ 2 (d), , 9., 14., Space for Rough Work, , 10., 15., , Mg ³ F ³ Mg 1 + µ 2, , 11., 16., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP04, , P-15, , 19. Which one of the following motions on a smooth plane, surface does not involve force?, (a) Accelerated motion in a straight line, (b) Retarded motion in a straight line, (c) Motion with constant momentum along a straight line, (d) Motion along a straight line with varying velocity, 20. A block A of mass m1 rests on a horizontal table. A light, string connected to it passes over a frictionless pulley at, the edge of table and from its other end another block B of, mass m2 is suspended. The coefficient of kinetic friction, between the block and the table is µk. When the block A is, sliding on the table, the tension in the string is, m1m 2 (1 + m k ) g, (m 2 – m k m1 ) g, (a), (b), (m1 + m 2 ), (m1 + m 2 ), (m 2 + m k m1 )g, m1m2 (1 – mk )g, (d), (c), (m1 + m 2 ), (m1 + m2 ), 21. The upper half of an inclined plane with inclination f 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 cos f (b) 2 sin f, (c) tan f, (d) 2 tan f, 22. A particle describes a horizontal circle in a conical funnel, whose inner surface is smooth with speed of 0.5 m/s. What, is the height of the plane of circle from vertex of the funnel?, (a) 0.25 cm (b) 2 cm, (c) 4 cm, (d) 2.5 cm, 23. You are on a frictionless horizontal plane. How can you get, off if no horizontal force is exerted by pushing against the, surface?, (a) By jumping, (b) By spitting or sneezing, (c) by rolling your body on the surface, (d) By running on the plane, 24. The coefficient of static and dynamic friction between a, body and the surface are 0.75 and 0.5 respectively. A force is, applied to the body to make it just slide with a constant, acceleration which is, g, g, 3g, (a), (b), (c), (d) g, 4, 2, 2, 25. In the system shown in figure, the pulley is smooth and, massless, the string has a total mass 5g, and the two, suspended blocks have masses 25 g and 15 g. The system, is released from state l = 0 and is studied upto stage l¢ = 0., During the process, the acceleration of block A will be, g, (a) constant at, 9, l, l', g, (b) constant at, A, 4, 25 g, (c) increasing by factor of 3, B, 15 g, (d) increasing by factor of 2, , RESPONSE, GRID, , 19., 24., 29., , 20., 25., 30., , 26., , The minimum force required to start pushing a body up, rough (frictional coefficient m) inclined plane is F1 while the, minimum force needed to prevent it from sliding down is F2., If the inclined plane makes an angle q from the horizontal, such that tan q = 2m then the ratio, (a) 1, , (b) 2, , (c) 3, , F1, is, F2, , (d) 4, , 27., , Two blocks are connected over a, massless pulley as shown in fig., A, The mass of block A is 10 kg and, the coefficient of kinetic friction is, B, 30º, 0.2. Block A slides down the incline, at constant speed. The mass of, block B in kg is, (a) 3.5, (b) 3.3, (c) 3.0, (d) 2.5, 28. Tension in the cable supporting an elevator, is equal to the, weight of the elevator. From this, we can conclude that the, elevator is going up or down with a, (a) uniform velocity, (b) uniform acceleration, (c) variable acceleration (d) either (b) or (c), 29. A particle tied to a string describes a vertical circular motion, of radius r continually. If it has a velocity 3 gr at the, highest point, then the ratio of the respective tensions in, the string holding it at the highest and lowest points is, (a) 4 : 3, (b) 5 : 4, (c) 1 : 4, (d) 3 : 2, 30. It is difficult to move a cycle with brakes on because, (a) rolling friction opposes motion on road, (b) sliding friction opposes motion on road, (c) rolling friction is more than sliding friction, (d) sliding friction is more than rolling friction, 31. A plumb line is suspended from a celling of a car moving, with horizontal acceleration of a. What will be the angle of, inclination with vertical?, (a) tan–1 (a/g), (b) tan–1 (g/a), –1, (c) cos (a/g), (d) cos–1 (g/a), 32. A cart of mass M has a block of mass m, attached to it as shown in fig. The, coefficient of friction between the block, M, m, and the cart is m. What is the minimum, acceleration of the cart so that the block, m does not fall?, (a) mg, (c) m/g, , 21., 26., 31., Space for Rough Work, , (b) g/m, (d) M mg/m, , 22., 27., 32., , 23., 28.
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t.me/Magazines4all, , DPP/ CP04, , P-16, , 33., , What is the maximum value of the force F such that the, block shown in the arrangement, does not move?, µ= 1, 2Ö 3, , 60°, , 40., , 41., , m= Ö3 kg, , (a) 20 N, (b) 10 N, (c) 12 N, (d) 15 N, A block has been placed on an inclined plane with the slope, angle q, block slides down the plane at constant speed. The, coefficient of kinetic friction is equal to, (a) sin q (b) cos q, (c) g, (d) tan q, 35. A block of mass m is connected to another block of mass M, by a spring (massless) of spring constant k. The block are, kept on a smooth horizontal plane. Initially the blocks are at, rest and the spring is unstretched. Then a constant force F, starts acting on the block of mass M to pull it. Find the force, of the block of mass m., mF, mF, MF, (a), (b), (c) ( M + m) F (d), M, (m + M ), (m + M ), m, 36. A block of mass m is placed on a surface with a vertical, 34., , 42., , 43., , x3, . If the coefficient of friction, 6, is 0.5, the maximum height above the ground at which the, block can be placed without slipping is:, , cross section given by y =, , 1, 2, 1, 1, m, m, (b) m, (c) m, (d), 6, 3, 3, 2, A ball of mass 10 g moving perpendicular to the plane of the, wall strikes it and rebounds in the same line with the same, velocity. If the impulse experienced by the wall is 0.54 Ns,, the velocity of the ball is, (a) 27 ms–1 (b) 3.7 ms–1 (c) 54 ms–1, (d) 37 ms–1, A block is kept on a inclined plane of inclination q of length l., The velocity of particle at the bottom of inclined is (the, coefficient of friction is m), , (a), , 37., , 38., , (a), 39., , [2gl(m cos q - sin q)]1 / 2 (b), , 2gl(sin q - m cos q), , 44., , 45., , (c), (d), 2gl(sin q + m cos q), 2gl(cos q + m sin q), A 100 g iron ball having velocity 10 m/s collides with a wall, at an angle 30° and rebounds with the same angle. If the, period of contact between the ball and wall is 0.1 second,, then the force experienced by the wall is, (a) 10 N (b) 100 N, (c) 1.0 N, (d) 0.1 N, , 33., 37., 42., , RESPONSE, GRID, , 34., 38., 43., , A bullet is fired from a gun. The force on the bullet is given by, F = 600 – 2 × 105 t where, F is in newton and t 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) 1.8 N-s (b) zero, (c) 9 N-s, (d) 0.9 N-s, Two stones of masses m and 2 m are whirled in horizontal, r, circles, the heavier one in radius and the lighter one in, 2, radius r. The tangential speed of lighter stone is n times that, of the value of heavier stone when they experience same, centripetal forces. The value of n is :, (a) 3, (b) 4, (c) 1, (d) 2, A 0.1 kg block suspended from a massless string is moved, first vertically up with an acceleration of 5ms–2 and then, moved vertically down with an acceleration of 5ms–2. If T1, and T2 are the respective tensions in the two cases, then, (a) T2 > T1, (b) T1 – T2 = 1 N, if g = 10ms–2, (c) T1 – T2 = 1kg f, (d) T1 – T2 = 9.8 N, if g = 9.8 ms–2, Three forces start acting simultaneously C, r, on a particle moving with velocity, v ., These forces are represented in magnitude, and direction by the three sides of a, triangle ABC. The particle will now move, with velocity, A, r, B, (a) less than v, r, (b) greater than v, (c) |v| in the direction of the largest force BC, r, (d) v , remaining unchanged, If in a stationary lift, a man is standing with a bucket full of, water, having a hole at its bottom. The rate of flow of water, through this hole is R0. If the lift starts to move up and, down with same acceleration and then the rates of flow of, water are Ru and Rd, then, (a) R0 > Ru > Rd, (b) Ru > R0 > Rd, (c) Rd > R0 > Ru, (d) Ru > Rd > R0, A stationary body of mass 3 kg explodes into three equal, pieces. Two of the pieces fly off in two mutually, perpendicular directions, one with a velocity of 3iˆ ms -1, and the other with a velocity of 4ˆj ms -1. If the explosion, occurs in 10–4 s, the average force acting on the third piece, in newton is, (a) (3iˆ + 4ˆj) ´ 10 - 4, (b) (3iˆ - 4ˆj) ´ 10 - 4, ˆ ´ 104, (c) (3iˆ - 4ˆj) ´ 10 4, (d) -(3iˆ + 4j), , 35., 39., 44., , 36., 40., 45., , 41., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP04 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45 Space for Qualifying, Score, Rough Work, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP05, , SYLLABUS : Work, Energy and Power, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , 4., , 5., , A spring of spring constant 5 × 103 N/m is stretched initially, by 5cm from the unstretched position. Then the work, required to stretch it further by another 5 cm is, (a) 12.50 Nm, (b) 18.75 Nm, (c) 25.00 Nm, (d) 6.25 Nm, A particle of mass 10 g moves along a circle of radius 6.4 cm, with a constant tangential acceleration. What is the, magnitude of this acceleration if the kinetic energy of the, particle becomes equal to 8 × 10–4 J by the end of the second, revolution after the beginning of the motion ?, (a) 0.1 m/s2 (b) 0.15 m/s2 (c) 0.18 m/s2 (d) 0.2 m/s2, 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, (a) t 3/4, (b) t 3/2, (c) t 1/4, (d) t 1/2, A ball is thrown vertically downwards from a height of 20 m, with an initial velocity v0. It collides with the ground and, loses 50% of its energy in collision and rebounds to the, same height. The initial velocity v0 is : (Take g = 10 ms–2), (a) 20 ms–1, (b) 28 ms–1, (c) 10 ms–1, (d) 14 ms–1, A cord is used to lower vertically a block of mass M,, a distance d at a constant downward acceleration of g/4., The work done by the cord on the block is, , RESPONSE, GRID, , 1., 6., , 2., 7., , 6., , d, d, d, (b) 3Mg, (c) -3Mg, (d) Mg d, 4, 4, 4, A rubber ball is dropped from a height of 5m on a plane,, where the acceleration due to gravity is not shown. On, bouncing it rises to 1.8 m. The ball loses its velocity on, bouncing by a factor of, , 7., , 16, 2, 3, 9, (b), (c), (d), 25, 5, 5, 25, A ball of mass m moving with a constant velocity strikes, against a ball of same mass at rest. If e = coefficient of, restitution, then what will be the ratio of velocity of two, balls after collision?, , 8., , 1- e, e -1, 1+ e, 2+e, (b), (c), (d), 1+ e, e +1, 1- e, e -1, A particle of mass m is driven by a machine that delivers a, constant power of k watts. If the particle starts from rest the, force on the particle at time t is, , (a), , Mg, , (a), , (a), , (a), , mk t –1/2, , (b), , 2mk t –1/2, , (c), , 1, mk t –1/2, 2, , (d), , mk –1/2, t, 2, , 3., 8., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP05, , P-18, , 9., , A body of mass 2 kg moving under a force has relation, , between displacement x and time t as x =, , 15., , t3, where x is in, 3, , metre and t is in sec. The work done by the body in first two, second will be, (a) 1.6 joule, (b) 16 joule, (c) 160 joule, (d) 1600 joule, 10. A sphere of mass 8m collides elastically (in one dimension), with a block of mass 2m. If the initial energy of sphere is E., What is the final energy of sphere?, (a) 0.8 E, (b) 0.36 E, (c) 0.08 E, (d) 0.64 E, 11. Two similar springs P and Q have spring constants KP and, KQ, such that KP > KQ. They are stretched, first by the same, amount (case a,) then by the same force (case b). The work, done by the springs WP and WQ are related as, in case (a), and case (b), respectively, (a) WP = WQ ; WP = WQ (b) WP > WQ ; WQ > WP, (c) WP < WQ ; WQ < WP (d) WP = WQ ; WP > WQ, 12. In the figure, the variation of potential energy of a particle, of mass m = 2 kg is represented w.r.t. its x-coordinate. The, particle moves under the effect of this conservative force, along the x-axis., , The relationship between the force F and position x of a, body is as shown in figure. The work done in displacing the, body form x = 1 m to x = 5 m will be, F(N), 10, 5, 0, , l, , 2 3, , 4, , 5, , 6, , x(m), , –5, –10, , (a) 30 J, (b) 15 J, (c) 25 J, (d) 20 J, A body is allowed to fall freely under gravity from a height, of 10m. If it looses 25% of its energy due to impact with the, ground, then the maximum height it rises after one impact is, (a) 2.5m, (b) 5.0m, (c) 7.5m, (d) 8.2m, 17. A block C of mass m is moving with velocity v0 and collides, elastically with block A of mass m and connected to another, block B of mass 2m through spring constant k. What is k if, x0 is compression of spring when velocity of A and B is, same?, 16., , C, , v0, , A, , B, , `, , U (in J), 20, , (a), , 15, 10, 5, , –5, , 2, , –10, , 10, , (c), , X (in meter), , mv02, , (b), , x 02, 2, 3 mv0, 2 x 02, , (d), , mv02, 2x 02, 2, 2 mv0, 3 x 02, , 18., , –12, –15, , 13., , 14., , If the particle is released at the origin then, (a) it will move towards positive x-axis, (b) it will move towards negative x-axis, (c) it will remain stationary at the origin, (d) its subsequent motion cannot be decided due to lack, of information, The potential energy of a certain spring when stretched, through distance S is 10 joule. The amount of work done (in, joule) that must be done on this spring to stretch it through, an additional distance s, will be, (a) 20, (b) 10, (c) 30, (d) 40, A force applied by an engine of a train of mass 2.05×106 kg, changes its velocity from 5m/s to 25 m/s in 5 minutes. The, power of the engine is, (a) 1.025 MW, (b) 2.05 MW, (c) 5 MW, (d) 6 MW, , RESPONSE, GRID, , 9., 14., 19., , 10., 15., 20., , Two springs of force constants 300 N/m, (Spring A) and 400 N/m (Spring B) are joined together in, series. The combination is compressed by 8.75 cm. The ratio, E, E, of energy stored in A and B is A . Then A is equal to :, EB, EB, 4, 16, 9, 3, (a), (b), (c), (d), 3, 9, 16, 4, 19. A body of mass 1 kg begins to move under the action of a, r, time dependent force F=(2tiˆ+3t 2 ˆj) N, where î and ĵ are, unit vectors alogn x and y axis. What power will be developed, by the force at the time t?, (a) (2t2 + 3t3)W, (b) (2t2 + 4t4)W, 3, 4, (c) (2t + 3t ) W, (d) (2t3 + 3t5)W, 20. A bullet of mass 20 g and moving with 600 m/s collides with, a block of mass 4 kg hanging with the string. What is the, velocity of bullet when it comes out of block, if block rises, to height 0.2 m after collision?, (a) 200 m/s (b) 150 m/s (c) 400 m/s, (d) 300 m/s, , 11., 16., Space for Rough Work, , 12., 17., , 13., 18., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP05, , P-19, , 21. A body of mass m kg is ascending on a smooth inclined, 1ö, æ, plane of inclination q ç sin q = ÷ with constant acceleration, è, xø, of a m/s2. The final velocity of the body is v m/s. The work, done by the body during this motion is, (Initial velocity of the body = 0), (a), , 1, mv2 (g + xa), 2, , (b), , mv2 æ g ö, ç + a÷, 2 è2 ø, , 2mv 2 x, mv 2, (d), ( a + gx ), ( g + xa ), a, 2ax, 22. A glass marble dropped from a certain height above the, horizontal surface reaches the surface in time t and then, continues to bounce up and down. The time in which the, marble finally comes to rest is, , (c), , é1 - e ù, é1+ e ù, tê, (d) t ê, ú, ú, ë1 + e û, ë1 - e û, 23. The potential energy of a 1 kg particle free to move along, , (a) en t, , (b) e2 t, , (c), , æ x4 x2 ö, ÷ J., the x-axis is given by V( x) = ç, ç 4, ÷, 2, è, ø, The total mechanical energy of the particle is 2 J. Then, the, maximum speed (in m/s) is, 3, , 1, , (d) 2, 2, 2, 24. Water falls from a height of 60 m at the rate of 15 kg/s to, operate a turbine. The losses due to frictional force are 10%, of energy. How much power is generated by the turbine?( g, = 10 m/s2), (a) 8.1 kW (b) 10.2 kW (c) 12.3 kW (d) 7.0 kW, 24. A car of mass m starts from rest and accelerates so that the, instantaneous power delivered to the car has a constant, magnitude P0. The instantaneous velocity of this car is, proportional to :, (a), , (b), , 2, , (c), , t, (a) t 2P 0, (b) t 1/2, (c) t –1/2, (d), m, 25. When a 1.0kg mass hangs attached to a spring of length 50, cm, the spring stretches by 2 cm. The mass is pulled down, until the length of the spring becomes 60 cm. What is the, amount of elastic energy stored in the spring in this, condition. if g = 10 m/s2., (a) 1.5 joule (b) 2.0 joule(c) 2.5 joule (d) 3.0 joule, 26. A block of mass m rests on a rough horizontal surface, (Coefficient of friction is µ). When a bullet of mass m/2, strikes horizontally, and get embedded in it, the block moves, a distance d before coming to rest. The initial velocity of the, bullet is k 2mgd , then the value of k is, , RESPONSE, GRID, , 21., 26., 31., , 22., 27., 32., , m/2, , m, , /////////////////////////////////////////////////////////////, d, , (a) 2, (b) 3, (c) 4, (d) 5, A force acts on a 30 gm particle in such a way that the, position of the particle as a function of time is given by x =, 3t – 4t2 + t3, where x is in metres and t is in seconds. The, work done during the first 4 seconds is, (a) 576mJ (b) 450mJ (c) 490mJ (d) 530mJ, 28. A particle of mass m1 moving with velocity v strikes with a, mass m2 at rest, then the condition for maximum transfer of, kinetic energy is, (a) m1 >> m2 (b) m2 >> m2 (c) m1 = m2 (d) m1 = 2m2, 29. A mass m is moving with velocity v collides inelastically, with a bob of simple pendulum of mass m and gets embedded, into it. The total height to which the masses will rise after, collision is, 27., , v2, 2v 2, v2, v2, (b), (c), (d), 4g, g, 2g, 8g, A 10 H.P. motor pumps out water from a well of depth 20 m, and fills a water tank of volume 22380 litres at a height of, 10 m from the ground. The running time of the motor to fill, the empty water tank is (g = 10ms–2), (a) 5 minutes, (b) 10 minutes, (c) 15 minutes, (d) 20 minutes, A particle of mass m1 is moving with a velocity v1 and, another particle of mass m2 is moving with a velocity v2., Both of them have the same momentum but their different, kinetic energies are E1 and E2 respectively. If m1 > m2 then, E1 m1, (d)E1 > E2, (a) E1 = E2 (b) E1 < E2 (c) E = m, 2, 2, A block of mass 10 kg, moving in x direction with a constant, speed of 10 ms–1, is subject to a retarding force F = 0.1 × J m, during its travel from x = 20 m to 30 m. Its final KE will be :, (a) 450 J, (b) 275 J, (c) 250 J, (d) 475 J, Identify the false statement from the following, (a) Work-energy theorem is not independent of Newton's, second law., (b) Work-energy theorem holds in all inertial frames., (c) Work done by friction over a closed path is zero., (d) No potential energy can be associated with friction., A one-ton car moves with a constant velocity of, 15 ms–1 on a rough horizontal road. The total resistance to, the motion of the car is 12% of the weight of the car. The, power required to keep the car moving with the same constant, velocity of 15ms–1 is [Take g = 10 ms–2], (a) 9 kW, (b) 18 kW (c) 24 kW, (d) 36 kW, A ball is released from the top of a tower. The ratio of work, done by force of gravity in first, second and third second of, the motion of the ball is, (a) 1 : 2 : 3 (b) 1 : 4 : 9, (c) 1 : 3 : 5, (d) 1 : 5 : 3, (a), , 30., , 31., , 32., , 33., , 34., , 35., , 23., 28., 33., Space for Rough Work, , 24., 29., 34., , 25., 30., 35.
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t.me/Magazines4all, , DPP/ CP05, , P-20, , 36., , 37., , 38., , 39., , 40., , Two spheres A and B of masses m1 and m2 respectively, collide. A is at rest initially and B is moving with velocity v, v, along x-axis. After collision B has a velocity in a direction, 2, perpendicular to the original direction. The mass A moves, after collision in the direction., (a) Same as that of B, (b) Opposite to that of B, (c) q = tan–1 (1/2) to the x-axis, (d) q = tan–1 (–1/2) to the x-axis, A 2 kg block slides on a horizontal floor with a speed of 4m/s., It strikes a uncompressed spring, and compresses it till the, block is motionless. The kinetic friction force is 15N and spring, constant is 10,000 N/m. The spring compresses by, (a) 8.5 cm, (b) 5.5 cm (c) 2.5 cm, (d) 11.0 cm, An engine pumps water through a hose pipe. Water passes, through the pipe and leaves it with a velocity of 2 m/s. The, mass per unit length of water in the pipe is 100 kg/m. What, is the power of the engine?, (a) 400 W, (b) 200 W (c) 100 W, (d) 800 W, A uniform chain of length 2 m is kept on a table such that 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) 12 J, (b) 3.6 J, (c) 7.2 J, (d) 1200 J, A mass ‘m’ moves with a velocity ‘v’ and collides inelastically, with another identical mass. After collision the lst mass moves, with velocity, , v, 3, , in a direction perpendicular to the initial, , direction of motion. Find the speed of the 2nd mass after, collision., m, m, A before, collision, 2, v, v, (a), (d), (c), 3 v (b) v, 3, 3, A spherical ball of mass 20 kg is stationary at the top of a hill, of height 100 m. It rolls down a smooth surface to the ground,, then climbs up another hill of height 30 m and finally rolls, down to a horizontal base at a height of 20 m above the, ground. The velocity attained by the ball is, v, , 41., , 3, Aafter, collision, , (a) 20 m/s (b) 40 m/s, , RESPONSE, GRID, , 36., 41., , (c), , 10 30 m/s (d) 10 m/s, , 37., 42., , 42., , A block of mass M is kept on a platform which is accelerated, upward with a constant acceleration 'a' during the time, interval T. The work done by normal reaction between the, block and platform is, M, , (a), , -, , MgaT 2, 2, , a, , (b), , 1, M (g + a) aT 2, 2, , 1, Ma 2 T, (d) Zero, 2, A spring lies along an x axis attached to a wall at one end, and a block at the other end. The block rests on a frictionless, surface at x = 0. A force of constant magnitude F is applied, to the block that begins to compress the spring, until the, block comes to a maximum displacement xmax., , (c), , 43., , Energy, of work, , 4, , 1, 2, 3, , x, , xmax, , During the displacement, which of the curves shown in the, graph best represents the kinetic energy of the block ?, (a) 1, (b) 2, (c) 3, (d) 4, 44. The K.E. acquired by a mass m in travelling a certain distance, d, starting form rest, under the action of a constant force is, directly proportional to, (a) m, (b), m, 1, (d) independent of m, (c), m, 45. A vertical spring with force constant k is fixed on a table. A, ball of mass m at a height h above the free upper end of the, spring falls vertically on the spring so that the spring is, compressed by a distance d. The net work done in the, process is, 1, 1, (a) mg(h + d) - kd 2, (b) mg(h - d) - kd 2, 2, 2, 1, 1, (c) mg(h - d) + kd 2, (d) mg(h + d) + kd2, 2, 2, , 38., 43., , 39., 44., , 40., 45., , DAILY PRA CTICE PROBLEM DPP CHAPTERWISE CP05 - PHYSICS, Total Ques tions, 45, Total Marks, Attempted, Correct, Incorrect, Net Score, Cut-off Score, 50, Qualifying Score, Succes s Gap = Net Score – Space, Qualifying, for RoughScore, Work, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP06, , SYLLABUS : System of Particles and Rotational Motion, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 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 center and perpendicular, to one of its faces is :, (a), , 2., , 3., , 4MR 2, , (b), , 4MR 2, , (c), , MR 2, , (d), , MR 2, , 9 3p, 3 3p, 32 2p, 16 2p, A hollow sphere is held suspended. Sand, is now poured into it in stages., The centre of mass of the sphere with, the sand, (a) rises continuously, (b) remains unchanged in the process, SAND, (c) first rises and then falls to the, original position, (d) first falls and then rises to the, original position, A body A of mass M while falling vertically downwards, 1, under gravity breaks into two parts; a body B of mass M, 3, 2, and a body C of mass, M. The centre of mass of bodies, 3, B and C taken together shifts compared to that of body A, towards, (a) does not shift, (b) depends on height of breaking, (c) body B, (d) body C, , RESPONSE GRID, , 1., 6., , 2., 7., , 4., , 5., , 6., , 7., , 8., , From a uniform wire, two circular loops are made (i) P of, radius r and (ii) Q of radius nr. If the moment of inertia of Q, about an axis passing through its centre and perpendicular, to its plane is 8 times that of P about a similar axis, the value, of n is (diameter of the wire is very much smaller than r or nr), (a) 8, (b) 6, (c) 4, (d) 2, A billiard ball of mass m and radius r, when hit in a horizontal, direction by a cue at a height h above its centre, acquired a, linear velocity v0. The angular velocity w0 acquired by the, ball is, 5v0 h, 2v0 r 2, 5v0 r 2, 2v0 h, (a), (b), (c), (d), 5h, 2r 2, 2h, 5r 2, Three bricks each of length L and Wall, mass M are arranged as shown, from the wall. The distance of the, centre of mass of the system from, L/4, L/2, the wall is, (a) L/4, (b) L/2, (c) (3/2)L L (d) (11/12)L, 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, (a) 2ml 2 (b), 3ml2 (c) 3ml 2 (d) ml 2, A loop of radius r and mass m rotating with an angular velocity, w0 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?, rw0, rw0, rw0, (a), (b), (c), (d) rw0, 3, 4, 2, , 3., 8., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP06, , P-22, , 9., , 10., , 11., , 12., , Two masses m1 and m2 are connected by a massless spring, of spring constant k and unstretched length l. The masses, are placed on a frictionless straight channel, which are, consider our x-axis. They are initially at x = 0 and x = l, respectively. At t = 0, a velocity v0 is suddenly imparted to, the first particle. At a later time t, the centre of mass of the, two masses is at :, m2 l, (a) x = m + m, 1, 2, , m2 v0t, m1l, (b) x = m + m + m + m, 1, 2, 1, 2, m2v0t, m2 l, m1v0 t, m2 l, +, (c) x =, (d) x = m + m + m + m, m1 + m1 m1 + m2, 1, 2, 1, 2, A body of mass 1.5 kg rotating about an axis with angular, velocity of 0.3 rad s–1 has the angular momentum of 1.8 kg, m2s–1. The radius of gyration of the body about an axis is, (a) 2 m, (b) 1.2 m, (c) 0.2 m, (d) 1.6 m, r, If F is the force acting on a particle having position, r, r, vector r and t be the torque of this force about the origin,, then:, r r, r r, (a) r . t > 0 and F . t < 0, r r, r r, (b) r . t = 0 and F . t = 0, r r, r r, (c) r . t = 0 and F . t ¹ 0, r r, r r, (d) r . t ¹ 0 and F . t = 0, A thin uniform rod of length l and mass m is swinging freely, about a horizontal axis passing through its end. Its maximum, angular speed is w. Its centre of mass rises to a maximum, height of, , 1 l 2w2, 1 l 2w 2, 1 l 2w 2, (c), (d), 3 g, 2 g, 6 g, A wheel is rolling straight on ground, without slipping. If the axis of the, P, q, wheel has speed v, the instantenous, velocity of a point P on the rim, defined, by angle q, relative to the ground will, be, , (a), 13., , (a), , 14., , 1 lw, 6 g, , 16., , 17., , 18., , (b), , æ1 ö, v cos ç q ÷, è2 ø, , (b), , 19., , æ1 ö, 2 v cos ç q ÷, è2 ø, , (c) v(1 + sin q), (d) v(1 + cos q), A solid sphere having mass m and radius r rolls down an, inclined plane. Then its kinetic energy is, (a), , 15., , 20., , 5, 2, rotational and translational, 7, 7, , RESPONSE, GRID, , 9., 14., 19., , 10., 15., , 2, 5, (b), rotational and translational, 7, 7, 2, 3, (c), rotational and translational, 5, 5, 1, 1, (d), rotational and translational, 2, 2, A ring of mass M and radius R is rotating about its axis with, angular velocity w. 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 :, Mm, m( M + 2m) 2 2, w2 R2, w R, (a), (b), (, M, +, m, ), M, ( M + m) M 2 2, Mm, w2 R 2, (c), (d) ( M + 2 m) w R, ( M + 2m), n F1, Acertain bicycle can go up a, Chai, gentle incline with constant speed, d, when the frictional force of R2, Roa, ground pushing the rear wheel is, R1, F2 = 4 N. With what force F1 must, 4N, the chain pull on the sprocket, F2 =, wheel if R1=5 cm and R2 = 30 cm?, Horizontal, 35, (a) 4 N, (b) 24 N, (c) 140 N (d), N, 4, A wooden cube is placed on a rough horizontal table, a, force is applied to the cube. Gradually the force is increased., Whether the cube slides before toppling or topples before, sliding is independent of :, (a) the position of point of application of the force, (b) the length of the edge of the cube, (c) mass of the cube, (d) Coefficient of friction between the cube and the table, From a circular ring of mass M and radius R, an arc, corresponding to a 90° sector is removed. The moment of, inertia of the ramaining part of the ring about an axis passing, through the centre of the ring and perpendicular to the plane, of the ring is k times MR2. Then the value of k is, (a) 3/4, (b) 7/8, (c) 1/4, (d) v1, 0, A mass m moves in a circle on a, smooth horizontal plane with, velocity v0 at a radius R0. The, m, mass is attached to string which, passes through a smooth hole in, the plane as shown., The tension in the string is increased gradually and finally, R, m moves in a circle of radius 0 . The final value of the, 2, kinetic energy is, 1, 1, mv02 (b) 2mv02 (c), mv02 (d) mv20, (a), 4, 2, A rod PQ of length L revolves in a horizontal plane about the, axis YY´. The angular velocity of the rod is w. If A is the area, of cross-section of the rod and r be its density, its rotational, kinetic energy is, , 11., 16., Space for Rough Work, , 12., 17., , 13., 18., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP06, , P-23, , 1, 1 3 2, AL3rw 2 `, AL rw, (b), 3, 2, 1, 1, AL3rw 2, AL3rw 2, (c), (d), 18, 24, A solid sphere of mass 2 kg rolls on a smooth horizontal, surface at 10 m/s. It then rolls up a smooth inclined plane of, inclination 30° with the horizontal. The height attained by, the sphere before it stops is, (a) 700 cm (b) 701 cm (c) 7.1 m, (d) 70 m, A hollow smooth uniform sphere A of mass m rolls without, sliding on a smooth horizontal surface. It collides head on, elastically with another stationary smooth solid sphere B of, the same mass m and same radius. The ratio of kinetic energy, of B to that of A just after the collision is, A, B, (a) 1 : 1, (b) 2 : 3, v, (c) 3 : 2, (d) 4 : 3, Two discs of same thickness but of different radii are made, of two different materials such that their masses are same., The densities of the materials are in the ratio of 1 : 3. The, moments of inertia of these discs about the respective axes, passing through their centres and perpendicular to their, planes will be in the ratio of, (a) 1 : 3, (b) 3 : 1, (c) 1 : 9, (d) 9 : 1, A pulley fixed to the ceiling carries a string with blocks of, mass m and 3 m attached to its ends. The masses of string, and pulley are negligible. When the system is released, its, centre of mass moves with what acceleration ?, (a) 0, (b) – g/4, (c) g/2, (d) – g/2, A ring of mass m and radius R has, four particles each of mass m, attached to the ring as shown in, figure. The centre of ring has a speed, v0. The kinetic energy of the system, is, , (a), , 21., , 22., , 0, , 23., , 24., , 25., , (a), , mv02, , (b), , 3mv02, , (c), , 5mv02, , (d), , 28., , 29., , 30., , 5, 1, 7, 2, Ma 2 (b), Ma 2 (c), Ma 2 (d), Ma 2, 6, 12, 12, 3, 27. A dancer is standing on a stool rotating about the vertical, axis passing through its centre. She pulls her arms towards, the body reducing her moment of inertia by a factor of n., The new angular speed of turn table is proportional to, (a) n, (b) n –1, (c) n 0, (d) n 2, , 31., , 20., 25., 30., , 21., 26., 31., , m1d, m1d, (d) m, (m1 + m 2 ), 2, A uniform bar of mass M and length L is horizontally, suspended from the ceiling by two vertical light cables as, shown. Cable A is connected 1/4th distance from, the left end of the bar. Cable, Cable B, B is attached at the far right Cable A, 1L, end of the bar. What is the 4, tension in cable A?, L, (a) 1/4 Mg (b) 1/3 Mg (c) 2/3 Mg (d) 3/4 Mg, A couple produces, (a) purely linear motion, (b) purely rotational motion, (c) linear and rotational motion, (d) no motion, Point masses 1, 2, 3 and 4 kg are lying at the point (0, 0, 0),, (2, 0, 0), (0, 3, 0) and (–2, –2, 0) respectively. The moment of, inertia of this system about x-axis will be, (a) 43 kgm2 (b) 34 kgm2 (c) 27 kgm2 (d) 72 kgm2, , (a), 32., , 33., , (a), , RESPONSE, GRID, , 55 2, 127 2, 111 2, mr (b), mr (c), mr (d) 55 mr2, 2, 2, 2, In a two-particle system with particle masses m1 and m2, the, first particle is pushed towards the centre of mass through, a distance d, the distance through which second particle, must be moved to keep the centre of mass at the same, position is, , (a), , 6mv02, , 26. 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 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. Then the moment of inertia about, y, y, the z-axis, (a) increases, Hole, Q, x, x, (b) decreases, (c) remains same, (d) changed in unpredicted manner., A circular turn table has a block of ice placed at its centre., The system rotates with an angular speed w about an axis, passing through the centre of the table. If the ice melts on, its own without any evaporation, the speed of rotation of, the system, (a) becomes zero, (b) remains constant at the same value w, (c) increases to a value greater than w, (d) decreases to a value less than w, Seven identical coins are rigidly arranged on a flat table in, the pattern shown below so that each coin touches it, neighbors. Each coin is a thin disc of mass m and radius r., The moment of inertia of the system of seven coins about, an axis that passes through point P and perpendicular to, the plane of the coin is :, , 34., , 22., 27., 32., Space for Rough Work, , m 2d, m1, , (b) d, , 23., 28., 33., , (c), , 24., 29., 34.
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t.me/Magazines4all, , DPP/ CP06, , P-24, , 35., , 36., , 37., , 38., , 39., , 40., , A solid sphere of mass M and, radius R is pulled horizontally on, a sufficiently rough surface as, shown in the figure., Choose the correct alternative., (a) The acceleration of the centre of mass is F/M, 2 F, (b) The acceleration of the centre of mass is, 3M, (c) The friction force on the sphere acts forward, (d) The magnitude of the friction force is F/3, The moment of inertia of a body about a given axis is, 1.2 kg m2. Initially, the body is at rest. In order to produce a, rotational kinetic energy of 1500 joule, an angular, acceleration of 25 radian/sec2 must be applied about that, axis for a duration of, (a) 4 sec (b) 2 sec, (c) 8 sec, (d) 10 sec, A gymnast takes turns with her arms and legs stretched., When she pulls her arms and legs in, (a) the angular velocity decreases, (b) the moment of inertia decreases, (c) the angular velocity stays constant, (d) the angular momentum increases, An equilateral triangle ABC formed from, A, a uniform wire has two small identical, g, beads initially located at A. The triangle is, set rotating about the vertical axis AO., Then the beads are released from rest, C, B, simultaneously and allowed to slide, O, down, one along AB and the other along AC as shown., Neglecting frictional effects, the quantities that are conserved, as the beads slide down, are, (a) angular velocity and total energy (kinetic and potential), (b) total angular momentum and total energy, (c) angular velocity and moment of inertia about the axis, of rotation, (d) total angular momentum and moment of inertia about, the axis of rotation, The moment of inertia of a uniform semicircular wire of mass, m and radius r, about an axis passing through its centre of, kö, æ, mass and perpendicular to its plane is mr 2 ç1 - 2 ÷ . Find, è p ø, the value of k., (a) 2, (b) 3, (c) 4, (d) 5, Initial angular velocity of a circular disc of mass M is w 1., Then two small spheres of mass m are attached gently to, diametrically opposite points on the edge of the disc. What, is the final angular velocity of the disc?, , 35., 40., 45., , RESPONSE, GRID, , 36., 41., , (a), , æ M + mö, çè, ÷ w1, M ø, , (b), , æ M + mö, çè, ÷ w1, m ø, , æ M ö, æ M ö, w., (d) ç, çè, ÷w, M + 4m ø 1, è M + 2m ÷ø 1, Two identical discs of mass m and radius r are //////////////////, arranged as shown in the figure. If a is the, angular acceleration of the lower disc and acm, is acceleration of centre of mass of the lower, disc, then relation between a cm ,, a and r is, (a) acm = a/r, (b) acm = 2ar, (c) acm = a r, (d) None of these, Five masses are placed in a plane as shown in figure. The, coordinates of the centre of mass are nearest to, , (c), , 41., , 42., , y, 2 3 kg, , (a) 1.2, 1.4, (b) 1.3, 1.1, , 1, , 4 kg, 5 kg, , (c) 1.1, 1.3, 0 1 kg, 0, , (d) 1.0, 1.0, 43., , 44., , 45., , 1, , 2 kg x, 2, , Three particles, each of mass m gram, are situated at the, vertices of an equilateral triangle ABC of side l cm (as shown, in the figure). The moment of inertia of the system about a, line AX perpendicular to AB and in the plane of ABC, in, gram-cm2 units will be, X, 3, 2, m, l, (a), m, C, 2, 3, l, l, ml 2, (b), 4, (c) 2 ml2, B, A, 5, m, l, m, ml 2, (d), 4, When a ceiling fan is switched on, it makes 10 rotations in, the first 3 seconds. Assuming a uniform angular, acceleration, how many rotation it will make in the next 3, seconds?, (a) 10, (b) 20, (c) 30, (d) 40, A solid sphere spinning about a horizontal axis with an, angular velocity w is placed on a horizontal surface., Subsequently it rolls without slipping with an angular, velocity of :, 2w, 7w, 2w, (a), (b), (c), (d) w, 5, 5, 7, , 37., 42., , 38., 43., , 39., 44., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP06 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Space for Rough Work, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP07, , SYLLABUS : Gravitation, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , The radius of a planet is 1/4th of Re and its acc. due to, gravity is 2g. What would be the value of escape velocity, on the planet, if escape velocity on earth is ve., ve, ve, (a), (b) v e 2, (c), 2 ve (d), 2, 2, A projectile is fired vertically from the Earth with a velocity, kve where ve is the escape velocity and k is a constant less, than unity. The maximum height to which projectile rises, as, measured from the centre of Earth, is, R, R, R, R, (d), (a), (b), (c), 2, k, k -1, 1- k, 1+ k2, A solid sphere of uniform, density and radius R applies a, A, gravitational force of attraction, equal to F1 on a particle placed, at A, distance 2R from the centre, of the sphere., R, R, , RESPONSE GRID, , 1., , 2., , 4., , A spherical cavity of radius R/2 is now made in the sphere, as shown in the figure. The sphere with cavity now applies, a gravitational force F2 on the same particle placed at A., The ratio F2/F1 will be, (a) 1/2, (b) 3, (c) 7, (d) 1/9, A geostationary satellite is orbiting the earth at a height of, 5R above that surface of the earth, R being the radius of the, earth. The time period of another satellite in hours at a height, of 2R from the surface of the earth is :, (a) 5, , 5., , (b) 10, , (c), , 6 2 (d), , 6, , 2, A satellite of mass m is orbiting around the earth in a circular, orbit with a velocity v. What will be its total energy?, (a) (3/4) mv2, (b) (1/2) mv2, 2, (c) mv, (d) – (1/2)m v2, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP07, , P-26, , 6., , The gravitational force of attraction between a uniform, sphere of mass M and a uniform rod of length l and mass m, oriented as shown is, m, , M, r, , 7., , 8., , GMm, r(r + l ), , l, , 13., , GM, , (c) Mmr2 + l (d) (r2 + l) mM, r2, If the gravitational force between two objects were, proportional to 1/R (and not as 1/R2) where R is separation, between them, then a particle in circular orbit under such a, force would have its orbital speed v proportional to, (a) 1/R2 (b) R0, (c) R1, (d) 1/R, 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), , (b), , 12., , æ gR 2 ö 1/ 2, gR, gR 2, ÷, (b), (c) gx (d) çç, (a), ÷, R, x, +, R-x, R+x, ø, è, 9., A body is projected up with a velocity equal to 3/4th of the, escape velocity from the surface of the earth. The height it, reaches from the centre of the earth is (Radius of the earth = R), 10R, 16R, 9R, 10R, (a), (b), (c), (d), 9, 7, 8, 3, 10. A Planet is revolving around the sun., B, , A, , C, S, , 14., , 15., , 16., , 1, changes and becomes an inverse cube law i.e. F µ, , but, r3, still remaining a central force. Then, (a) Kepler’s law of area still holds, (b) Kepler’s law of period still holds, (c) Kepler’s law of area and period still holds, (d) neither the law of area nor the law of period still holds, 17. Four equal masses (each of mass M) are placed at the corners, of a square of side a. The escape velocity of a body from the, centre O of the square is, (a), , D, , 11., , Which of the following is correct option?, (a) The time taken in travelling DAB is less than that for, BCD, (b) The time taken in travelling DAB is greater than that, for BCD, (c) The time taken in travelling CDA is less than that for, ABC, (d) The time taken in travelling CDA is greater than that, for ABC, The acceleration due to gravity on the planet A is 9 times, the acceleration due to gravity on planet B. A man jumps to, a height of 2m on the surface of A. What is the height of, jump by the same person on the planet B?, 2, 2, m, m, (a), (b), (c) 18 m, (d) 6 m, 3, 9, , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , If suddenly the gravitational force of attraction between the, earth and a satellite revolving around it becomes zero, then, the satellite will, (a) continue to move in its orbit with same speed, (b) move tangentially to the original orbit with same speed, (c) become stationary in its orbit, (d) move towards the earth, Mass M is divided into two parts xM and (1 – x )M. For a, given separation, the value of x for which the gravitational, attraction between the two pieces becomes maximum is, 3, 1, (a), (b), (c) 1, (d) 2, 5, 2, The potential energy of a satellite, having mass m and, rotating at a height of 6.4 × 106 m from the earth surface, is, (a) – mgRe, (b) – 0.67 mgRe, (c) – 0.5 mgRe, (d) – 0.33 mgRe, If the radius of the earth were to shrink by 1%, with its mass, remaining the same, the acceleration due to gravity on the, earth’s surface would, (a) decrease by 1%, (b) decrease by 2%, (c) increase by 1%, (d) increase by 2%, Suppose the law of gravitational attraction suddenly, , 18., , 19., , 4, , 2GM, a, , (b), , 8 2 GM, a, , (c), , 4GM, a, , (d), , 4 2 GM, a, , If the gravitational force had varied as r –5/2 instead of r–2;, the potential energy of a particle at a distance ‘r’ from the, centre of the earth would be directly proportional to, (a) r -1, (b) r - 2, (c) r -3 / 2, (d) r -5 / 2, A particle of mass ‘m’ is kept at rest at a height 3R from the, surface of earth, where ‘R’ is radius of earth and ‘M’ is mass, of earth. The minimum speed with which it should be, projected, so that it does not return back, is (g is acceleration, due to gravity on the surface of earth), 1, , (a), , 8., 13., 18., Space for Rough Work, , æ GM ö 2, ç, ÷, è R ø, , 1, , æ GM ö 2, (b) ç, ÷, è 2R ø, , 9., 14., 19., , 1, , æ gR ö 2, (c) ç, ÷, è 4 ø, , 10., 15., , 1, , æ 2g ö 2, (d) ç, ÷, è 4 ø, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP07, , P-27, , 20. The ratio between the values of acceleration due to gravity, at a height 1 km above and at a depth of 1 km below the, Earth’s surface is (radius of Earth is R), , R–2, R, R–2, (b), (c), (d) 1, R –1, R –1, R, 21. The weight of an object in the coal mine, sea level and at the, top of the mountain, are respectively W1, W2 and W3 then, (a) W1< W2 > W3, (b) W1= W2 = W3, (c) W1< W2 < W3, (d) W1> W2 > W3, 22. The period of moon's rotation around the earth is nearly 29, days. If moon's mass were 2 fold its present value and all, other things remain unchanged, the period of moon's rotation, would be nearly, , O, , (c), , 1, rad sec –1, 800, 1, 1, rad sec–1, (d) 8 rad sec –1, (c), 80, A body weighs 72 N on the surface of the earth. What is the, gravitational force on it due to earth at a height equal to half, the radius of the earth from the surface?, (a) 32 N (b) 28 N, (c) 16 N, (d) 72 N, A body weighs W newton at the surface of the earth. Its, weight at a height equal to half the radius of the earth, will be, W, 2W, 4W, 8W, (a), (b), (c), (d), 2, 3, 9, 27, A shell of mass M and radius R has a point mass m placed at, a distance r from its centre. The graph of gravitational, potential energy U(r) vs distance r will be, , 25., , 26., , 27., , (a), , O, , 29., , 31., , r, , (b) O, , 20., 25., 30., , 21., 26., 31., , r, , The largest and the shortest distance of the earth from the, sun are r1 and r2. Its distance from the sun when it is at, perpendicular to the major-axis of the orbit drawn from the sun, (a) (r1 + r2)/4, (b) (r1 + r2)/(r1 – r 2), (c) 2r1 r2 /(r1 + r2), (d) (r1 + r2)/3, A planet is moving in an elliptical orbit around the sun. If T,, V, E and L stand respectively for its kinetic energy,, gravitational potential energy, total energy and magnitude, of angular momentum about the centre of force, then which, of the following is correct ?, (a) T is conserved, (b) V is always positive, (c) E is always negative, (d) L is conserved but direction of vector L changes, continuously, The earth is assumed to be sphere of radius R. A platform is, arranged at a height R from the surface of Earth. The escape, velocity of a body from this platform is kv, where v is its, escape velocity from the surface of the earth. The value of k is, 1, 1, 1, (a), (b), (c), (d), 2, 3, 2, 2, m, A solid sphere of mass M and radius R is, h, A, surrounded by a spherical shell of same, mass M and radius 2R as shown. A small, B, particle of mass m is released from rest, R, from a height h [ << R] above the shell., There is a hole in the shell., 2R, What time will it enter the hole at A ?, , (a), , 2, , hR 2, GM, , (b), , 2hR 2, GM, , 3hR 2, hR 2, (d), GM, GM, 32. A body starts from rest from a point distance R0 from the, centre of the earth. The velocity acquired by the body when, it reaches the surface of the earth will be (R represents radius, of the earth)., , (c), , U (r ), , U (r ), , RESPONSE, GRID, , 30., , (b), , r, , O, , U (r ), , U (r ), , 28., , (a), , (a) Zero, , (d), , GMm, –, R, , (a), , (b) 29 / 2 days, 29 2 days, (c) 29 × 2 days, (d) 29 days, 23. The mean radius of earth is R, its angular speed on its own, axis is w and the acceleration due to gravity at earth's surface, is g. What will be the radius of the orbit of a geostationary, satellite ?, (a) (R2g / w2)1/3, (b) (Rg / w2)1/3, 2, 2, 1/3, (c) (R w / g), (d) (R2g / w)1/3, 24. In order to make the effective acceleration due to gravity, equal to zero at the equator, the angular velocity of rotation, of the earth about its axis should be (g = 10 ms–2 and radius, of earth is 64000 km), , r, , (a), , æ1, 1 ö, ÷, 2 G M çç ÷, R, R, 0ø, è, , (b), , æ 1, 1ö, 2 G M çç, - ÷÷, R, R, ø, è 0, , (c), , æ1, 1 ö, ÷, G M çç ÷, R, R, 0ø, è, , (d), , æ1, 1 ö, ÷, 2 G M çç ÷, è R R0 ø, , 22., 27., 32., Space for Rough Work, , 23., 28., , 24., 29.
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t.me/Magazines4all, , DPP/ CP07, , P-28, , 33., , A satellite of mass M is moving in a circle of radius R under a, centripetal force given by (–k/R2), where k is a constant. Then, k, (a) The kinetic energy of the particle is, R, 12, (b) The total energy of the particle is æç - k ö÷, 2R, è, , ø, , 39., , 40., , æ k ö, , (c), (a), (b), gR, 2.5gR (d), 7.1gR, R, 37. A satellite revolves around the earth of radius R in a circular, orbit of radius 3R. The percentage increase in energy, required to lift it to an orbit of radius 5R is, (a) 10 %, (b) 20 %, (c) 30 %, (d) 40 %, 38. A (nonrotating) star collapses onto itself from an initial radius, Ri with its mass remaining unchanged. Which curve, a, in figur e best gives th e, gravitational acceleration ag on, d, the surface of the star as a, b, function of the radius of the, a, star during the collapse, g, , (a) a, , (b) b, , 33., 38., 43., , RESPONSE, GRID, , (c) c, , (d) d, , 34., 39., 44., , Ri, , 2 pMR 2, 4 pMR 2, pMR 3, MR 2 p, (b), (c), (d), 5T, 5T, T, T, A satellite is launched into a circular orbit of radius R around, the earth. A second satellite is launched into an orbit of, radius 1.01 R. The period of second satellite is larger than, the first one by approximately, (a) 0.5%, (b) 1.0%, (c) 1.5%, (d) 3.0%, A uniform spherical shell gradually shrinks maintaining, its shape. The gravitational potential at the centre, (a) increases, (b) decreases, (c) remains constant, (d) cannot say, The depth d at which the value of acceleration due to gravity, 1, becomes times the value at the surface of the earth, is, n, [R = radius of the earth], , (a), , æ kö, (c) The kinetic energy of the particle is ç - ÷, è Rø, , (d) The potential energy of the particle is ç ÷, è 2R ø, 34. 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, earth. When both ‘d’ and ‘h’ are much smaller than the radius, of earth, then which one of the following is correct?, h, 3h, (a) d =, (b) d =, (c) d = h, (d) d =2 h, 2, 2, 35. Two identical geostationary satellites are moving with equal, speeds in the same orbit but their sense of rotation brings, them on a collision course. The debris will, (a) fall down, (b) move up, (c) begin to move from east to west in the same orbit, (d) begin to move from west to east in the same orbit, 36. A diametrical tunnel is dug across the Earth. A ball is dropped, into the tunnel from one side. The velocity of the ball when, it reaches the centre of the Earth is .... (Given : gravitational, 3 GM, potential at the centre of Earth = –, ), 2 R, , If the earth is treated as a sphere of radius R and mass M;, its angular momentum about the axis of its rotation with, period T, is, , 41., , 42., , æ n -1 ö, R, R, æ n ö, (b) R ç, (c), (d) R ç, ÷, ÷, 2, n, è n ø, è n +1ø, n, Radius of moon is 1/4 times that of earth and mass is 1/81, times that of earth. The point at which gravitational field, due to earth becomes equal and opposite to that of moon, is, (Distance between centres of earth and moon is 60R, where, R is radius of earth), (a) 5.75 R from centre of moon, (b) 16 R from surface of moon, (c) 53 R from centre of earth, (d) 54 R from centre of earth, If earth is supposed to be a sphere of radius R, if g30 is, value of acceleration due to gravity at lattitude of 30° and g, at the equator, the value of g – g30 is, 3 2, 1 2, (a) 1 w2 R (b), (d), w R (c) w 2 R, w R, 4, 2, 4, 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?, , (a), 43., , 44., , 45., , (a), , R, , 35., 40., 45., , 5GmM, (b), 6R, , 36., 41., , 2GmM, (c), 3R, , GmM, (d), 2R, , 37., 42., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP07 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score –Space, Qualifying, for RoughScore, Work, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , GmM, 2R, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP08, , SYLLABUS : Mechanical Properties of Solids, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , Two wires A and B are of the same material. Their lengths, are in the ratio 1 : 2 and the diameter are in the ratio 2 : 1. If, they are pulled by the same force, then increase in length, will be in the ratio, (a) 2 : 1, (b) 1 : 4, (c) 1 : 8, (d) 8 : 1, The load versus elongation graphs for four wires of same, length and made of the same material are shown in the figure., The thinnest wire is represented by the line, Load, , (a) OA, , 5., , D, C, B, , (b) OC, , 6., , A, , (c) OD, 3., , 4., , O Elongation, (d) OB, A spring of force constant 800 N/m has an extension of 5, cm. The work done in extending it from 5 cm to 15 cm is, (a) 16 J, (b) 8 J, (c) 32 J, (d) 24 J, A metal wire of length L1 and area of cross-section A is, attached to a rigid support. Another metal wire of length L2, and of the same cross-sectional area is attached to the free, end of the first wire. A body of mass M is then suspended, from the free end of the second wire. If Y1 and Y2 are the, Young’s moduli of the wires respectively, the effective force, constant of the system of two wires is, , RESPONSE, GRID, , 1., 6., , 2., 7., , 7., , (a), , (Y1Y2 ) A, 2(Y1 L2 + Y2 L1 ), , (b), , (c), , (Y1Y2 ) A, Y1L2 + Y2 L1, , (d), , (Y1Y2 ) A, ( L1 L2 )1/2, , (Y1Y2 )1/2 A, , ( L2 L1 )1/2, The approximate depth of an ocean is 2700 m. The, compressibility of water is 45.4 × 10–11 Pa–1 and density of, water is 103 kg/m3.What fractional compression of water, will be obtained at the bottom of the ocean ?, (a) 1.0 × 10–2, (b) 1.2 × 10–2, –2, (c) 1.4 × 10, (d) 0.8 × 10–2, The Young's modulus of steel is twice that of brass. Two, wires of same length and of same area of cross section, one, of steel and another of brass are suspended from the same, roof. If we want the lower ends of the wires to be at the same, level, then the weights added to the steel and brass wires, must be in the ratio of :, (a) 2 : 1, (b) 4 : 1, (c) 1 : 1, (d) 1 : 2, Choose the wrong statement., (a) The bulk modulus for solids is much larger than for, liquids., (b) Gases are least compressible., (c) The incompressibility of the solids is due to the tight, coupling between neighbouring atoms., (d) The reciprocal of the bulk modulus is called, compressibility., , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP08, , P-30, , 8., , A copper wire of length 1.0 m and a steel wire of length, 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) 1.75 mm (b) 2.0 mm (c) 1.50 mm (d) 1.25 mm, 9., A cube at temperature 0ºC is compressed equally from all, sides by an external pressure P. By what amount should its, temperature be raised to bring it back to the size it had before, the external pressure was applied. The bulk modulus of the, material of the cube is B and the coefficient of linear, expansion is a., (a) P/B a (b) P/3 B a (c) 3 p a/B (d) 3 B/P, 10. The diagram below shows the change in the length X of a, thin uniform wire caused by the application of stress F at, two different temperatures T1 and T2. The variation shown, suggests that, (a) T1 > T2, , F, , (a), , 16., , T1, , (c) T2 > T1, , 17., , X, (d) T1 ³ T2, If the ratio of lengths, radii and Young's moduli 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), (b), (c), (d), , 3c, 2ab2, , Steel, , 2a 2 c, b, , M, , 2M, , 2b2c, 2ac, , b2, 12. The Young’s modulus of brass and steel are respectively, 1010 N/m2. and 2 × 1010 N/m2. A brass wire and a steel wire, of the same length are extended by 1 mm under the same, force, the radii of brass and steel wires are RB and RS, respectively. Then, (a), , 18., , Brass, , 3a, , RS = 2 R B, , (b), , 19., , 20., , 8., 13., 18., , 9., 14., 19., , (0.8) 4, , (b), , (0.8) 2, , (8.02) 2 - ( 7.98) 2, (0.8) 2, , (0.8) 2, , (d), (8.02) 4 - (7.98) 4, (8.02) 3 - (7.98) 2, Two wires are made of the same material and have the 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 Dx on applying force F, how much force is needed to, stretch wire 2 by the same amount?, (a) 4 F, (b) 6 F, (c) 9 F, (d) F, In materials like aluminium and copper, the correct order of, magnitude of various elastic modului is:, (a) Young’s modulus < shear modulus < bulk modulus., (b) Bulk modulus < shear modulus < Young’s modulus, (c) Shear modulus < Young’s modulus < bulk modulus., (d) Bulk modulus < Young’s modulus < shear modulus., What per cent of length of wire increases by applying a, stress of 1 kg weight/mm2 on it?, (Y = 1 × 1011 N/m2 and 1 kg weight = 9.8 newton), (a) 0.0067%, (b) 0.0098%, (c) 0.0088%, (d) 0.0078%, An elastic string of unstretched length L and force constant, k is stretched by a small length x. It is further stretched by, another small length y. The work done in the second, stretching is :, (a), , 1 2, ky, 2, , (b), , 1, k ( x2 + y 2 ), 2, , (c), , 1, k ( x + y)2, 2, , (d), , 1, ky (2 x + y ), 2, , Two, spring P and Q of force constants k p and, kp ö, æ, kQ ç kQ =, 2 ÷ø are stretched by applying forces of equal, è, , RS = R B / 2, , (c) R S = 4R B, (d) R S = R B / 4, 13. Steel ruptures when a shear of 3.5 × 108 N m–2 is applied., The force needed to punch a 1 cm diameter hole in a steel, sheet 0.3 cm thick is nearly:, , RESPONSE, GRID, , (8.02) 4 - (7.98) 4, , (c), , T2, , (b) T1 < T2, , 11., , (a) 1.4 × 104 N, (b) 2.7 × 104 N, 4, (c) 3.3 × 10 N, (d) 1.1 × 104 N, 14. A ball falling in a lake of depth 400 m has a decrease of 0.2%, in its volume at the bottom. The bulk modulus of the material, of the ball is (in N m–2), (a) 9.8 × 109, (b) 9.8 × 1010, 10, (c) 1.96 × 10, (d) 1.96 × 109, 15. A circular tube of mean radius 8 cm and thickness 0.04 cm is, melted up and recast into a solid rod of the same length. The, ratio of the torsional rigidities of the circular tube and the, solid rod is, , magnitude. If the energy stored in Q is E, then the energy, stored in P is, (a) E, (b) 2 E, (c) E/2, (d) E/4, , 10., 15., 20., Space for Rough Work, , 11., 16., , 12., 17., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP08, , P-31, , 21. 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 100ºC is:, , 29., , (For steel Young’s modulus is 2 ´ 1011 Nm -2 and coefficient, of thermal expansion is 1.1 ´ 10-5 K -1 ), , 22., , 23., , 24., , 25., , 26., , 2.2 ´ 108 Pa, , (b), , 30., , 2.2 ´ 10 9 Pa, , (c) 2.2 ´ 10 7 Pa, (d) 2.2 ´ 10 6 Pa, A steel ring of radius r and cross sectional area A is fitted, onto a wooden disc of radius R (R > r). If the Young’s modulus, of steel is Y, then the force with which the steel ring is, expanded is, (a) A Y (R/r), (b) A Y (R – r)/r, (c) (Y/A)[(R – r)/r], (d) Y r/A R, Two wires A and B of same material and of equal length with, the radii in the ratio 1 : 2 are subjected to identical loads. If, the length of A increases by 8 mm, then the increase in, length of B is, (a) 2 mm (b) 4 mm (c) 8 mm (d) 16 mm, A material has 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, The upper end of a wire of diameter 12mm and length 1m is, clamped and its other end is twisted through an angle of, 30°. The angle of shear is, (a) 18°, (b) 0.18°, (c) 36°, (d) 0.36°, The pressure on an object of bulk modulus B undergoing, hydraulic compression due to a stress exerted by, 2, æ DV ö, surrounding fluid having volume strain ç, is, è V ÷ø, , æ DV ö, (a) B2 ç, è V ÷ø, 1 æ DV ö, (c) B ç, è V ÷ø, , æ DV ö, (b) B ç, è V ÷ø, æ DV ö, (d) B ç, è V ÷ø, , RESPONSE, GRID, , 21., 26., 31., , (c) 1/Y, , (b), , l 1T2 + l 2 T1, T1 + T2, , (c), , l 1T2 - l 2 T1, T2 - T1, , (d), , T1T2 l 1l 2, , 31., , For the same cross-sectional area and for a given load, the, ratio of depressions for the beam of a square cross-section, and circular cross-section is, (a) 3 : p, (b) p : 3, (c) 1 : p, (d) p : 1, 32. The bulk moduli of ethanol, mercury and water are given as, 0.9, 25 and 2.2 respectively in units of 109 Nm–2. For a given, value of pressure, the fractional compression in volume is, DV, DV, . Which of the following statements about, for, V, V, these three liquids is correct ?, (a) Ethanol > Water > Mercury, (b) Water > Ethanol > Mercury, (c) Mercury > Ethanol > Water, (d) Ethanol > Mercury > Water, 33. The graph given is a stress-strain curve for, 1.0, 0.5, 0, , d, , (b) Y, , l1 + l 2, 2, , 2, , 27. A structural steel rod has a radius of 10 mm and length of, 1.0 m. A 100 kN force stretches it along its length. Young’s, modulus of structural steel is 2 × 1011 Nm–2. The percentage, strain is about, (a) 0.16% (b) 0.32% (c) 0.08% (d) 0.24%, 28. A beam of metal supported at the two edges is loaded at the, centre. The depression at the centre is proportional to, , (a) Y 2, , (a), , Stress (N/ m2), , (a), , When a 4 kg mass is hung vertically on a light spring that, obeys Hooke’s law, the spring stretches by 2 cms. The work, required to be done by an external agent in stretching this, spring by 5 cms will be (g = 9.8 m/sec2), (a) 4.900 joule, (b) 2.450 joule, (c) 0.495 joule, (d) 0.245 joule, The length of a metal is l1 when the tension in it is T1 and is, l2 when the tension is T2. The original length of the wire is, , (d) 1/Y 2, , 22., 27., 32., , 0.5, , 1.0, Strain, , (a) elastic objects, (b) plastics, (c) elastomers, (d) None of these, 34. A metal rod of Young's modulus 2 × 1010 N m–2 undergoes, an elastic strain of 0.06%. The energy per unit volume stored, in J m–3 is, (a) 3600, (b) 7200, (c) 10800 (d) 14400, 35. Two wires of the same material and same length but diameters, in the ratio 1 : 2 are stretched by the same force. The potential, energy per unit volume of the two wires will be in the ratio, (a) 1 : 2, (b) 4 : 1, (c) 2 : 1, (d) 16 : 1, , 23., 28., 33., Space for Rough Work, , 24., 29., 34., , 25., 30., 35.
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t.me/Magazines4all, , DPP/ CP08, , P-32, , The length of an elastic string is a metre when the, longitudinal tension is 4 N and b metre when the longitudinal, tension is 5 N. The length of the string in metre when the, longitudinal tension is 9 N is, (a) a – b, , 1, 4, , (b) 5b – 4a (c) 2b – a (d) 4a – 3b, , A force of 103 newton, stretches the length of a hanging, wire by 1 millimetre. The force required to stretch a wire of, same material and length but having four times the diameter, by 1 millimetre is, (a) 4 × 103 N, (b) 16 × 103 N, 1, 1, ´ 103 N, ´ 103 N, (c), (d), 16, 4, 38. A steel wire of length l and cross sectional area A is stretched, by 1 cm under a given load. When the same load is applied, to another steel wire of double its length and half of its, cross section area, the amount of stretching (extension) is, (a) 0.5 cm (b) 2 cm, (c) 4 cm, (d) 1.5 cm, 39. The adjacent graph shows the extension (D l) of a wire of, length 1 m suspended from the top of a roof at one end with, a load W connected to the other end. If the cross-sectional, area of the wire is 10–6 m2, calculate the Young’s modulus, of the material of the wire :, 37., , Dl(x10 )m, -4, , 4, 3, 2, , 41., , (a) 19.6 ´ 10 -8 N / m 2, , I., , (a) 2 × 1011 N/m2, , (b) 2 × 10–11 N/m2, , (c) 3 × 10–12 N/m2, (d) 2 × 10–13 N/m2, If a rubber ball is taken at the depth of 200 m in a pool, its, volume decreases by 0.1%. If the density of the water is, 1 × 103 kg/m3 and g = 10m/s2, then the volume elasticity in, N/m2 will be, (a) 108, (b) 2 × 108 (c) 109, (d) 2 × 109, , 36., 41., , RESPONSE, GRID, , It will be easier to compress, this rubber than expand it, , 37., 42., , Force, , II., , Rubber does not return to its original length after it is, stretched, III. The rubber band will get heated if it is stretched and, released, Which of these can be deduced from the graph:, (a) III only (b) II and III (c) I and III (d) I only, 43. The Poisson’s ratio of a material is 0.5. If a force is applied to a, wire of this material, there is a decrease in the cross-sectional, area by 4%. The percentage increase in the length is:, (a) 1%, (b) 2%, (c) 2.5%, (d) 4%, 44. Copper of fixed volume ‘V; is drawn into wire of length ‘l’., When this wire is subjected to a constant force ‘F’, the, extension produced in the wire is ‘Dl’. Which of the following, graphs is a straight line?, (a) Dl versus, , 20 40 60 80 W(N), , (b) 19.6 ´ 1010 N / m 2, , (d) 19.6 ´108 N / m 2, (c) 19.6 ´10 -10 N / m 2, 42. The diagram shows a forceextension graph for a rubber, band. Consider the following, statements :, , 1, , 40., , A ball is falling in a lake of depth 200 m creates a decrease, 0.1 % in its volume at the bottom. The bulk modulus of the, material of the ball will be, , Extension, , 36., , 1, l, , (b) Dl versus l2, , 1, (d) Dl versus l, (c) Dl versus 2, l, 45. When a 4 kg mass is hung vertically on a light spring that, obeys Hooke’s law, the spring stretches by 2 cms. The work, required to be done by an external agent in stretching this, spring by 5 cm will be (g = 9.8 m/sec2), (a) 4.900 joule, (b) 2.450 joule, (c) 0.495 joule, (d) 0.245 joule, , 38., 43., , 39., 44., , 40., 45., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP08 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP09, , SYLLABUS : Mechanical Properties of Fluids, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , The density of water at the surface of ocean is r. If the bulk, modulus of water is B, what is the density of ocean water at, a depth where the pressure is nP 0 , where P 0 is the, atmospheric pressure ?, (a), , rB, B - ( n - 1) P0, , (b), , rB, B + ( n - 1) P0, , (c), , rB, B - nP0, , (d), , rB, B + nP0, , 4., , A ball of radius r and density r falls freely under, gravity through a distance h before entering, water. Velocity of ball does not change even on, entering water. If viscosity of water is h the value, of h is given by, 2 2 æ 1- r ö, r ç, (a), ÷g, 9 è h ø, , (b), , 2 2 æ r -1 ö, r ç, ÷g, 81 è h ø, , (d), , 2 4 æ r -1 ö, r ç, ÷ g, 9 è h ø, , 2 4 æ r -1 ö, r ç, ÷ g, 81 è h ø, , RESPONSE GRID, , 1., , h, , 5., , 2, , 2, , (c), , 3., , 2., , Two parallel glass plates are dipped partly in the liquid of, density 'd' keeping them vertical . If the distance between, the plates is 'x', surface tension for liquids is T and angle of, contact is q, then rise of liquid between the plates due to, capillary will be, 2T cos q, 2T, T cos q, T cos q, (a), (b), (c), (d), xdg, xdg cos q, xdg, xd, A liquid is allowed to flow into a tube of truncated cone, shape. Identify the correct statement from the following, (a) The speed is high at the wider end and high at the, narrow end, (b) The speed is low at the wider end and high at the, narrow end, (c) The speed is same at both ends in a streamline flow, (d) The liquid flows with uniform velocity in the tube, A wide vessel with a small hole at the bottom is filled with, water (density r1, height h1) and kerosene (density r2, height, h2). Neglecting viscosity effects, the speed with which water, flows out is :, (a) [2g(h1 + h2)]1/2, (b) [2g(h1r1 + h2r2)]1/2, (c) [2g(h1 + h2(r2/r1))]1/2 (d) [2g (h1 + h2(r1/r2))]1/2, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP09, , P-34, , 6., , 7., , A capillary tube of radius r is immersed vertically in a liquid, such that liquid rises in it to height h (less than the length of, the tube). Mass of liquid in the capillary tube is m. If radius, of the capillary tube is increased by 50%, then mass of liquid, that will rise in the tube, is, 3, 2, 4, 9, m, m, m, m, (a), (b), (c), (d), 2, 3, 9, 4, A lead shot of 1 mm diameter falls through a long column of, glycerine. The variation of its velocity v with distance, covered is represented by, v, , v, , (a), , (b), , Distance, covered, , v, , v, , (c), 8., , 9., , 10., , 11., , 12., , Distance, covered, , (d), , Distance, covered, , Distance, covered, , Two mercury drops (each of radius ‘r’) merge to form bigger, drop. The surface energy of the bigger drop, if T is the, surface tension, is :, (a) 4pr 2T, (b) 2pr 2T, (c) 28/ 3 pr 2T, (d) 25/ 3 pr 2T, Wax is coated on the inner wall of a capillary tube and the, tube is then dipped in water. Then, compared to the unwaxed, capillary, the angle of contact q and the height h upto which, water rises change. These changes are :, (a) q increases and h also increases, (b) q decreases and h also decreases, (c) q increases and h decreases, (d) q decreases and h increases, A rain drop of radius 0.3 mm has a terminal velocity in air =, 1 m/s. The viscosity of air is 8 × 10–5 poise. The viscous, force on it is, (a) 45.2 × 10–4 dyne, (b) 101.73×10–5 dyne, –4, (c) 16.95 × 10 dyne, (d) 16.95 × 10–5 dyne, A water tank of height 10m, completely filled with water is, placed on a level ground. It has two holes one at 3 m and the, other at 7 m from its base. The water ejecting from, (a) both the holes will fall at the same spot, (b) upper hole will fall farther than that from the lower hole, (c) upper hole will fall closer than that from the lower hole, (d) more information is required, Two capillary of length L and 2L and of radius R and 2R are, connected in series. The net rate of flow of fluid through, them will be (given rate to the flow through single capillary,, X=, , 7, 9, 8, 5, X, (b), (c), (d), (a), X, X, X, 5, 8, 9, 7, 13. A candle of diameter d is floating on a liquid in a cylindrical, container of diameter D (D >> d) as shown in figure. If it is, burning at the rate of 2 cm/hour then the top of the candle, will, L, (a) remain at the same height, (b) fall at the rate of 1 cm/hour, L, (c) fall at the rate of 2 cm/hour, d, D, (d) go up at the rate of 1 cm/hour, 14. An isolated and charged spherical soap bubble has a radius, r and the pressure inside is atmospheric. T is the surface, , tension of soap solution. If charge on drop is X pr 2rTe 0, then find the value of X., (a) 8, (b) 9, (c) 7, (d) 2, 15. A thread is tied slightly loose to a wire frame as in figure and, the frame is dipped into a soap solution and taken out. The, frame is completely covered with the film. When the portion, A is punctured with a pin, the thread, (a) becomes concave towards A, Frame, A, (b) becomes convex towards A, B, (c) remains in the initial position, Thread, (d) either (a) or (b) depending, on the size of A w.r. t. B, 16. Which of the following expressions represents the excess, of pressure inside the soap bubble?, (a) Pi – Po = s, (b) Pi – Po = 2s, r, r, 2s, 4s, (c) Pi – Po =, + hrg, (d) Pi – Po =, r, r, 17. A spherical solid ball of volume V is made of a material of, density r1. It is falling through a liquid of density, r1 (r2< r1). Assume that the liquid applies a viscous force on, the ball that is proportional to the square of its speed, v, i.e., Fviscous = –kv2 (k > 0). The terminal speed of the ball, is, (a), , pPR 4, ), 8hL, , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , (b), , Vg r1, k, , (d), , Vg r1, k, , Vg (r1 – r2 ), k, Select the correct statements from the following., (a) Bunsen burner and sprayers work on Bernoulli's, principle, (b) Blood flow in arteries is explained by Bernoulli's, principle, (c) A siphon works on account of atmospheric pressure., (d) All are correct, , (c), , 18., , Vg (r1 – r2 ), k, , 8., 13., 18., Space for Rough Work, , 9., 14., , 10., 15., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP09, , P-35, , 19. The wetability of a surface by a liquid depends primarily on, (a) surface tension, (b) density, (c) angle of contact between the surface and the liquid, (d) viscosity, 20. The relative velocity of two parallel layers of water is, 8 cm/sec. If the perpendicular distance between the layers, is 0.1 cm, then velocity gradient will be, (a) 80/sec (b) 60 /sec (c) 50/sec, (d) 40/sec, 21. Choose the correct statement, (a) Terminal velocities of rain drops are proportional to, square of their radii, (b) Water proof agents decrease the angle of contact, between water and fibres, (c) Detergents increase the surface tension of water, (d) Hydraulic machines work on the principle of Torricelli's, law, 22. When a ball is released from rest in a very long column of, viscous liquid, its downward acceleration is ‘a’ (just after, release). Its acceleration when it has acquired two third of, the maximum velocity is a/X. Find the value of X., (a) 2, (b) 3, (c) 4, (d) 5, 23. A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm, external diameter. It is supported horizontally from the pan, of a balance, so that it comes in contact with the water in a, glass vessel. If an extra 3.97. If is required to pull it away, from water, the surface tension of water is, (a) 72 dyne cm–1, (b) 70.80 dyne cm–1, (c) 63.35 dyne cm–1, (d) 60 dyne cm–1, 24. Which of the following graph represents the variation of, surface tension with temperature over small temperature, ranges for water?, (a), , one -half that of a freely falling body ? (The densities of, metal and of liquid are r and s respectively, and the viscosity, of the liquid is h)., (a), , (a) = H, , 30., , (d) S.T., , S.T., , T, , T, , 25. When a large air bubble rises from the bottom of a lake to, the surface, its radius doubles. If the atmospheric pressure, is equal to that of a column of water of height H, then depth, of the lake is, (a) H, (b) 2H, (c) 7H, (d) 8H, 26. What is the velocity v of a metallic ball of radius r falling in, a tank of liquid at the instant when its acceleration is, , RESPONSE, GRID, , 19., 24., 29., , 20., 25., 30., , H, , (b) > H, , 31., (c), , r2 g, (2r - s), 9h, , r2g, 2r 2 g, (r - s), ( r - s), (d), 9h, 9h, 27. Two pieces of metals are suspended from the arms of a, balance and are found to be in equilibrium when kept, immersed in water. The mass of one piece is 32 g and its, density 8 g cm–3. The density of the other is 5 g per cm3., Then the mass of the other is, (a) 28 g, (b) 35 g, (c) 21 g, (d) 33.6 g, 28. A block of material of specific gravity 0.4 is held submerged, at a depth of 1m in a vessel filled with water. The vessel is, accelerated upwards with acceleration of ao = g/5. If the, block is released at t = 0, neglecting viscous effects, it will, reach the water surface at t equal to (g = 10 m/s2) :, (a) 0.60 s (b) 0.33 s, (c) 3.3 s, (d) 1.2 s, 29. Figure shows a capillary rise H. If the air is blown through, the horizontal tube in the direction as shown then rise in, capillary tube will be, , T, , T, , (b), , (c), , (b) S.T., , S.T., , r2 g, (r - 2s), 9h, , 32., , (c) < H, (d) zero, A small spherical ball falling through a viscous medium of, negligible density has terminal velocity v. Another ball of, the same mass but of radius twice that of the earlier falling, through the same viscous medium will have terminal velocity, (a) v, (b) v/4, (c) v/2, (d) 2v, Two non-mixing liquids of densities r and n r, (n > 1) are put in a container. The height of each liquid is h., A solid cylinder of length L and density d is put in this, container. The cylinder floats with its axis vertical and length, pL(p < 1) in the denser liquid. The density d is equal to :, (a) {1 + (n + 1)p}r, (b) {2 + (n + 1)p}r, (c) {2 + (n – 1)p}r, (d) {1 + (n – 1)p}r, A thin liquid film formed between a U-shaped, wire and a light slider supports a weight of, Film, 1.5 × 10–2 N (see figure). The length of the, slider is 30 cm and its weight negligible. The, surface tension of the liquid film is, (a) 0.0125 Nm–1, (b) 0.1 Nm–1, W, (c) 0.05 Nm–1, (d) 0.025 Nm–1, , 21., 26., 31., Space for Rough Work, , 22., 27., 32., , 23., 28.
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t.me/Magazines4all, , DPP/ CP09, , P-36, , 33., , Two liquids of densities d1 and d2 are flowing in identical, capillary tubes uder the same pressure difference. If t1 and, t2 are time taken for the flow of equal quantities (mass) of, liquids, then the ratio of coefficient of viscosity of liquids, must be, t, d2 t2, d1 t 1, (d), (c), (b) 1, t, d1 t 1, d2t2, 2, Let T1 be surface tension between T, liquid air, solid and air, T2 be the surface, q, tension between solid and liquid and, T2, Solid, T be the surface tension between T1, liquid and air. Then in equilibrium, for a drop of liquid on, a clean glass plate, the correct relation is (q is angle of, contact), , (a), , 34., , 35., , 36., , 37., , 38., , 39., , 40., , d1 t 1, d2t 2, , (a) cos q =, , T, T1 + T2, , (b) cos q =, , T, T1 - T2, , (c) cos q =, , T1 + T2, T, , (d) cos q =, , T1 - T2, T, , A uniform rod of density r is placed in a wide tank containing, a liquid of density r0(r0 > r). The depth of liquid in the, tank is half the length of the rod. The rod is in equilibrium,, with its lower end resting on the bottom of the tank. In this, position the rod makes an angle q with the horizontal, 1 r0, 1, r0 / r, (a) sin q =, (b) sin q = ., 2 r, 2, (c) sin q = r / r0, (d) sin q = r0 / r, A spherical ball of iron of radius 2 mm is falling through a, column of glycerine. If densities of glycerine and iron are, respectively 1.3 × 103 kg/m3 and 8 × 103 kg/m3. h for glycerine, = 0.83 Nm–2 sec, then the terminal velocity is, (a) 0.7 m/s (b) 0.07 m/s (c) 0.007 m/s (d) 0.0007 m/s, A water film is formed between two straight parallel wires of, 10 cm length 0.5 cm apart. If the distance between wires is, increased by 1 mm. What will be the work done ?, (surface tension of water = 72 dyne/cm), (a) 36 erg (b) 288 erg (c) 144 erg (d) 72 erg, A waterproofing agent changes the angle of contact, (a) from obtuse to acute., (b) from acute to obtuse., (c) from obtuse to p/2., (d) from acute to p/ 2., A thin metal disc of radius r floats on water surface and, bends the surface downwards along the perimeter making, an angle q with vertical edge of the disc. If the disc displaces, , 33., 38., 43., , RESPONSE, GRID, , 34., 39., 44., , 41., , a weight of water W and surface tension of water is T, then, the weight of metal disc is:, (a) 2 prT + W, (b) 2 prT cos q – W, (c) 2 prT cos q + W, (d) W – 2 prT cos q, A tank has a small hole at its botom of area of cross-section, a. Liquid is being poured in the tank at the rate Vm3/s,, the maximum level of liquid in the container will be (Area of, tank = A), V2, V2, V, V, (a), (b), (c), (d), 2gAa, gAa, 2gaA, gaA, A jar is filled with two non-mixing liquids 1 and, 2 having densities r1 and, r2 respectively. A, r, solid ball, made of a material of density r3 , is, dropped in the jar. It comes to equilibrium in, the position shown in the figure.Which of the following is, true for r1, r2and r3?, (a) r3 < r1 < r2, (b) r1 > r3 > r2, (c) r1 < r2 < r3, (d) r1 < r3 < r2, 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 with the bottom of the vessel., R, If r << R and the surface tension of water, is T, value of r just before bubbles, 2r, detach is: (density of water is rw), r1, , 3, , 42., , r g, 2rw g, 3r w g, r g, (b) R 2 w (c) R 2 w (d) R 2, 6T, 3T, T, T, The lift of an air plane is based on, (a) Torricelli's theorem, (b) Bernoulli's theorem, (c) Law of gravitation, (d) conservation of linear momentum, The cylindrical tube of a spray pump has radius, R, one end, of which has n fine holes, each of radius r. If the speed of the, liquid in the tube is V, the speed of the ejection of the liquid, through the holes is :, , (a) R 2, , 43., , 44., , 45., , VR 2, , VR 2, V 2R, (d), nr, nr 2, n3r 2, n2r2, Drops of liquid of density r are floating half immersed in a, liquid of density s. If the surface tension of liquid is T, the, radius of the drop will be, , (a), , (a), , 35., 40., 45., , (b), , 3T, (b), g (3r - s), , 36., 41., , VR 2, , (c), , 6T, (c), g (2r - s ), , 3T, 3T, (d), g (4r - 3s), g (2r - s ), , 37., 42., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP09 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, Space, for, Rough Work Score, 45, Qualifying, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP10, , P-37, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP10, , SYLLABUS : Thermal Properties of Matter, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , The total radiant energy per unit area, normal to the direction, of incidence, received at a distance R from the centre of a, star of radius r, whose outer surface radiates as a black, body at a temperature T K is given by: (s is Stefan's, constant), (a), (c), , 2., , s r 2T 4, , (b), , R2, s r 4T 4, , (d), , s r 2T 4, , 4., , 4p r 2, 4 p s r 2T 4, , r4, R2, Three rods of same dimensions are arranged as shown in, figure, have thermal conductivities K1, K2 and K3 . The points, P and Q are maintained at different temeratures for the heat, to flow at the same rate along PRQ and PQ. Then which of, the following option is correct?, , RESPONSE GRID, , 1., , 5., , K1, , P, , 3, , K2, , K3, , 2., , The sprinkling of water slightly reduces the temperature, of a closed room because, (a) temperature of water is less than that of the room, (b) specific heat of water is high, (c) water has large latent heat of vaporisation, (d) water is a bad conductor of heat, The value of molar heat capacity at constant temperature is, (a) zero, (b) infinity, (c) unity, (d) 4.2, The specific heat capacity of a metal at low temperature (T), is given as, æ T ö, C p (kJK -1kg -1 ) = 32 ç, è 400 ÷ø, A 100 gram vessel of this metal is to be cooled from 20ºK to, 4ºK by a special refrigerator operating at room temperature, (27°C). The amount of work required to cool the vessel is, (a) greater than 0.148 kJ, (b) between 0.148 kJ and 0.028 kJ, (c) less than 0.028 kJ, (d) equal to 0.002 kJ, , R, , 1, (a) K 3 = ( K1 + K 2 ), 2, (b) K3 = K1 + K2, , K1K 2, (c) K3 =, K1 + K2, (d) K3 = –2(K1 + K2), , 3., , Q, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP10, , P-38, , 7., , 8., , 9., , 10., , The emissive power of a black body at T = 300K is 100 Watt/, m2. Consider a body B of area A = 10m2 coefficient of, reflectivity r = 0.3 and coefficient of transmission t = 0.5. Its, temperature is 300K. Then which of the following is incorrect?, (a) The emissive power of B is 20 W/m2, (b) The emissive power of B is 200 W/m2, (c) The power emitted by B is 200 Watts, (d) The emissivity of B is 0.2, A solid cube and a solid sphere of the same material have, equal surface area. Both are at the same temperture 120°C ,, then, (a) both the cube and the sphere cool down at the same rate, (b) the cube cools down faster than the sphere, (c) the sphere cools down faster than the cube, (d) whichever is having more mass will cool down faster, The density of water at 20°C is 998 kg/m3 and at 40°C 992, kg/m3. The coefficient of volume expansion of water is, (a) 10–4/°C, (b) 3 × 10–4/°C, (c) 2 × 10–4/°C, (d) 6 × 10–4/°C, A metallic rod l cm long, A square cm in cross-section is, heated through tºC. If Young’s modulus of elasticity of the, metal is E and the mean coefficient of linear expansion is a per, degree celsius, then the compressional force required to, prevent the rod from expanding along its length is, (a) E A a t, (b) E A a t/(1 + at), (c) E A a t/(1 – a t), (d) E l a t, If liquefied oxygen at 1 atmospheric pressure is heated from, 50 K to 300 K by supplying heat at constant rate. The graph, of temperature vs time will be, , 13., , 14., , 15., , (b), , 16., , 17., , 18., , T, t, , t, , (c), , (d), , T, , T, , t, , 11., , 12., , 19., , t, , If a bar is made of copper whose coefficient of linear, expansion is one and a half times that of iron, the ratio of, force developed in the copper bar to the iron bar of identical, lengths and cross-sections, when heated through the same, temperature range (Young’s modulus of copper may be taken, to be equal to that of iron) is, (a) 3/2, (b) 2/3, (c) 9/4, (d) 4/9, A piece of ice falls from a height h so that it melts completely., Only one-quarter of the heat produced is absorbed by the, ice and all energy of ice gets converted into heat during its, fall. The value of h is :, [Latent heat of ice is 3.4 × 105 J/kg and g = 10 N/kg], (a) 34 km (b) 544 km (c) 136 km (d) 68 km, , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , r1 r2, (r2 - r1 ), C, B, A block of steel heated to 100ºC is left, in a room to cool. Which of the curves, A, shown in fig., represents the correct, Time, behaviour?, (a) A, (b) B, (c) C, (d) None of these, Which of the following will expand the most for same rise, in temperature?, (a) Aluminium, (b) Glass, (c) Wood, (d) All will expand same, The plots of intensity versus, T3, wavelength for three black bodies, T2, at temperatures T 1 , T 2 and T 3, I T1, respectively are as shown. Their, temperature are such that, l, (a) T1 > T2 > T3, (b) T1 > T3 > T2, (c) T2 > T3 > T1, (d) T3 > T2 > T1, When the temperature of a rod increases from t to t + Dt, its, moment of inertia increases from I to I + DI. If a be the, coefficient of linear expansion of the rod, then the value of, DI, is, I, aDt, Dt, (a) 2aDt, (b) aDT, (c), (d), 2, a, Two rods, one of aluminum and the other made of steel,, having initial length l1 and l2 are connected together to, form a single rod of length l1 + l2. The coefficients of linear, expansion for aluminum and steel are aa and as and, respectively. If the length of each rod increases by the same, amount when their temperature are raised by t0C, then find, the ratio l1/(l1 + l2), (a) as/aa (b) aa/as (c) a s/(aa + as) (d) aa/(aa + as), , (c), , T, , (a), , A body of mass 5 kg falls from a height of 20 metres on the, ground and it rebounds to a height of 0.2 m. If the loss in, potential energy is used up by the body, then what will be the, temperature rise?, (specific heat of material = 0.09 cal gm–1 ºC–1), (a) 0ºC, (b) 4ºC, (c) 8ºC, (d) None of these, Two straight metallic strips each of thickness t and length l, are rivetted together. Their coefficients of linear expansions, are a1 and a2 . If they are heated through temperature DT,, the bimetallic strip will bend to form an arc of radius, (a) t/{a1 + a2)DT}, (b) t/{(a2 – a1)DT}, (c) t(a1 – a2)DT, (d) t(a2 – a1)DT, The figure shows a system of two, concentric spheres of radii r1 and r2 are, kept at temperatures T 1 and T 2 ,, r1, respectively. The radial rate of flow of, T1, heat in a substance between the two, r2, T2, concentric spheres is proportional to, (r2 - r1 ), æ ö, (b), (a) ln ç r2 ÷, (r1 r2 ), è r1 ø, , 20., , 8., 13., 18., Space for Rough Work, , ( r2 - r1 ), , (d), , Temperature, , 6., , 9., 14., 19., , 10., 15., 20., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP10, , P-39, , 21. A polished metal plate with a rough black spot on it is heated, to about 1400 K and quickly taken into a dark room. Which, one of the following statements will be true?, (a) The spot will appear brighter than the plate, (b) The spot will appear darker than the plate, (c) The spot and plate will appear equally bright, (d) The spot and the plate will not be visible in the dark room, 22. On observing light from three different stars P, Q and R, it, was found that intensity of violet colour is maximum in the, spectrum of P, the intensity of green colour is maximum in, the spectrum of R and the intensity of red colour is maximum, in the spectrum of Q. If TP, TQ and TR are the respective, absolute temperature of P, Q and R, then it can be concluded, from the above observations that, (a) TP > TR > TQ, (b) TP < TR < TQ, (c) TP < TQ < TR, (d) TP > TQ > TR, 23. A partition wall has two layers of different materials A and B, in contact with each other. They have the same thickness, but the thermal conductivity of layer A is twice that of layer, B. At steady state the temperature difference across the, layer B is 50 K, then the corresponding difference across, the layer A is, (a) 50 K, (b) 12.5 K (c) 25 K, (d) 60 K, 24. Which of the following statements is/are false about, mode of heat transfer?, (a) In radiation, heat is transfered from one medium to, another without affecting the intervening medium, (b) Radiation and convection are possible in vaccum, while conduction requires material medium., (c) Conduction is possible in solids while convection, occurs in liquids and gases., (d) All are correct, 25. In a vertical U-tube containing a liquid,, the two arms are maintained at, t1, different temperatures t1 and t2. The, t, liquid columns in the two arms have, 2, l1, heights l 1 and l 2 respectively. The, l, coefficient of volume expansion of the 2, liquid is equal to, l1 – l2, l1 + l2, l1 – l2, l1 + l2, (a), (b), (c), (d), l1t1 – l2 t2, l2 t1 + l1t2, l2 t1 – l1t2, l1t1 + l2 t2, 26. The top of an insulated cylindrical container is covered by, a disc having emissivity 0.6 and conductivity 0.167 WK–, 1m–1 and thickness 1 cm. The temperature is maintained by, circulating oil as shown in figure., Find the radiation loss to the, surroundin g in Jm –2 s –1 if Oil out, temperature of the upper surface, of the disc is 27°C and temperature, Oil in, of the surrounding is 27°C., (a) 595 Jm–2s–1, (b) 545 Jm–2s–1, –2, –1, (c) 495 Jm s, (d) None of these, 27. Wien's law is concerned with, (a) relation between emissivity and absorptivity of a, radiating surface, , RESPONSE, GRID, , 21., 26., 31., , 22., 27., 32., , 28., , (b) total radiation, emitted by a hot surface, (c) an expression for spectral distribution of energy of a, radiation from any source, (d) a relation between the temperature of a black body and, the wavelength at which there is maximum radiant, energy per unit wavelength, If a piece of metal is heated to temperature q and then allowed, to cool in a room which is at temperature q0, the graph, between the temperature T of the metal and time t will be, closest to, (a), , T, , O, , (c), 29., , 30., , 31., , 32., , 33., , (b), t, , T, q0, O, , t, , O, , (d), t, , T, q0, , T, q0, t, , O, , Two rods of same length and, transfer a given amount of heat, 12 second, when they are joined, as shown in figure (i). But when, they are joined as shwon in figure, (ii), then they will transfer same, heat in same conditions in, (a) 24 s, (b) 13 s, (c) 15 s, (d) 48 s, Consider a compound slab consisting of two different, materials having equal thicknesses and thermal, conductivities K and 2K, respectively. The equivalent thermal, conductivity of the slab is, 4, 2, K, K, (a), (b), (c), 3 K (d) 3 K, 3, 3, The coefficient of thermal conductivity of copper, mercury, and glass are respectively Kc, Km and Kg such that Kc > Km, > Kg. If the same quantity of heat is to flow per sec per unit, area of each and corresponding temperature gradients are, Xc, Xm and Xg then, (a) Xc = Xm = Xg, (b) Xc > Xm > Xg, (c) Xc < Xm < Xg, (d) Xm < Xc < Xg, The radiation energy density per unit wavelength at a, temperature T has a maximum at a wavelength l0. At, temperature 2T, it will have a maximum wavelength, l0, l0, (a) 4l 0 (b) 2l 0, (c), (d), 2, 4, Assuming the Sun to be a spherical body of radius R at a, temperature of TK, evaluate the total radiant powerd incident, of Earth at a distance r from the Sun, 4, 4, 2 2 T, 2 2 T, (a) 4pr0 R s 2, (b) pr0 R s 2, r, r, 4, 4, T, T, 2, 2 2, (c) r0 R s, (d) R s 2, r, 4pr 2, , 23., 28., 33., Space for Rough Work, , 24., 29., , 25., 30.
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t.me/Magazines4all, , DPP/ CP10, , P-40, , 36., , 37., , 38., , 39., , 34., 39., 44., , RESPONSE, GRID, , 35., 40., 45., , 40., , 41., , 42., , 43., , 44., , 45., , In a surrounding medium of temperature 10°C, a body takes, 7 min for a fall of temperature from 60°C to 40°C. In what time, the temperature of the body will fall from 40°C to 28°C?, (a) 7 min (b) 11 min, (c) 14 min, (d) 21 min, Two rods of same length and area of cross-section A1 and, A2 have their ends at the same temperature. If K 1 and K2 are, their thermal conductivities, c1 and c2 are their specific heats, and d1 and d2 are their densities, then the rate of flow of, heat is the same in both the rods if, A1, k c d, A1 - k 1, =, = 1 1 1, (a), (b), A2, k2, A 2 k 2 c2 d 2, A1, k c d, A1 k 2, = 2 1 1, =, (c), (d), A 2 c 2 d 2 k1, A 2 k1, A student takes 50gm wax, 250, (specific heat = 0.6 kcal/kg°C), 200, and heats it till it boils. The, 150, graph between temperature and, 100, time is as follows. Heat supplied, 50, 0, to the wax per minute and, 1 2 3 4 567 8, boiling point are respectively, Time (Minute), (a) 500 cal, 50°C, (b) 1000 cal, 100°C, (c) 1500 cal, 200°C, (d) 1000 cal, 200°C, Consider two identical iron spheres , one, which lie on a thermally insulating plate,, while the other hangs from an insulatory, thread. Equal amount of heat is supplied, to the two spheres, then, (a) temperature of A will be greater than B, (b) temperature of B will be greater than A, (c) their temperature will be equal, (d) can’t be predicted, Steam at 100°C is passed into 20 g of water at 10°C. When, water acquires a temperature of 80°C, the mass of water, present will be: [Takespecific heat of water = 1 cal g– 1 °C– 1, and latent heat of steam = 540 cal g– 1], (a) 24 g, (b) 31.5 g (c) 42.5 g (d) 22.5 g, Two solid spheres, of radii R1 and R2 are made of the same, material and have similar surfaces. The spheres are raised to, the same temperature and then allowed to cool under, identical conditions. Assuming spheres to be perfect, conductors of heat, their initial rates of loss of heat are, Temperature (°C), , 35., , A metal ball immersed in alcohol weighs W1 at 0ºC and W2 at, 59ºC. The coefficient of cubical expansion of the metal is less, than that of alcohol. Assuming that the density of the metal is, large compared to that of alcohol, it can be shown that, (a) W1 > W2, (b) W1 = W2, (c) W1 < W2, (d) W1 = (W2/2), One end of a thermally insulated rod is kept at a temperature, T1 and the other at T2. The rod is composed of two sections, of length l 1 and l 2 and thermal T1 l1, l2, T2, conductivities K 1 and K 2, respectively. The temperature at the, interface of the two sections is, K1, K2, ( K 2l2T1 + K1l1T2 ), (, K, l, T, +, K, l, T, ), 1, 1, 1, 2, 2, 2, (a), (b), ( K1l1 + K 2 l2 ), ( K1l1 + K 2l2 ), ( K1l2T1 + K 2l1T2 ), ( K 2l1T1 + K1l2T2 ), (c), (d), ( K1l2 + K 2 l1 ), ( K 2 l1 + K1l2 ), Two spheres of different materials one with double the radius, and one-fourth wall thickness of the other are filled with ice., If the time taken for complete melting of ice in the larger, sphere is 25 minute and for smaller one is 16 minute, the, ratio of thermal conductivities of the materials of larger, spheres to that of smaller sphere is, (a) 4 : 5, (b) 5 : 4, (c) 25 : 8, (d) 8 : 25, A black body has maximum wavelength lm at temperature, 2000 K. Its corresponding wavelength at temperature 3000, K will be, 4, 3, 2, 9, lm, lm, lm, lm, (a), (b), (c), (d), 9, 2, 3, 4, A solid material is supplied with, E, heat at constant rate and the, C, temperature of the material changes, D, A, CD = 2AB, as shown. From the graph, the, B, FALSE conclusion drawn is, O, Heat Input, (a) AB and CD of the graph represent phase changes, (b) AB represents the change of state from solid to liquid, (c) latent heat of fusion is twice the latent heat of, vaporization, (d) CD represents change of state from liquid to vapour, 10 gm of ice cubes at 0°C are released in a tumbler (water, equivalent 55 g) at 40°C. Assuming that negligible heat is, taken from the surroundings, the temperature of water in, the tumbler becomes nearly (L = 80 cal/g), (a) 31°C, (b) 22°C, (c) 19°C, (d) 15°C, Temperature, , 34., , (a), , 36., 41., , R12 / R22 (b), , 37., 42., , R1 / R 2 (c), , R 2 / R1 (d), , 38., 43., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP10 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score – Qualifying Score, for Rough Work, Net Score = Space, (Correct, × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , R22 / R12, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP11, , SYLLABUS : Thermodynamics, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , The relation between U, P and V for an ideal gas in an, adiabatic process is given by relation U = a + bP V. Find the, value of adiabatic exponent (g) of this gas, b +1, b +1, a +1, a, (b), (c), (d), b, a, b, a+b, Carbon monoxide is carried around P, a closed cycle abc in which bc is an P2, b, isothermal process as shown in the, figure. The gas absorbs 7000 J of, c, heat as its temperture increases from P1 a, 300 K to 1000 K in going from a to b., V, O, V1, V2, The quantity of heat rejected by the, gas during the process ca is, (a) 4200 J (b) 5000 J (c) 9000 J (d) 9800 J, A Carnot engine, having an efficiency of h = 1/10 as heat, engine, is used as a refrigerator. If the work done on the, system is 10 J, the amount of energy absorbed from the, reservoir at lower temperature is, (a) 100 J, (b) 99 J, (c) 90 J, (d) 1 J, In a thermodynamic process, fixed mass of a gas is changed, in such a manner that the gas release 20 J of heat and 8 J of, work was done on the gas. If the initial internal energy of, the gas was 30 J. Then the final internal energy will be, (a) 2 joule (b) 18 joule (c) 42 joule (d) 58 joule, , 5., , (a), 2., , 3., , 4., , RESPONSE, GRID, , 1., 6., , 2., 7., , A closed gas cylinder is divided into two parts by a piston, held tight. The pressure and volume of gas in two parts, respectively are (P, 5V) and (10P, V). If now the piston is left, free and the system undergoes isothermal process, then the, volumes of the gas in two parts respectively are, 10, 20, V,, V, 11 11, The efficiency of an ideal gas with adiabatic exponent ‘g’ for, the shown cyclic process would be, , (a) 2V, 4V, , 6., , (a), , 7., , (b) 3V, 3V, , ( 2 ln 2 - 1), g / (g - 1), , (c) 5V, V, , (d), , v, , (b), , (1 - 2 ln 2), g /( g - 1), , 2V0, , (c), , ( 2 ln 2 + 1), g /( g - 1), , V0, , (d), , ( 2 ln 2 - 1), g /( g + 1), , C, A, , B, T0, , 2T0, , T, , A mass of diatomic gas (g = 1.4) at a pressure of 2, atmospheres is compressed adiabatically so that its, temperature rises from 27°C to 927°C. The pressure of the, gas in final state is, (a) 28 atm (b) 68.7 atm (c) 256 atm (d) 8 atm, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP11, , P-42, , 8., , 9., , A diatomic ideal gas is used in a Carnot engine as the working, substance. If during the adiabatic expansion part of the cycle, the volume of the gas increases from V to 32 V, the efficiency, of the engine is, (a) 0.5, (b) 0.75, (c) 0.99, (d) 0.25, The P-V diagram of a gas system undergoing cyclic process, is shown here. The work done during isobaric compression, A, D, is, (a) 100 J, , 2 × 102, P(N m–2), 102, , B, , 15., , C, , 16., , (b) 200 J, (c) 600 J, 0, , 1, , 2, , 3, , –3, , V(m ), (d) 400 J, 10. During an adiabatic process of an ideal gas, if P is, 1, proportional to 1.5 , then the ratio of specific heat, V, capacities at constant pressure to that at constant volume, for the gas is, (a) 1.5, (b) 0.25, (c) 0.75, (d) 0.4, 11. The work of 146 kJ is performed in order to compress one, kilo mole of gas adiabatically and in this process the, temperature of the gas increases by 7°C. The gas is, (R = 8.3 J mol–1 K–1), (a) diatomic, (b) triatomic, (c) a mixture of monoatomic and diatomic, (d) monoatomic, 12. Consider a spherical shell of radius R at temperature T. The, black body radiation inside it can be considered as an ideal, U, gas of photons with internal energy per unit volume u =, V, 1æUö, 4, µ T and pressure p = ç ÷ . If the shell now undergoes, 3è V ø, an adiabatic expansion the relation between T and R is :, 1, 1, (b) T µ 3 (c) T µ e–R (d) T µ e–3R, (a) T µ, R, R, 13. The specific heat capacity of a metal at low temperature (T), 3, , æ T ö, is given as C (kJK –1kg –1 ) = 32 ç, ÷ . A 100 g vessel of, è 400 ø, 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 in vessel is, (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, 14. 5.6 litre of helium gas at STP is adiabatically compressed to, 0.7 litre. Taking the initial temperature to be T1, the work, done in the process is, , RESPONSE, GRID, , 8., 13., 18., , 9., 14., 19., , 15, 9, 9, 3, RT, RT1 (d), (a), RT1 (b), RT1 (c), 2 1, 8, 2, 8, Four curves A, B, C and D are drawn in the figure for a given, amount of a gas. The curves which represent adiabatic and, isothermal changes are, (a) C and D respectively, B, P, C, (b) D and C respectively, A, D, (c) A and B respectively, (d) B and A respectively, V, 2, In an adiabatic process, the pressure is increased by % ., 3, 3, If g = , then the volume decreases by nearly, 2, , 4, 2, 9, %, %, %, (b), (c) 1%, (d), 9, 3, 4, 17. A reversible engine converts one-sixth of the heat input, into work. When the temperature of the sink is reduced by, 62ºC, the efficiency of the engine is doubled. The, temperatures of the source and sink are, (a) 99ºC, 37ºC, (b) 80ºC, 37ºC, (c) 95ºC, 37ºC, (d) 90ºC, 37ºC, 1, 18. A diatomic ideal gas is compressed adiabatically to, of, 32, its initial volume. If the initial temperature of the gas is Ti (in, Kelvin) and the final temperature is aTi, the value of a is, (a) 8, (b) 4, (c) 3, (d) 5, 19. When the state of a gas adiabatically changed from an, equilibrium state A to another equilibrium state B an amount, of work done on the stystem is 35 J. If the gas is taken from, state A to B via process in which the net heat absorbed by, the system is 12 cal, then the net work done by the system, is (1 cal = 4.19 J), (a) 13.2 J (b) 15.4 J (c) 12.6 J (d) 16.8 J, 20. Calculate the work done when 1 mole of a perfect gas is, compressed adiabatically. The initial pressure and volume, of the gas are 105 N/m2 and 6 litre respectively. The final, volume of the gas is 2 litres. Molar specific heat of the gas at, constant volume is 3R/2. [Given (3)5/3 = 6.19], (a) –957 J (b) +957 J (c) – 805 J (d) + 805 J, 21. A Carnot engine whose efficiency is 40%, receives heat at, 500K. If the efficiency is to be 50%, the source temperature, for the same exhaust temperature is, (a) 900 K (b) 600 K (c) 700 K (d) 800 K, 22. 1 gm of water at a pressure of 1.01 × 105 Pa is converted into, steam without any change of temperature. The volume of 1, g of steam is 1671 cc and the latent heat of evaporation is, 540 cal. The change in internal energy due to evaporation of, 1 gm of water is, (a) » 167 cal (b) 500 cal (c) 540 cal (d) 581 cal, , (a), , 10., 15., 20., Space for Rough Work, , 11., 16., 21., , 12., 17., 22., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP11, , P-43, , 23. One mole of an ideal gas at temperature T was cooled, P, . Then, by, n, an isobaric process, the gas was restored to the initial, temperature. The net amount of heat absorbed by the gas in, the process is, RT, (a) nRT, (b), n, (c) RT (1 – n–1), (d) RT (n – 1), 1, 24. A Carnot engine, having an efficiency of h =, as heat, 10, engine, is used as a refrigerator. If the work done on the, system is 10 J, the amount of energy absorbed from the, reservoir at lower temperature is, (a) 99 J, (b) 90 J, (c) 1 J, (d) 100 J, 25. The volume of an ideal gas is 1 litre and its pressure is equal, to 72 cm of mercury column. The volume of gas is made, 900 cm3 by compressing it isothermally. The stress of the, gas will be, (a) 8 cm of Hg, (b) 7 cm of Hg, (c) 6 cm of Hg, (d) 4 cm of Hg, 26. An ideal gas is taken through the cycle A ® B ® C ® A , as, shown in figure. If the net heat supplied to the gas in the cycle, is 5 J, the work done by the gas in the process C ® A is, , isochorically till the gas pressure fell from P to, , (a) – 5 J, 2, , (b) – 10 J, , 10, , 33., , 35., , (d) – 20 J, 27. An ideal gas undergoing adiabatic change has the following, pressure-temperature relationship, (a) P g -1T g = constant, (b) P g T g -1 = constant, g 1-g, (c) P T, = constant, (d) P1-g T g = constant, 28. In a thermodynamic process, fixed mass of a gas is changed, in such a manner that the gas release 20 J of heat and 8 J of, work was done on the gas. If the initial internal energy of, the gas was 30 J, the final internal energy will be, (a) 2 joule (b) 18 joule (c) 42 joule (d) 58 joule, 29. The coefficient of performance of a refrigerator is 5. If the, inside temperature of freezer is –20°C, then the temperature, of the surroundings to which it rejects heat is, (a) 41°C, (b) 11°C, (c) 21°C, (d) 31°C, 30. Monatomic, diatomic and polyatomic ideal gases each, undergo slow adiabatic expansions from the same initial, volume and same initial pressure to the same final volume., The magnitude of the work done by the environment on the, gas is, (a) the greatest for the polyatomic gas, (b) the greatest for the monatomic gas, (c) the greatest for the diatomic gas, , 24., 29., 34., , p0, , A, , B, , C, , D, v0, , 2v0, , v, , 32. For an ideal gas graph is shown for three processes. Process, Work done (magnitude), 1, 2 and 3 are respectively., , A, 5, P ( N / m2 ), , æ 13 ö, ç ÷ p0 v0 2p0, è2ø, , æ 11 ö, ç ÷ p0 v0, è2ø, (d) 4p0v0, (c), , B, , V (in m3 ), 1, , 23., 28., 33., , (b), , 34., , C, , (c) – 15 J, , RESPONSE, GRID, , (d) the question is irrelevant, there is no meaning of slow, adiabatic expansion, 31. The given p-v diagram represents the thermodynamic cycle, of an engine, operating with an ideal monatomic gas. The, amount of heat, extracted from the source in a single cycle is, p, (a) p 0 v 0, , 36., , (a) Isobaric, adiabatic, isochoric, 3, 2, (b) Adiabatic, isobaric, isochoric, (c) Isochoric, adiabatic, isobaric, 1, DT, (d) Isochoric, isobaric, adiabatic, Temperature change, During an adiabatic process an object does 100J of work, and its temperature decreases by 5K. During another, process it does 25J of work and its temperature decreases, by 5K. Its heat capacity for 2nd process is, (a) 20 J/K (b) 24 J/K (c) 15 J/K (d) 100 J/K, A refrigerator works between 4°C and 30°C. It is required to, remove 600 calories of heat every second in order to keep, the temperature of the refrigerated space constant. The, power required is: (Take 1 cal = 4.2 joule), (a) 2.365 W (b) 23.65 W (c) 236.5 W (d) 2365 W, A perfect gas goes from a state A to another state B by, absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external, work. It is now transferred between the same two states in, another process in which it absorbs 105 J of heat. In the, second process, (a) work done by gas is 105 J, (b) work done on gas is 105 J, (c) work done by gas is 0.5 × 105 J, (d) work done on the gas is 0.5 × 105 J, One mole of a diatomic ideal gas, undergoes a cyclic process ABC as, B, shown in figure. The process BC is, 800 K, adiabatic. The temperatures at A, B, and C are 400 K, 800 K and 600 K, P, respectively. Choose the correct, statement:, 600 k, C, (a) The change in internal energy, A, 400 K, in whole cyclic process is 250, R., V, (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., , 25., 30., 35., Space for Rough Work, , 26., 31., 36., , 27., 32.
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t.me/Magazines4all, , DPP/ CP11, , P-44, , 37. Two Carnot engines A and B are operated in series. The, engine A receives heat from the source at temperature T 1, and rejects the heat to the sink at temperature T. The second, engine B receives the heat at temperature T and rejects to, its sink at temperature T2. For what value of T the efficiencies, of the two engines are equal?, T1 + T2, T -T, (b) 1 2 (c) T 1T 2, (d), T1T2, 2, 2, 38. An ideal gas is initially at P1, V1 is expanded to P2, V2 and, then compressed adiabatically to the same volume V1 and, pressure P3. If W is the net work done by the gas in complete, process which of the following is true ?, (a) W > 0 ; P3 > P1, (b) W < 0 ; P3 > P1, (c) W > 0 ; P3 < P1, (d) W < 0 ; P3 < P1, 39. Which of the following statements is correct for any, thermodynamic system ?, (a) The change in entropy can never be zero, (b) Internal energy and entropy are state functions, (c) The internal energy changes in all processes, (d) The work done in an adiabatic process is always zero., 40. One mole of an ideal gas goes from an initial state A to final, state B via two processes : It first undergoes isothermal, expansion from volume V to 3V and then its volume is, reduced from 3V to V at constant pressure. The correct P-V, diagram representing the two processes is :, (a), (b), B, , (a), , A, , P, , P, , A, , B, , V, , V, , A, , (c), , 2, , V, , 3V, , (b) Final pressure will be three times more than initial, pressure., (c) Change in pressure will be more than three times the, initial pressure., (d) Change in pressure will be less than three times the, initial pressure., 42. A gas is compressed isothermally to half its initial volume., The same gas is compressed separately through an adiabatic, process until its volume is again reduced to half. Then :, (a) Compressing the gas isothermally will require more work, to be done., (b) Compressing the gas through adiabatic process will, require more work to be done., (c) Compressing the gas isothermally or adiabatically will, require the same amount of work., (d) Which of the case (whether compression through, isothermal or through adiabatic process) requires more, work will depend upon the atomicity of the gas., 43. An ideal gas goes from state A to state B via three different, processes as indicated in the P-V diagram :, If Q1, Q2, Q3 indicate the heat a absorbed, A, 1, 2, by the gas along the three processes and P, B, DU1, DU2, DU3 indicate the change in, 3, internal energy along the three processes, respectively, then, V, (a) Q1 > Q2 > Q3 and DU1 = DU2 = DU3, (b) Q3 > Q2 > Q1 and DU1= DU2 = DU3, (c) Q1 = Q2 = Q3 and DU1 > DU2 > DU3, (d) Q3 > Q2 > Q1 and DU1> DU2 > DU3, 44. In P-V diagram shown in figure ABC is a semicircle. The, work done in the process ABC is, , 3V, , (a) 4 J, , A, , (d), , P, B, , (b), P, B, , V, , V, , 3V, , (c), V, , V, , 3V, , 41. What will be the final pressure if an ideal gas in a cylinder, 1, is compressed adiabatically to rd of its volume?, 3, (a) Final pressure will be three times less than initial, pressure., , RESPONSE, GRID, , 37., 42., , 38., 43., , P(N/m ), A, , 3, , -p, J, 2, p, J, 2, , B, 1, , C, 3, , 0, , V(m ), (d) zero, 1, 2, 45. For an isothermal expansion of a perfect gas, the value of, , DP, is equal to, P, DV, 1/ 2 D V, (b) –, (a) – g, V, V, , 39., 44., , 40., 45., , (c), , –g, , DV, (d), V, , 41., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP11 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = Space, (Correct, × 4) Work, – (Incorrect × 1), for Rough, Space for Rough Work, , 180, , 60, , – g2, , DV, V, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP12, , SYLLABUS : Kinetic Theory, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat, capacity of the gas at constant volume is 5.0JK–1. If the, speed of any quantity x in this gas at NTP is 952 ms–1, then, the heat capacity at constant pressure is (Take gas constant, R = 8.3 JK–1 mol–1), (a) 7.5 JK–1 mol–1, (b) 7.0 JK–1 mol–1, (c) 8.5 JK–1 mol–1, (d) 8.0 JK–1 mol–1, A fixed mass of gas at constant pressure occupies a volume, V. The gas undergoes a rise in temperature so that the root, mean square velocity of its molecules is doubled. The new, volume will be, (a) V/2, , 3., , (b), , (d) 4 V, V / 2 (c) 2 V, A gaseous mixture consists of 16 g of helium and 16 g of, oxygen. The ratio, (a) 1.62, , RESPONSE, GRID, , Cp, Cv, , 5., , 6., , of the mixture is, , (b) 1.59, , 1., 6., , 4., , (c) 1.54, , (d) 1.4, , 2., , Air is pumped into an automobile tube upto a pressure of, 200 kPa in the morning when the air temperature is 22°C., During the day, temperature rises to 42°C and the tube, expands by 2%. The pressure of the air in the tube at this, temperature, will be approximately, (a) 212 kPa (b) 209 kPa (c) 206 kPa (d) 200 kPa, The rms speed of the particles of fume of mass 5 × 10–17 kg, executing Brownian motion in air at N.T.P. is (k = 1.38 ×, 10–23 J/K), (a) 1.5 m/s (b) 3.0 m/s (c) 1.5 cm/s (d) 3 cm/s, One mole of an ideal monoatomic gas requires 207 J heat to, raise the temperature by 10 K when heated at constant, pressure. If the same gas is heated at constant volume to, raise the temperature by the same 10 K, the heat, required is [Given the gas constant R = 8.3 J/ mol. K], (a) 198.7 J (b) 29 J, (c) 215.3 J (d) 124 J, , 3., , Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP12, , P-46, , 7., , Figure shows the variation in temperature (DT) 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 scales for the two axes coincide. Ignoring, vibrational degrees of freedom, the lines a, b and c, respectively correspond to, a, (a) P, M and D, b, Q, (b) M, D and P, , 8., , 9., , 10., , 11, , 12., , 13., , (d) D, M and P, DT, 1 mole of a monatomic and 2 mole of a diatomic gas are, mixed. The resulting gas is taken through a process in which, molar heat capacity was found 3R. Polytropic constant in, the process is, (a) –1/5 (b) 1/5, (c) 2/5, (d) –2/5, The density of a gas is 6 × 10–2 kg/m3 and the root mean, square velocity of the gas molecules is 500 m/s. The pressure, exerted by the gas on the walls of the vessel is, (a) 5 × 103 N/m2, (b) 1.2 × 10–4 N/m2, –4, 2, (c) 0.83 × 10 N/m, (d) 30 N/m2, The absolute temperature of a gas is increases 3 times. The, root mean square velocity of the moelcules will be, (a) 3 times, (b) 9 times, (c) 1/3 times, (d) Ö3 times, Consider an ideal gas confined in an isolated closed chamber., As the gas undergoes an adiabatic expansion, the average, time of collision between molecules increases as Vq, where, Cp ö, æ, V is the volume of the gas. The value of q is : ç g =, ÷, Cv ø, è, 3g + 5, 3g - 5, g +1, g -1, (a), (b), (c), (d), 6, 6, 2, 2, One kg of a diatomic gas is at a pressure of 8 × 104N/m2. The, density of the gas is 4kg/m3. What is the energy of the gas, due to its thermal motion?, (a) 5 × 104 J, (b) 6 × 104 J, 4, (c) 7 × 10 J, (d) 3 × 104 J, A thermally insulated vessel contains an ideal gas of molecular, mass M and ratio of specific heats g. It is moving with speed v, and it suddenly brought to rest. Assuming no heat is lost to, the surroundings, its temperature increases by, ( g –1), gMv 2, Mv 2 K, K, (a), (b), 2gR, 2R, , RESPONSE, GRID, , 14., , c, , (c) P, D and M, , 7., 12., 17., , 8., 13., 18., , (g –1), ( g – 1), Mv 2 K, Mv 2 K, (d), 2(, g, +, 1), R, 2R, Figure shows a parabolic graph between T and 1/V for a, mixture of a gases undergoing an adiabatic process. What, is the ratio of Vrms of molecules and speed of sound in mixture?, , (c), , 15., , 16., , 17., , 18., , 19., , (a), , 3/ 2, , (b), , 2, , (c), , 2/3, , (d), , 3, , T, , 2T 0, T0, 1/V0 4/V0, , 1/V, , The work of 146 kJ is performed in order to compress one, kilomole of gas adiabatically and in this process the, temperature of the gas increases by 7°C. The gas is, (R = 8.3 J mol–1 K–1), (a) diatomic, (b) triatomic, (c) a mixture of monatomic and diatomic, (d) monatomic, At what temperature is root mean square velocity of gaseous, hydrogen molecules equal to that of oxygen molecules at, 47°C ?, (a) 40K, (b) 80K, (c) – 73K (d) 3K, The kinetic theory of gases states that the average squared, velocity of molecules varies linearly with the mean molecular, weight of the gas. If the root mean square (rms) velocity of, oxygen molecules at a certain temperature is 0.5 km/sec., The rms velocity for hydrogen molecules at the same, temperature will be :, (a) 2 km/sec (b) 4 km/sec (c) 8 km/sec (d) 16 km/sec, If 2 moles of an ideal monatomic gas at temperature T0 is, mixed with 4 moles of another ideal monatomic gas at, temperature 2T0, then the temperature of the mixture is, 5, 3, 4, 5, T0, T0, T0, T0, (a), (b), (c), (d), 3, 2, 3, 4, From the following statements, concerning ideal gas at any, given temperature T, select the incorrect one(s), (a) The coefficient of volume expansion at constant, pressure is same for all ideal gas, (b) The average translational kinetic energy per molecule, of oxygen gas is 3 KT (K being Boltzmann constant), (c) In a gaseous mixture, the average translational kinetic, energy of the molecules of each component is same, (d) The mean free path of molecules increases with, decrease in pressure, , 9., 14., 19., Space for Rough Work, , 10., 15., , 11., 16., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP12, , P-47, , 20. The adjoining figure shows graph of pressure and volume, of a gas at two tempertures T1 and T2. Which of the following, inferences is correct?, P, (a) T1 > T2, (b) T1 = T2, T2, (c) T1 < T2, T1, (d) None of these, V, 21. The molecules of a given mass of gas have a root mean, square velocity of 200m s–1 at 27°C and 1.0 × 105 N m–2, pressure. When the temperature is 127°C and the pressure, 0.5 × 105 Nm–2, the root mean square velocity in ms–1, is, , 26., , 27., , 28., , 400, , 100, 100 2, (b) 100 2 (c), (d), 3, 3, 3, 22. A graph is plotted with PV/T on y-axis and mass of the gas, along x-axis for different gases. The graph is, (a) a straight line parallel to x-axis for all the gases, (b) a straight line passing through origin with a slope, having a constant value for all the gases, (c) a straight line passing through origin with a slope, having different values for different gases, (d) a straight line parallel to y-axis for all the gases, 23. At identical temperatures, the rms speed of hydrogen, molecules is 4 times that for oxygen molecules. In a mixture, of these in mass ratio H2 : O2 = 1:8, the rms speed of all molecules, is n times the rms speed for O2 molecules, where n is, (a) 3, (b) 4/3, (c) (8/3)1/2 (d) (11)1/2, 24. Work done by a system under isothermal change from a, volume V1 to V2 for a gases which obeys Vander Waal's, , (a), , æ, an 2 ö, equation (V - bn) ç P +, ÷ = nRT is, ç, V ÷ø, è, æ V2 - nb ö, 2 æ V1 - V2 ö, (a) nRT log e ç V - nb ÷ + an ç V V ÷, è 1, ø, è 1 2 ø, (b), , æ V - nb ö, 2 æ V1 - V2 ö, nRT log10 ç 2, ÷ + an ç, ÷, è V1 - nb ø, è V1V2 ø, , (c), , æ V - nb ö, 2 æ V1 - V2 ö, nRT loge ç 2, ÷ + bn ç, ÷, è V1 - nb ø, è V1V2 ø, , 20., 25., 30., , 21., 26., 31., , 29., , 30., , 31., , (a) Boyle’s law, (b) Charle’s law, (c) Avogadro’s law, (d) Gaylussac’s law, 32., , æ V1 - nb ö, 2 æ V1V2 ö, (d) nRT log e ç V - nb ÷ + an ç V – V ÷, è 2, ø, è 1 2ø, 25. Two vessels separately contain two ideal gases A and B at, the same temperature. The pressure of A being twice that of, B. Under such conditions, the density of A is found to be, 1.5 times the density of B. The ratio of molecular weight of A, and B is :, , RESPONSE, GRID, , 3, 2, 1, (b) 2, (c), (d), 3, 4, 2, The temperature of the mixture of one mole of helium and, one mole of hydrogen is increased from 0°C to 100°C at, constant pressure. The amount of heat delivered will be, (a) 600 cal (b) 1200 cal (c) 1800 cal (d) 3600 cal, If the intermolecular forces vanish away, the volume, occupied by the molecules contained in 4.5 g water at, standard temperature and pressure will be, (a) 5.6 litre (b) 4.5 litre (c) 11.2 litre (d) 6.5 litre, A gas mixture consists of 2 moles of oxygen and 4 moles of, Argon at temperature T. Neglecting all vibrational moles,, the total internal energy of the system is, (a) 4 RT, (b) 15 RT, (c) 9 RT, (d) 11RT, A vessel has 6g of hydrogen at pressure P and temperature, 500K. A small hole is made in it so that hydrogen leaks out., How much hydrogen leaks out if the final pressure is P/2, and temperature falls to 300 K ?, (a) 2g, (b) 3g, (c) 4g, (d) 1g, For a gas if ratio of specific heats at constant pressure and, volume is g then value of degrees of freedom is, 3g –1, 2, 9, 25, ( g –1) (d), (g –1), (a), (b), (c), 2g –1, g –1, 2, 2, The given P-V curve is predicted by, (a), , 33., , 1.4, 1.2, 1.0, P 0.8, 0.6, 0.4, 0.2, 0, , T1 > T2 > T3, T1, T2, T3, 20 60, , 100 140 160 200, V, , Three perfect gases at absolute temperatures T1, T2 and T3, are mixed. The masses of molecules are m1, m2 and m3 and, the number of molecules are n1, n2 and n3 respectively., Assuming no loss of energy, the final temperature of the, mixture is :, (a), , n1T1 + n2T2 + n3T3, n1 + n2 + n3, , (b), , (c), , n12T12 + n22T22 + n32T32, n1T1 + n2T2 + n3T3, , (d), , n1T12 + n2T22 + n3T32, n1T1 + n2T2 + n3T3, , (T1 + T2 + T3 ), 3, , A gas is enclosed in a cube of side l. What will be the, change in momentum of the molecule, if it suffers an, elastic collision with the plane wall parallel to yz-plane and, rebounds with the same velocity ?, [(Vx, Vy & Vz) initial velocities of the gas molecules], (a) mvx, (b) zero, (c) – mvx (d) – 2mvx., , 22., 27., 32., Space for Rough Work, , 23., 28., 33., , 24., 29.
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t.me/Magazines4all, , DPP/ CP12, , P-48, , 34., , 35., , 36., , What will be the ratio of number of molecules of a, monoatomic and a diatomic gas in a vessel, if the ratio of, their partial pressures is 5 : 3 ?, (a) 5 : 1, (b) 3 : 1, (c) 5 : 3, (d) 3 : 5, The average transitional energy and the rms speed of, molecules in a sample of oxygen gas at 300 K are 6.21 × 10–21 J, and 484 m/s respectively. The corresponding values at 600, K are nearly (assuming ideal gas behaviour), (a) 12.42 × 10–21 J, 968 m/s (b) 8.78 × 10–21 J, 684 m/s, (c) 6.21 × 10–21 J, 968 m/s (d) 12.42 × 10–21 J, 684 m/s, At 10° C the value of the density of a fixed mass of an ideal, gas divided by its pressure is x. At 110°C this ratio is:, 383, 10, 283, x, (c), x (d), x, 283, 110, 383, If the potential energy of a gas molecule is, U = M/r6 – N/r12, M and N being positive constants. Then, the potential energy at equilibrium must be, (a) zero, (b) M2/4N (c) N2/4M (d) MN2/4, , (a) x, 37., , 40., , (a), , 41., , 42., , (b), , Consider a gas with density r and c as the root mean, square velocity of its molecules contained in a volume. If, the system moves as whole with velocity v, then the pressure, exerted by the gas is, 1, 1 2, r( c + v ) 2, rc, (a), (b), 3, 3, 1, 1, r(c – v) 2, r(c –2 – v ) 2, (c), (d), 3, 3, 39. How is the mean free path (l) in a gas related to the, interatomic distance?, (a) l is 10 times the interatomic distance, (b) l is 100 times the interatomic distance, (c) l is 1000 times the interatomic distance, , 43., , 38., , (d) l is, , 44., , 34., 39., 44., , 35., 40., 45., , (d) 3 3, 54 / 4 (b), 54 / 2 (c) 3.5, If R is universal gas constant, the amount of heat needed to, raise the temperature of 2 moles of an ideal monoatomic gas, from 273 K to 373 K, when no work is done, is, (a) 100 R, (b) 150 R, (c) 300 R (d) 500 R, N molecules, each of mass m, of gas A and 2 N molecules, each, of mass 2 m, of gas B are contained in the same vessel which, is maintained at a temperature T. The mean square velocity of, molecules of B type is denoted by V2 and the mean square, V, velocity of A type is denoted by V1, then 1 is, V2, (a) 2, (b) 1, (c) 1/3, (d) 2/3, The root mean square value of the speed of the molecules in, a fixed mass of an ideal gas is increased by increasing, (a) the volume while keeping the temperature constant, (b) the pressure while keeping the volume constant, (c) the temperature while keeping the pressure constant, (d) the pressure while keeping the temperature constant, The P-V diagram of a diatomic gas is a straight line passing, through origin. The molar heat capacity of the gas in the, process will be, 4R, 3, For a gas, difference between two specific heats is 5000 J/, mole°C. If the ratio of specific heats is 1.6, the two specific, heats in J/mole-°C are, (a) CP = 1.33 × 104, CV = 2.66 × 104, (b) CP = 13.3 × 104, CV = 8.33 × 103, (c) CP = 1.33 × 104, CV = 8.33 × 103, (d) CP = 2.6 × 104, CV = 8.33 × 104, , (a) 4 R, , 45., , 1, times of the interatomic distance, 10, , RESPONSE, GRID, , Four molecules have speeds 2 km/sec, 3 km/sec, 4 km/sec, and 5 km/sec. The root mean square speed of these, molecules (in km/sec) is, , 36., 41., , (b) 2.5 R, , 37., 42., , (c) 3 R, , (d), , 38., 43., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP12 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP13, , SYLLABUS : Oscillations, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , 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?, (a) aT/x, (b) aT + 2pv, (c) aT/v, (d) a2T2 + 4p2v2, A mass is suspended separately by two different springs in, successive order, then time periods is t1 and t2 respectively., It is connected by both springs as shown in fig. then time, period is t0. The correct relation is, (a), , t 02 = t12 + t 22, , (b), , t 0-2, , (c), , t 0-1 = t1-1 + t -2 1, , =, , t 1-2, , +, , t 2-2, , k1, , 4., , k2, , 1., , 1, 2 2, ma n, (d) 4p2ma2n2, 4, A mass M attached to a spring oscillates with a period of 2s., If the mass is increased by 2 kg, then the period increases, by 2s. Find the initial mass M assuming that Hooke's law is, obeyed., , (c), m, , 5., , (d) t 0 = t1 + t 2, A rod of length l is in motion such that its ends A and B are, moving along x-axis and y-axis respectively. It is given that, dq, = 2 rad/sec always. P is a fixed point on the rod. Let M, dt, , RESPONSE GRID, , be the projection of P on x-axis. For the time interval in, p, which q changes from 0 to , the correct statement is, 2, (a) The acceleration of M is always directed towards right, (b) M executes SHM, (c) M moves with constant speed, (d) M moves with constant acceleration, A particle of mass m executes simple harmonic motion with, amplitude a and frequency n. The average kinetic energy, during its motion from the position of equilibrium to the end, is, (a) 2p2ma2n2, (b) p2ma2n2, , 2., , (a), , 3., Space for Rough Work, , 2, kg, 3, , (b), , 4., , 1, kg, 3, , (c), , 1, kg, 2, , 5., , (d) 1 kg
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t.me/Magazines4all, , DPP/ CP13, , P-50, , rd, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , æ 1ö, The amplitude of a damped oscillator becomes ç ÷ in 2, è 3ø, 1, seconds. If its amplitude after 6 seconds is, times the, n, original amplitude, the value of n is, , (a) 32, (b) 33, (c) 3 3, (d) 23, Assume the earth to be perfect sphere of uniform density. If, a body is dropped at one end of a tunnel dug along a diameter, of the earth (remember that inside the tunnel the force on, the body is – k times the displacement from the centre, k, being a constant), it (body) will, (a) reach the earth’s centre and stay there, (b) go through the tunnel and comes out at the other end, (c) oscillate simple harmonically in the tunnel, (d) stay somewhere between the earth’s centre and one of, the ends of tunnel., A particle undergoes simple harmonic motion having time, period T. The time taken in 3/8th oscillation is, 5, 3, 5, 7, T, T, T, T, (a), (b), (c), (d), 8, 8, 12, 12, A particle is executing simple harmonic motion with amplitude, A. When the ratio of its kinetic energy to the potential energy, 1, is , its displacement from its mean position is, 4, 2, 3, 1, 3, A (b), A, A, A (c), (d), (a), 5, 4, 4, 2, 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%, The time period of a mass suspended from a spring is T. If, the spring is cut into four equal parts and the same mass is, suspended from one of the parts, then the new time period, will be, T, T, (a) 2T, (b), (c) 2, (d), 4, 2, Two simple harmonic motions act on a particle. These, harmonic motions are x = A cos (wt + d ), y = A cos (wt + a), p, when d = a + , the resulting motion is, 2, (a) a circle and the actual motion is clockwise, (b) an ellipse and the actual motion is counterclockwise, (c) an elllipse and the actual motion is clockwise, (d) a circle and the actual motion is counter clockwise, A point mass oscillates along the x-axis according to the law, x = x0 cos(wt – p/4). If the acceleration of the particle is, written as a = A cos(wt – d), then, (a) A = x0w2, d = 3p/4, (b) A = x0, d = –p/4, (c) A = x0w2, d = p/4, (d) A = x0w2, d = –p/4, , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , 14., , 15., , A mass M is suspended from a spring of negligible mass., The spring is pulled a little and then released so that the, mass executes SHM of time period T. If the mass is increased, m, 5T, by m, the time period becomes, . Then the ratio of, M, 3, is, 25, 16, 5, 3, (a), (c), (d), (b), 9, 9, 3, 5, A body oscillates with a simple harmonic motion having, amplitude 0.05 m. At a certain instant of time, its displacement, is 0.01 m and acceleration is 1.0 m/s2. The period of, oscillation is, p, p, s, s, (d), 10, 5, The particle executing simple harmonic motion has a kinetic, energy K0 cos2 wt. The maximum values of the potential, energy and the total energy are respectively, (a) K0/2 and K0, (b) K0 and 2K0, (c) K0 and K0, (d) 0 and 2K0, A simple pendulum attached to the ceiling of a stationary, lift has a time period T. The distance y covered by the lift, moving upwards varies with time t as y = t2 where y is in, metres and t in seconds. If g = 10 m/s2, the time period of, pendulum will be, , (a) 0.1 s, , 16., , 17., , (b) 0.2 s, , (c), , 4, 5, 5, 6, T, T, T (d), T, (b), (c), 5, 6, 4, 5, A particle moves with simple harmonic motion in a straight, line. In first ts, after starting from rest it travels a distance a,, and in next t s it travels 2a, in same direction, then:, (a) amplitude of motion is 3a, (b) time period of oscillations is 8t, (c) amplitude of motion is 4a, (d) time period of oscillations is 6t, Two simple harmonic motions are represented by the, p, equations y1 = 0.1 sin æç100pt + ö÷ and y 2 = 0.1 cos pt ., è, 3ø, The phase difference of the velocity of particle 1 with respect, to the velocity of particle 2 is, p, -p, p, -p, (a), (b), (c), (d), 6, 6, 3, 3, Masses MA and MB hanging from the ends of strings of, lengths LA and LB are executing simple harmonic motions. If, their frequencies are fA = 2fB, then, (a) LA = 2LB and MA = MB/2, (b) LA = 4LB regardless of masses, (c) LA = LB/4 regardless of masses, (d) LA = 2LB and MA = 2MB, , (a), 18., , 19., , 20., , 8., 13., 18., Space for Rough Work, , 9., 14., 19., , 10., 15., 20., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP13, , P-51, , 21. In damped oscillations, the amplitude of oscillations is, reduced to one-third of its inital value a0 at the end of 100, oscillations. When the oscillator completes 200 oscillations,, its amplitude must be, (a) a0/2, (b) a0/4, (c) a0/6, (d) a0/9, 22. The spring constant from the adjoining combination of, springs is, 2K, (a) K, (b) 2 K, m, (c) 4 K, K, K, (d) 5 K/2, 23. A body executes simple harmonic motion. The potential, energy (P.E), the kinetic energy (K.E) and total energy (T.E), are measured as a function of displacement x. Which of the, following statements is true ?, (a) K.E. is maximum when x = 0, (b) T.E is zero when x = 0, (c) K.E is maximum when x is maximum, (d) P.E is maximum when x = 0, 24. A simple harmonic wave having an amplitude a and time, period T is represented by the equation y = 5sin p (t + 4)m ., Then the value of amplitude (a) in (m) and time period (T) in, second are, (a) a = 10, T = 2, (b) a = 5, T = 1, (c) a = 10, T = 1, (d) a = 5, T = 2, 25. A particle moves such that its acceleration ‘a’ is given by a, = – zx where x is the displacement from equilibrium position, and z is constant. The period of oscillation is, (a) 2 p/z, (b) 2 π / z (c), 2 p / z (d) 2 p / z, 26. The displacement of an object attached to a spring and, executing simple harmonic motion is given by x = 2 × 10–2, cos pt metre.The time at which the maximum speed first, occurs is, (a) 0.25 s (b) 0.5 s, (c) 0.75 s (d) 0.125 s, 27. A tunnel has been dug through the centre of the earth and, a ball is released in it. It executes S.H.M. with time period, (a) 42 minutes, (b) 1 day, (c) 1 hour, (d) 84.6 minutes, 28. The displacement equation of a particle is, , x = 3sin2t + 4cos2t . The amplitude and maximum, velocity will be respectively, (a) 5, 10, (b) 3, 2, (c) 4, 2, (d) 3, 4, 29. A body of mass 0.01 kg executes simple harmonic motion, about x = 0 under the influence of a force as shown in figure., F(N), The time period of SHM is, (a) 1.05 s, , 31., , 32., , 33., , 34., , Two oscillators are started simultaneously in same phase., After 50 oscillations of one, they get out of phase by p, that, is half oscillation. The percentage difference of frequencies, of the two oscillators is nearest to, (a) 2%, (b) 1%, (c) 0.5%, (d) 0.25%, The length of a second’s pendulum at the surface of earth is, 1 m. The length of second’s pendulum at the surface of, moon where g is 1/6th that at earth’s surface is, (a) 1/6 m, (b) 6 m, (c) 1/36 m (d) 36 m, A simple spring has length l and force constant K. It is cut, into two springs of lengths l1 and l2 such that l1 = n l2, (n = an integer). The force constant of spring of length l1 is, (a) K (1 + n), (b) (K/n) (1 + n), (c) K, (d) K/(n + 1), The displacement of a particle from its mean position (in, metre) is given by y = 0.2 sin (10p t + 1.5p) cos (10pt + 1.5 p), The motion of particle is, (a) periodic but not SHM, (b) non-periodic, (c) simple harmonic motion with period 0.1s, (d) simple harmonic motion with period 0.2s., A point particle of mass 0.1 kg is executing S.H.M. of, amplitude 0.1 m. When the particle passes through the mean, position, its kinetic energy is 8 × 10–3 joule. Obtain the, equation of motion of this particle, if the initial phase of, oscillation is 45º., (a), , –0.2, , E, , 21., 26., 31., , PE, , (b), , d, , KE, d, , PE, , E, , KE, , PE, , x(m), , (c), , 22., 27., 32., , E, , KE, , (a), , –80, , (d) 0.03 s, , pö, æ, y = 0.2sin ç ±4t + ÷, è, 4ø, , pö, æ, p, y = 0.1sin ç ±2t + ÷ (d) y = 0.2sin æç ±2t + ö÷, è, è, 4ø, 4ø, 35. 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 scale), , E, , 0.2, , (c) 0.25 s, , pö, æ, y = 0.1sin ç ±4t + ÷ (b), è, 4ø, , (c), , 80, , (b) 0.52 s, , RESPONSE, GRID, , 30., , 23., 28., 33., Space for Rough Work, , PE, d, , 24., 29., 34., , (d), , KE, , 25., 30., 35., , d
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t.me/Magazines4all, , DPP/ CP13, , P-52, , 36., , 37., , The equation of a simple harmonic wave is given by, p, y = 3sin (50t - x), 2, Where x and y are in meters and t is in seconds. The ratio of, maximum particle velocity to the wave velocity is, 3, 2, p, p, (a) 2p, (b), (c) 3p, (d), 2, 3, If the mass shown in figure is slightly displaced and then let, go, then the system shall oscillate with a time period of, (a), , 2p, , m, 3k, , (b), , 2p, , 3m, 2k, , (c), , 2p, , 2m, 3k, , 38., , 39., , 40., , k, , 42., k, , k, , 3k, m, m, A hollow sphere is filled with water. It is hung by a long, thread. As the water flows out of a hole at the bottom, the, period of oscillation will, (a) first increase and then decrease, (b) first decrease and then increase, (c) go on increasing, (d) go on decreasing, The figure shows a position time graph of a particle, executing SHM. If the time period of SHM is 2 sec, then the, equation of SHM is, x, (a) x = 10 cos pt, 10, pö, æ, (b) x = 5sin ç pt + ÷, 5, è, 3ø, t, pö, æ, (c) x = 10sin ç pt + ÷, è, 3 ø –10, pö, æ, (d) x = 10sin ç pt + ÷, è, 6ø, A coin is placed on a horizontal platform which undergoes, vertical simple harmonic motion of angular frequency w. The, amplitude of oscillation is gradually increased. The coin will, , (d), , 41., , 2p, , RESPONSE, GRID, , 36., 41., , 37., 42., , 43., , leave contact with the platform for the first time, (a) at the mean position of the platform, g, (b) for an amplitude of 2, w, g2, (c) for an amplitude of 2, w, (d) at the highest position of the platform, The bob of a simple pendulum executes simple harmonic, motion in water with a period t, while the period of oscillation, of the bob is t0 in air. Neglecting frictional force of water and, given that the density of the bob is (4/3) × 1000 kg/m3. The, relationship between t and t0 is, (a) t = 2t0 (b) t = t0/2 (c) t = t0, (d) t = 4t0, Starting from the origin a body oscillates simple harmonically, with a period of 2 s. After what time will its kinetic energy be, 75% of the total energy?, 1, 1, 1, 1, s, s, s, s, (a), (b), (c), (d), 6, 3, 4, 12, A body executes simple harmonic motion under the action, 4, of a force F1 with a time period s. If the force is changed, 5, 3, to F2, it executes S.H.M. with time period s. If both the, 5, forces F1 and F2 act simultaneously in the same direction, on the body, its time period in second is, 12, 7, 24, 5, (b), (c), (d), 25, 5, 25, 7, A block connected to a spring oscillates vertically. A damping, force Fd, acts on the block by the surrounding medium. Given, as Fd = –bV, b is a positive constant which depends on :, (a) viscosity of the medium, (b) size of the block, (c) shape of the block, (d) All of these, If a simple pendulum of length l has maximum angular, displacement q, then the maximum K.E. of bob of mass m is, , (a), , 44., , 45., , 1, ml / g, 2, (c) mgl (1 – cos q), , (a), , 38., 43., , 39., 44., , (b) mg / 2l, (d) mgl sin q/2, , 40., 45., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP13 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score – Qualifying Score, for Rough, Net Score = Space, (Correct, × 4) Work, – (Incorrect × 1), Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP14, , P-53, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP14, , SYLLABUS : Waves, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , Where should the two bridges be set in a 110cm long wire, so that it is divided into three parts and the ratio of the, frequencies are 3 : 2 : 1 ?, (a) 20cm from one end and 60cm from other end, (b) 30cm from one end and 70cm from other end, (c) 10cm from one end and 50cm from other end, (d) 50cm from one end and 40cm from other end, When a wave travel in a medium, the particle displacement, is given by the equation y = a sin 2p (bt– cx) where a, b and, c are constants. The maximum particle velocity will be twice, the wave velocity if, 1, 1, (b) c = pa (c) b = ac (d) b =, pa, ac, The wave described by y = 0.25 sin (10px – 2pt),, where x and y are in meters and t in seconds, is a wave, travelling along the:, (a) –ve x direction with frequency 1 Hz., (b) +ve x direction with frequency p Hz and wavelength l, = 0. 2 m., (c) +ve x direction with frequency 1 Hz and wavelength l, = 0.2 m, (d) –ve x direction with amplitude 0.25 m and wavelength l, = 0.2 m, , (a), , 3., , 4., , é xù, 4p êt - ú . When it is reflected at a rigid support, its, ë 2û, , 2, of its previous value. The equation, 3, of the reflected wave is, , amplitude becomes, , c=, , RESPONSE GRID, , 1., , 2., , The equation of a plane progressive wave is y = 0.9 sin, , (a), , é xù, y = 0 .6 sin 4 p ê t + ú, 2û, ë, , (b), , é xù, y = - 0.6 sin 4 p ê t + ú, ë 2û, , (c), , é xù, y = - 0.9 sin 8p ê t - ú, ë 2û, , é xù, y = - 0.6 sin 4 p ê t + ú, ë 2û, A person carrying a whistle emitting continuously a note of, 272 Hz is running towards a reflecting surface with a speed, of 18 km h–1. The speed of sound in air is 345 m s–1. The, number of beats heard by him is, (a) 4, (b) 6, (c) 8, (d) zero, , (d), , 5., , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP14, , P-54, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , A closed organ pipe (closed at one end) is excited to support, the third overtone. It is found that air in the pipe has, (a) three nodes and three antinodes, (b) three nodes and four antinodes, (c) four nodes and three antinodes, (d) four nodes and four antinodes, A wave disturbance in a medium is described by, pö, æ, y( x , t ) = 0.02 cos ç 50pt + ÷ cos(10px ) where x and y are, 2ø, è, in metre and t is in second. Which of the following is correct?, (a) A node occurs at x = 0.15 m, (b) An antinode occurs at x = 0.3 m, (c) The speed wave is 5 ms–1, (d) The wavelength is 0.3 m, In a resonance column, first and second resonance are, obtained at depths 22.7 cm and 70.2 cm. The third resonance, will be obtained at a depth, (a) 117.7 cm, (b) 92.9 cm, (c) 115.5 cm, (d) 113.5 cm, 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 seconds. 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 (c) 60 m/s, (d) 30 m/s, Two identical piano wires kept under the same tension T, have a fundamental frequency of 600 Hz. The fractional, increase in the tension of one of the wires which will lead to, occurrence of 6 beats/s when both the wires oscillate, together would be, (a) 0.02, (b) 0.03, (c) 0.04, (d) 0.01, Two sound sources emitting sound each of wavelength l, are fixed at a given distance apart. A listener moves with a, velocity u along the line joining the two sources. The number, of beats heard by him per second is, u, 2u, u, u, (a), (b), (c), (d), 2l, l, l, 3l, An observer moves towards a stationary source of sound,, with a velocity one-fifth of the velocity of sound. What is, the percentage increase in the apparent frequency ?, (a) 0.5%, (b) zero, (c) 20 %, (d) 5 %, Velocity of sound in air is 320 m s–1. A pipe closed at one, end has a length of 1 m. Neglecting end correction, the air, column in the pipe cannot resonate with sound of frequency, (a) 80 Hz, (b) 240 Hz (c) 320 Hz (d) 400 Hz, The driver of a car travelling with speed 30 m/sec towards, a hill sounds a horn of frequency 600 Hz. If the velocity of, sound in air is 330 m/s, the frequency of reflected sound as, heard by driver is, (a) 555.5 Hz (b) 720 Hz (c) 500 Hz (d) 550 Hz, , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , 15., , What will be the frequency of beats formed from the, superposition of two harmonic waves shown below?, y, 1.0, 0, , t(s), , –1.0, , (a), , y, 1.0, 0, , t(s), , –1.0, , 16., , 17., , 18., , (b), (a) 20 Hz, (b) 11 Hz (c) 9 Hz, (d) 2 Hz, What is the effect of increase in temperature on the frequency, of sound produced by an organ pipe?, (a) increases, (b) decreases, (c) no effect, (d) erratic change, A cylinderical tube open at both ends, has a fundamental, frequency f in air. The tube is dipped vertically in water so, that half of it is in water. The fundamental frequency of air, column is now, (a) f / 2, (b) f, (c) 3f / 4, (d) 2 f, The transverse displacement y (x, t) of a wave on a string is, given by y( x, t ) = e, This represents a:, , (, , - ax2 + bt 2 + 2 abxt, , )., , (a) wave moving in –x direction with speed, (b) standing wave of frequency, (c) standing wave of frequency, , b, 1, b, , (d) wave moving in + x direction with speed, 19., , 20., , b, a, , A longitudinal wave is represented by, , a, b, , xö, æ, x = x 0 sin 2 pç nt - ÷, l, ø, è, The maximum particle velocity will be four times the wave, velocity if, px 0, (b) l = 2px 0, (a) l =, 4, px, (c) l = 0, (d) l = 4px 0, 2, Two tones of frequencies n 1 and n2 are sounded together., The beats can be heard distinctly when, (a) 10 < (n1 – n 2) < 20, (b) 5 < (n1 – n 2) > 20, (c) 5 < (n1 – n 2) < 20, (d) 0 < (n1 – n 2) < 10, , 8., 13., 18., Space for Rough Work, , 9., 14., 19., , 10., 15., 20., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP14, , P-55, , 21. A pipe of length 85 cm is closed from one end. Find the, number of possible natural oscillations of air column in the, pipe whose frequencies lie below 1250 Hz. The velocity of, sound in air is 340 m/s., (a) 12, (b) 8, (c) 6, (d) 4, 22. A vehicle, with a horn of frequency n is moving with a velocity, of 30 m/s 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. Then (if the sound, velocity in air is 300 m/s), (a) n1 = 10n, (b) n1 = 0, (c) n1 = 0.1n, (d) n1 = – 0.1n, 23. A source of sound gives 5 beats per second, when sounded, with another source of frequency 100/sec. The second, harmonic of the source, together with a source of frequency, 205/sec gives 5 beats per second. What is the frequency of, the source?, (a) 95 sec–1, (b) 100 sec–1, (c) 105 sec–1, (d) 205 sec–1, 24. If we study the vibration of a pipe open at both ends, then, which of the following statements is not true ?, (a) Odd harmonics of the fundamental frequency will be, generated, (b) All harmonics of the fundamental frequency will be, generated, (c) Pressure change will be maximum at both ends, (d) Antinode will be at open end, 25. 41 forks are so arranged that each produces 5 beats per sec, when sounded with its near fork. If the frequency of last, fork is double the frequency of first fork, then the frequencies, (in Hz) of the first and the last fork are respectively., (a) 200, 400, (b) 205, 410, (c) 195, 390, (d) 100, 200, 26. Two points are located at a distance of 10 m and 15 m from, the source of oscillation. The period of oscillation is 0.05, sec and the velocity of the wave is 300 m/sec. What is the, phase difference between the oscillations of two points?, p, 2p, p, (a), (b), (c) p, (d), 3, 3, 6, 27. A sound absorber attenuates the sound level by 20 dB. The, intensity decreases by a factor of, (a) 100, (b) 1000, (c) 10000 (d) 10, 28. A wave travelling along the x-axis is described by the, equation y(x, t) = 0.005 cos (a x – bt). If the wavelength and, the time period of the wave are 0.08 m and 2.0s, respectively,, then a and b in appropriate units are, 0.08, 2.0, ,b =, (a) a = 25.00 p , b = p, (b) a =, p, p, 0.04, 1.0, p, ,b =, (c) a =, (d) a = 12.50p, b =, p, p, 2.0, , RESPONSE, GRID, , 21., 26., 31., , 22., 27., 32., , 29., , The equation Y = 0.02 sin (500pt) cos (4.5 x) represents, (a) progressive wave of frequency 250 Hz along x-axis, (b) a stationary wave of wavelength 1.4 m, (c) a transverse progressive wave of amplitude 0.02 m, (d) progressive wave of speed of about 350 m s–1, 30. Which of the following statements is/are incorrect about, waves ?, (a) Waves are patterns of disturbance which move without, the actual physical transfer of flow of matter as a whole., (b) Waves cannot transport energy., (c) The pattern of disturbance in the form of waves carry, information that propagate from one point to another., (d) All our communications essentially depend on, transmission of signals through waves., 31. An organ pipe P1, closed at one end vibrating in its first, harmonic and another pipe P2, open at both ends vibrating, in its third harmonic, are in resonance with a given tuning, fork. The ratio of the lengths of P1 and P2 is :, 8, 1, 1, 1, (b), (c), (d), 3, 6, 3, 2, Two vibrating tuning forks producing waves given by, y1 = 27 sin 600pt and y2 = 27 sin 604 pt are held near the ear, of a person, how many beats will be heard in three seconds, by him ?, (a) 4, (b) 2, (c) 6, (d) 12, A source of sound A emitting waves of frequency 1800 Hz, is falling towards ground with a terminal speed v. The, observer B on the ground directly beneath the source, receives waves of frequency 2150 Hz. The source A receives, waves, reflected from ground of frequency nearly: (Speed, of sound = 343 m/s), (a) 2150 Hz (b) 2500 Hz (c) 1800 Hz (d) 2400 Hz, Consider the three waves z 1, z2 and z3 as, z1 = A sin (kx – wt), z2 = A sin (kx + wt), z3 = A sin (ky – wt), Which of the following represents a standing wave?, (a) z1 + z2, (b) z2 + z3, (c) z3 + z1, (d) z1 + z2 + z3, A sonometer wire supports a 4 kg load and vibrates in, fundamental mode with a tuning fork of frequency 416 Hz., The length of the wire between the bridges is now doubled., In order to maintain fundamental mode, the load should be, changed to, (a) 1 kg, (b) 2 kg, (c) 4 kg, (d) 16 kg, , (a), , 32., , 33., , 34., , 35., , 23., 28., 33., Space for Rough Work, , 24., 29., 34., , 25., 30., 35.
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t.me/Magazines4all, , DPP/ CP14, , P-56, , 36., , The vibrations of a string of length 60 cm fixed at both the, , æ 4p x ö, ends are represented by the equation y = 2 sin ç, cos, è 15 ÷ø, (96pt) where x and y are in cm. The maximum number of, loops that can be formed in it is, (a) 4, (b) 16, (c) 5, (d) 15, 37. If n1, n2 and n3 are the fundamental frequencies of three, segments into which a string is divided, then the original, fundamental frequency n of the string is given by, (a) n = n1 + n2 + n3, 1 1, 1, 1, (b) n = n + n + n, 1, 2, 3, (c), (d), 38., 39., , 40., , 1, n, , =, , 1, n1, , +, , 1, n2, , +, , p, 3, (c) speed of propagation = 5, (d) period = p, 15, If the intensities of two interfering waves be I1 and I2, the, contrast between maximum and minimum intensity is, maximum, when, (a) I1 > > I2, (b) I1 << I2, (c) I1 = I2, (d) either I1 or I2 is zero, The fundamental frequency of a closed organ pipe of length, 20 cm is equal to the second overtone of an organ pipe open, at both the ends. The length of organ pipe open at both the, ends is, (a) 100 cm (b) 120 cm (c) 140 cm (d) 80 cm, The equation of a travelling wave is y = 60 cos (180 t – 6x), where y is in mm, t in second and x in metres. The ratio of, maximum particle velocity to velocity of wave propagation is, (a) 3.6 × 10–2, (b) 3.6 × 10–4, –6, (c) 3.6 × 10, (d) 3.6 × 10–11, , (b) wavelength = 4, , 43., , 37., 42., , A whistle of frequency 1000 Hz is sounded on a car travelling, towardsacliff with velocity of 18 m s–1 normal to the cliff. If, velocity of sound (v) = 330 m s–1 , then the apparent, frequency of the echo as heard by the car driver is nearly, (a) 1115 Hz (b) 115 Hz, (c) 67 Hz, (d) 47.2 Hz, The transverse wave represented by the equation, æ pö, y = 4 sin ç ÷ sin (3x - 15t ) has, è6ø, (a) amplitude = 4, , n3, , An echo repeats two syllables. If the velocity of sound is, 330 m/s, then the distance of the reflecting surface is, (a) 66.0 m (b) 33.0 m, (c) 99.0 m, (d) 16.5 m, What is the effect of humidity on sound waves when, humidity increases?, (a) Speed of sound waves is more, (b) Speed of sound waves is less, (c) Speed of sound waves remains same, (d) Speed of sound waves becomes zero, If the ratio of maximum to minimum intensity in beats is 49,, then the ratio of amplitudes of two progressive wave trains, is, (a) 7 : 1, (b) 4 : 3, (c) 49 : 1, (d) 16 : 9, , 36., 41., , 42., , 1, , n = n1 + n2 + n3, , RESPONSE, GRID, , 41., , 44., , 45., , 38., 43., , 39., 44., , 40., 45., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP14 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP15, , SYLLABUS : Electric Charges and Fields, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , The surface charge density of a thin charged disc of radius, R is s. The value of the electric field at the centre of the disc, s, is, . With respect to the field at the centre, the electric, 2 Î0, field along the axis at a distance R from the centre of the disc, reduces by, (a) 70.7% (b) 29.3% (c) 9.7%, (d) 14.6%, A solid conducting sphere of radius a has, a net positive charge 2Q. A conducting, spherical shell of inner radius b and outer, a, radius c is concentric with the solid sphere, b, and has a net charge – Q. The surface, c, charge density on the inner and outer, surfaces of the spherical shell will be, respectively, 2Q, Q, Q, Q, ,, ,, (a) (b) 4 pb 2 4 pc 2, 4 pb 2 4 pc 2, , 3., , 0,, , Q, , 4., , (d), , 2, , RESPONSE GRID, , 1., , Q, , (b) remains same, (c) increases, 5., , A, , (d) decreases, ABC is an equilateral triangle. Charges +q are placed at, each corner as shown in fig. The electric intensity at centre, +q A, O will be, (a), , Q, , 2., , (b), , (a) becomes zero, , ,0, 4pc, 4pc 2, Two equally charged, identical metal spheres A and B repel, each other with a force ‘F’. The spheres are kept fixed with, a distance ‘r’ between them. A third identical, but uncharged, sphere C is brought in contact with A and then placed at the, mid point of the line joining A and B. The magnitude of the, net electric force on C is, (c), , F, F, 3F, (c), (d), 2, 4, 4, In the figure, the net electric flux through the area A is, r r, f = E × A when the system is in air. On immersing the system, in water the net electric flux through the area, , (a) F, , (b), , 1 q, 4 p Îo r, , r, O, , r, , 1, q, 4 p Îo r 2, , +q, , +q, C, , B, , 1 3q, 4 p Îo r 2, (d) zero, , (c), , 3., Space for Rough Work, , 4., , r, , 5.
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t.me/Magazines4all, , DPP/ CP15, , P-58, , 6., , 7., , 8., , An electric dipole is placed in a uniform electric field. The, dipole will experience, (a) a force that will displace it in the direction of the field, (b) a force that will displace it in a direction opposite to the, field., (c) a torque which will rotate it without displacement, (d) a torque which will rotate it and a force that will displace, it, An uniform electric field E exists along positive x-axis. The, work done in moving a charge 0.5 C through a distance 2 m, along a direction making an angle 60° with x-axis is 10 J., Then the magnitude of electric field is, (a) 5 Vm–1 (b) 2 Vm–1 (c), 5 Vm –1 (d) 20 Vm–1, Which one of the following graphs represents the variation, of electric field with distance r from the centre of a charged, spherical conductor of radius R?, E, , r=R, , E, , qE, æ, çg+ m, ç, è g, , E, , (c), , (d), r=R, , r, , r=R, , 15., , r, , A hollow cylinder has a charge q coulomb within it. If f is, the electric flux in units of voltmeter associated with the, curved surface B, the flux linked with the plane surface A in, units of voltmeter will be, B, (a), , q, 2e 0, , (b), , f, 3, , ö, 1æ q, - f÷, ç, 2 è e0, ø, 10. If Ea be the electric field strength of a short dipole at a, point on its axial line and Ee that on the equatorial line at, the same distance, then, (a) Ee= 2Ea, (b) Ea = 2Ee, (c) Ea = Ee, (d) None of the above, 11. Three positive charges of equal value q are placed at vertices, of an equilateral triangle. The resulting lines of force should, be sketched as in, , (c), , q, - f (d), e0, , (a), 12., , (c), , (b), , 16., , A, , C, , (d), , 17., , 18., , 19., , Three point charges Q1, Q2, Q3 in the order are placed equally, spaced along a straight line. Q2 and Q3 are equal in magnitude, but opposite in sign. If the net force on Q3 is zero. The value, of Q1 is, , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , Q = 2(Q 3 ), , 100Q, 100Q, 10Q, Q, (b), (c), (d), pe 0, e0, pe 0, e0, 14. A small sphere carrying a charge ‘q’ is hanging in between, two parallel plates by a string of length L. Time period of, pendulum is T0. When parallel, plates are charged, the time, L, m, period changes to T. The ratio, T/T0 is equal to, , r, , r=R, , (b), , (a), , (b), r, , Q1 = 4(Q 3 ), , (c) Q1 = 2 (Q 3 ), (d) Q1 =| Q 3 |, 13. Electric charge is uniformly distributed along a long straight, wire of radius 1 mm. The charge per cm length of the wire is, Q coulomb. Another cylindrical surface of radius 50 cm and, length 1 m symmetrically encloses the wire. The total electric, flux passing through the cylindrical surface is, , E, , (a), , 9., , (a), , 1/ 2, , ö, ÷, ÷, ø, , 3/ 2, , æ g ö, qE ÷, (a), (b) ç, çg+, ÷, m ø, è, 1/ 2, 5/2, æ g ö, (c) ç, (d) æ g ö, ÷, ç, qE, qE ÷, çg+, ÷, çg+, ÷, m ø, è, m ø, è, An electric dipole, consisting of two opposite charges of, 2 ´ 10 -6 C each separated by a distance 3 cm is placed in, an electric field of 2 ´ 10 5 N/C. Torque acting on the dipole is, (b) 12 ´ 10 - 2 N - m, (a) 12 ´ 10 -1 N - m, (c) 12 ´ 10 - 3 N - m, (d) 12 ´ 10 - 4 N - m, The electric field in a certain region is acting radially outward, and is given by E = Ar. A charge contained in a sphere of, radius 'a' centred at the origin of the field, will be given by, (a) A e0 a2 (b) 4 pe0 Aa3 (c) e0 Aa3 (d) 4 pe0 Aa2, The spatial distribution of electric field due to charges (A, B), is shown in figure. Which one of the following statements is, correct?, (a) A is +ve and B –ve, |A| > |B|A, B, (b) A is –ve and B +ve, |A| = |B|, (c) Both are +ve but A > B, (d) Both are –ve but A > B, Point charges + 4q, –q and +4q are kept on the X-axis at, points x = 0, x =a and x = 2a respectively., (a) only – q is in stable equilibrium, (b) none of the charges is in equilibrium, (c) all the charges are in unstable equilibrium, (d) all the charges are in stable equilibrium., Figure shows some of the electric field, A, B, C, lines corresponding to an electric field., The figure suggests that, (a) EA > EB > EC, (b) EA = EB = EC, (c) EA = EC > EB, (d) EA = EC < EB, , 8., 13., 18., Space for Rough Work, , 9., 14., 19., , 10., 15., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP15, , P-59, , 20. For distance far away from centre of dipole the change in, magnitude of electric field with change in distance from the, centre of dipole is, (a) zero., (b) same in equatorial plane as well as axis of dipole., (c) more in case of equatorial plane of dipole as compared, to axis of dipole., (d) more in case of axis of dipole as compared to equatorial, plane of dipole., 21. Two charge q and –3q are placed fixed on x–axis, separated by distance d. Where should a third charge 2q, be placed such that it will not experience any force ?, q, –3q, d, , A, , B, , d + 3d, d - 3d, d + 3d, d - 3d, (c), (b), (d), 2, 2, 2, 2, 22. A charge Q is placed at each of the opposite corners of a, square. A charge q is placed at each of the other two corners., If the net electrical force on Q is zero, then Q/q equals:, 1, (d) -2 2, (a) –1, (b) 1, (c) 2, 23. Identify the wrong statement in the following. Coulomb's, law correctly describes the electric force that, (a) binds the electrons of an atom to its nucleus, (b) binds the protons and neutrons in the nucleus of an, atom, (c) binds atoms together to form molecules, (d) binds atoms and molecules together to form solids, 24. An oil drop of radius r and density r is held stationary in a, uniform vertically upwards electric field 'E'. If r0 (< r) is the, density of air and e is quanta of charge, then the drop has–, , (a), , (a), (b), , 4pr3 (r - r0 ) g, excess electrons, 3eE, 4 p r 2 (r - r0 ) g, eE, , (c) deficiency of, (d) deficiency of, , p, 3p, (c) p, (d), 2, 2, 27. Which of the following statements is incorrect?, (a) The charge q on a body is always given by q = ne,, where n is any integer, positive or negative., (b) By convention, the charge on an electron is taken to, be negative., (c) The fact that electric charge is always an integral, multiple of e is termed as quantisation of charge., (d) The quatisation of charge was experimentally, demonstrated by Newton in 1912., 28. Two positive ions, each carrying a charge q, are separated, by a distance d. If F is the force of repulsion between the, ions, the number of electrons missing from each ion will be, (e being the charge of an electron), , (a) zero, , (a), , (c), , 4pe0 Fd 2, , (b), , e2, , 4pe0 Fe 2, d2, 4 pe0 Fd 2, , 4pe0 Fd 2, , 9.1 ´10-31 kg and 1.6 ´10 -19 C .), , 4pr3 (r - r0 ) g, electrons, 3eE, 4 pr 2 (r - r0 ) g, electrons, eE, , 20., 25., 30., , (b), , (d), q2, e2, 29. Two small similar metal spheres A and B having charges 4q, and – 4q, when placed at a certain distance apart, exert an, electric force F on each other. When another identical, uncharged sphere C, first touched with A then with B and, then removed to infinity, the force of interaction between A, and B for the same separation will be, (a) F/2, (b) F/8, (c) F/16, (d) F/32, 30. The electric field intensity just sufficient to balance the, earth’s gravitational attraction on an electron will be: (given, mass and charge of an electron respectively are, , excess electrons, , (a), , 21., 26., 31., , - 5.6 ´10 -11 N / C, , (b), , - 4.8 ´10 -15 N / C, , (c) - 1.6 ´10 -19 N / C, (d) - 3.2 ´10 -19 N / C, An electric dipole is placed at an angle of 30° with an electric, field of intensity 2 × 105 NC–1, It experiences a torque of 4, Nm. Calculate the charge on the dipole if the dipole length is, 2 cm., (a) 8 mC, (b) 4 mC, (c) 8 mC, (d) 2 mC, 32. A particle of mass m and charge q is placed at rest in a, uniform electric field E and then released. The kinetic energy, attained by the particle after moving a distance y is, (a) qEy2, (b) qE2 y (c) qEy, (d) q2 Ey, 33. There is an electric field E in x-direction. If the work done on, moving a charge of 0.2 C through a distance of 2 m along a, line making an angle 60° with x-axis is 4 J, then what is the, value of E?, (a) 3 N/C (b) 4 N/C (c) 5 N/C (d) 20 N/C, 31., , 25. A square surface of side L meter in the, E, plane of the paper is placed in a uniform, electric field E (volt/m) acting along the, q, same plane at an angle q with the, horizontal side of the square as shown, in Figure. The electric flux linked to the, surface, in units of volt. m, is, (a) EL2, (b) EL2 cos q, 2, (c) EL sin q, (d) zero, ur, 26. An electric dipole of moment p placed in a uniform electric, field E has minimum potential energy when the angle, , RESPONSE, GRID, , between P and E is, , 22., 27., 32., Space for Rough Work, , 23., 28., 33., , 24., 29.
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t.me/Magazines4all, , DPP/ CP15, , P-60, , 34., , r, A surface has the area vector A = 2$i + 3 $j m 2 . The flux of an, , (, , ), , r, , electric field through it if the field is E = 4$i, 35., , V, :, m, , 39., , (a) 8 V-m (b) 12 V-m (c) 20 V-m (d) zero, There exists a non!-uniform electric field along x-axis as, shown in the figure below. The field increases at a uniform, rate along +ve x-axis. A dipole is placed inside the field as, shown. Which one of the following is correct for the dipole?, –q, a, , +q, , q, , (a) zero, , x-axis, , (a) Dipole moves along positive x-axis and undergoes a, clockwise rotation, (b) Dipole moves along negative x-axis and undergoes a, clockwise rotation, (c) Dipole moves along positive x-axis and undergoes a, anticlockwise rotation, (d) Dipole moves along negative x-axis and undergoes a, anticlockwise rotation, 36. A square surface of side L metres is in, the plane of the paper. A uniform, uur, electric field E (volt /m), also in the, E, plane of the paper, is limited only to, the lower half of the square surface (see figure). The electric, flux in SI units associated with the surface is, (a) EL2/2 (b) zero, (c) EL2, (d) EL2 / (2e0), 37. Among two discs A and B, first have radius 10 cm and charge, 10 -6 µC and second have radius 30 cm and charge 10 -5 C., When they are touched, charge on both q A and q B, respectively will, be, (a) q A =2.75μC, q B =3.15μC, (b), , 40., , q A =1.09 μC, q B =1.53μC, , 34., 39., 44., , (c) q/2eo, (d) 2q/eo, If the electric flux entering and leaving a closed surface are, 6 × 106 and 9 × 106 S.I. units respectively, then the charge, inside the surface of permittivity of free space e0 is, (a) e0 × 106, (b) – e0 × 106, (c) – 2e0 × 106, (d) 3e0 × 106, 42. Two particle of equal mass m and charge q are placed at a, distance of 16 cm. They do not experience any force. The, q, value of, is, m, G, pe0, 4pe0 G, (c), (d), 4pe 0, G, A rod of length 2.4 m and radius 4.6 mm carries a negative, charge of 4.2 × 10–7 C spread uniformly over it surface., The electric field near the mid–point of the rod, at a point, on its surface is, (a) –8.6 × 105 N C–1, (b) 8.6 × 104 N C–1, 5, –1, (c) –6.7 × 10 N C, (d) 6.7 × 104 N C–1, A hollow insulated conduction sphere is given a positive, charge of 10 mC. What will be the electric field at the centre, of the sphere if its radius is 2 m?, (a) Zero, (b) 5 mCm–2, –2, (c) 20 mCm, (d) 8 mCm–2, A charge Q is enclosed by a Gaussian spherical surface of, radius R. If the radius is doubled, then the outward electric, flux will, (a) increase four times, (b) be reduced to half, (c) remain the same, (d) be doubled, , (a) l, , 43., , 44., , 45., , 35., 40., 45., , (b) q/eo, , 41., , (c) q A = q B =5.5μC, (d) None of these, 38. The total electric flux emanating from a closed surface, enclosing an a-particle (e-electronic charge) is, , RESPONSE, GRID, , 2e, e0 e, e, (a), (b), (c) ee0, (d), e0, e0, 4, Which of the following is a wrong statement?, (a) The charge of an isolated system is conserved, (b) It is not possible to create or destroy charged, particles, (c) It is possible to create or destroy charged particles, (d) It is not possible to create or destroy net charge, A charge q is placed at the centre of the open end of a, cylindrical vessel. The flux of the electric field through the, surface of the vessel is, , 36., 41., , (b), , 37., 42., , 38., 43., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP15 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP16, , P-61, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP16, , SYLLABUS : Electrostatic Potential & Capacitance, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , If n drops, each charged to a potential V, coalesce to form a, single drop. The potential of the big drop will be, V, V, (a), (b), (c) Vn1/3, (d) Vn2/3, 2/3, 1/ 3, n, n, The capacitance of a parallel plate capacitor is Ca (Fig. a). A, dielectric of dielectric constant K is inserted as shown in, fig. (b) and (c). If Cb and Cc denote the capacitances in fig., (b) and (c), then, d/2, , d, , 4., , A parallel plate condenser is immersed in an oil of dielectric, constant 2. The field between the plates is, (a) increased, proportional to 2, 1, 2, (c) increased, proportional to – 2, , (b) decreased, proportional to, , 1, 2, What is the effective capacitance between points X and Y?, , (d) decreased, proportional to 5., , K, (a), , Ca, , Cb, , C1=6m F, (b), , X, K, , d, , (c), , 6., , Cc, , 3., , (a) both Cb, Cc > Ca, (b) Cc > Ca while Cb > Ca, (c) both Cb, Cc < Ca, (d) Ca = Cb = Cc, The electric potential V(x) in a region around the origin is, given by V(x) = 4x2 volts. The electric charge enclosed in a, cube of 1 m side with its centre at the origin is (in coulomb), (a) 8e0, (b) – 4e0, (c) 0, (d) – 8e0, , RESPONSE, GRID, , 1., 6., , 2., , C3 =6m F C5=20m F C2=6m F, Y, B, A, C, D, C4=6m F, , (a) 24 mF (b) 18 mF (c) 12 mF (d) 6 mF, Two identical particles each of mass m and having charges, – q and +q are revolving in a circle of radius r under the, influence of electric attraction. Kinetic energy of each, æ, 1 ö, particle is ç k =, 4pe 0 ÷ø, è, , (a) kq2/4r, , 3., Space for Rough Work, , (b) kq2/2r, , 4., , (c) kq2/8r, , 5., , (d) kq2/r
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t.me/Magazines4all, , DPP/ CP16, , P-62, , 7., , Four metallic plates each with a surface area of one side A,, are placed at a distance d from each other. The two outer, plates are connected to one point A and the two other inner, plates to another point B as shown in the figure. Then the, capacitance of the system is, , (c), , (d), V, , V, , S, , S, , 13., A, , B, , e0A, 2e 0 A, 3e 0 A, 4e 0 A, (b), (c), (d), d, d, d, d, A parallel plate condenser with a dielectric of dielectric, constant K between the plates has a capacity C and is, charged to a potential V volt. The dielectric slab is slowly, removed from between the plates and then reinserted. The, net work done by the system in this process is, 1, (a) zero, (b), ( K - 1) CV 2, 2, , (a), , 8., , CV 2 ( K - 1), (d) ( K - 1) CV 2, K, 9., If a slab of insulating material 4 × 10–5 m thick is introduced, between the plates of a parallel plate capacitor, the distance, between the plates has to be increased by 3.5 × 10–5 m to, restore the capacity to original value. Then the dielectric, constant of the material of slab is, (a) 8, (b) 6, (c) 12, (d) 10, 10. A unit charge moves on an equipotential surface from a, point A to point B, then, (a) VA – VB = + ve, (b) VA – VB = 0, (c) VA – VB = – ve, (d) it is stationary, 11. Identify the false statement., (a) Inside a charged or neutral conductor, electrostatic, field is zero, (b) The electrostatic field at the surface of the charged, conductor must be tangential to the surface at any, point, (c) There is no net charge at any point inside the, conductor, (d) Electrostatic potential is constant throughout the, volume of the conductor, 12. In a hollow spherical shell, potential (V) changes with respect, to distance (s) from centre as, , 14., , 15., , (c), , (a), , (b), V, , V, , S, , S, , RESPONSE, GRID, , 7., 12., 17., , 8., 13., 18., , 16., , 17., , 18., , The 1000 small droplets of water each of radius r and charge, Q, make a big drop of spherical shape. The potential of big, drop is how many times the potential of one small droplet ?, (a) 1, (b) 10, (c) 100, (d) 1000, The work done in carrying a charge q once around a circle, of radius r with a charge Q placed at the centre will be, (a) Qq(4pe0r2), (b) Qq/(4pe0r), (c) zero, (d) Qq2/(4pe0r), A parallel plate condenser is filled with two, dielectrics as shown. Area of each plate is, A m2 and the separation is t m. The, k1 k2, dielectric constants are k 1 and k 2, respectively. Its capacitance in farad will be, eo A, eo A k1 + k 2, ., (a), (k1 + k2), (b), t, t, 2, eo A k1 - k2, 2e o A, ., (c), (k1 + k2), (d), t, t, 2, Two metal pieces having a potential difference of 800 V are, 0.02 m apart horizontally. A particle of mass 1.96 × 10–15 kg, is suspended in equilibrium between the plates. If e is the, elementary charge, then charge on the particle is, (a) 8, (b) 6, (c) 0.1, (d) 3, A one microfarad capacitor of a TV is subjected to 4000 V, potential difference. The energy stored in capacitor is, (a) 8 J, (b) 16 J, (c) 4 × 10–3 J, (d) 2 × 10–3 J, An unchanged parallel plate capacitor filled with a dielectric, constant K is connected to an air filled identical parallel, capacitor charged to potential V1. If the common potential, is V2, the value of K is, V1 - V2, V1, (b), (a), V1, V1 - V2, , V2, V1 - V2, , V1 - V2, V2, 19. In the circuit given below, the, 2mF, 3mF, charge in mC, on the capacitor, e, having 5 mF is, 5mF, f, (a) 4.5, 4mF, (b) 9, +, a, (c) 7, 6V, (d) 15, , (c), , 9., 14., 19., Space for Rough Work, , (d), , 10., 15., , 11., 16., , d, c, b, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP16, , P-63, , 20. Two concentric, thin metallic spheres of radii R1 and R2, (R1 > R2) bear charges Q1 and Q2 respectively. Then the, potential at distance r between R 1 and R 2 will be, æ Q1 Q 2 ö, æ Q1 + Q2 ö, +, (a) k ç, (b) k ç, ÷, è, ø, è r R 2 ÷ø, r, æ Q2 Q1 ö, æ Q1 Q2 ö, +, (c) k ç r + R ÷, (d) k ç, è, è R1 R 2 ÷ø, 1ø, 21. Charge Q on a capacitor varies y, A, with voltage V as shown in the, figure, where Q is taken along the, X-axis and V along the Y-axis. The, V, area of triangle OAB represents, (a) capacitance, (b) capacitive reactance, x, (c) magnetic field between the plates O, B, Q, (d) energy stored in the capacitor, 22. An alpha particle is accelerated through a potential difference, of 106 volt. Its kinetic energy will be, (a) 1 MeV (b) 2 MeV (c) 4 MeV (d) 8 MeV, 23. Four point charges –Q, –q, 2q and 2Q are placed, one at, each corner of the square. The relation between Q and q for, which the potential at the centre of the square is zero is :, 1, 1, (a) Q = – q (b) Q = – (c) Q = q (d) Q =, q, q, 24. 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 one with area 1/3A,, dielectric constant 2K and another with area 2/3A and, dielectric constant K. If the capacitance of this new capacitor, C, is C then, is, C0, (a) 1, (b) 4/3, (c) 2/3, (d) 1/3, 25. Two conden sers, on e of, capacity C and other of, capacity C/2 are connected to, V, a V-volt battery, as shown. The, C, C/2, work done in charging fully, both the condensers is, 1, 3, 1, CV 2 (b), CV 2 (c), CV 2 (d) 2CV2., (a), 4, 4, 2, 26. A, B and C are three points, A, in a uniform electric field., ®, B, E, The electric potential is, (a) maximum at B, C, (b) maximum at C, (c) same at all the three points A, B and C, (d) maximum at A, , RESPONSE, GRID, , 20., 25., 30., , 21., 26., 31., , 27., , 28., , 29., , 30., , 31., , Three capacitors are connected in the, arms of a triangle ABC as shown in, figure 5 V is applied between A and B., The voltage between B and C is, , 33., , 2µF, , 3µF, , A, B, (a) 2 V, (b) 1 V, 2µF, (c) 3 V, (d) 1.5 V, Two parallel metal plates having charges + Q and –Q face, each other at a certain distance between them. If the plaves, are now dipped in kerosene oil tank, the electric field between, the plates will, (a) remain same, (b) become zero, (c) increases, (d) decrease, An air capacitor C connected to a battery of e.m.f. V acquires, a charge q and energy E. The capacitor is disconnected, from the battery and a dielectric slab is placed between the, plates. Which of the following statements is correct ?, (a) V and q decrease but C and E increase, (b) V remains unchange, but q, E and C increase, (c) q remains unchanged, C increases, V and E decrease, (d) q and C increase but V and E decrease., Choose the wrong statement about equipotential surfaces., (a) It is a surface over which the potential is constant, (b) The electric field is parallel to the equipotential surface, (c) The electric field is perpendicular to the equipotential, surface, (d) The electric field is in the direction of steepest decrease, of potential, Two spherical conductors A and B of radii a, b, B, and b (b>a) are placed concentrically in air., A, a, The two are connected by a copper wire as, shown in figure. Then the equivalent, capacitance of the system is, , (a), 32., , C, , 4pe 0, , ab, b-a, , (b) 4pe0(a + b), , (c) 4pe0b, (d) 4pe0a, A capacitor is charged to store an energy U. The charging, battery is disconnected. An identical capacitor is now, connected to the first capacitor in parallel. The energy in, each of the capacitors is, (a) U / 2, (b) 3U / 2 (c) U, (d) U / 4, Equipotentials at a great distance from a collection of charges, whose total sum is not zero are approximately, (a) spheres, (b) planes, (c) paraboloids, (d) ellipsoids, , 22., 27., 32., Space for Rough Work, , 23., 28., 33., , 24., 29.
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t.me/Magazines4all, , DPP/ CP16, , P-64, , 34., , Which of the following figure shows the correct, equipotential surfaces of a system of two positive charges?, (a), , +, , (b), , +, , +, , +, , 40., (c), 35., , +, , (d), , +, , +, , +, , Two identical metal plates are given positive charges Q1, and Q2 ( < Q1) respectively. If they are now brought close, together to form a parallel plate capacitor with capacitance, C, the potential difference between them is, (a), , Q – Q2, Q – Q2, Q1 + Q 2, Q + Q2, (b) 1, (c) 1, (d) 1, C, 2C A, 2C, C, , 2, A1, The capacitance of the capacitor of plate, areas A1 and A2 (A1 < A2) at a distance d,, as shown in figure is, Î0 (A1 + A 2 ), Î0 A 2, (b), (a), 2d, d, Î, 0 A1, (c) Î0 A1A 2, (d), d, d, d, 37. In a given network the, C1, equivalent, capacitance, C2, between A and B is [C1 = C4 =, A, B, 1 mF, C2 = C3 = 2mF], C3, (a) 3 mF, (b) 6 mF, C4, (c) 4.5 mF (d) 2.5 mF, 38. A parallel plate air capacitor is charged to a potential, difference of V volts. After disconnecting the charging, battery the distance between the plates of the capacitor is, increased using an insulating handle. As a result the, potential difference between the plates, (a) does not change, (b) becomes zero, (c) increases, (d) decreases, +Q, 39. Figure shows three circular arcs,, each of radius R and total charge as, 45°, indicated. The net electric potential, 30°, –2Q, at the centre of curvature is, R, , 36., , 41., , 42., , 43., , 44., , 45., , 34., 39., 44., , 35., 40., 45., , Q, 2pe 0 R, , (b), , Q, 4pe 0 R, , (c), , 2Q, pe0 R, , (d), , Q, pe0 R, , r, -1 exists in a region of, $, An electric field E = (25i$ + 30j)NC, space. If the potential at the origin is taken to be zero then, the potential at x = 2 m, y = 2 m is :, (a) –110 V (b) –140 V (c) –120 V (d) –130 V, If a unit positive charge is taken from one point to another, over an equipotential surface, then, (a) work is done on the charge, (b) work is done by the charge, (c) work done is constant, (d) no work is done, Three large plates A, B and C are placed, parallel to each other and charges are, given as shown. The charge that appears, on the left surface of plate B is, (a) 5C, (b) 6C, (c) 3C, (d) –3 C, Three charges 2 q, – q and – q are located at the vertices of, an equilateral triangle. At the centre of the triangle, (a) the field is zero but potential is non-zero, (b) the field is non-zero, but potential is zero, (c) both field and potential are zero, (d) both field and potential are non-zero, If a charge – 150 nC is given to a concentric spherical shell, and a charge +50 nC is placed at its centre then the charge, on inner and outer surface of the shell is, (a) –50 nC, –100 nC, (b) +50 nC, –200 nC, (c) –50 nC, –200 nC, (d) 50 nC, 100 nC, Two capacitors of capacitances C1 and C2 are connected in, parallel across a battery. If Q1 and Q2 respectively be the, Q, charges on the capacitors, then 1 will be equal to, Q2, (a), , +3Q, , RESPONSE, GRID, , (a), , 36., 41., , C2, C1, , (b), , 37., 42., , C1, C2, , (c), , C12, , (d), , C 22, , 38., 43., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP16 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , C 22, C12, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP17, , SYLLABUS : Current Electricity, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , When 5V potential difference is applied across a wire of, length 0.1 m, the drift speed of electrons is 2.5 × 10 –4 ms–1. If, the electron density in the wire is 8 × 10 28 m–3, the resistivity, of the material is close to :, (a) 1.6 × 10–6 Wm, (b) 1.6 × 10–5 Wm, –8, (c) 1.6 × 10 Wm, (d) 1.6 × 10–7 Wm, Variation of current passing through a conductor as the, voltage applied across its ends is varied as shown in the, adjoining diagram. If the resistance (R) is determined at the, points A, B, C and D, we will find that, V, (a) RC = RD, C D, B, (b) RB > RA, A, , (a), (c), 4., , 3., , RESPONSE GRID, , 1., 6., , 2., , 30 ( E - 0.5i ), 100, , 5., , 6., , (b), , 30 E, (100 - 0.5), , (d), , 30 E, 100, , The masses of the three wires of copper are in the ratio of, 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio, of their electrical resistance is, (a) 1 : 3 : 5, , (c) RC > RB, , i, (d) RA > RB, The length of a wire of a potentiometer is 100 cm, and the e., m.f. of its standard cell is E volt. It is employed to measure, the e.m.f. of a battery whose internal resistance is 0.5W. If, the balance point is obtained at l = 30 cm from the positive, end, the e.m.f. of the battery is, , 30 E, 100.5, , (b) 5 : 3 : 1, , (c) 1 : 25 : 125, (d) 125 : 15 : 1, n equal resistors are first connected in series and then, connected in parallel. What is the ratio of the maximum to, the minimum resistance ?, (a) n, (b) 1/n2, (c) n2, (d) 1/ n, A battery is charged at a potential of 15V for 8 hours when, the current flowing is 10A. The battery on discharge, supplies a current of 5A for 15 hours. The mean terminal, voltage during discharge is 14V. The “watt-hour” efficiency, of the battery is, (a) 87.5% (b) 82.5% (c) 80%, (d) 90%, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP17, , P-66, , 7., , (b) 6 V, , R, , 55W, , + –, 12 V, 8W, , (a) 12 V, , Shown in the figure below is a meter-bridge set up with null, deflection in the galvanometer., , 6W, , (c) 4 V, , E, , (d) 2 V, + –, 16. Find emf E of the cell as shown in figure., , G, 20 cm, , 12V, , 2W, , I, , D, , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , The value of the unknown resistor R is, (a) 13.75 W (b) 220 W (c) 110 W (d) 55 W, In the equation AB = C, A is the current density, C is the, electric field, Then B is, (a) resistivity, (b) conductivity, (c) potential difference, (d) resistance, The Kirchhoff's first law (Si = 0) and second law (SiR = SE),, are respectively based on, (a) conservation of charge, conservation of momentum, (b) conservation of energy, conservation of charge, (c) conservation of momentum, conservation of charge, (d) conservation of charge, conservation of energy, You are given a resistance coil and a battery. In which of, the following cases the largest amount of heat generated ?, (a) When the coil is connected to the battery directly, (b) When the coil is divided into two equal parts and both, the parts are connected to the battery in parallel, (c) When the coil is divided into four equal parts and all, the four parts are connected to the battery in parallel, (d) When only half the coil is connected to the battery, The resistance of the coil of an ammeter is R. The shunt, required to increase its range n-fold should have a resistance, R, R, R, (a), (b), (c), (d) nR, n -1, n, n +1, On increasing the temperature of a conductor, its resistance, increases because the, (a) relaxation time increases, (b) mass of electron increases, (c) electron density decreases, (d) relaxation time decreases, An electric current is passed through a circuit containing, two wires of the same material, connected in parallel. If the, 4, 2, lengths and radii are in the ratio of and , then the ratio, 3, 3, of the current passing through the wires will be, (a) 8/9, (b) 1/3, (c) 3, (d) 2, In a meter bridge experiment null point is obtained at 20 cm., from one end of the wire when resistance X is balanced against, another resistance Y. If X < Y, then where will be the new, position of the null point from the same end, if one decides to, balance a resistance of 4 X against Y, (a) 40 cm (b) 80 cm (c) 50 cm (d) 70 cm, In the circuit shown, the current through 8 ohm is same, before and after connecting E. The value of E is, , RESPONSE, GRID, , 7., 12., 17., , 10 W, , 8., 13., 18., , 1W, , E, , A, , C, B, , 1A, , F, , E, 2A, , (a) 15V, (b) 10V, (c) 12V, (d) 5V, 17. A torch bulb rated as 4.5 W, 1.5 V is connected as shown in, fig. The e.m.f. of the cell, needed to make the bulb glow at, full intensity is, 4.5 W,, 1.5V, , 2E/9, E/9, , 0.33 W, , E/3, , E, r = 2.67 W, , (a) 4.5 V, (b) 1.5 V, (c) 2.67 V (d) 13.5 V, 18. In a given network, each resistance has value of 6W. The, point X is connected to point A by a copper wire of negligible, resistance and point Y is connected to point B by the same, wire. The effective resistance between X and Y will be, 6W, , X, , 19., , 6W A6W, , Y, , (a) 18W, (b) 6W, (c) 3W, (d) 2W, If N, e, t and m are representing electron density, charge,, relaxation time and mass of an electron respectively, then, the resistance of wire of length l and cross-sectional area A, is given by, , Ne 2 t A, Ne 2 A, (d), 2ml, 2m tl, Ne 2 l, Ne 2 At, Cell having an emf e and internal resistance r is connected, across a variable external resistance R. As the resistance R, is increased, the plot of potential difference V across R is, given by :, (a) e, (b) e, 2ml, , (a), , 20., , B, , (b), , 2mtA, , (c), , V, , V, , (c), , 0, , Space for Rough Work, , R, , (d), , e, V, 0, , 9., 14., 19., , 0, , R, , V, , 0, , R, , 10., 15., 20., , R, , 11., 16., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP17, , P-67, , 21. If voltage across a bulb rated 220 Volt-100 Watt drops by, 2.5% of its rated value, the percentage of the rated value by, which the power would decrease is :, (a) 20%, (b) 2.5%, (c) 5%, (d) 10%, 22. If specific resistance of a potentiometer wire is 10–7 Wm, the, current flow through it is 0.1 A and the cross-sectional area, of wire is 10–6 m2 then potential gradient will be, (a) 10–2 volt/m, (b) 10–4 volt/m, (c) 10–6 volt/m, (d) 10–8 volt/m, 23. Two resistances R1 and R2 are made of different materials., The temperature coefficient of the material of R1 is a and, that of material of R2 is – b. The resistance of the series, combination of R1 and R2 will not change with temperature, R1, equal to, R2, a, a 2 + b2, b, a +b, (a), (c), (b), (d), 2ab, a, b, a -b, Five cells each of emf E and internal resistance r send the, same amount of current through an external resistance R, whether the cells are connected in parallel or in series. Then, R, the ratio æç ö÷ is, è rø, 1, 1, (c), (d) 1, (a) 2, (b), 2, 5, The length of a given cylindrical wire is increased by 100%., Due to the consequent decrease in diameter the change in, the resistance of the wire will be, (a) 200% (b) 100% (c) 50%, (d) 300%, Potentiometer wire of length 1 m is connected in series with, 490W resistance and 2 V battery. If 0.2 mV/cm is the potential, gradient, then resistance of the potentiometer wire is, (a) 4.9 W (b) 7.9 W (c) 5.9 W, (d) 6.9 W, See the electric circuit shown in the figure. R, Which of the following, equations is a correct equation, for it?, r1, i1 e 1, (a) e2 – i2 r2 – e1 – i1 r1 = 0, (b) - e2 – (i1 + i2) R+ i2 r2 = 0, (c) e1 – (i1 + i2) R + i1 r1 = 0, i2, e2, r2, (d) e1 – (i1 + i2) R– i1 r1 = 0, In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100, W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of, electric mains is 220 V. The minimum capacity of the main, fuse of the building will be:, (a) 8 A, (b) 10 A, (c) 12 A, (d) 14 A, Two sources of equal emf are connected to an external, resistance R. The internal resistance of the two sources are, R1 and R2 (R1 > R1). If the potential difference across the, source having internal resistance R2 is zero, then, (a) R = R2 – R1, (b) R = R2 × (R1 + R2)/(R2 – R1), (c) R = R1R2/(R2 – R1), , 30., , 31., , 32., , if, , 24., , 25., , 26., , 27., , 28., , 29., , (d) R = R1R2 /( R1 - R2 ), , RESPONSE, GRID, , 21., 26., 31., 36., , 22., 27., 32., , 33., , 34., , The resistance of the series combination of two resistances, is S. when they are joined in parallel the total resistance is P., If S = nP then the minimum possible value of n is, (a) 2, (b) 3, (c) 4, (d) 1, If an ammeter is to be used in place of a voltmeter, then we, must connect with the ammeter a, (a) low resistance in parallel, (b) high resistance in parallel, (c) high resistance in series, (d) low resistance in series., A d.c. main supply of e.m.f. 220 V is connected across a, storage battery of e.m.f. 200 V through a resistance of 1W., The battery terminals are connected to an external resistance, ‘R’. The minimum value of ‘R’, so that a current passes, through the battery to charge it is:, (a) 7 W, (b) 9 W, (c) 11 W, (d) Zero, In the given circuit diagram when the current reaches steady, state in the circuit, the charge on the capacitor of capacitance, C will be :, r2, (a) CE, (r + r2 ), r1, (b) CE, (r1 + r), r2, (c) CE, r, ( + r1 ), r, (d) CE 1, (r2 + r), Suppose the drift velocity nd in a material varied with the, applied electric field E as nd µ E . Then V – I graph for a, wire made of such a material is best given by :, V, , V, , (a), , (b), I, , I, , V, , V, , (c), , (d), I, , I, , 35. In a neon gas discharge tube Ne+ ions moving through a, cross-section of the tube each second to the right is 2.9 ×, 1018, while 1.2 × 1018 electrons move towards left in the same, time; the electronic charge being 1.6 × 10–19 C, the net electric, current is, (a) 0.27 A to the right, (b) 0.66 A to the right, (c) 0.66 A to the left, (d) zero, 36. Two rods are joined end to end, as shown. Both have a, cross-sectional area of 0.01 cm2. Each is 1 meter long. One, rod is of copper with a resistivity of 1.7 × 10–6 ohm-centimeter,, the other is of iron with a resistivity of 10–5 ohm-centimeter., How much voltage is required to produce a current of 1, V, ampere in the rods?, (a) 0.117 V, (b) 0.00145 V, (c) 0.0145 V, Cu Fe, (d) 1.7 × 10–6 V, , 23., 28., 33., Space for Rough Work, , 24., 29., 34., , 25., 30., 35.
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t.me/Magazines4all, , DPP/ CP17, , P-68, , 37., , 38., , 39., 40., , 41., , An energy source will supply a constant current into the, load if its internal resistance is, (a) very large as compared to the load resistance, (b) equal to the resistance of the load, (c) non-zero but less than the resistance of the load, (d) zero, The resistance of a wire at room temperature 30°C is found, to be 10 W. Now to increase the resistance by 10%, the, temperature of the wire must be [ The temperature coefficient, of resistance of the material of the wire is 0.002 per °C], (a) 36°C, (b) 83°C, (c) 63°C, (d) 33°C, If current flowing in a conductor changes by 1% then power, consumed will change by, (a) 10%, (b) 2%, (c) 1%, (d) 100%, In the circuit shown in figure, the 5W resistance, develops 20.00 cal/s due to the current flowing through it., The heat developed in 2 W resistance (in cal/s) is, , (a) 23.8, (b) 14.2, (c) 11.9, (d) 7.1, In a Wheatstone's bridge, three resistances P, Q and R, connected in the three arms and the fourth arm is formed by, two resistances S1 and S2 connected in parallel. The, condition for the bridge to be balanced will be, P R (S1 + S2 ), P, 2R, =, =, (a), (b), Q, S1S2, Q S1 + S2, , 37., 42., , RESPONSE, GRID, , 38., 43., , (c), 42., , 43., , 44., , 45., , P R (S1 + S2 ), =, Q, 2S1S2, , (d), , P, R, =, Q S1 + S2, , The electric resistance of a certain wire of iron is R. If its, length and radius are both doubled, then, (a) the resistance and the specific resistance, will both, remain unchanged, (b) the resistance will be doubled and the specific, resistance will be halved, (c) the resistance will be halved and the specific resistance, will remain unchanged, (d) the resistance will be halved and the specific resistance, will be doubled, A car battery has e.m.f. 12 volt and internal resistance 5 × 10–2, ohm. If it draws 60 amp current, the terminal voltage of the, battery will be, (a) 15 volt (b) 3 volt (c) 5 volt (d) 9 volt, A conducting wire of cross-sectional area 1 cm 2 has, 3 × 1023 charge carriers per m3. If wire carries a current of 24, mA, then drift velocity of carriers is, (a) 5 × 10–2 m/s, (b) 0.5 m/s, (c) 5 × 10–3 m/s, (d) 5 × 10–6 m/s, In the series combination of n cells each cell having emf, e and internal resistance r. If three cells are wrongly, connected, then total emf and internal resistance of this, combination will be, (a) ne, (nr – 3r), (b) (ne – 2e) nr, (c) (ne – 4e), nr, (d) (ne – 6e), nr, , 39., 44., , 40., 45., , 41., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP17 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP18, , SYLLABUS : Moving Charges and Magnetism, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , An insulating rod of length l carries a charge q distributed, uniformly on it. The rod is pivoted at its mid point and is, rotated at a frequency f about a fixed axis perpendicular to, rod and passing through the pivot. The magnetic moment, 1, pqfl 2 . Find the value of a., of the rod system is, 2a, (a) 6, (b) 4, (c) 5, (d) 8, A portion of a conductive wire is bent in the form of a, semicircle of radius r as shown below in fig. At the centre of, semicircle, the magnetic induction will be, i, , r, , i, , 3., , 4., , O, , (a) zero, (b) infinite, μ 0 2π i, μ0 π i, (d) 4 π . r, (c) 4 π . r, A closely wound solenoid of 2000 turns and area of crosssection 1.5 × 10–4 m2 carries a current of 2.0 A. It suspended, through its centre and perpendicular to its length, allowing, it to turn in a horizontal plane in a uniform magnetic field, , RESPONSE GRID, , 1., , 2., , 5., , 5 × 10–2 tesla making an angle of 30° with the axis of the, solenoid. The torque on the solenoid will be:, (a) 3 × 10–2 N-m, (b) 3 × 10–3 N-m, –3, (c) 1.5 × 10 N-m, (d) 1.5 × 10–2 N-m, An alternating electric field, of frequency v, is applied across, the dees (radius = R) of a cyclotron that is being used to, accelerate protons (mass = m). The operating magnetic field, (B) used in the cyclotron and the kinetic energy (K) of the, proton beam, produced by it, are given by :, (a) B = mn and K = 2mp2n2R2, e, 2, (b) B = pmn and K = m2pnR2, e, 2, p, (c) B = mn and K = 2mp2n2R2, e, mn, (d) B =, and K = m2pnR2, e, A galvanometer of 50 ohm resistance has 25 divisions. A, current of 4 × 10–4 ampere gives a deflection of one per, division. To convert this galvanometer into a voltmeter, having a range of 25 volts, it should be connected with a, resistance of, (a) 2450 W in series, (b) 2500 W in series., (c) 245 W in series., (d) 2550 W in series., , 3., Space for Rough Work, , 4., , 5.
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P-70, , 6., , 7., , 8., , 9., , 10., , If we double the radius of a coil keeping the current through, it unchanged, then the magnetic field at any point at a large, distance from the centre becomes approximately, (a) double, (b) three times, (c) four times, (d) one-fourth, A particle of mass m, charge Q and kinetic energy T enters a, r, transverse uniform magnetic field of induction B . After 3, seconds, the kinetic energy of the particle will be:, (a) 3T, (b) 2T, (c) T, (d) 4T, A 10 eV electron is circulating in a plane at right angles to a, uniform field at magnetic induction 10–4 Wb/m2 (= 1.0, gauss). The orbital radius of the electron is, (a) 12 cm, (b) 16 cm, (c) 11 cm, (d) 18 cm, A uniform electric field and a uniform magnetic field exist in, a region in the same direction. An electron is projected with, velocity pointed in the same direction. The electron will, (a) turn to its right, (b) turn to its left, (c) keep moving in the same direction but its speed will, increase, (d) keep moving in the same direction but its speed will, decrease, Proton, deuteron and alpha particle of same kinetic energy, are moving in circular trajectories in a constant magnetic, field. The radii of proton, deuteron and alpha particle are, respectively rp, rd and ra. Which one of the following relation, is correct?, (a), , 11., , 12., , 13., , 14., , ra = rp = rd, , (b), , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , DPP/ CP18, , r, (b) – F, r, r, (c) – 3 F, (d) F, A straight section PQ of a circuit lies along the X-axis from, a, a, x = - to x =, and carries a steady current i. The, 2, 2, magnetic field due to the section PQ at a point X = + a will be, (a) proportional to a, (b) proportional to a2, (c) proportional to 1/a, (d) zero, A and B are two, A, conductors carrying a, current i in the same, direction. x and y are, B, two electron beams, moving in the same, x, direction. Then, y, (a) there will be repulsion betwen A and B, attraction between, x and y, (b) there will be attraction between A and B, repulsion, between x and y, (c) there will be repulsion between A and B and also x and y, (d) there will be attraction between A and B and also x and y, A galvanometer of resistance, G is shunted by a resistance, S ohm. To keep the main current in the circuit unchanged,, the resistance to be put in series with the galvanometer is, G, SG, S2, G2, (a), (b), (c), (d), (S, +, G), (S, +, G), (S + G), (S + G), A current I flows in an infinitely long wire with cross section, in the form of a semi-circular ring of radius R. The magnitude, of the magnetic induction along its axis is:, r, (a) 3 F, , 15., , 16., , 17., , 18., , ra = rp < rd, , rd > rp, (c) ra > rd > rp, (d) ra =, A moving coil galvanometer has 150 equal divisions. Its, current sensitivity is 10-divisions per milliampere and voltage, sensitivity is 2 divisions per millivolt. In order that each, division reads 1 volt, the resistance in ohms needed to be, connected in series with the coil will be, (a) 105, (b) 103, (c) 9995, (d) 99995, A 2 µC charge moving around a circle with a frequency of, 6.25 × 1012 Hz produces a magnetic field 6.28 tesla at the, centre of the circle. The radius of the circle is, (a) 2.25 m (b) 0.25 m (c) 13.0 m (d) 1.25 m, A charged particle with charge q enters a region, ofuconstant,, ur, r, uniform and mutually orthogonalurfields uE, and, B with a, r, r, velocity v perpendicular to both E and B , and comes out, r, without any change in magnitude or direction of v . Then, r ur ur 2, r ur ur 2, (a) v = B ´ E / E, (b) v = E ´ B / B, r ur ur 2, r ur ur, (c) v = B ´ E / B, (d) v = E ´ B / E 2, A square current carrying loop is suspended in a uniform, magnetic field acting in the, r plane of the loop. If the force on, one arm of the loop is F , the net force on the remaining, three arms of the loop is, , t.me/Magazines4all, , (a), 19., , m0 I, 2, , 2p R, , (b), , m0 I, 2pR, , (c), , m0 I, 4pR, , Two equal electric currents are flowing, perpendicular to each other as shown, in the figure. AB and CD are, perpendicular to each other and, symmetrically placed with respect to, the current flow. Where do we expect, the resultant magnetic field to be zero?, , (d), , m0 I, , 2, Ip R, , C, , A, O, , B, , I, D, , (a) On AB, (b) On CD, (c) On both AB and CD (d) On both OD and BO, 20. A closed loop PQRS carrying a current is placed in a uniform, magnetic field., Q, If the magnetic forces on segments PS,, SR, and RQ are F1 , F2 and F3 respectively, and are in the plane of the paper and along P, the directions shown, the force on the, F3, segment QP is, (a) F3 – F1– F2, F1, (b), , (F3 – F1 )2 + F22, , S, , (c), (F3 – F1 ) – F22, (d) F3 – F1+F2, 2, , 8., 13., 18., Space for Rough Work, , 9., 14., 19., , R, F2, , 10., 15., 20., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP18, , P-71, , 21. A long solenoid carrying a current produces a magnetic, field B along its axis. If the current is double and the number, of turns per cm is halved, the new value of the magnetic, field is, (a) 4B, (b) B/2, (c) B, (d) 2B, 22. A particle of charge q and mass m moves in a circular orbit, of radius r with angular speed w . The ratio of the magnitude, of its magnetic moment to that of its angular momentum, depends on, (a) w and q, (b) w , q and m, (c) q and m, (d) w and m, 23. A current loop in a magnetic field, (a) can be in equilibrium in one orientation, (b) can be in equilibrium in two orientations, both the equilibrium states are unstable, (c) can be in equilibrium in two orientations, one stable, while the other is unstable, (d) experiences a torque whether the field is uniform or, non-uniform in all orientations, 24. Two long parallel wires P and Q are held perpendicular to, the plane of paper with distance of 5 m between them. If P, and Q carry current of 2.5 amp. and 5 amp. respectively in, the same direction, then the magnetic field at a point halfway between the wires is, (a) m 0 / 17, (b), 3 m0 / 2 p, (c) m 0 / 2 p, (d), 25. A very long straight wire, carries a current I. At the, instant when a charge + Q, r, at point P has velocity v ,, as shown, the force on the, charge is, , 3 m0 / 2 p, , O, , 2A, , 1A, , 30., , 31., , 32., , 33., , Y, , X, , (a) along OY, (b) opposite to OY, (c) along OX, (d) opposite to OX, 26. Two wires with currents 2 A and 1 A are enclosed in a circular, loop. Another wire with current 3 A is situated outside the, r r, loop as shown. The Ñò B.dl around the loop is, (a) µ0, , 29., , (c) v > u at y > 0, 34., , 3A, , (b) 3µ0, (c) 6µ0, , 35., , (d) 2µ0, 27. If in a circular coil A of radius R, current I is flowing and in, another coil B of radius 2R a current 2I is flowing, then the, ratio of the magnetic fields BA and BB, produced by them, will be, (a) 1, (b) 2, (c) 1/2, (d) 4, 28. A charged particle moves through a magnetic field, perpendicular to its direction. Then, (a) kinetic energy changes but the momentum is constant, (b) the momentum changes but the kinetic energy is, constant, , RESPONSE, GRID, , 21., 26., 31., , 22., 27., 32., , (c) both momentum and kinetic energy of the particle are, not constant, (d) both momentum and kinetic energy of the particle are, constant, The deflection in a galvanometer falls from 50 division to 20, when a 12 ohm shunt is applied. The galvanometer resistance, is, (a) 18 ohm (b) 36 ohm (c) 24 ohm (d) 30 ohm, When a long wire carrying a steady current is bent into a, circular coil of one turn, the magnetic induction at its centre, is B. When the same wire carrying the same current is bent, to form a circular coil of n turns of a smaller radius, the, magnetic induction at the centre will be, (a) B/n, (b) nB, (c) B/n2, (d) n2B, The magnetic field due to a current carrying circular loop of, radius 3 cm at a point on the axis at a distance of 4 cm from, the centre is 54 mT. What will be its value at the centre of, loop ?, (a) 125 mT, (b) 150 mT, (c) 250 mT, (d) 75 mT, A charge moving with velocity v in X-direction is subjected, to a field of magnetic induction in negative X-direction. As, a result, the charge will, (a) remain unaffected, (b) start moving in a circular path Y–Z plane, (c) retard along X-axis, (d) move along a helical path around X-axis, An electron travelling with a speed u along the positive, x-axis enters into a region of magnetic field where, B = –B0 k̂ (x > 0). It comes out of the region with speed v, then, ×B, y, (a) v = u at y > 0, e– u, x®, (b) v = u at y < 0, , (d) v > u at y < 0, If an ammeter is to be used in place of a voltmeter, then we, must connect with the ammeter a, (a) low resistance in parallel, (b) high resistance in parallel, (c) high resistance in series, (d) low resistance in series, An infinite straight conductor carrying current 2 I is split, into a loop of radius r as shown in fig. The magnetic field at, the centre of the coil is, I, m 0 2 (p + 1), (a), 4p, r, 2I, 2I, O, m 0 2 ( p - 1), (b), 4p, r, m 0 ( p + 1), 4p r, (d) zero, , (c), , 23., 28., 33., Space for Rough Work, , 24., 29., 34., , I, , 25., 30., 35.
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t.me/Magazines4all, , DPP/ CP18, , P-72, , 36., , 37., , 38., , A parallel plate capacitor of area 60 cm2 and separation 3 mm, is charged initially to 90 mC. 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) 2.5 × 10–8 T, (b) 2.0 × 10–7 T, –11, (c) 1.63 × 10 T, (d) Zero, Four wires, each of length 2.0 m, are bent into four loops P,, Q, R and S and then suspended in a uniform, magnetic field. If the same, current is passed in each, then, Q, S, P, R, the torque will be maximum on, the loop, (a) P, (b) Q, (c) R ur, (d) S, ˆ N / C and, A certain region has an electric field E = (2iˆ - 3j), ur, ˆ T . The force, a uniform magnetic field B = (5iˆ + 3jˆ + 4k), , 42., , 43., , experienced by a charge 1C moving with velocity (iˆ + 2ˆj), ms–1 is, ˆ, ˆ, (a) (10iˆ - 7ˆj - 7k), (b) (10iˆ + 7ˆj + 7k), , ˆ, ˆ, (c) (-10iˆ + 7ˆj + 7k), (d) (10iˆ + 7ˆj - 7k), 39. A galvanometer of resistance 100 W gives a full scale, deflection for a current of 10–5 A. To convert it into a ammeter, capable of measuring upto 1 A, we should connect a, resistance of, (a) 1 W in parallel, (b) 10–3 W in parallel, 5, (c) 10 W in series, (d) 100 W in series, I1, 40. A square loop, carrying a steady, d, current I, is placed in a horizontal plane, I, near a long straight conductor carrying, a steady current I1 at a distance d from, the conductor as shown in figure. The, loop will experience, I, (a) a net repulsive force away from the conductor, (b) a net torque acting upward perpendicular to the, horizontal plane, (c) a net torque acting downward normal to the horizontal, plane, (d) a net attractive force towards the conductor, 41. Two coaxial solenoids of different radius carry current I in, uur, the same direction. F1 be the magnetic force on the inner, uur, solenoid due to the outer one and F2 be the magnetic force, , RESPONSE, GRID, , 36., 41., , 37., 42., , on the outer solenoid due to the inner one. Then :, uur, uur, (a) F1 is radially inwards and F2 = 0, uur, uur, (b) F1 is radially outwards and F2 = 0, uur uur, (c) F1 = F2 = 0, uur, uur, (d) F1 is radially inwards and F2 is radially outwards, A beam of electrons is moving with constant velocity in a, region having simultaneous perpendicular electric and, magnetic fields of strength 20 Vm–1 and 0.5 T respectively, at right angles to the direction of motion of the electrons., Then the velocity of electrons must be, (a) 8 m/s, (b) 20 m/s (c) 40 m/s (d) 1 m / s, 40, The magnetic flux density B at a distance r from a long, straight wire carrying a steady current varies with r as, (b) B, (a) B, , r, , r, , (c), , 44., , 45., , (d), , B, , B, , r, , r, , The AC voltage across a resistance can be measured, using a :, (a) hot wire voltmeter, (b) moving coil galvanometer, (c) potential coil galvanometer, (d) moving magnet galvanometer, r, When a charged particle moving with velocityv is subjected, ur, to a magnetic field of induction B , the force on it is nonzero. This implies that, ur, r, (a) angle between v and B is necessarily 90°, ur, r, (b) angle between v and B can have any value other, than 90°, r, ur, (c) angle between v and B can have any value other, than zero and 180°, ur, r, (d) angle between v and B is either zero or 180°, , 38., 43., , 39., 44., , 40., 45., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP18 - PHYSICS, Total Questions, , 45, , Total Marks, , Attempted, , Correct, , Incorrect, , Net Score, , Cut-off Score, , 45, , Qualifying Score, , Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP19, , SYLLABUS : Magnetism and Matter, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , Two identical magnetic dipoles of magnetic moments, 1.0 A-m2 each, placed at a separation of 2 m with their axis, perpendicular to each other. The resultant magnetic field at, point midway between the dipole is, (a) 5 × 10–7 T, (b), 5 × 10–7 T, (c) 10–7 T, (d) 2 × 10–7 T, Two identical thin bar magnets each, N, of length l and pole strength m are, placed at right angles to each other,, with north pole of one touching, south pole of the other, then the, S, magnetic moment of the system is, 1, , 5., , (a) er = 0.5, mr = 1.5, (b) er = 1.5, mr = 0.5, (c) er = 0.5, mr = 0.5, (d) er = 1.5, mr = 1.5, If the period of oscillation of freely suspended bar magnet, in earth’s horizontal field H is 4 sec. When another magnet, is brought near it, the period of oscillation is reduced to 2s., The magnetic field of second bar magnet is, , 6., , 3H, (a) 4 H, (b) 3 H, (c) 2 H, (d), Three identical bars A, B and C are made of different magnetic, materials. When kept in a uniform magnetic field, the field, lines around them look as follows:, , 2, , 1, , 3., , 4., , N2, , S2, , A, , 2 ml (d) ml/2, The magnetic lines of force inside a bar magnet, (a) are from north-pole to south-pole of the magnet, (b) do not exist, (c) depend upon the area of cross-section of the bar magnet, (d) are from south-pole to north-pole of the magnet, Relative permittivity and permeability of a material er and mr,, respectively. Which of the following values of these, quantities are allowed for a diamagnetic material?, (a) 1 ml, , RESPONSE, GRID, , (b) 2 ml, , 1., 6., , (c), , 2., , C, , B, , Make the correspondence of these bars with their material, being diamagnetic (D), ferromagnetic (F) and paramagnetic, (P):, (a) A « D, B « P, C « F, (b) A « F, B « D, C « P, (c) A « P, B « F, C « D, (d) A « F, B « P, C « D, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP19, , P-74, , 8., , 9., , Curie temperature is the temperature above which, (a) a ferromagnetic material becomes paramagnetic, (b) a paramagnetic material becomes diamagnetic, (c) a ferromagnetic material becomes diamagnetic, (d) a paramagnetic material becomes ferromagnetic, A watch glass containing some powdered substance is placed, between the pole pieces of a magnet. Deep concavity is, observed at the centre. The substance in the watch glass is, (a) iron, (b) chromium (c) carbon (d) wood, A coil in the shape of an equilateral triangle of side l is, suspended®between the pole pieces of a permanent magnet, such that B is in the plane of the coil. If due to a current i in, the triangle a torque t acts on it, the side l of the triangle is, 1, , 1, , 10., , 11., , 12., , 13., , (a), , 2 æ t ö2, ç ÷, 3 è B.i ø, , (b), , (c), , 2 æ t ö, ç, ÷, 3 è B.i ø, , (d), , æ t ö2, ÷÷, 2çç, è 3B.i ø, 1 t, 3 B.i, , A compass needle whose magnetic moment is 60 Am2, is, directed towards geographical north at any place, experiencing moment of force of 1.2 × 10–3 Nm. At that place, the horizontal component of earth field is 40 micro W/m2., What is the value of dip angle at that place?, (a) 30°, (b) 60°, (c) 45°, (d) 15°, The materials suitable for making electromagnets should, have, (a) high retentivity and low coercivity, (b) low retentivity and low coercivity, (c) high retentivity and high coercivity, (d) low retentivity and high coercivity, The length of a magnet is large compared to its width and, breadth. The time period of its oscillation in a vibration, magnetometer is 2s. The magnet is cut along its length into, three equal parts and these parts are then placed on each, other with their like poles together. The time period of this, combination will be, 2, 2, s, s, (a) 2 3 s (b), (c) 2 s, (d), 3, 3, Hysteresis loops for two magnetic materials A and B are, given below :, B, , These materials are used to make magnets for elecric, generators, transformer core and electromagnet core. Then, it is proper to use :, (a) A for transformers and B for electric generators., (b) B for electromagnets and transformers., (c) A for electric generators and trasformers., (d) A for electromagnets and B for electric generators., 14. Which of the following is responsible for the earth’s magnetic, field?, (a) Convective currents in earth’s core., (b) Diversive current in earth’s core., (c) Rotational motion of earth., (d) Translational motion of earth., 15. In a vibration megnetometer, the time period of a bar magnet, oscillating in horizontal component of earth's magnetic field, is 2 sec. When a magnet is brought near and parallel to it, the, time period reduces to 1 sec. The ratio H/F of the horizontal, component H and the field F due to magnet will be, 16., , 17., , 18., , 19., , 20., , (a) 3, (b) 1/3, (c), (d) 1/ 3, 3, Let V and H be the vertical and horizontal components of, earth's magnetic field at any point on earth. Near the north, pole, (a) V >> H (b) V << H (c) V = H, (d) V = H = 0, A thin circular wire carrying a current I has a magnetic, moment M. The shape of the wire is changed to a square, and it carries the same current. It will have a magnetic, moment, 4, 4, p, M (c), M, M, (a) M, (b), (d), 2, p, 4, p, A bar magnet of magnetic moment M is placed at right angles, to a magnetic induction B. If a force F is experienced by, each pole of the magnet, the length of the magnet will be, (a) F/MB (b) MB/F (c) BF/M (d), MF/B, If the susceptibility of dia, para and ferro magnetic materials, are cd, cp, cf respectively, then, (a) cd < cp < cf, (b) cd < cf < cp, (c) cf < cd < cp, (d) cf < cp < cd, The basic magnetization curve for a ferromagnetic material, is shown in figure. Then, the value of relative permeability, is highest for the point, 1.5, , R, , D, , H, , H, (A), , B(Tesla), , 7., , 1.0, 0.5, , (B), , RESPONSE, GRID, , 7., 12., 17., , 8., 13., 18., , Q, P, , O 0 1, , (a) P, , 9., 14., 19., Space for Rough Work, , S, , 2, , 3 4 5 6 7, 3, H (× 10 A/m), , (b) Q, , 10., 15., 20., , (c) R, , (d) S, , 11., 16., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP19, , P-75, , configuration corresponds to the lowest potential energy, among all the configurations shown ?, , 21. A magnetic needle suspended by a silk thread is vibrating, in the earth’s magnetic field. if the temperature of the needle, is increased by 700ºC, then, (a) time period decreases, (b) time period increases, (c) time period remains unchanged, (d) the needle stops vibrating, 22. Torques t1 and t2 are required for a magnetic needle to, remain perpendicular to the magnetic fields at two different, places. The magnetic fields at those places are B1 and B2, respectively; then ratio, (a), , t2, t1, , B1, is, B2, , (b), , Q4, , (a) PQ3, (b) PQ4, , t1, t2, , (c), , 23., , 24., , 25., , 26., , 27., , 28., , RESPONSE, GRID, , 21., 26., 31., , 22., 27., 32., , P, , Q1, , Q2, , (c) PQ5, 29., , t1 + t 2, t1 - t 2, (d), t1 - t 2, t1 + t 2, A bar magnet has a length 8 cm. The magnetic field at a, point at a distance 3 cm from the centre in the broad side-on, position is found to be 4×10–6 T. The pole strength of the, magnet is., (a) 6 × 10–5 Am, (b) 5 × 10–5 Am, (c) 2 × 10–4 Am, (d) 3 × 10–4 Am, A vibration magnetometer consists of two identical bar magnets, placed one over the other such that they are perpendicular, and bisect each other. The time period of oscillation in a, horizontal magnetic field is 25/4 seconds. One of the magnets, is removed and if the other magnet oscillates in the same field,, then the time period in seconds is, (a) 21/4, (b) 21/2, (c) 2, (d) 23/4, A magnetic needle is kept in a non-uniform magnetic field. It, experiences, (a) neither a force nor a torque, (b) a torque but not a force, (c) a force but not a torque, (d) a force and a torque, The angle of dip at a place is 37° and the vertical component, of the earth’s magnetic field is 6 × 10–5T. The earth’s, magnetic field at this place is (tan 37° = 3/4), (a) 7 × 10–5 T, (b) 6 × 10–5 T, –5, (c) 5 × 10 T, (d) 10–4 T, Needles N1, N2 and N3 are made of a ferromagnetic, a, paramagnetic and a diamagnetic substance respectively. A, magnet when brought close to them will, (a) attract N1 and N2 strongly but repel N3, (b) attract N1 strongly, N2 weakly and repel N3 weakly, (c) attract N1 strongly, but repel N2 and N3 weakly, (d) attract all three of them, The figure shows the various positions (labelled by, subscripts) of small magnetised needles P and Q. The arrows, show the direction of their magnetic moment. Which, , Q5, , 30., , 31., , (d) PQ6, Q6, A dip needle lies initially in the magnetic meridian when it, shows an angle of dip q at a place. The dip circle is rotated, through an angle x in the horizontal plane and then it shows, an angle of dip q'., tan q ', Then, is, tan q, 1, 1, 1, (a), (b), (c), (d) cos x, cos x, sin x, tan x, Two tangent galvanometers having coils of the same radius, are connected in series. A current flowing in them produces, deflections of 60º and 45º respectively. The ratio of the number, of turns in the coils is, 3 +1, 3 +1, 3, (a) 4/3, (b), (c), (d), 3 -1, 1, 1, Following figures show the arrangement of bar magnets in, different configurations., Each magnet has magnet ic dipole, r, moment m . Which configuration has highest net magnetic, dipole moment?, N, , (a), , (b), S S, , S, N, , N, S, , N, , N, , N, , (c), 32., , 33., , S, , 30º, N, , (d), S, , 60º, N, , A compass needle which is allowed to move in a horizontal, plane is taken to a geomagnetic pole. It :, (a) will become rigid showing no movement, (b) will stay in any position, (c) will stay in north-south direction only, (d) will stay in east-west direction only, If a magnetic dipole of moment M situated in the direction of, a magnetic field B is rotated by 180°, then the amount of, work done is, (a) MB, , 23., 28., 33., Space for Rough Work, , (b) 2MB, , 24., 29., , (c), , MB, 2, , 25., 30., , (d) 0
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t.me/Magazines4all, , DPP/ CP19, , P-76, , 34., , 35., , 36., 37., , 38., , A bar magnet is oscillating in the earth’s magnetic field with, a period T. What happens to its period of motion, if its mass, is quadrupled, (a) motion remains simple harmonic with new period = T/2, (b) motion remains simple harmonic with new period = 2 T, (c) motion remains simple harmonic with new period = 4T, (d) motion remains simple harmonic and the period stays, nearly constant, The magnetic field of earth at the equator is approximately 4, × 10–5 T. The radius of earth is 6.4 × 106 m. Then the dipole, moment of the earth will be nearly of the order of:, (a) 1023 A m2 (b) 1020 A m2 (c) 1016 A m2 (d) 1010 A m2, The relative permeability of a medium is 0.075. What is its, magnetic susceptibility?, (a) 0.925, (b) – 0.925 (c) 1.075, (d) –1.075, A dip circle is so set that its needle moves freely in the, magnetic meridian. In this position, the angle of dip is 40º., Now the dip circle is rotated so that the plane in which the, needle moves makes an angle of 30º with the magnetic, meridian. In this position, the needle will dip by an angle, (a) 40º (b) 30º (c) more than 40º (d) less than 40º, The mid points of two small magnetic dipoles of length d in, end-on positions, are separated by a distance x, (x > > d)., The force between them is proportional to x–n where n is:, N, , S, , N, , 41., , 42., , 43., , 44., , S, , 45., , x, , 39., , 40., , (a) 1, (b) 2, (c) 3, (d) 4, At a temperatur of 30°C, the susceptibility of a ferromagnetic, material is found to be c. Its susceptibility at 333°C is, (a) c, (b) 0.5c, (c) 2c, (d) 11.1c, , 34., 39., 44., , RESPONSE, GRID, , 35., 40., 45., , The susceptibility of annealed iron at saturation is 5500., Find the permeability of annealed iron at saturation., (a) 6.9 × 10–3 (b) 5.1 × 10–2 (c) 5 × 102 (d) 3.2 × 10–5, A short magnet oscillates in an oscillation magnetometer, with a time period of 0.10s where the earth¢s horizontal, magnetic field is 24 mT. A downward current of 18 A is, established in a vertical wire placed 20 cm east of the magnet., Find the new time period., (a) 0.076 s (b) 0.5 s, (c) 0.1 s, (d) 0.2 s, A permanent magnet in the shape of a thin cylinder of, length 10 cm has magnetisation (M) = 106 A m–1. Its, magnetization current IM is, (a) 105 A (b) 106 A (c) 107 A (d) 108 A, The earth’s magnetic field lines resemble that of a dipole at, the centre of the earth. If the magnetic moment of this dipole, is close to 8 × 1022 Am2, the value of earth’s magnetic field, near the equator is close to (radius of the earth = 6.4 × 106 m), (a) 0.6 Gauss, (b) 1.2 Gauss, (c) 1.8 Gauss, (d) 0.32 Gauss, The coercivity of a small magnet where the ferromagnet, gets demagnetized 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 demagnetized when inside the, solenoid, is:, (a) 30 mA (b) 60 mA (c) 3 A, (d) 6 A, A thin bar magnet of length 2 l and breadth 2 b pole strength, m and magnetic moment M is divided into four equal parts, with length and breadth of each part being half of original, magnet., Then, the magnetic moment of each part is, (a) M/4, (b) M, (c) M/2, (d) 2 M, , 36., 41., , 37., 42., , 38., 43., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP19 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP20, , SYLLABUS : Electromagnetic Induction, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , A metal disc of radius 100 cm is rotated at a constant angular, speed of 60 rad/s in a plane at right angles to an external, field of magnetic induction 0.05 Wb/m2. The emf induced, between the centre and a point on the rim will be, (a) 3 V, (b) 1.5 V, (c) 6 V, (d) 9 V, In a coil of resistance 100 W , 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), , 4., , 250 Wb, 5., , (b) 275 Wb, (c) 200 Wb, 3., , (d) 225 Wb, A 10-meter wire is kept in east-west direction. It is falling, down with a speed of 5.0 meter/second, perpendicular to, the horizontal component of earth's magnetic field of, 0.30 ×10-4 weber/meter2. The momentary potential difference, induced between the ends of the wire will be, (a) 0.0015 V, (b) 0.015 V, (c) 0.15 V, (d) 1.5 V, , RESPONSE GRID, , 1., 6., , 2., 7., , 6., , 7., , The figure shows certain wire segments, c, d, joined together to form a coplanar loop., a, b, The loop is placed in a perpendicular, magnetic field in the direction going into, the plane of the figure., The magnitude of the field increases with time. I1 and I2 are, the currents in the segments ab and cd. Then,, (a) I1 > I2, (b) I1 < I2, (c) I1 is in the direction ba and I2 is in the direction cd, (d) I1 is in the direction ab and I2 is in the direction dc, Two solenoids of equal number of turns have their lengths, and the radii in the same ratio 1 : 2. The ratio of their self, inductances will be, (a) 1 : 2, (b) 2 : 1, (c) 1 : 1, (d) 1 : 4, A metal conductor of length 1 m rotates vertically about one, of its ends at angular velocity 5 radians per second. If the, horizontal component of earth’s magnetic field is 0.2×10–4T,, then the e.m.f. developed between the two ends of the, conductor is, (a) 5 mV, (b) 50 mV (c) 5 mV, (d) 50mV, Eddy currents do not produce, (a) heat, (b) a loss of energy, (c) spark, (d) damping of motion, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP20, , P-78, , 8., , 9., , A conducting square frame of side ‘a’ and a long staight, wire carrying current I are located in the same plane as shown, in the figure. The frame moves to the right with a constant, velocity ‘V’. The emf induced in the frame will be proportional, to, (a), , 1, (2x – a)2, , (b), , 1, (2x + a)2, , (c), , 1, (2x – a)(2x + a), , 10., , 11., , 12., , 13., , 14., , X, , 16., , l, V, , 1, a, (d), x2, Which of the following figure correctly depicts the Lenz’s, law. The arrows show the movement of the labelled pole of, a bar magnet into a closed circular loop and the arrows on, the circle show the direction of the induced current, , (a), , N, , (b), , N, , (c), , S, , (d), , 8., 13., 18., , 9., 14., 19., , 17., 18., , (a), , -, , ( W1 - W2 ), Rnt, , (b), , -, , n ( W2 - W1 ), 5 Rt, , (c), , -, , ( W2 - W1 ), 5 Rnt, , (d), , -, , n ( W2 - W1 ), Rt, , S, , The magnetic flux (in weber) linked with a coil of resistance, 10 W is varying with respect to time t as f = 4t2 + 2t + 1. Then, the current in the coil at time t = 1 second is, (a) 0.5 A, (b) 2 A, (c) 1.5 A, (d) 1 A, Two coaxial solenoids are made by winding thin insulated, wire over a pipe of cross-sectional area A = 10 cm2 and, length = 20 cm. If one of the solenoid has 300 turns and the, other 400 turns, their mutual inductance is, (m0 = 4p × 10 –7 Tm A–1), (a) 2.4p × 10–5 H, (b) 4.8p × 10–4 H, (c) 4.8p × 10–5 H, (d) 2.4p × 10–4 H, When the current changes from +2 A to –2A in 0.05 second,, an e.m.f. of 8 V is induced in a coil. The coefficient of self induction of the coil is, (a) 0.2 H, (b) 0.4 H, (c) 0.8 H, (d) 0.1 H, A long solenoid has 500 turns. When a current of 2 ampere, is passed through it, the resulting magnetic flux linked with, each turn of the solenoid is 4 ×10–3 Wb. The self- inductance, of the solenoid is, (a) 2.5 henry, (b) 2.0 henry, (c) 1.0 henry, (d) 40 henry, A metallic square loop ABCD is moving, A, B, in its own plane with velocity v in a, uniform magnetic field perpendicular to, v, its plane as shown in the figure. An, D, C, electric field is induced, , RESPONSE, GRID, , 15., , (a) in AD, but not in BC, (b) in BC, but not in AD, (c) neither in AD nor in BC (d) in both AD and BC, In an AC generator, a coil with N turns, all of the same area, A and total resistance R, rotates with frequency w in a, magnetic field B. The maximum value of emf generated in, the coil is, (a) N.A.B.R.w, (b) N.A.B., (c) N.A.B.R., (d) N.A.B.w, In an inductor of self-inductance L = 2 mH, current changes, with time according to relation i = t2e–t. At what time emf is, zero?, (a) 4s, (b) 3s, (c) 2s, (d) 1s, Choke coil works on the principle of, (a) transient current, (b) self induction, (c) mutual induction, (d) wattless current, A coil having n turns and resistance R W is connected with, a galvanometer of resistance 4R W. This combination is, moved in time t seconds from a magnetic field W1 weber to, W2 weber. The induced current in the circuit is, , 19., , A thin circular ring of area A is held perpendicular to a, 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, , B2 A, AB, BR, (b), (c) ABR, (d), R, A, R2, A boat is moving due east in a region where the earth's, magnetic field is 5.0 × 10–5 NA–1 m–1 due 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) 0.75 mV (b) 0.50 mV (c) 0.15 mV (d) 1mV, In a coil of area 10 cm2 and 10 turns with magnetic field, directed perpendicular to the plane and is changing at the, rate of 108 Gauss/second. The resistance of the coil is 20W., The current in the coil will be, (a) 0.5 A, (b) 5 A, (c) 50 A, (d) 5 × 108 A, A horizontal straight wire 20 m long extending from east to, west falling with a speed of 5.0 m/s, at right angles, to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb/m2. The instantaneous value of the e.m.f., induced in the wire will be, (a) 3 mV, (b) 4.5 mV (c) 1.5 mV (d) 6.0 mV, (a), , 20., , 21., , 22., , 10., 15., 20., Space for Rough Work, , 11., 16., 21., , 12., 17., 22., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP20, , P-79, , 23. The self inductance of a long solenoid cannot be increased, by, (a) increasing its area of cross section, (b) increasing its length, (c) changing the medium with greater permeability, (d) increasing the current through it, 24. A metallic rod of length ‘l’ is tied to a string of length 2l and, made to rotate with angular speed w on a horizontal, table with one end of the string fixed., If there is a vertical magnetic field ‘B’, in the region, the e.m.f. induced across, the ends of the rod is, , 2 Bwl2, 4 Bwl2, 5 Bwl 2, 3Bwl 2, (b), (c), (d), 2, 2, 2, 2, 25. Lenz’s law gives, (a) the magnitude of the induced e.m.f., (b) the direction of the induced current, (c) both the magnitude and direction of the induced current, (d) the magnitude of the induced current, 26. A metal ring is held horizontally and bar magnet is dropped, through the ring with its length along the axis of the ring., The acceleration of the falling magnet, (a) is equal to g, (b) is less than g, (c) is more than g, (d) depends on the diameter of ring and length of magnet, (a), , 27. The pointer of a dead-beat galvanometer gives a steady, deflection because, (a) eddy currents are produced in the conducting frame, over which the coil is wound., (b) its magnet is very strong., (c) its pointer is very light., (d) its frame is made of ebonite., 28. A metal rod of length l cuts across a uniform magnetic field, B with a velocity v. If the resistance of the circuit of which, the rod forms a part is r, then the force required to move the, rod is, Blv, B2 l 2 v, B2 lv, B2 l 2 v 2, (b), (c), (d), r, r, r, r, 29. In an A.C. generator, when the plane of the armature is, perpendicular to the magnetic field, (a) both magnetic flux and emf are maximum, (b) both magnetic flux and emf are zero, (c) both magnetic flux and emf are half of their respective, maximum values, (d) magnetic flux is maximum and emf is zero, , 30., , 31., , 32., , 33., , 34., , 35., , 36., , (a), , RESPONSE, GRID, , 23., 28., 33., , 24., 29., 34., , 37., , A copper disc of radius 0.1 m rotated about its centre with, 10 revolutions per second in a uniform magnetic field of 0.1, tesla with its plane perpendicular to the field. The e.m.f., induced across the radius of disc is, p, 2p, (b), (a), volt, volt, 10, 10, –2, (c) p × 10 volt, (d) 2p × 10–2 volt, The mutual inductance of a pair of coils, each of N turns, is, M henry. If a current of I ampere in one of the coils is, brought to zero in t second, the emf induced per turn in the, other coil, in volt, will be, MN, MI, MI, NMI, (a), (b), (c), (d), It, t, t, Nt, Consider the situation shown in figure. If the switch is closed, and after some time it is opened again, the closed loop will, show, (a) a clockwise current, (b) an anticlockwise current, (c) an anticlockwise current and then clockwise, (d) a clockwise current and then an anticlock wise current., A magnet is moved towards a coil (i) quickly (ii) slowly,, then the induced e.m.f. is, (a) larger in case (i), (b) smaller in case (i), (c) equal to both the cases, (d) larger or smaller depending upon the radius of the coil, A circular wire of radius r rotates about its own axis with, angular speed w in a magnetic field B perpendicular to its, plane, then the induced e.m.f. is, 1, Brw 2 (b) Brw2, (a), (c) 2Brw2 (d) zero, 2, A conducting ring of radius l m kept in a uniform magnetic, field B of 0.01 T, rotates uniformly with an angular velocity, 100 rad s–1 with its axis of rotation perpendicular to B. The, maximum induced emf in it is, (a) 1.5pV (b) pV, (c) 2pV, (d) 0.5pV, A magnetic field of 2 × 10–2 T acts at right angles to a coil of, area 100 cm2, with 50 turns. The average e.m.f. induced in, the coil is 0.1 V, when it is removed from the field in t sec., The value of t is, (a) 10 s, (b) 0.1 s, (c) 0.01 s, (d) 1 s, The magnetic flux through a circuit of resistance R changes, by an amount Df in a time Dt. Then the total quantity of, electric charge Q that passes any point in the circuit during, the time Dt is represented by, Df, 1 Df, (a) Q = R ., (b) Q = ., Dt, R Dt, Df, Df, (c) Q =, (d) Q =, Dt, R, , 25., 30., 35., Space for Rough Work, , 26., 31., 36., , 27., 32., 37.
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t.me/Magazines4all, , DPP/ CP20, , P-80, , 38., , Fig shown below represents an area A = 0.5 m 2 situated in a, uniform magnetic field B = 2.0 weber/m2 and making an angle, of 60º with respect to magnetic field., , 42., , B, 60, , The value of the magnetic flux through the area would be, equal to, (a) 2.0 weber, , (b), , 3 weber, , (c), , 39., , 40., , 41., , (d) 0.5 weber, 3 / 2 weber, As a result of change in the magnetic flux, linked to the closed loop shown in the fig, an, e.m.f. V volt is induced in the loop. The work, done (joule) in taking a charge Q coulomb, once along the loop is, (a) QV, (b) 2QV, (c) QV/2, (d) Zero, Two coils are placed close to each other. The mutual, inductance of the pair of coils depends upon, (a) the rates at which currents are changing in the two, coils, (b) relative position and orientation of the two coils, (c) the materials of the wires of the coils, (d) the currents in the two coils, When current i passes through an inductor of self, inductance L, energy stored in it is 1/2. L i 2. This is stored in, the, , 38., 43., , RESPONSE, GRID, , 39., 44., , (a) current, (b) voltage, (c) magnetic field, (d) electric field, A conducting wire frame is placed in a magnetic field which, is directed into the paper. The magnetic field is increasing at, a constant rate. The directions of induced current in wires, AB and CD are, , (a) B to A and D to C, (b) A to B and C to D, (c) A to B and D to C, (d) B to A and C to D, 43. Two different wire loops are concentric, and lie in the same plane. The current in, the outer loop (I) is clockwise and I, increases with time. The induced current, in the inner loop, (a) is clockwise, (b) is zero, (c) is counter clockwise, (d) has a direction that depends on the ratio of the loop, radii., 44. When current in a coil changes from 5 A to 2 A in 0.1 s,, average voltage of 50 V is produced. The self - inductance, of the coil is :, (a) 6 H, (b) 0.67 H (c) 3 H, (d) 1.67 H, 45. Two conducting circular loops of radii R1 and R2 are placed, in the same plane with their centres coinciding. If, R1>>R2, the mutual inductance M between them will be, directly proportional to, (a) R1/R2, , 40., 45., , (b) R2/R1, , 41., , (c), , R12 / R2 (d), , 42., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP20 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 70, , R22 / R1, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP21, , SYLLABUS : Alternating Current, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , 4., , In a series resonant LCR circuit, the voltage across R is 100, volts and R = 1 kW with C = 2mF. The resonant frequency w, is 200 rad/s. At resonance, the voltage across L is, (a) 2.5 × 10–2 V, (b) 40 V, (c) 250 V, (d) 4 × 10–3 V, An alternating voltage V = V0 sin wt is applied across a, circuit. As a result, a current I = I0 sin (wt – p/2) flows in it., The power consumed per cycle is, (a) zero, (b) 0.5 V0I0, (c) 0.707 V0I0, (d) 1.414 V0I0, For the circuit shown in the fig., the current through the, inductor is 0.9 A while the current through the condenser is, 0.4 A. Then, C, (a) current drawn from, generator I = 1.13 A, L, (b) w = 1/(1.5 L C), (c) I = 0.5 A, ~, (d) I = 0.6 A, V = V0 sin wt, A capacitor has capacity C and reactance X. If capacitance, and frequency become double, then reactance will be, (a) 4X, (b) X/2, (c) X/4, (d) 2X, , 1., RESPONSE GRID 6., , 2., 7., , 5., , 6., , 7., , A coil of inductance 300 mH and resistance 2W is connected, to a source of voltage 2V. The current reaches half of its, steady state value in, (a) 0.1 s, (b) 0.05 s (c) 0.3 s, (d) 0.15 s, In an A.C. circuit, a resistance of R ohm is connected in, series with an inductance L. If phase angle between voltage, and current be 45°, the value of inductive reactance will be, (a) R/4, (b) R/2, (c) R, (d) R/5, A bulb is rated at 100 V, 100 W, it can be treated as a resistor., Find out the inductance of an inductor (called choke coil), that should be connected in series with the bulb to operate, the bulb at its rated power with the help of an ac source of, 200 V and 50 Hz., , 8., , p, , 2, 3, H (b) 100 H (c), H (d), H, 3, p, p, An ac source of angular frequency w is fed across a resistor, r and a capacitor C in series. The current registered is I. If, now the frequency of source is changed to w/3 (but, maintaining the same voltage), the current in the circuit is, found to be halved. The ratio of reactance to resistance at, the original frequency w is, (a), , (a), , 3., 8., Space for Rough Work, , 3, 5, , (b), , 4., , 2, 5, , (c), , 1, 5, , 5., , (d), , 4, 5
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t.me/Magazines4all, , DPP/ CP21, , P-82, , 9., , Large transformers, when used for some time, become hot, and are cooled by circulating oil. The heating of transformer, is due to, (a) heating effect of current alone, (b) hysteresis loss alone, (c) both the hysteresis loss and heating effect of current, (d) none of the above, , An inductor of inductance L = 400 mH, and resistors of resistance R1 = 2W E, and R2 = 2W are connected to a battery, L, of emf 12 V as shown in the figure., R, The internal resistance of the battery, is negligible. The switch S is closed at, R, t = 0. The potential drop across L as a, function of time is, S, `, 12 -3t, e V, (a), (b) 6(1 – e–t/0.2)V, t, (c) 12e–5t V, (d) 6e–5t V, 11. An ideal coil of 10H is connected in series with a resistance, of 5W and a battery of 5V. 2second after the connection is, made, the current flowing in ampere in the circuit is, (a) (1 – e–1) (b) (1 – e) (c) e, (d) e–1, 12. In an A.C. circuit, the current flowing in inductance is, I = 5 sin (100 t – p/2) amperes and the potential difference is, V = 200 sin (100 t) volts. The power consumption is equal to, (a) 1000 watt, (b) 40 watt, (c) 20 watt, (d) Zero, 13. In an oscillating LC circuit the maximum charge on the, capacitor is Q. The charge on the capacitor when the energy, is stored equally between the electric and magnetic field is, , 16., , O, , 17., , 10., , 1, , 2, , Q, Q, Q, (a), (b), (c), (d) Q, 3, 2, 2, 14. A fully charged capacitor C with initial charge q0 is connected, to a coil of self inductance L at t = 0. The time at which the, energy is stored equally between the electric and the, magnetic fields is:, p, , 15., , LC (b) 2p LC (c), (d) p LC, (a), LC, 4, For an LCR series circuit with an A.C. source of angular, frequency w, , (a) circuit will be capacitive if w >, (b) circuit will be inductive if w =, , 1, LC, 1, , LC, (c) power factor of circuit will be unity if capacitive, reactance equals inductive reactance, (d) current will be leading voltage if w >, , RESPONSE, GRID, , 9., 14., 19., , 1, LC, , 10., 15., 20., , The r.m.s. value of potential V, difference V shown in the, V0, figure is, , 18., , T/2, , t, , T, , (d) V0 / 3, (a) V0, (b) V0 / 2 (c) V0/2, Which of the following statements is/are incorrect?, (a) If the resonance is less sharp, not only is the, maximum current less, the circuit is close to resonance, for a larger range Dw of frequencies and the tuning, of the circuit will not be good., (b) Less sharp the resonance less is the selectivity of, the circuit or vice–versa., (c) If quality factor is large, i.e., R is low or L is large,, the circuit is more selective., (d) Below resonance, voltage leads the current while, above it, current leads the voltage., A lamp consumes only 50% of peak power in an a.c. circuit., What is the phase difference between the applied voltage, and the circuit current?, , p, p, p, p, (b), (c), (d), 6, 3, 4, 2, 19. A step down transformer reduces 220 V to 110 V. The primary, draws 5 ampere of current and secondary supplies 9 ampere., The efficiency of transformer is, (a) 20%, (b) 44%, (c) 90%, (d) 100%, 20. The voltage time (V-t) graph for triangular wave having, peak value V0 is as shown in +V, 0, figure.The rms value of V in, time interval from t = 0 to T/4 is, T/2, t, , (a), , V0, x, , 0, , then find the value of x., , T/4, , T, , –V0, , (a) 5, (b) 4, (c) 7, (d) 3, The tuning circuit of a radio receiver has a resistance of, 50 W , an inductor of 10 mH and a variable capacitor. A, 1 MHz radio wave produces a potential difference of, 0.1 mV. The values of the capacitor to produce resonance is, (Take p2 = 10), (a) 2.5 pF (b) 5.0 pF, (c) 25 pF (d) 50 pF, 22. In an alternating current circuit in which an inductance and, capacitance are joined in series, current is found to be, maximum when the value of inductance is 0.5 henry and the, value of capacitance is 8µF. The angular frequency of applied, alternating voltage will be, (a) 5000 rad/sec, (b) 4000 rad/sec, (c) 2 × 105 rad/sec, (d) 500 rad/sec, 23. A coil has resistance 30 ohm and inductive reactance 20, ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz,, is connected across the coil, the current in the coil will be, 20, A (d) 2.0 A, (a) 4.0 A, (b) 8.0 A, (c), 13, 21., , 11., 16., 21., Space for Rough Work, , 12., 17., 22., , 13., 18., 23., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP21, , P-83, , V, V, 24. In the figure shown, three AC, voltmeters are connected. At, R, L, C, resonance, (a) V2 = 0, (b) V1 = 0, V, (c) V3 = 0, (d) V1 = V2 ¹ 0, ~, 25. A.C. power is transmitted from a power house at a high, voltage as, (a) the rate of transmission is faster at high voltages, (b) it is more economical due to less power loss, (c) power cannot be transmitted at low voltages, (d) a precaution against theft of transmission lines, 26. A transformer has an efficiency of 80%. It works at 4 kW, and 100 V. If secondary voltage is 240 V, the current in primary, coil is, (a) 0.4 A, (b) 4 A, (c) 10 A, (d) 40 A, 27. A 12 W resistor and a 0.21 henry inductor are connected in, series to an a.c. source operating at 20 volt, 50 cycle. The, phase angle between the current and source voltage is, (a) 30º, (b) 40º, (c) 80º, (d) 90º, 28. In LCR series circuit fed by a DC source, how does the, amplitude of charge oscillations vary with time during, discharge ?, 2, , 1, , 3, , q, , (a), , (c), , (b), , qo, , t, , O, , t, , O, , t, , O, , t, , O, , 29. The primary and secondary coil of a transformer have 50, and 1500 turns respectively. If the magnetic flux f linked, with the primary coil is given by f = f0 + 4t, where f is in, webers, t is time in seconds and f0 is a constant, the output, voltage across the secondary coil is, (a) 120 volts, (b) 220 volts, (c) 30 volts, (d) 90 volts, 30. The primary winding of a transformer has 100 turns and its, secondary winding has 200 turns. The primary is connected, to an A.C. supply of 120 V and the current flowing in it is 10, A. The voltage and the current in the secondary are, (a) 240 V, 5 A, (b) 240 V, 10 A, (c) 60 V, 20 A, (d) 120 V, 20 A, 31. The resistance in the following circuit is increased at a, 10mH, particular instant. At this instant, the value of resistance is 10W. The i, current in the circuit will be now, (a) i = 0.5 A (b) i > 0.5 A, 32., , p, p, (c), (d) 0, 2, 4, 34. What is the value of inductance L for which the current is, maximum in a series LCR circuit with, C = 10 µF and w = 1000s– 1 ?, (a) 1 mH, (b) cannot be calculated unless R is known, (c) 10 mH, (d) 100 mH, 35. In the circuit of Fig, the bulb will become suddenly bright if, , 5V, , (c) i < 0.5 A, , +, , 36., , 37., , (a), , RESPONSE, GRID, , (b), , 24., 29., 34., , (c), , 3, s, ln2, , 25., 30., 35., , K, , (a) contact is made or broken, (b) contact is made, (c) contact is broken, (d) won't become bright at all, The voltage of an ac source varies with time according to, the equation V = 100 sin 100 pt cos 100 pt where t is in, seconds and V is in volt. Then, (a) the peak voltage of the source is 100 volt, (b) the peak voltage of the source is 50 volt, (c) the peak voltage of the source is 100 / 2 volt, (d) the frequency of the source is 50 Hz, The current (I) in the inductance is varying with time, according to the plot shown in figure., , T/2, t, , T, , Which one of the following is the correct variation of voltage, with time in the coil?, V, , V, , Rh, , (a), , (d) i = 0, , (d), , –, , I, , t, T, , T/2, , (b), , (c), , 4, s, ln2, , 26., 31., 36., Space for Rough Work, , T/2, , t, , T, , V, , V, , state value in 4s. The time constant of this circuit is, 2, s, ln2, , B, , B, , 3, The current in a LR circuit builds up to th of its steady, 4, 1, s, ln2, , (b), , L, , qo, , (d), , qo, , An LCR series circuit is connected to a source of alternating, current. At resonance, the applied voltage and the current, flowing through the circuit will have a phase difference of, (a) p, , q, , q, , q, , 33., , T/2, , t T, , 27., 32., 37., , (d), , t, T/2, , 28., 33., , T
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t.me/Magazines4all, , DPP/ CP21, , P-84, , 38., , Using an A.C. voltmeter the potential difference in the, electrical line in a house is read to be 234 volt. If the line, frequency is known to be 50 cycles/second, the equation, for the line voltage is, , (a) V = 165 sin (100 p t), (b) V = 331 sin (100 p t), (c) V = 220 sin (100 p t), (d) V = 440 sin (100 p t), 39. In the circuit shown, when the switch is closed, the capacitor, charges with a time constant, R, C, (a) RC, (b) 2RC, (c), , 1, RC, 2, , 42., , An AC generator of 220 V having internal resistance r = 10W, and external resistance R = 100W. What is the power, developed in the external circuit?, (a) 484 W (b) 400 W (c) 441 W, (d) 369 W, 43. What is increased in step-down transformer?, (a) Voltage, (b) Current, (c) Power, (d) Current density, K, V, 44. In the circuit shown below, the key, K is closed at t = 0. The current, L, R, through the battery is, R, VR1R2, V, (a), at t = 0 and R at t = ¥, 2, R 2 + R2, 1, , 2, , 1, , +, , (b), , B, , 40., , 41., , (d) RC ln 2, A 100 mF capacitor in series with a 40W resistance is, connected to a 110 V, 60 Hz supply., What is the maximum current in the circuit?, (a) 3.24 A (b) 4.25 A (c) 2.25 A, (d) 5.20 A, The core of any transformer is laminated so as to, (a) reduce the energy loss due to eddy currents, (b) make it light weight, (c) make it robust and strong, (d) increase the secondary voltage, , (c), (d), 45., , V ( R1 + R2 ), V, at t = 0 and, at t = ¥, R1 R2, R2, VR1R2, V, at t = 0 and, at t = ¥, R2, R12 + R22, V ( R1 + R2 ), V, at t = 0 and, at t = ¥, R1 R2, R2, , The inductance between A and D is, , A, , (a) 3.66 H, , 38., 43., , RESPONSE, GRID, , 39., 44., , 2, , 40., 45., , 3H, , (b) 9 H, , 41., , 3H, , 3H, , D, , (c) 0.66 H, , (d) 1 H, , 42., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP21 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP22, , SYLLABUS : Electromagnetic Waves, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , 4., , An electromagnetic wave in vacuum has the electric and, r, r, magnetic field E and B , which are always perpendicular to, r, each other. The direction of polarization is given by X and, r, that of wave propagation by k . Then, r r r, r r, r r, r r r, (a) X || B and k || B ´ E (b) X || E and k || E ´ B, r r r, r, r r, r r, r r, (c) X || B and k || E ´ B (d) X || E and k || B ´ E, The rms value of the electric field of the light coming from, the Sun is 720 N/C. The average total energy density of the, electromagnetic wave is, (a) 4.58 × 10–6 J/m3, (b) 6.37 × 10–9 J/m3, (c) 81.35 × 10–12 J/m3, (d) 3.3 × 10–3 J/m3, In order to establish an instantaneous displacemet current, of 1 mA in the space between the plates of 2mF parallel, plate capacitor, the potential difference need to apply is, (a) 100 Vs–1 (b) 200 Vs–1 (c) 300 Vs–1 (d) 500 Vs–1, During the propagation of electromagnetic waves in a, medium:, , (b) Electric energy density is half of the magnetic energy, density., (c) Electric energy density is equal to the magnetic energy, density., 5., , (d) Both electric and magnetic energy densities are zero., An electromagnetic wave with frequency w and wavelength, l travels in the + y direction. Its magnetic field is along + xaxis. The vector equation for the associated electric field (of, amplitudeE0) is, (a), (b), , (c), , (d), , (a) Electric energy density is double of the magnetic, energy density., , RESPONSE GRID, , 1., , 2., , 3., Space for Rough Work, , ®, , 2p, æ, E = - E0 cos ç wt +, è, l, , ®, , 2p, æ, E = E0 cos ç wt è, l, , ö, y ÷ xˆ, ø, , ö, y ÷ xˆ, ø, , ®, , 2p ö, æ, E = E0 cos ç wt y÷ zˆ, è, l ø, , ®, , 2p ö, æ, E = - E0 cos ç wt +, y÷ zˆ, è, l ø, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP22, , P-86, , 6., , 7., , An electromagnetic wave of frequency n = 3.0 MHz, passes from vacuum into a dielectric medium with, permittivity Î = 4.0 . Then, (a) wavelength is halved and frequency remains, unchanged, (b) wavelength is doubled and frequency becomes half, (c) wavelength is doubled and the frequency remains, unchanged, (d) wavelength and frequency both remain unchanged., The average electric field of electromagnetic waves in certain, region of free space is 9 × 10–4 NC–1. Then the average, magnetic field in the same region is of the order of, (a) 27 × 10–4 T, , 8., , 9., , 10., , 15., , 16., , (b) 3 × 10–12 T, , æ 1ö, –12, T, (c) ç ÷ ´ 10, (d) 3 × 1012 T, è 3ø, The electric field of an electromagnetic wave travelling, through vaccum is given by the equation, E = E0 sin (kx – wt). The quantity that is independent of, wavelength is, k, (c) k2 w, (d) w, w, The electric and the magnetic field associated with an E.M., wave, propagating along the +z-axis, can be represented by, r r, r, r, r, ˆ B = B ˆjù, (a) éë E = E 0 i,, (b) éë E = E 0 k, B = B0ˆi ùû, 0 û, r, r, r, r, (c) éë E = E 0 ˆj, B = B0 iˆ ùû (d) éë E = E 0 ˆj,B = B0 kˆ ùû, , (a) kw, , 14., , 17., , (b), , 18., , E2, B2, +, 2e 0 2µ 0, , (b), , 19., , 1, 1, e0E 2 + µ0B2, 2, 2, , E 2 + B2, 1, B2, (d), e0E 2 +, c, 2, 2µ0, A plane electromagnetic wave is incident on a plane surface, of area A, normally and is perfectly reflected. If energy E, strikes the surface in time t then average pressure exerted, on the surface is (c = speed of light), , (c), 11., , 12., , 13., , (a) zero, (b) E/Atc (c) 2E/Atc (d) E/c, An electromagnetic wave travels along z-axis. Which of the, following pairs of space and time varying fields would, generate such a wave ?, (a) Ex, By (b) Ey, Bx (c) Ez, Bx (d) Ey, Bz, The magnetic field in a travelling electromagnetic wave has, a peak value of 20 nT. The peak value of electric field strength, is :, , RESPONSE, GRID, , 6., 11., 16., , 7., 12., 17., , 2E, E, 2E, E, (b), (c), (d), C, C2, C2, C, Match List - I (Electromagnetic wave type) with List - II (Its, association/application) and select the correct option from, the choices given below the lists:, List 1, List 2, 1. Infrared waves, (i) To treat muscular strain, 2. Radio waves, (ii) For broadcasting, 3. X-rays, (iii) To detect fracture of bones, 4. Ultraviolet rays (iv) Absorbed by the ozone layer, of the atmosphere, 1, 2, 3, 4, (a) (iv), (iii), (ii), (i), (b) (i), (ii), (iv), (iii), (c) (iii), (ii), (i), (iv), (d) (i), (ii), (iii), (iv), A plane electromagnetic wave travels in free space along, r, X-direction. If the value of B (in tesla) at a particular point, r, in space and time is 1.2 × 10–8 k̂. The value of E (in Vm–1), at that point is, , (a), , The energy of electromagnetic wave in vacuum is given by, the relation, (a), , (a) 3 V/m (b) 6 V/m (c) 9 V/m (d) 12 V/m, Microwave oven acts on the principle of :, (a) giving rotational energy to water molecules, (b) giving translational energy to water molecules, (c) giving vibrational energy to water molecules, (d) transferring electrons from lower to higher energy, levels in water molecule, Displacement current is, (a) continuous when electric field is changing in the circuit, (b) continuous when magnetic field is changing in the, circuit, (c) continuous in both types of fields, (d) continuous through wires and resistance only, The electric field associated with an e.m. wave in vacuum is, r, given by E = iˆ 40 cos (kz – 6 × 108t), where E, z and t are in, volt/m, meter and seconds respectively. The value of wave, vector k is, (a) 2 m–1 (b) 0.5 m–1 (c) 6 m–1 (d) 3 m–1, The charge on a parallel plate capacitor varies as q = q0, cos 2put. The plates are very large and close together, (area = A, separation = d). The displacement current, through the capacitor is, (a) q0 2pu sinput, (b) – q0 2pu sin2put, (c) q0 2p sinput, (d) q0 pu sin2put, A radiation of energy ‘E’ falls normally on a perfectly, reflecting surface. The momentum transferred to the surface, is (C = Velocity of light), , 20., , (a) 1.2 ˆj, , 8., 13., 18., Space for Rough Work, , (b) 3.6 kˆ, , 9., 14., 19., , (c), , 1.2 kˆ, , 10., 15., 20., , (d), , 3.6 ˆj, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP22, , P-87, , 21. If vs, vx and vm are the speed of soft gamma rays, X-rays, and microwaves respectively in vacuum, then, (a) vs > vx > vm, , (b) vs < vx < vm, , (c) vs > vx < vm, (d) vs = vx = vm, 22. Photons of an electromagnetic radiation has an energy, 11 keV each. To which region of electromagnetic spectrum, does it belong ?, (a) X-ray region, (b) Ultra violet region, (c) Infrared region, (d) Visible region, 23. A plane electromagnetic wave travels in free space along, x-axis. At a particular point in space, the electric field along, y-axis is 9.3 V m–1. The magnetic induction (B) along z-axis, is, (a) 3.1 × 10–8 T, (b) 3 × 10–5 T, (c) 3 × 10–6 T, (d) 9.3 × 10–6 T, 24. The ratio of amplitude of magnetic field to the amplitude of, electric field for an electromagnetic wave propagating in, vacuum is equal to :, (a) the speed of light in vacuum, (b) reciprocal of speed of light in vacuum, (c) the ratio of magnetic permeability to the electric, susceptibility of vacuum, (d) unity, 25. A plane electromagnetic wave is incident on a material, surface. If the wave delivers momentum p and energy E,, then, (a) p = 0, E = 0, (c), , p ¹ 0, E = 0, , (b), , p ¹ 0, E ¹ 0, , (d), , p = 0, E ¹ 0, , (a), , I = cm 0 B20 / 2, , (b), , I = ce 0 B20 / 2, , (c), , I = B20 / cm 0, , (d), , I = E 02 / 2ce 0, , 27. The decreasing order of wavelength of infrared, microwave,, ultraviolet and gamma rays is, (a) microwave, infrared, ultraviolet, gamma rays, (b) gamma rays, ultraviolet, infrared, micro-waves, (c) microwaves, gamma rays, infrared, ultraviolet, (d) infrared, microwave, ultraviolet, gamma rays, 28. Which radiation in sunlight, causes heating effect ?, (a) Ultraviolet, (b) Infrared, (c) Visible light, (d) All of these, , 21., 26., 31., 36., , 22., 27., 32., , 30., , 31., , The speed of electromagnetic wave in vacuum depends, upon the source of radiation. It, (a) increases as we move from g-rays to radio waves, (b) decreases as we move from g-rays to radio waves, (c) is same for all of them, (d) None of these, When an electromagnetic waves enter the ionised layer of, ionosphere, the motion of electron cloud produces a space, current and the electric field has its own capacitative, displacement current, then, (a) the space current is in phase of displacement current, (b) the space current lags behind the displacement current, by a phase 180°., (c) the space current lags behind the displacement current, by a phase 90°., (d) the space current leads the displacement current by a, phase 90°., The displacement current is, (a), , e o d f E / dt, , (b), , eo, d f E / dt, R, , (c) e o E / R, (d) e o q C / R, 32. Electromagnetic radiation of highest frequency is, (a) infrared radiations, (b) visible radiation, (c) radio waves, (d) g-rays, 33. A point source of electromagnetic radiation has an average, power output of 1500 W. The maximum value of electric field, at a distance of 3m from this sources in Vm–1 is, 500, 250, (d), 3, 3, 15, 34. Frequency of a wave is 6 × 10 Hz. The wave is, (a) radiowave, (b) microwave, (c) x-ray, (d) ultraviolet, 35. The electromagnetic waves do not transport, (a) energy, (b) charge, (c) momentum, (d) information, 36. Which of the following statement is false for the properties, of electromagnetic waves?, (a) Both electric and magnetic field vectors attain the, maxima and minima at the same place and same time., (b) The energy in electromagnetic wave is divided equally, between electric and magnetic vectors, (c) Both electric and magnetic field vectors are parallel to, each other and perpendicular to the direction of, propagation of wave, (d) These waves do not require any material medium for, propagation., , (a) 500, , 26. Intensity of electromagnetic wave will be, , RESPONSE, GRID, , 29., , 23., 28., 33., Space for Rough Work, , (b) 100, , 24., 29., 34., , (c), , 25., 30., 35.
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t.me/Magazines4all, , DPP/ CP22, , P-88, , 37., , 38., 39., , 40., , 41., , Which of the following electromagnetic waves has minimum, frequency ?, (a) Microwaves, (b) Audible waves, (c) Ultrasonic wave, (d) Radiowaves, The wave impendance of free space is, (a) zero (b) 376.6 W (c) 33.66 W (d) 3.76 W, A plane electromagnetic, wave in a non-magnetic dielectric, ur ur, medium is given by E = E 0 (4 ´ 10 -7 x - 50t ) with distance, being in meter and time in seconds. The dielectric constant, of the medium is :, (a) 2.4, (b) 5.8, (c) 8.2, (d) 4.8, We consider the radiation emitted by the human body., Which of the following statements is true?, (a) the radiation emitted lies in the ultraviolet region and, hence is not visible., (b) the radiation emitted is in the infra-red region., (c) the radiation is emitted only during the day., (d) the radiation is emitted during the summers and, absorbed during the winters., In a plane electromagnetic wave propagating in space has, an electric field of amplitude 9 × 103 V/m, then the amplitude, of the magnetic field is, (a) 2.7 × 1012 T, (b) 9.0 × 10–3 T, –4, (c) 3.0 × 10 T, (d) 3.0 × 10–5 T, , 37., 42., , RESPONSE, GRID, , 38., 43., , 42., , 43., , 44., , 45., , Out of the following options which one can be used to, produce a propagating electromagnetic wave ?, (a) A charge moving at constant velocity, (b) A stationary charge, (c) A chargeless particle, (d) An accelerating charge, Radio waves of constant amplitude can be generated with, (a) rectifier, (b) filter, (c) F.E.T., (d) oscillator, In an electromagnetic wave, (a) power is transmitted along the magnetic field, (b) power is transmitted along the electric field, (c) power is equally transferred along the electric and, magnetic fields, (d) power is transmitted in a direction perpendicular to, both the fields, If c is the speed of electromagnetic waves in vacuum, its, speed in a medium of dielectric constant K and relative, permeability µr is, (a), , v=, , (c), , v=, , 39., 44., , 1, mr K, c, mr K, , 40., 45., , (b), , v = c mr K, , (d), , v=, , K, mr C, , 41., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP22 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP23, , SYLLABUS : Ray Optics and Optical Instruments, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , A double convex lens is made of glass which has its, refractive index 1.45 for violet rays and 1.50 for red rays. If, the focal length for violet ray is 20cm, the focal length for, red ray will be, (a) 9 cm, (b) 18 cm (c) 20 cm (d) 22 cm, A, and, 2, the angle of prism is A, then angle of minimum deviation is, , If the refractive index of the material of a prism is cot, , p, p, - 2A (d), -A, 2, 2, If two + 5 diopter lenses are mounted at some distance apart,, the equivalent power will always be negative if the distance, is, (a) greater than 40 cm, (b) equal to 40 cm, (c) equal to 10 cm, (d) less than 10 cm, Refraction of light from air to glass and from air to water, are shown in figure (i) and figure (ii) below. The value of, the angle q in the case of refraction as shown in figure, (iii) will be, , (a) p - 2A, 3., , 4., , (i), , Glass, Air 60°, , (b) p - A, , RESPONSE GRID, , 1., 6., , (iii), 41°, , (c), , 2., , 5., , 60°, , (ii), , Air, Water, 41°, , q, Glass, Water, , (a) 30°, (b) 35°, (c) 60°, (d) 41°, A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of, water is, , 6., , 35°, , 4, and the fish is 12 cm below the surface, the, 3, , radius of this circle in cm is, (a) 36 5, (b) 4 5, (c) 36 7, (d) 36/ 7, If fV and fR are the focal lengths of a convex lens for violet and, red light respectively and FV and FR are the focal lengths of, concave lens for violet and red light respectively, then we have, (a) fV < fR and FV > FR, (b) fV < fR and FV < FR, (c) fV > fR and FV > FR, (d) fV > fR and FV < FR, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP23, , P-90, , 7., , Spherical aberration in a lens :, (a) is minimum when most of the deviation is at first surface, (b) is minimum when most of the deviation is at the second, surface, (c) is minimum when the total deviation is equally distributed, over the two surfaces, (d) does not depend on the above considerations, 8., A rod of length 10 cm lies along the principal axis of a, concave mirror of focal length 10 cm in such a way that its, end closer to the pole is 20 cm away from the mirror. The, length of the image is :, (a) 10 cm, (b) 15 cm, (c) 2.5 cm, (d) 5 cm, 9., A telescope consists of two thin lenses of focal lengths, 0.3, m and 3 cm respectively. It is focused on moon which, subtends an 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°, (d) 0.35°, 10. The layered lens as shown is made of, two types of transparent materials-one, indicated by horizontal lines and the, other by vertical lines. The number of, images formed of an object will be, (a) 1, (b) 2, (c) 3, (d) 6, 11. A man’s near point is 0.5 m and far point is 3 m. Power of, spectacle lenses required for (i) reading purposes, (ii) seeing, distant objects, respectively, are, (a) –2 D and + 3 D, (b) +2 D and –3 D, (c) +2 D and –0.33 D, (d) –2 D and + 0.33 D, 12. A ray of light falls on a transparent glass slab of refractive, index 1.62. If the reflected ray and the refracted ray are, mutually perpendicular, the angle of incidence is, (a), , tan -1 (1.62), , (b), , æ 1 ö, tan -1 ç, è 1.62 ÷ø, , tan -1 (1.33), , RESPONSE, GRID, , (b) R, (c) 2R, (d) 3R, 17., , 18., , 19., , 20., , 21., , æ 1 ö, tan -1 ç, è 1.33 ÷ø, 13. A telescope has an objective of focal length 100 cm and an, eyepiece of focal length 5 cm. What is the magnifying power, of the telescope when the final image is formed at the least, distance of distinct vision ?, (a) 20, (b) 24, (c) 28, (d) 32, 14. Which light rays undergoes two internal reflection inside, a raindrop, which of the rainbow is formed?, (a) Primary rainbow, (b) Secondary rainbow, (c) Both (a) and (b), (d) Can’t say, 15. When a plane mirror is placed horizontally on a level ground, at a distance of 60 m from the foot of a tower, the top of the, tower and its image in the mirror subtend an angle of 90° at, the eye. The height of the tower will be, , (c), , (a) 30 m, (b) 60 m, (c) 90 m, (d) 120 m, 16. A parallel beam of light is incident on the surface of a, transparent hemisphere of radius R and refractive index 2.0, as shown in figure. The position of the image formed by, refraction at the first surface is :, (a) R/2, , (d), , 7., 12., 17., 22., , 8., 13., 18., , 22., , A lens made of glass whose index of refraction is 1.60 has a, focal length of + 20 cm in air. Its focal length in water, whose, refractive index is 1.33, will be, (a) three times longer than in air, (b) two times longer than in air, (c) same as in air, (d) None of these, A compound microscope has an eye piece of focal length 10, cm and an objective of focal length 4 cm. Calculate the, magnification, if an object is kept at a distance of 5 cm from, the objective so that final image is formed at the least, distance vision (20 cm) :, (a) 12, (b) 11, (c) 10, (d) 13, For a prism kept in air it is found that for an angle of incidence, 60°, the angle of Prism A, angle of deviation d and angle of, emergence ‘e’ become equal. Then the refractive index of, the prism is, (a) 1.73, (b) 1.15, (c) 1.5, (d) 1.33, A person can see clearly only upto a distance of 30 cm. He, wants to read a book placed at a distance of 50 cm from his, eyes. What is the power of the lens of his spectacles ?, (a) –1.0 D, (b) –1.33 D (c) –1.67 D (d) –2.0 D, An object is placed at a distance of 40 cm in front of a, concave mirror of focal length 20 cm. The image produced is, (a) real, inverted and smaller in size, (b) real, inverted and of same size, (c) real and erect, (d) virtual and inverted, A vessel of depth x is half filled with oil of refractive index, m1 and the other half is filled with water of refractive index, m2. The apparent depth of the vessel when viewed from, above is, (a), , x (m1 + m 2 ), 2m1m2, , (c), , xm1m2, (m1 + m 2 ), , 9., 14., 19., Space for Rough Work, , 10., 15., 20., , xm1m 2, 2(m1 + m2 ), 2 x (m1 + m2 ), (d), m1m2, , (b), , 11., 16., 21., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP23, , P-91, , 23. The following figure, shows refraction of light, at the interface of three, media Correct the order of, optical density (d) of the, media is, , 24., , 25., , 26., , 27., , 29., Medium 1, , i1, , Medium 2, r1, , r2 Medium 3, , r1, , (a) d1 > d2 > d3, (b) d2 > d1 > d3, (c) d3 > d3 > d2, (d) d2 > d3 > d1, Light travels in two media A and B with speeds 1.8 ×, 108 m s–1 and 2.4 × 108 m s–1 respectively. Then the critical, angle between them is, -1 æ 2 ö, -1 æ 3 ö, (a) sin ç ÷, (b) tan ç ÷, 3, è ø, è 4ø, æ 2ö, æ 3ö, (c) tan -1 ç ÷, (d) sin -1 ç ÷, è 3ø, è 4ø, The refractive index of a glass is 1.520 for red light and, 1.525 for blue light. Let D1 and D2 be angles of minimum, deviation for 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, Which of the following is not due to total internal reflection?, (a) Working of optical fibre, (b) Difference between apparent and real depth of pond, (c) Mirage on hot summer days, (d) Brilliance of diamond, A body is located on a wall. Its image of equal size is to be, obtained on a parallel wall with the help of a convex lens., The lens is placed at a distance 'd' ahead of second wall,, then the required focal length will be, , 30., , 31., , 32., , 33., , A ray of light passes through an equilateral prism such that, the angle of incidence is equal to the angle of emergence, 3, and the latter is equal to th of angle of prism. The angle, 4, of deviation is, (a) 25°, (b) 30°, (c) 45°, (d) 35°, The power of a biconvex lens is 10 dioptre and the radius of, curvature of each surface is 10 cm. Then the refractive index, of the material of the lens is, 5, 3, 4, 9, (a), (b), (c), (d), 2, 3, 8, 3, A microscope is focussed on a mark on a piece of paper and, then a slab of glass of thickness 3 cm and refractive index, 1.5 is placed over the mark. How should the microscope be, moved to get the mark in focus again ?, (a) 4.5 cm downward, (b) 1 cm downward, (c) 2 cm upward, (d) 1 cm upward, What causes chromatic aberration?, (a) Marginal rays, (b) Central rays, (c) Difference in radii of curvature of its surfaces, (d) Variation of focal length of lens with colour, The graph between angle of deviation (d) and angle of, incidence (i) for a triangular prism is represented by, (a), , o, , (c), , d, 4, d, (b) only, 2, , (a) only, , 28., , RESPONSE, GRID, , 23., 28., 33., , 24., 29., 34., , (d), , i, , d, , o, , i, , d, , o, , d, d, (c) more than, but less than, 4, 2, d, (d) less than, 4, A concave mirror forms the image of an object on a screen., If the lower half of the mirror is covered with an opaque, card, the effect would be to make the, (a) image less bright., (b) lower half of the image disappear., (c) upper half of the image disappear., (d) image blurred., , (b), , d, , i, , d, , o, , i, , 34., , The ratio of thickness of plates of two transparent medium, A and B is 6 : 4. If light takes equal time in passing through, them, then refractive index of A with respect to B will be, (a) 1.33, (b) 1.75, (c) 1.4, (d) 1.5, 35. A rectangular block of glass is placed on a mark made on the, surface of the table and it is viewed from the vertical position, of eye. If refractive index of glass be m and its thickness d,, then the mark will appear to be raised up by, (a), , 25., 30., 35., Space for Rough Work, , (m + 1)d, (m - 1)d, (m + 1), (m - 1) m, (b), (c), (d), m, m, md, d, , 26., 31., , 27., 32.
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t.me/Magazines4all, , DPP/ CP23, , P-92, , 36., , 37., , 38., 39., , If a glass prism is dipped in water, its dispersive power, (a) increases, (b) decreases, (c) does not change, (d) may increase or decrease depending on whether the, angle of the prism is less than or greater than 60º, A planoconcave lens is placed, Radius of, on a paper on which a flower is curvature =20cm, drawn. How far above its actual, Air, position does the flower appear, µ=3/2, t=20cm, to be?, (a) 10 cm (b) 15 cm, Paper, (c) 50 cm (d) None of these, To get three images of a single object, one should have two, plane mirrors at an angle of, (a) 60º, (b) 90º, (c) 120º, (d) 30º, Light propagates with speed of 2.2 ´ 108 m / s and, 2.4 ´ 108 m / s in the media P and Q respectively. The critical, angle of incidence for light undergoing reflection from P, and Q is, æ1ö, 11, (b) sin -1 æç ö÷, sin -1 ç ÷, 12, è 11 ø, è ø, -1 æ 5 ö, -1 æ 5 ö, (c) sin ç ÷, (d) sin ç ÷, è 11 ø, è 12 ø, A thin convergent glass lens (mg = 1.5) has a power of, + 5.0 D. When this lens is immersed in a liquid of refractive, index m, it acts as a divergent lens of focal length 100 cm., The value of m must be, (a) 4/3, (b) 5/3, (c) 5/4, (d) 6/5, A ray of light travelling inside a rectangular glass block of, refractive index 2 is incident on the glass-air surface at, an angle of incidence of 45º. The refractive index of air is, one. Under these conditions the ray will, (a) emerge into the air without any deviation, (b) be reflected back into the glass, (c) be absorbed, (d) emerge into the air with an angle of refraction equal, to 90º, , 42., , 3 cm, , 4 cm, , coin, , 43., , (a), , 40., , 41., , 36., 41., , RESPONSE, GRID, , 37., 42., , A small coin is resting on the bottom of a beaker filled with, liquid. A ray of light from the coin travels upto the surface of, the liquid and moves along its surface. How fast is the light, travelling in the liquid?, , 44., , 45., , 2.4 × 108 m/s, 1.2 × 108 m/s, , (a), (b) 3.0 × 108 m/s, (c), (d) 1.8 × 108 m/s, A, A ray PQ incident on the refracting, face BA is refracted in the prism, 60°, R, BAC as shown in the figure and, Q, emerges from the other refracting, S, face AC as RS such that AQ = AR., If the angle of prism A = 60° and the P B, C, refractive index of the material of prism is 3 , then the, angle of deviation of the ray is, (a) 60°, (b) 45°, (c) 30°, (d) None of these, When a biconvex lens of glass having refractive index 1.47, is dipped in a liquid, it acts as a plane sheet of glass. This, implies that the liquid must have refractive index., (a) equal to that of glass, (b) less then one, (c) greater than that of glass, (d) less then that of glass, If a thin prism of glass is dipped in water then minimum, deviation (with respect to air) of light produced by prism, æ, , 3, 2, , 4ö, 3ø, , will be ç w µg = , aµ w = ÷, è, , 1, (a), 5, , 38., 43., , (b), , 1, 4, , 39., 44., , (c), , 1, 2, , (d), , 40., 45., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP23 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 60, , 1, 3, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP24, , SYLLABUS : Wave Optics, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , In young’s double-slit experiment, the intensity of light at a, point on the screen where the path difference is l is I, l, being the wavelength of light used. The intensity at a point, l, where the path difference is, will be, 4, I, I, (b), (c) I, (d) zero, (a), 2, 4, A beam of light is incident on a A, B, glass slab (m = 1.54) in a direction, as shown in the figure. The, reflected light is analysed by a, 33°, 33°, polaroid prism. On rotating the, Glass slab, polaroid, (tan 57° = 1.54), (a) the intensity remains unchanged, (b) the intensity is reduced to zero and remains at zero, (c) the intensity gradually reduces to zero and then again, increase, (d) the intensity increases continuously, Two sources of light of wavelengths 2500 Å and 3500 Å are, used in Young’s double slit expt. simultaneously. Which, orders of fringes of two wavelength patterns coincide?, (a) 3rd order of 1st source and 5th of the 2nd, (b) 7th order of 1st and 5th order of 2nd, , RESPONSE GRID, , 1., , 2., , 4., , (c) 5th order of 1st and 3rd order of 2nd, (d) 5th order of 1st and 7th order of 2nd, Figure shows behavior of a wavefront when it passes, through a prism., , A, Incident, wavefront, , A’, Refracted, wavefront, B’, , B, , 5., , Which of the following statements is/are correct ?, (a) Lower portion of wavefront (B’) is delayed resulting, in a tilt., (b) Time taken by light to reach A’ is equal to the time, taken to reach B’ from B., (c) Speed of wavefront is same everywhere., (d) A particle on wavefront A’ B’ is in phase with a, particle on wavefront AB., When the angle of incidence is 60° on the surface of a, glass slab, it is found that the reflected ray is completely, polarised. The velocity of light in glass is, (a), (c), , 3., Space for Rough Work, , 2 ´ 108 ms -1, 2 ´ 108 ms -1, , 4., , (b), (d), , 3 ´ 108 ms -1, 3 ´ 108 ms -1, , 5.
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t.me/Magazines4all, , DPP/ CP24, , P-94, , 6., , 7., , 8., , Figure shows two coherent sources S1 and S2 vibrating in, same phase. AB is an irregular wire lying at a far distance, l, -3, from the sources S1 and S2. Let = 10 and ÐBOA = 0.12°., d, How many bright spots will be seen on the wire, including, points A and B?, A, S1, (a) 5, (b) 4, d O, (c) 2, S2, B, (d) 7, Two identical light waves, propagating in the same direction,, have a phase difference d. After they superpose, the intensity, of the resulting wave will be proportional to, (a) cos d, (b) cos (d/2), (c) cos2 (d/2), (d) cos2 d, In YSDE, both slits are covered by transparent slab. Upper, slit is covered by slab of R.I. 1.5 and thickness t and lower is, covered by R.I., , 4, and thickness 2t, then central maxima, 3, , (a) shifts in +ve y-axis direction, , 9., , 10., , 11., , 6., 11., 16., , 13., , 7., 12., 17., , On a rainy day, if there is an oil drop on tar road coloured, rings are seen around this drop. This is due to, (a) total internal reflection of light, (b) polarisation, (c) diffraction pattern, (d) interference pattern produced due to oil film, In a Young’s double slit experiment, the intensity at a point, where the path difference, , l, (l – is wavelength of the light), 6, , is I. If I0 denotes the maximum intensity, then, , I, is equal to, I0, , 1, 1, 3, 3, (b), (d), (c), 2, 2, 4, 2, According to Huygens, medium through which light, waves travel is, (a) vacuum only, (b) luminiferous ether, (c) liquid only, (d) solid only, If we observe the single slit Fraunhofer diffraction with, wavelength l and slit width b, the width of the central, maxima is 2q. On decreasing the slit width for the same l, (a) q increases, (b) q remains unchanged, (c) q decreases, (d) q increases or decreases depending on the intensity of, light, Aperture of the human eye is 2 mm. Assuming the mean, wavelength of light to be 5000 Å, the angular resolution, limit of the eye is nearly, (a) 2 minute, (b) 1 minute, (c) 0.5 minute, (d) 1.5 minute, Unpolarised light is incident on a dielectric of refractive, , (a), , 14., , 15., , y, , (b) shifts in –ve y-axis direction, x, (c) remains at same position, (d) may shift in upward, or downward depending upon wavelength of light, A beam of light of l = 600 nm from a distant source falls on, a single slit 1 mm wide and the resulting diffraction pattern, is observed on a screen 2 m away. The distance between, first dark fringes on either side of the central bright fringe is, (a) 1.2 cm, (b) 1.2 mm, (c) 2.4 cm, (d) 2.4 mm, A parallel beam of light of wavelength l is incident normally, on a narrow slit. A diffraction pattern is formed on a screen, placed perpendicular to the direction of the incident beam., At the second minimum of the diffraction pattern, the phase, difference between the rays coming from the two edges of, slit is, (a) pl, (b) 2p, (c) 3p, (d) 4p, The diffraction effects in a microscopic specimen become, important when the separation between two points is, (a) much greater than the wavelength of light used., (b) much less than the wavelength of light used., (c) comparable to the wavelength of light used., (d) independent of the wavelength of light used., , RESPONSE, GRID, , 12., , 16., , 17., , index, , 18., , 3 . What is the angle of incidence if the reflected, beam is completely polarised?, (a) 30°, (b) 45°, (c) 60°, (d) 75°, The figure shows the, Central bright, interference pattern obtained in, fringe, a double-slit experiment using, light of wavelength 600nm. 1, 2,, 3, 4 and 5 are marked on five, fringes., 12 3 4 5, The third order bright fringe is, (a) 2, (b) 3, (c) 4, (d) 5, , 8., 13., 18., Space for Rough Work, , 9., 14., , 10., 15., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP24, , P-95, , 19. Which of the following diagrams represent the variation of, electric field vector with time for a circularly polarised, light ?, (a), , (b), |E |, , (c), |E |, , t, (a), , (d), , 27., , |E |, , |E |, , t, (b), , t, , t, , (d), , 20. With a monochromatic light, the fringe-width obtained in a, Young’s double slit experiment is 0.133 cm. The whole setup is immersed in water of refractive index 1.33, then the, new fringe-width is, (a) 0.133 cm, (b) 0.1 cm, (c) 1.33 cm, (d) 0.2 cm, 21. The condition for obtaining secondary maxima in the, diffraction pattern due to single slit is, (a), , a sin q = ( 2n - 1), , 28., , 29., , l, 2, , nl, 2, In double slit experiment, the angular width of the fringes is, , (c), 22., , (b), , a sin q = nl, , respectively. What is the actual distance of separation?, (a) 12.5 cm (b) 12 cm (c) 13 cm (d) 14 cm, , a sin q = ( 2n - 1) l, , (d), , a sin q =, , 0.20° for the sodium light (l = 5890Å ) . In order to increase, the angular width of the fringes by 10%, the necessary, change in wavelength is, (a) zero, (b) increased by 6479 Å, (c) decreased by 589 Å (d) increased by 589 Å, 23. In Young's double slit experiment with sodium vapour lamp, of wavelength 589 nm and the slits 0.589 mm apart, the half, angular width of the central maximum is, (a) sin–1 (0.01), (b) sin –1 (0.0001), –1, (c) sin (0.001), (d) sin–1 (0.1), 24. The adjacent figure shows Fraunhoffer’s diffraction due to, a single slit. If first minimum is obtained in the direction, shown, then the path difference between rays 1 and 3 is, (a) 0, 1, (b) l / 4, 2, (c) l / 2, 3, (d) l, 25. A YDSE is conducted in water (µ1) as shown in figure. A, glass plate of thickness t and refractive index µ2 is placed in, the path of S2. The optical path difference at O is, (a) (m 2 - 1)t, S1, water µ1, (b) (m1 - 1)t, (c), , æ m2, ö, - 1÷ t, ç, è m1, ø, (m 2 – m1 )t, , (a), , 30., , 31., , 32., , 33., , µ, S2 2, t, , 19., 24., 29., , 20., 25., 30., , (b) p radian, , p, p, radian, radian, (d), 8, 4, The central fringe of the interference pattern produced by, light of wavelength 6000Å is found to shift to the position, of 4th bright fringe after a glass plate of refractive index 1.5, is introduced in front of one of slits in Young's experiment., The thickness of the glass plate will be, (a) 4.8 µm, (b) 8.23 µm, (c) 14.98 µm, (d) 3.78 µm, , Sodium light (l = 6 ´10 -7 m) is used to produce, interference pattern. The observed fringe width is 0.12 mm., The angle between two interfering wave trains, is, (a) 1 ´ 10 -3 rad, , (d), Screen, 26. In a Fresnel biprism experiment, the two positions of lens, give separation between the slits as 16 cm and 9 cm, , RESPONSE, GRID, , p, radian, 2, , (c), , O, , S, , pö, æ, If two waves represented by y1 = 4 sin wt and y2 = ç wt + ÷, 3ø, è, interfere at a point, then the amplitude of the resulting wave, will be about, (a) 7, (b) 6, (c) 5, (d) 3.5, In Young’s double slit experiment, the separation between, the slits is halved and the distance between the slits and, screen is doubled. The fringe width will, (a) be halved, (b) be doubled, (c) be quadrupled, (d) remain unchanged, At the first minimum adjacent to the central maximum of a, single-slit diffraction pattern, the phase difference between, the Huygen's wavelet from the edge of the slit and the, wavelet from the midpoint of the slit is :, , (b) 1´10 -2 rad, , (c) 5 ´ 10 -3 rad, (d) 5 ´ 10 -2 rad, The Young’s double slit experiment is performed with blue, and with green light of wavelengths 4360Å and 5460Å, respectively. If x is the distance of 4th maxima from the central, one, then, (a) x (blue) = x (green), (b) x (blue) > x (green), x (blue) 5460, =, (c) x (blue) < x (green), (d), x (green) 4360, If yellow light emitted by sodium lamp in Young’s double, slit experiment is replaced by a monochromatic blue light of, the same intensity, (a) fringe width will decrease, (b) finge width will increase, (c) fringe width will remain unchanged, (d) fringes will become less intense, , 21., 26., 31., Space for Rough Work, , 22., 27., 32., , 23., 28., 33.
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t.me/Magazines4all, , DPP/ CP24, , P-96, , 34., , 35., , 36., , 37., , 38., , When unpolarised light is incident on a plane glass plate at, Brewster’s angle, then which of the following statements is, correct?, (a) Reflected and refracted rays are completely polarised, with their planes of polarization parallel to each other, (b) Reflected and refracted rays are completely polarised, with their planes of polarization perpendicular to each, other, (c) Reflected light is plane polarised but transmitted light, is partially polarised, (d) Reflected light is partially polarised but refracted light, is plane polarised, The maximum number of possible interference maxima for, slit- separation equal to twice the wavelength in Young¢s, double-slit experiment is, (a) infinite (b) five, (c) three, (d) zero, In the figure shown if a parallel, beam of white light is incident, on the plane of the slits then, d 2d/3, O, the distance of the nearest, white spot on the screen from, O is d/A. Find the value of A., D, (assume d << D, l << d], (a) 3, (b) 5, (c) 6, (d) 4, Two light waves superimposing at the mid-point of the, screen are coming from coherent sources of light with, phase difference 3p rad. Their amplitudes are 1 cm each., The resultant amplitude at the given point will be., (a) 5 cm, (b) 3 cm, (c) 2 cm, (d) zero, Spherical wavefronts, emanating from a point source, strike, a plane reflecting surface. What will happen to these wave, fronts, immediately after reflection?, (a) They will remain spherical with the same curvature,, both in magnitude and sign., (b) They will become plane wave fronts., (c) They will remain spherical, with the same curvature,, but sign of curvature reversed., (d) They will remain spherical, but with different, curvature, both in magnitude and sign., , RESPONSE GRID, , 34., 39., 44., , 35., 40., 45., , 39., , 40., , 41., , 42., , 43., , Two coherent point sources S1 and S2 are separated by a, small distance d as shown. The fringes obtained on the, vertical screen will be :, (a) points, d, (b) straight bands, S2, S1, screen, (c) concentric circles, D, (d) semicircles, In the phenomena of diffraction of light, when blue light is, used in the experiment in spite of red light, then, (a) fringes will become narrower, (b) fringes will become broader, (c) no change in fringe width, (d) None of these, 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, the Huygens', principle leads us to conclude that as it travels, the light, beam :, (a) bends downwards, (b) bends upwards, (c) becomes narrower, (d) goes horizontally without any deflection, If I 0 is the intensity of the principal maximum in the single, slit diffraction pattern, then what will be its intensity when, the slit width is doubled?, I0, (a) 4 I 0, (b) 2 I 0, (c), (d) I 0, 2, Conditions of diffraction is, (a), , 44., , 45., , a, =1, l, , (b), , a, >> 1, l, , (c), , a, << 1, l, , (d) None of these, In Fresnel’s biprism expt., a mica sheet of refractive index 1.5, and thickness 6 × 10–6 m is placed in the path of one of, interfering beams as a result of which the central fringe gets, shifted through 5 fringe widths. The wavelength of light, used is, (a) 6000 Å (b) 8000 Å (c) 4000 Å (d) 2000 Å, Two nicols are oriented with their principal planes making, an angle of 60º. Then the percentage of incident unpolarised, light which passes through the system is, (a) 100, (b) 50, (c) 12.5, (d) 37.5, , 36., 41., , 37., 42., , 38., 43., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP24 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP25, , SYLLABUS : Dual Nature of Radiation and Matter, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , A particle of mass 1 mg has the same wavelength as an, electron moving with a velocity of 3×106 ms–1. The velocity, of the particle is:, (a) 2.7× 10–18 ms–1, , (b) 9 × 10–2 ms–1, , (c) 3 × 10–31 ms–1, , (d) 2.7×10–21 ms–1, , The maximum kinetic energy of the electrons hitting a target, so as to produce X-ray of wavelength 1 Å is, (a) 1.24 keV, (b) 12.4 keV, (c) 124 keV, (d) None of these, , 5., , An X-ray tube is operated at 15 kV. Calculate the upper limit, of the speed of the electrons striking the target., , An electron of mass m and a photon have same energy E., The ratio of de-Broglie wavelengths associated with them, is :, 1, , (a), , (c), 3., , 4., , 6., , 1, , æ E ö2, (b) ç, è 2m ÷ø, , 1æ E ö2, ç, ÷, c è 2m ø, , 1, , 1, c(2mE) 2, , (d), , 1 æ 2m ö 2, ç, ÷, cè E ø, , All electrons ejected from a surface by incident light of, wavelength 200nm can be stopped before travelling 1m in, the direction of uniform electric field of 4N/C. The work, function of the surface is, (a) 4 eV, (b) 6.2 eV, (c) 2 eV, (d) 2.2 eV, , RESPONSE GRID, , (a) 7.26 × 107 m/s, , 1., 6., , 2., 7., , 7., , (b) 7.62 × 109 m/s, , (c) 7.62 × 107 cm/s, (d) 7.26 × 109 m/s, A and B are two metals with threshold frequencies, 1.8 × 1014 Hz and 2.2 × 1014 Hz. Two identical photons of, energy 0.825 eV each are incident on them. Then, photoelectrons are emitted in (Take h = 6.6 × 10–34 Js), (a) B alone, , (b) A alone, , (c) neither A nor B, , (d) both A and B., , If E1, E2, E3 are the respective kinetic energies of an electron,, an alpha-particle and a proton, each having the same, de-Broglie wavelength, then, (a) E1 > E3 > E2, , (b) E2 > E3 > E1, , (c) E1 > E2 > E3, , (d) E1 = E2 = E3, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP25, , P-98, , 8., , Which of the following when falls on a metal will emit, photoelectrons ?, (a) UV radiations, (b) Infrared radiation, (c) Radio waves, (d) Microwaves, 9., The stopping potential (V 0 ) versus V0, frequency (v) plot of a substance is 2, shown in figure, the threshold, wavelength is, 1, (a) 5 × 1014m, (b) 6000 Å, 4 5 6 7 8, (c) 5000 Å, v × 1014 Hz, (d) Cannot be estimated from given data, 10. A material particle with a rest mass m0 is moving with speed, of light c. The de-Broglie wavelength associated is given, by, h, m0c, (c) zero, (d) ¥, (b), m0c, h, A 200 W sodium street lamp emits yellow light of wavelength, 0.6 µm. Assuming it to be 25% efficient in converting, electrical energy to light, the number of photons of yellow, light it emits per second is, (a) 1.5 × 1020, (b) 6 × 1018, 20, (c) 62 × 10, (d) 3 × 1019, A proton has kinetic energy E = 100 keV which is equal to, that of a photon. The wavelength of photon is l2 and that, of proton is l1. The ratio of l2/l1 is proportional to, (a) E2, (b) E1/2, (c) E–1, (d) E–1/2, In photoelectric effect the work function of a metal is 3.5 eV., The emitted electrons can be stopped by applying a potential, of –1.2 V. Then, (a) the energy of the incident photon is 4.7 eV, (b) the energy of the incident photon is 2.3 eV, (c) if higher frequency photon be used, the photoelectric, current will rise, (d) when the energy of photon is 3.5 eV, the photoelectric, current will be maximum, The threshold frequency for a metallic surface corresponds, to an energy of 6.2 eV and the stopping potential for a, radiation incident on this surface is 5 V. The incident radiation, lies in, (a) ultra-violet region, (b) infra-red region, (c) visible region, (d) X-ray region, When photons of energy hn fall on an aluminium plate (of, work function E0), photoelectrons of maximum kinetic energy, K are ejected. If the frequency of the radiation is doubled,, the maximum kinetic energy of the ejected photoelectrons, will be, (a) 2K, (b) K, (c) K + hn, (d) K + E0, , 16., , 17., , (a), , 12., , 13., , 14., , 15., , RESPONSE, GRID, , 8., 13., 18., 23., , 9., 14., 19., 24., , 2( hc + lf ), ml, , (b), , 2(hc + lf ), ml, , 2(hl - f), 2( hc - lf ), (d), m, ml, If the kinetic energy of a free electron doubles, it’s deBroglie, wavelength changes by the factor, , (c), 18., , (a), , 11., , Which metal will be suitable for a photoelectric cell using, light of wavelength 4000Å. The work functions of sodium, and copper are respectively 2.0 eV and 4.0 eV., (a) Sodium, (b) Copper, (c) Both, (d) None of these, The maximum velocity of an electron emitted by light of, wavelength l incident on the surface of a metal of workfunction f is, , 1, 1, (c), (d), 2, 2, 2, Radiations of two photon’s energy, twice and ten times the, work function of metal are incident on the metal surface, successsively. The ratio of maximum velocities of, photoelectrons emitted in two cases is, (a) 1 : 2, (b) 1 : 3, (c) 1 : 4, (d) 1 : 1, The cathode of a photoelectric cell is changed such that the, work function changes from W1 to W2 (W2 > W1). If the, current before and after changes are I1 and I2, all other, conditions remaining unchanged, then (assuming hn > W2), (a) I1 = I2, (b) I1 < I2, (c) I1 > I2, (d) I1 < I2 < 2 I1, Monochromatic radiation emitted when electron on, hydrogen atom jumps from first excited to the ground state, irradiates a photosensitive material. The stopping potential, is measured to be 3.57 V. The threshold frequency of the, materials is :, (a) 4 × 1015 Hz, (b) 5 × 1015 Hz, (c) 1.6 × 1015 Hz, (d) 2.5 × 1015 Hz, Photoelectric work function of a metal is 1eV. Light of, wavelength l = 3000 Å falls on it. The photo electrons come, out with velocity, (a) 10 metres/sec, (b) 102 metres/sec, 4, (c) 10 metres/sec, (d) 106 metres/sec, When the energy of the incident radiation is incredased by, 20%, the kinetic energy of the photoelectrons emitted from, a metal surface increased from 0.5 eV to 0.8 eV. The work, function of the metal is :, (a) 0.65 eV (b) 1.0 eV (c) 1.3 eV (d) 1.5 eV, The maximum distance between interatomic lattice planes is, 15 Å. The maximum wavelength of X-rays which are, diffracted by this crystal will be, (a) 15 Å, (b) 20 Å, (c) 30 Å, (d) 45 Å, , (a) 2, , 19., , 20., , 21., , 22., , 23., , 24., , 10., 15., 20., Space for Rough Work, , (b), , 11., 16., 21., , 12., 17., 22., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP25, , P-99, , 25. In photoelectric effect, stopping potential for a light of, frequency n 1 is V1. If light is replaced by another having a, frequency n 2 then its stopping potential will be, (a), , V1 -, , h, ( n 2 - n1 ), e, , (b), , V1 +, , h, ( n 2 + n1 ), e, , h, h, ( n 2 - 2n1 ), (d) V1 + ( n 2 - n1 ), e, e, 26. The maximum kinetic energy of the photoelectrons ejected, from a photocathode when it is irradiated with light of, wavelength 440nm is 1eV. If the threshold energy of the, surface is 1.9eV, then which of the following statement, is/are incorrect?, (a) The threshold frequency for photo sensitive metal is, 4.6 × 1014Hz, (b) The minimum wavelength of incident light required for, photoemission is 6513 Å., (c) The maximum wavelength of incident light required for, photoemission is 6513 Å., (d) The energy of incident photon is 2.9 eV., 27. The work functions of metals A and B are in the raio 1 : 2., If light of frequencies f and 2f are incident on the surfaces of, A and B respectively, the ratio of the maximum kinetic, energies of photoelectrons emitted is (f is greater than, threshold frequency of A, 2f is greater than threshold, frequency of B), (a) 1 : 1, (b) 1 : 2, (c) 1 : 3, (d) 1 : 4, 28. Which one of the following graphs represents the variation, of maximum kinetic energy (EK) of the emitted electrons, with frequency u in photoelectric effect correctly ?, , (c), , V1 +, , (a), , EK, , (b), , EK, , 30. X-rays are produced in X-ray tube operating at a given, accelerating voltage. The wavelength of the continuous, X-rays has values from, (a) 0 to ¥, (b) lmin to ¥, where lmin > 0, (c) 0 to lmax, where lmax < ¥, (d) lmin to lmax, where 0 < lmin < lmax < ¥, 31. Electrons used in an electron microscope are accelerated by, a voltage of 25 kV. If the voltage is increased to 100kV then, the de–Broglie wavelength associated with the electrons, would, (a) increase by 2 times, (b) decrease by 2 times, (c) decrease by 4 times, (d) increase by 4 times, 32. In the Davisson and Germer experiment, the velocity of, electrons emitted from the electron gun can be increased by, (a) increasing the potential difference between the anode, and filament, (b) increasing the filament current, (c) decreasing the filament current, (d) decreasing the potential difference between the anode, and filament, 33. Two radiations of photons energies 1 eV and 2.5 eV,, successively illuminate a photosensitive metallic surface of, work function 0.5 eV. The ratio of the maximum speeds of, the emitted electrons is :, (a) 1 : 4, (b) 1 : 2, (c) 1 : 1, (d) 1 : 5, 34. Photoelectric emission is observed from a metallic surface, for frequencies v1 and v2 of the incident light rays (v1 > v2)., If the maximum values of kinetic energy of the photoelectrons, emitted in the two cases are in the ratio of 1 : k, then the, threshold frequency of the metallic surface is, (a), , u, , u, , (c), , (d), , EK, , u, , u0, , u, , 29. The potential difference that must be applied to stop the, fastest photoelectrons emitted by a nickel surface, having, work function 5.01 eV, when ultraviolet light of 200 nm falls, on it, must be:, (a) 2.4 V, (b) – 1.2 V (c) – 2.4 V (d) 1.2 V, , RESPONSE, GRID, , 25., 30., 35., , 26., 31., 36., , (b), , kv1 - v 2, k -1, , kv2 - v1, v2 - v1, (d), k -1, k, Which of the following is/are false regarding cathode, rays?, (a) They produce heating effect, (b) They don’t deflect in electric field, (c) They cast shadow, (d) They produce fluorescence, The ratio of the respective de Broglie wavelengths, associated with electrons accelerated from rest with the, voltages 100 V, 200 V and 300 V is, , (c), 35., , EK, , v1 - v 2, k -1, , 36., , (a) 1 : 2 : 3 (b) 1 : 4 : 9, , 27., 32., Space for Rough Work, , 28., 33., , (c) 1:, , 1, 2, , :, , 1, , 29., 34., , 3, , (d) 1:, , 1 1, :, 2 3
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t.me/Magazines4all, , DPP/ CP25, , P-100, , 37. A 5 watt source emits monochromatic light of wavelength, 5000 Å. When placed 0.5 m away, it liberates photoelectrons, from a photosensitive metallic surface. When the source is, moved to a distance of 1.0 m, the number of photoelectrons, liberated will be reduced by a factor of, (a) 8, (b) 16, (c) 2, (d) 4, 38. In the photoeletric effect, electrons are emitted, (a) at a rate that is proportional to the amplitude of the, incident radiation, (b) with a maximum velocity proportional to the frequency, of the incident radiation, (c) at a rate that is independent of the emitter, (d) only if the frequency of the incident radiations is above, a certain threshold value, 39. The threshold frequency for a photosensitive metal is 3.3 ×, 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this, metal, the cut-off voltage for the photoelectric emission is, nearly, (a) 2 V, (b) 3 V, (c) 5 V, (d) 1 V, 40. In an experiment on photoelectric effect, a student plots, stopping potential V0 against reciprocal, of the wavelength l of the incident V, Metal A, Metal B, light for two different metals A and, B. These are shown in the figure., , 41., , 42., 43., , 44., , 45., , 1/l, , Looking at the graphs, you can most appropriately say that:, (a) Work function of metal B is greater than that of metal A, (b) For light of certain wavelength falling on both metal,, maximum kinetic energy of electrons emitted from A, will be greater than those emitted from B., , 37., 42., , 38., 43., , 30.8, , 3.08, , 0.308, , Å (d), , 0.0308, , Å, T, T, T, T, Which of the following cannot be explained on the basis, of photoelectric theory?, (a) Instantaneous emission of photoelectrons, (b) Existence of threshold frequency, (c) Sufficiently intense beam of radiation can emit, photoelectrons, (d) Existence of stopping potential, , (a), , 0, , RESPONSE, GRID, , (c) Work function of metal A is greater than that of metal B, (d) Students data is not correct, White X-rays are called white due to the fact that, (a) they are electromagnetic radiations having nature, same as that of white light., (b) they are produced most abundantly in X ray tubes., (c) they have a continuous wavelength range., (d) they can be converted to visible light using coated, screens and photographic plates are affected by, them just like light., The wavelength associated with an electron, accelerated, through a potential difference of 100 V, is of the order of, (a) 1000 Å (b) 100 Å (c) 10.5 Å, (d), 1.2 Å, Monochromatic light of frequency 6.0 × 1014 Hz is produced, by a laser. The power emitted is 2 × 10–3 w. The number of, photons emitted, on the average, by the sources per second, is, (a) 5 × l 016 (b) 5 × 1017 (c) 5 × 1014 (d) 5 × 1015, The de-Broglie wavelength of neutron in thermal, equilibrium at temperature T is, , 39., 44., , Å (b), , 40., 45., , Å (c), , 41., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP25 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 45, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 60, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP26, , SYLLABUS : Atoms, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , The potential energy associated with an electron in the orbit, (a) increases with the increases in radii of the orbit, (b) decreases with the increase in the radii of the orbit, (c) remains the same with the change in the radii of the orbit, (d) None of these, The diagram shows the energy levels for an electron in a, certain atom. Which transition shown represents the, emission of a photon with the most energy?, n =4, n =3, , 4., , (a) 15 × 106, , 3., , II, , III, , IV, , 5., , 1., , 2., , 3, × 106, 5, , 5, × 106, (d) None of these, 9, In the Bohr 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 :, , n =1, , (a), , (a) IV, (b) III, (c) II, (d) I, Electrons in a certain energy level n = n1, can emit 3 spectral, lines. When they are in another energy level, n = n 2. They, can emit 6 spectral lines. The orbital speed of the electrons, in the two orbits are in the ratio of, , RESPONSE GRID, , (b), , (c), , n =2, , I, , (a) 4 : 3, (b) 3 : 4, (c) 2 : 1, (d) 1 : 2, In Rutherford scattering experiment, the number of, a-particles scattered at 60° is 5 × 106. The number of, a-particles scattered at 120° will be, , (c), , 3., Space for Rough Work, , æ e n2 h ö, ç, ÷, ç 2m 2p ÷, è, ø, e, nh, æ, ö, ç, ÷, è 2m ø 2p, , 4., , (b), , æ e ö nh, ç ÷, è m ø 2p, , (d), , 2, æ e ön h, ç ÷, è m ø 2p, , 5.
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t.me/Magazines4all, , DPP/ CP26, , P-102, , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , A 12.5 eV electron beam is used to bombard gaseous, hydrogen at room temperature. It will emit :, (a) 2 lines in the Lyman series and 1 line in the Balmar, series, (b) 3 lines in the Lyman series, (c) 1 line in the Lyman series and 2 lines in the Balmar, series, (d) 3 lines in the Balmer series, A Hydrogen atom and a Li++ ion are both in the second, excited state. If lH and lLi are their respective electronic, angular momenta, and EH and ELi their respective energies,, then, (a) lH > lLi and |EH| > |ELi| (b) lH = lLi and |EH| < |ELi|, (c) lH = lLi and |EH| > |ELi| (d) lH < lLi and |EH| < |ELi|, The radius of hydrogen atom in its ground state is, 5.3 × 10–11 m. After collision with an electron it is found to, have a radius of 21.2 × 10–11 m. What is the principal quantum, number n of the final state of the atom, (a) n = 4 (b) n = 2, (c) n = 16, (d) n = 3, When hydrogen atom is in its first excited level, its radius is, (a) four times its ground state radius, (b) twice, (c) same, (d) half, Consider 3rd orbit of He+ (Helium), using non-relativistic, approach, the speed of electron in this orbit will be [given K, = 9 × 109 constant, Z = 2 and h (Plank's Constant), = 6.6 × 10–34 J s], (a) 1.46 × 106 m/s, (b) 0.73 × 106 m/s, 8, (c) 3.0 × 10 m/s, (d) 2.92 × 106 m/s, An electron in the hydrogen atom jumps from excited state, n to the ground state. The wavelength so emitted illuminates, a photosensitive material having work function 2.75 eV. If, the stopping potential of the photoelectron is 10 V, the value, of n is, (a) 3, (b) 4, (c) 5, (d) 2, The electron in a hydrogen atom makes a transition from an, excited state to the ground state. Which of the following, statements is true?, (a) Its kinetic energy increases and its potential energy, decreases., (b) Its kinetic energy decreases, potential energy increases., (c) Its kinetic and its potential energy increases., (d) Its kinetic, potential energy decrease., An energy of 24.6 eV is required to remove one of the, electrons from a neutral helium atom. The energy in (eV), required to remove both the electrons from a neutral helium, atom is, (a) 38.2, (b) 49.2, (c) 51.8, (d) 79.0, , RESPONSE, GRID, , 6., 11., 16., 21., , 7., 12., 17., 22., , One of the lines in the emission spectrum of Li2+ has the, same wavelength as that of the 2nd line of Balmer series in, hydrogen spectrum. The electronic transition corresponding, to this line is n = 12 ® n = x. Find the value of x., (a) 8, (b) 6, (c) 7, (c) 5, 257, 15. If the atom 100Fm follows the Bohr model and the radius, of 100Fm257 is n times the Bohr radius, then find n., (a) 100, (b) 200, (c) 4, (d) 1/4, 16. The energy of He+ in the ground state is – 54.4 eV, then the, energy of Li++ in the first excited state will be, (a) – 30.6 eV, (b) 27.2 eV, (c) – 13.6 eV, (d) – 27.2 eV, 17. If the angular momentum of an electron in an orbit is J then, the K.E. of the electron in that orbit is, 14., , 18., , 19., , 20., , 21., 22., , J2, , Jv, J2, J2, (c), (d), r, 2m, 2p, 2mr 2, Suppose an electron is attracted towards the origin by a, k, force where ‘k’ is a constant and ‘r’ is the distance of the, r, electron from the origin. By applying Bohr model to this, system, the radius of the nth orbital of the electron is found, to be ‘rn’ and the kinetic energy of the electron to be ‘Tn’., Then which of the following is true?, 1, 2, (a) Tn µ 2 , rn µ n, (b) Tn independent of n, rn µ n, n, 1, 1, (c) Tn µ , rn µ n, (d) Tn µ , rn µ n 2, n, n, In Hydrogen spectrum, the wavelength of Ha line is 656 nm,, whereas in the spectrum of a distant galaxy, Ha line, wavelength is 706 nm. Estimated speed of the galaxy with, respect to earth is, (a) 2 × 108 m/s, (b) 2 × 107m/s, (c) 2 × 106 m/s, (d) 2 × 105 m/s, In the hydrogen atom, an electron makes a transition from, n = 2 to n = 1. The magnetic field produced by the circulating, electron at the nucleus, (a) decreases 16 times, (b) increases 4 times, (c) decreases 4 times, (d) increases 32 times, What is the radius of iodine atom (At. no. 53, mass no. 126), (a) 2.5 × 10–11 m, (b) 2.5 × 10–9 m, (c) 7 × 10–9 m, (d) 7 × 10–6 m, When an a-particle of mass 'm' moving with velocity 'v', bombards on a heavy nucleus of charge 'Ze', its distance of, closest approach from the nucleus depends on m as :, 1, 1, 1, (a), (b), (c), (d) m, m, m, m2, , (a), , 8., 13., 18., Space for Rough Work, , (b), , 9., 14., 19., , 10., 15., 20., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP26, , P-103, , 23. The ionization energy of the electron in the hydrogen atom, in its ground state is 13.6 eV. The atoms are excited to higher, energy levels to emit radiations of 6 wavelengths. Maximum, wavelength of emitted radiation corresponds to the, transition between, (a) n = 3 to n = 1 states, (b) n = 2 to n = 1 states, (c) n = 4 to n = 3 states, (d) n = 3 to n = 2 states, 24. The wavelengths involved in the spectrum of deuterium, , ( ), , 30., , 31., , 2, 1D, , 25., , 26., , 27., , 28., , 29., , are slightly different from that of hydrogen spectrum,, because, (a) the size of the two nuclei are different, (b) the nuclear forces are different in the two cases, (c) the masses of the two nuclei are different, (d) the attraction between the electron and the nucleus is, differernt in the two cases, An electron in hydrogen atom makes a transition n1 ® n2, where n1 and n2 are principal quantum numbers of the two, states. Assuming Bohr’s model to be valid the time period, of the electron in the initial state is eight times that in the, final state. The possible values of n1 and n2 are, (a) n1 = 4 and n2 = 2, (b) n1 = 6 and n2 = 2, (c) n1 = 8 and n2 = 1, (d) n1 = 8 and n2 = 2, Ina hydrogen like atom electronmake 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 :, 1, 1, 1, 1, (a), (b), (c), (d), 2, 3, n, n, n, n3, 2, The spectrum obtained from a sodium vapour lamp is an, example of, (a) band spectrum, (b) continuous spectrum, (c) emission spectrum, (d) absorption spectrum, Ionization potential of hydrogen atom is 13.6eV. Hydrogen, atoms in the ground state are excited by monochromatic, radiation of photon energy 12.1 eV. According to Bohr’s, theory, the spectral lines emitted by hydrogen will be, (a) three (b) four, (c) one, (d) two, The Bohr model of atoms, (a) predicts the same emission spectra for all types of, atoms, (b) assumes that the angular momentum of electrons is, quantised, , RESPONSE, GRID, , 23., 28., 33., , 24., 29., 34., , 32., 33., , 34., , 35., , 36., , 37., , (c) uses Einstein’s photoelectric equation, (d) predicts continuous emission spectra for atoms, 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, A doubly ionised Li atom is excited from its ground state(n, = 1) to n = 3 state. The wavelengths of the spectral lines are, given by l32, l31 and l21. The ratio l32/l31 and l21/l31, are, respectively, (a) 8.1, 0.67, (b) 8.1, 1.2, (c) 6.4, 1.2, (d) 6.4, 0.67, In Rutherford scattering experiment, what will be the correct, angle for a-scattering for an impact parameter, b = 0 ?, (a) 90°, (b) 270°, (c) 0°, (d) 180°, Consider 3rd orbit of He+ (Helium), using non-relativistic, approach, the speed of electron in this orbit will be [given, K = 9 × 109 constant, Z = 2 and h (Plank's Constant), = 6.6 × 10–34 J s], (a) 1.46 × 106 m/s, (b) 0.73 × 106 m/s, (c) 3.0 × 108 m/s, (d) 2.92 × 106 m/s, The ionization energy of hydrogen atom is 13.6 eV. Following, Bohr’s theory, the energy corresponding to a transition, between 3rd and 4th orbit is, (a) 3.40 eV (b) 1.51 eV (c) 0.85 eV (d) 0.66 eV, The transition from the state n = 3 to n = 1 in a hydrogen like, atom results in ultraviolet radiation. Infrared radiation will, be obtained in the transition from :, (a) 2 ® 1 (b) 3 ® 2, (c) 4 ® 2, (d) 4 ® 3, Given the value of Rydberg constant is 107m–1, the wave, number of the last line of the Balmer series in hydrogen, spectrum will be :, (a) 0.025 × 104 m–1, (b) 0.5 × 107 m–1, 7, –1, (c) 0.25 × 10 m, (d) 2.5 × 107 m–1, Which of the plots shown in the figure represents speed, (vn) of the electron in a hydrogen atom as a function of the, principal quantum number (n)?, A, , C, , vn, , D, , B, , (a) B, , 25., 30., 35., Space for Rough Work, , o 1 2, , (b) D, , 26., 31., 36., , 3 4, , n, , (c) C, , (d) A, , 27., 32., 37.
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t.me/Magazines4all, , DPP/ CP26, , P-104, , 38., , 39., 40., , 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 will be (according to Bohr’s, theory), (a) 10, (b) 8, (c) 6, (d) 4, The energy of hydrogen atom in nth orbit is En, then the, energy in nth orbit of single ionised helium atom will be, (a) 4En, (b) En/4, (c) 2En, (d) En/2, In the Rutherford experiment, a-particles are scattered from, a nucleus as shown. Out of the four paths, which path is not, possible?, , 42., , 43., , A, , o, , B, , 44., C, D, , 41., , (a) D, (b) B, (c) C, (d) A, An electron changes its position from orbit n = 2 to the orbit, n = 4 of an atom. The wavelength of the emitted radiations is, (R = Rydberg’s constant), 16, 16, 16, 16, (a), (b), (c), (d), R, 3R, 5R, 7R, , 38., 43., , RESPONSE, GRID, , In a Rutherford scattering experiment when a projectile of, charge Z1 and mass M1approaches a target nucleus of, charge Z2 and mass M2, the distance of closest approach, is r0. The energy of the projectile is, (a) directly proportional to Z1 Z2, (b) inversely proportional to Z1, (c) directly proportional to mass M1, (d) directly proportional to M1 × M2, The wavelength of the first spectral line in the Balmer series, of hydrogen atom is 6561 A°. The wavelength of the second, spectral line in the Balmer series of singly-ionized helium, atom is, , 39., 44., , (c), 45., , o, , o, , 1, 1, 1, =, +, u2 u1 u3, , (d), , 1, 1, 1, =, +, u1 u2 u3, , In a hypothetical Bohr hydrogen atom, the mass of the electron, is doubled. The energy E¢0 and radius r¢0 of the first orbit will, be (r0 is the Bohr radius), (a) –11.2 eV (b) –6.8 eV, (c) –13.6 eV (d) –27.2 eV, , 40., 45., , 41., , 42., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP26 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , o, , (a) 1215 A (b) 1640 A, (c) 2430 A (d) 4687 A, If u1 is the frequency of the series limit of Lyman series,, u2 is the frequency of the first line of Lyman series and, u3 is the frequency of the series limit of the Balmer series, then, (a) u1 - u2 = u3, (b) u1 = u2 - u3, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP27, , SYLLABUS : Nuclei, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , The mass of a 37 Li nucleus is 0.042 u less than the sum of, the masses of all its nucleons. The binding energy per, nucleon of 37 Li nucleus is nearly, (a) 46 MeV, (b) 5.6 MeV, (c) 3.9 MeV, (d) 23 MeV, In the nuclear decay given below:, A, A– 4, A, ® Z + 1Y ¾¾, ® Z-1 B*, Z X ¾¾, , 3., 4., , 5., , 6., , (a) (R1T1 – R2T2), (c) (R1 – R2)/T, , A -4, , ¾¾, ® Z -1B,, the particles emitted in the sequence are, (a) g, b, a, (b) b, g, a, (c) a, b, g, (d) b, a, g, If the nuclear radius of 27Al is 3.6 Fermi, the approximate, nuclear radius of 64Cu in Fermi is :, (a) 2.4, (b) 1.2, (c) 4.8, (d) 3.6, Which of the following statements is true for nuclear forces?, (a) they obey the inverse square law of distance, (b) they obey the inverse third power law of distance, (c) they are short range forces, (d) they are equal in strength to electromagnetic forces., A radioactive sample at any instant has its disintegration, rate 5000 disintegrations per minute. After 5 minutes, the, rate is 1250 disintegrations per minute. Then, the decay, constant (per minute) is, (a) 0.4 ln 2 (b) 0.2 ln 2 (c) 0.1 ln 2(d) 0.8 ln 2, , RESPONSE GRID, , 1., 6., , The radioactivity of a sample is R1 at a time T1 and R2 at a, time T2. If the half-life of the specimen is T, the number of, atoms that have disintegrated in the time (T1 – T2 ) is, proportional to, , 2., 7., , (b) (R1 – R2), (d) (R1 – R2) T, , 7., , In the reaction, 12 H + 13 H, , 4, 1, , if the binding, 2 He + 0 n, , 8., , energies of 12 H , 13 H and 42 He are respectively, a, b and c, (in MeV), then the energy (in MeV) released in this reaction, is, (a) a + b + c, (b) a + b – c, (c) c – a – b, (d) c + a – b, If M (A; Z), Mp and Mn denote the masses of the nucleus, A, Z X,, , proton and neutron respectively in units of u ( 1u =, 931.5 MeV/c2) and BE represents its bonding energy in, MeV, then, (a) M (A, Z) = ZMp + (A – Z) Mn –BE/c2, (b) M (A, Z) = ZMp+ ( A–Z) Mn + BE, (c) M (A, Z) = ZMp + (A – Z) Mn – BE, (d) M (A, Z) = ZMp + (A – Z)Mn + BE/c2, , 3., 8., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP27, , P-106, , 9., , 10., , How does the binding energy per nucleon vary with the, increase in the number of nucleons?, (a) Increases continuously with mass number, (b) Decreases continuously with mass number, (c) First decreases and then increases with increase in mass, number, (d) First increases and then decreases with increase in mass, number, The energy spectrum of b-particles [Number N(E) as a, function of b-energy E] emitted from a radioactive source is, , 16., , (a), , 17., , (b), N(E), E, , 13., , 14., , 15., , N(E), , E0, , (b), , 9, 1, 4 Be + 1H, , (c), , 11, 1, 5 Be + 1H, , (d), , 7, 4, ® 11H + 104 B, 3 Li + 2 He ¾¾, , ¾¾, ® 63 Li + 42 He, , ¾¾, ® 94 Be + 42 He, , The ratio of half-life times of two elements A and B is, , E, , b, , E0, , a, , E, , 18., , g, , ® A1 ¾¾, ® A 2 ¾¾, ® A 3 ¾¾, ® A4, A ¾¾, If the mass number and atomic number of ‘A’ are 180 and 72, respectively, then what are these numbers for A4, (a) 172 and 69, (b) 174 and 70, (c) 176 and 69, (d) 176 and 70, The activity of a radioactive sample is measured as 9750, counts per minute at t = 0 and as 975 counts per minute at, t = 5 minutes. The decay constant is approximately, (a) 0.922 per minute, (b) 0.691 per minute, (c) 0.461 per minute, (d) 0.230 per minute, Actinium 231, 231 AC89, emit in succession two b particles,, four a-particles, one b and one a plus several g rays. What, is the resultant isotope?, (a) 221 Au 79, (b) 211 Au 79, 221, (c), Pb 82, (d) 211 Pb82, Fusion reactions take place at high temperature because, (a) atoms are ionised at high temperature, (b) molecules break up at high temperature, (c) nuclei break up at high temperature, (d) kinetic energy is high enough to overcome repulsion, between nuclei, , If MO is the mass of an oxygen isotope 8 O17 ,MP and MN, are the masses of a proton and a neutron respectively, the, nuclear binding energy of the isotope is, (a) (MO –17MN)c2, (b) (MO – 8MP)c2, (c) (MO– 8MP –9MN)c2 (d) MOc 2, , 9., 14., 19., , lA, , is, lB, , TA, ., TB, , (b) TA / TB, , TA + TB, TA - TB, (d), TA, TA, Two radioactive materials X1 and X2 have decay constants, 10l and l respectively. If initially they have the same number, of nuclei, then the ratio of the number of nuclei of X1 to that, of X2 will be 1/e after a time, (a) 1/10l, (b) 1/11l, (c) 11/10l, (d) 1/9l, In a radioactive material the activity at time t1 is R1 and at a, later time t2, it is R2. If the decay constant of the material is, l, then, , (c), , A radioactive nucleus undergoes a series of decay according, to the scheme, , RESPONSE, GRID, , 20, C ¾¾, ® 10, Ne + 42 He, , (a) TB / TA, , (d), N(E), , 12., , 12, 12, 6C+ 6, , The ratio of respective decay constant, , E, , E0, , (c), , a, , (a), , N(E), E0, , 11., , Which of the following nuclear reactions is not possible?, , 10., 15., 20., , 19., , (a), , R1 = R2 e l (t1 -t2 ), , (c) R1 = R2, , (b), , R1 = R2 e(t2 / t1 ), , (d), , R1 = R2 e -l (t1 -t2 ), , 20. The correct relation between t av = average life and, t 1/2 = half life for a radioactive nuclei., 1, t, 2 1/2, (c) 0.693 t av = t 1/2, (d) t av = 0.693 t 1/2, If the nuclear force between two protons, two neutrons and, between proton and neutron is denoted by Fpp, Fnn and Fpn, respectively, then, , (a) t av = t 1/2, , 21., , (a), , (b) t av =, , Fpp » Fnn » Fpn, , (b), , Fpp ¹ Fnn and Fpp = Fnn, , (c) Fpp = Fnn = Fpn, (d) Fpp ¹ Fnn ¹ Fpn, 22. Which one is correct about fission?, (a) Approx. 0.1% mass converts into energy, (b) Most of energy of fission is in the form of heat, (c) In a fission of U 23 5 about 200 eV energy is, released, (d) On an average, one neutron is released per, fission of U235, , 11., 16., 21., Space for Rough Work, , 12., 17., 22., , 13., 18., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , DPP/ CP27, , P-107, , the fission of a single U235, , 23. If 200 MeV energy is released in, nucleus, the number of fissions required per second to, produce 1 kilowatt power shall be (Given 1eV = 1.6 × 10–19 J), (a) 3.125 × 1013, (b) 3.125 × 1014, 15, (c) 3.125 × 10, (d) 3.125 × 1016, 24. In any fission process, the ratio of, , 25., , 26., , 27., , mass of fission products, is, mass of parent nucleus, (a) equal to 1, (b) greater than 1, (c) less than 1, (d) depends on the mass of the parent nucleus, In an a-decay the kinetic energy of a-particle is 48 MeV and, Q-value of the reaction is 50 MeV. The mass number of the, mother nucleus is X. Find value of X/25., (Assume that daughter nucleus is in ground state), (a) 2, (b) 4, (c) 6, (d) 8, A sample of radioactive element has a mass of 10gm at an, instant t=0. The approximate mass of this element in the, sample after two mean lives is, (a) 6.30 gm, (b) 1.35 gm, (c) 2.50 gm, (d) 3.70 gm, Consider a radioactive material of half-life 1.0 minute. If one, of the nuclei decays now, the next one will decay, (a) after 1 minute, , 30., , 31., , 32., , (c), , 33., , (a) c, 34., , 35., , 36., , (d), , Ta + Tb, , 23., 28., 32., 37., , (, , 1, Ta + Tb, 2, , 24., 28., 33., , ), , N, , (d), , (c), t, , t, , t, , A nucleus of mass M + Dm is at rest and decays into two, M, daughter nuclei of equal mass, each. Speed of light is c., 2, The speed of daughter nuclei is, Dm, M + Dm, , (b), , c, , 2 Dm, M, , (c), , c, , Dm, M, , (d) c, , Dm, M + Dm, , Atomic weight of Boron is 10.81 and it has two isotopes, 10 and, , 5B, , 11, , .Then the ratio 5 B10 :5 B11 in nature would, , be, (a) 19 : 81, (b) 10 : 11 (c) 15 : 16 (d) 81 : 19, A nucleus ruptures into two nuclear parts, which have their, velocity ratio equal to 2:1. What will be the ratio of their, nuclear size (nuclear radius)?, (a) 21/3 : 1, (b) 1 : 21/3 (c) 31/2 : 1 (d) 1 : 31/2, A nucleus of uranium decays at rest into nuclei of thorium, and helium. Then :, (a) the helium nucleus has less momentum than the thorium, nucleus., (b) the helium nucleus has more momentum than the, thorium nucleus., (c) the helium nucleus has less kinetic energy than the, thorium nucleus., (d) the helium nucleus has more kinetic energy than the, thorium nucleus., If radius of the, radius of, , Ta Tb, , RESPONSE, GRID, , (b), t, , 37., , (b) Ta + Tb, , Ta + Tb, , N, , N, , (a), , 5B, , 1, (c) after, minute, where N is the number of nuclei present, N, at that moment, (d) after any time, 28. The mass of a-particle is, (a) less than the sum of masses of two protons and two, neutrons, (b) equal to mass of four protons, (c) equal to mass of four neutrons, (d) equal to sum of masses of two protons and two neutron, 29. The decay constants of a radioactive substance for a and b, emission are la and lb respectively. If the substance emits, a and b simultaneously, then the average half life of the, material will be, 2Ta Tb, , If the end A of a wire is irradiated with a-rays and the other, end B is irradiated with b-rays. Then, (a) a current will flow from A to B, (b) a current will flow from B to A, (c) there will be no current in the wire, (d) a current will flow from each end to the mid-point of, the wire, A radioactive nucleus of mass M emits a photon of frequency, n and the nucleus recoils. The recoil energy will be, (a) Mc2 – hn, (b) h2n2 / 2Mc2, (c) zero, (d) hn, Radioactive element decays to form a stable nuclide. The, rate of decay of reactant is correctly depicted by, N, , 1, (b) after log 2 minute, e, , (a), , t.me/Magazines4all, , (a), , 25., 29., 34., Space for Rough Work, , 125, 53 Te, , 5, R Al, 3, , 27, 12 Al, , nucleus is taken to be RAl, then the, , nucleus is nearly:, , (b), , 26., 30., 35., , 1/3, 1/3, 3, 13, 53, R Al (c) æç ö÷ R Al (d) æç ö÷ R Al, 5, è 53 ø, è 13 ø, , 27., 31., 36.
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t.me/Magazines4all, , DPP/ CP27, , P-108, , 38., , 39., , 40., , 41., , M n an d M p represent mass of neutron and proton, respectively. If an element having atomic mass M has Nneutron and Z-proton, then the correct relation will be, (a) M < [NMn + ZMp], (b) M > [NMn + ZMp], (c) M = [NMn + ZMp], (d) M = N[Mn + Mp], After 300 days, the activity of a radioactive sample is 5000, dps (disintegrations per sec). The activity becomes 2500, dps after another 150 days. The initial activity of the sample, in dps is, (a) 20,000, (b) 10,000, (c) 7,000, (d) 25,000, Order of magnitude of density of uranium nucleus is, (mp = 1.67 × 10–27 kg), (a) 1020 kg / m3, (b) 1017 kg / m3, 14, 3, (c) 10 kg / m, (d) 1011 kg / m3, The electrons cannot exist inside the nucleus because, (a) de-Broglie wavelength associated with electron in bdecay is much less than the size of nucleus, (b) de-Broglie wavelength associated with electron in bdecay is much greater than the size of nucleus, (c) de-Broglie wavelength associated with electron in bdecay is equal to the size of nucleus, (d) negative charge cannot exist in the nucleus, , 38., 43., , RESPONSE, GRID, , 39., 44., , 42., , If the total binding energies of, , 2, 4, 56, 235, 1 H, 2 He, 26 Fe & 92 U, , nuclei are 2.22, 28.3, 492 and 1786 MeV respectively, identify, the most stable nucleus of the following., , 43., , 44., , (a), , 56, 26 Fe, , (b), , 2, 1H, , (c), , 235, 92 U, , (d), , 4, 2 He, , At a specific instant emission of radioactive compound is, deflected in a magnetic field. The compound cannot emit, (a) electrons, (b) protons, (c) He2+, (d) neutrons, A nuclear reaction is given by, ZX, , A, , ® Z+1Y A + -1 e 0 + n , represents, , (a) fission, (b) b-decay, (c) µ-decay, (d) fusion, 45. Radioactive material 'A' has decay constant '8 l' and material, 'B' has decay constant 'l'. Initially they have same number, of nuclei. After what time, the ratio of number of nuclei of, 1, material 'B' to that 'A' will be ?, e, 1, 1, 1, 1, (a), (b), (c), (d), 7l, 8l, 9l, l, , 40., 45., , 41., , 42., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP27 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP - Daily Practice Problems, Chapter-wise Sheets, Date :, , Start Time :, , End Time :, , PHYSICS, , CP28, , SYLLABUS : Semiconductor Electronics: Materials, Devices and Simple Circuits, , Max. Marks : 180, , Marking Scheme : (+4) for correct & (–1) for incorrect answer, , Time : 60 min., , INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct., Darken the correct circle/ bubble in the Response Grid provided on each page., , 1., , 2., , 3., , 4., , 5., , A change of 8.0 mA in the emitter current bring a change of, 7.9 mA in the collector current. The values of parameters a, and b are respectively, (a) 0.99, 90 (b) 0.96,79 (c) 0.97,99 (d) 0.99,79, A pure semiconductor has equal electron and hole, concentration of 1016 m–3. Doping by indium increases, number of hole concentration n h to 5 × 1022 m–3. Then, the, value of number of electron concentration n e in the doped, semiconductor is, (a) 106/m3, (b) 1022/m3, (c) 2 × 106/m3, (d) 2 × 109/m3, For LED’s to emit light in visible region of electromagnetic, light, it should have energy band gap in the range of:, (a) 0.1 eV to 0.4 eV, (b) 0.5 eV to 0.8 eV, (c) 0.9 eV to 1.6 eV, (d) 1.7 eV to 3.0 eV, A common emitter amplifier has a voltage gain of 50, an input, impedance of 100W and an output impedance of 200W. The, power gain of the amplifier is, (a) 1000, (b) 1250, (c) 100, (d) 500, Which logic gate with inputs A and B performs the same, operation as that performed by the following circuit?, , RESPONSE GRID, , 1., 6., , 2., 7., , A, B, V, Lamp, , 6., , 7., , (a) NAND gate, (b) OR gate, (c) NOR gate, (d) AND gate, In an unbiased p-n junction, holes diffuse from the p-region, to n-region because of, (a) the potential difference across the p-n junction, (b) the attraction of free electrons of n-region, (c) the higher hole concentration in p-region than that in, n-region, (d) the higher concentration of electrons in the n-region, than that in the p-region, A silicon diode has a threshold voltage of 0.7 V. If an input, voltage given by 2 sin(pt) is supplied to a half wave rectifier, circuit using this diode, the rectified output has a peak, value of, (a) 2 V, (b) 1.4 V, (c) 1.3 V, (d) 0.7 V, , 3., Space for Rough Work, , 4., , 5.
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t.me/Magazines4all, , DPP/ CP28, , P-110, , 8., , 9., , 10., , The current gain for a transistor working as common-base, amplifier is 0.96. If the emitter current is 7.2 mA, then the, base current is, (a) 0.29 mA (b) 0.35 mA (c) 0.39 mA (d) 0.43 mA, In a npn transistor 1010 electrons enter the emitter in, 10–6 s. 4% of the electrons are lost in the base. The current, transfer ratio will be, (a) 0.98, (b) 0.97, (c) 0.96, (d) 0.94, Assuming that the silicon diode having resistance of, 20 W, the current through the diode is (knee voltage 0.7 V), R =180W, , 11., , 12., , 13., , 14., , 15., , 16., 17., , 2V, 0V, (a) 0 mA (b) 10 mA, (c) 6.5 mA (d) 13.5 mA, Transfer characteristics [output, V I, II, voltage (V0) vs input voltage (Vi)] 0, III, for a base biased transistor in CE, configuration is as shown in the, figure. For using transistor as a, switch, it is used, Vi, (a) in region III, (b) both in region (I) and (III), (c) in region II, (d) in region (I), A half-wave rectifier is being used to rectify an alternating, voltage of frequency 50 Hz. The number of pulses of rectified, current obtained in one second is, (a) 50, (b) 25, (c) 100, (d) 2000, A diode having potential difference 0.5 V across its junction, which does not depend on current, is connected in series, with resistance of 20W across source. If 0.1 A current passes, through resistance then what is the voltage of the source?, (a) 1.5 V (b) 2.0 V, (c) 2.5 V, (d) 5 V, In common emitter amplifier, the current gain is 62. The collector, resistance and input resistance are 5 kW an 500W respectively., If the input voltage is 0.01V, the output voltage is, (a) 0.62 V (b) 6.2 V, (c) 62 V, (d) 620 V, On doping germanium with donor atoms of density, 1017 cm–3 its conductivity in mho/cm will be, [Given : me = 3800 cm2/V–s and ni = 2.5 × 1013 cm–13], (a) 30.4, (b) 60.8, (c) 91.2, (d), 121.6, The voltage gain of an amplifier with 9% negative feedback, is 10. The voltage gain without feedback will be, (a) 90, (b) 10, (c) 1.25, (d) 100, A system of four gates is set up as shown. The ‘truth table’, corresponding to this system is :, A, , Y, , B, , RESPONSE, GRID, , 8., 13., 18., , 9., 14., 19., , (a), , A B Y, 0 0 1, 0 1 0, 1 0 0, 1 1 1, , (b), , A B Y, 0 0 0, 0 1 0, 1 0 1, 1 1 0, , (c), , (d), A B Y, A B Y, 0 0 1, 0 0 1, 0 1 1, 0 1 0, 1 0 0, 1 0 1, 1 1 0, 1 1 0, 18. The intrinsic conductivity of germanium at 27° is 2.13 mho, m–1 and mobilities of electrons and holes are 0.38 and 0.18, m2V–1s–1 respectively. The density of charge carriers is, (a) 2.37 × 1019 m–3, (b) 3.28 × 1019 m–3, 19, –3, (c) 7.83 × 10 m, (d) 8.47 × 1019 m–3, 19. The logic circuit shown below has the input waveforms, ‘A’ and ‘B’ as shown. Pick out the correct output, waveform, A, , Y, , B, Input A, Input B, , Output is, (a), (b), (c), (d), 20., , Pure Si at 500K has equal number of electron (ne) and hole (nh), concentrations of 1.5 × 1016 m–3. Doping by indium increases, nh to 4.5 × 1022 m–3. The doped semiconductor is of, (a) n–type with electron concentration n e = 5 × 1022 m–3, (b) p–type with electron concentration ne = 2.5 ×1010 m–3, (c) n–type with electron concentration n e = 2.5 × 1023 m–3, (d) p–type having electron concentration n e = 5 × 109 m–3, 21. Which of the following statements is incorrect?, (a) The resistance of intrinsic semiconductors decrease, with increase of temperature, (b) Doping pure Si with trivalent impurities give p-type, semiconductors, (c) The majority carriers in n-type semiconductors are holes, (d) A p-n junction can act as a semiconductor diode, , 10., 15., 20., Space for Rough Work, , 11., 16., 21., , 12., 17., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP28, , P-111, , 22. The relation between number of free electrons (n) in a, semiconductor and temperature (T) is given by, (a) n µ T (b) n µ T2, (c) n µ T (d) n µ T3/2, 23. If a PN junction diode of depletion layer width W and barrier, height V0 is forward biased, then, (a) W increases, V0 decreases, (b) W decreases, V0 increases, (c) both W and V0 increase, (d) both W and V0 decrease, 24. The circuit has two oppositively connected ideal diodes in, parallel. The current flowing in the circuit is, 4W, D1, 12V, , 3W, , D2, 2W, , (a) 1.71 A (b) 2.00 A (c) 2.31 A (d) 1.33 A, 25. For a transistor amplifier in common emitter configuration, for load impedanceof 1kW (hfe = 50 and h0e = 25) the current, gain is, (a) – 24.8 (b) – 15.7 (c) – 5.2, (d) – 48.78, 26. A PN-junction has a thickness of the order of, (a) 1 cm, (b) 1mm, (c) 10–6 m (d) 10–12 cm, 27. A working transistor with its three legs marked P, Q and R is, tested using a multimeter. No conduction is found between, P and Q. By connecting 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 true for the transistor?, (a) It is an npn transistor with R as base, (b) It is a pnp transistor with R as base, (c) It is a pnp transistor with R as emitter, (d) It is an npn transistor with R as collector, 28. If in a p-n junction, a square input signal of 10 V is applied, as shown, then the output across RL will be, +5V, RL, –5V, 10V, , (a), , (b), –5V, , (c) number of electrons and holes remain same, (d) number of electrons and holes increases equally., 30. The ratio of electron and hole currents in a semiconductor, is 7/4 and the ratio of drift velocities of electrons and holes, is 5/4, then the ratio of concentrations of electrons and holes, will be, (a) 5/7, (b) 7/5, (c) 25/49, (d) 49/25, 31. C and Si both have same lattice structure, having 4 bonding, electrons in each. However, C is insulator whereas Si is, intrinsic semiconductor. This is because :, (a) In case of C the valence band is not completely filled at, absolute zero temperature., (b) In case of C the conduction band is partly filled even at, absolute zero temperature., (c) The four bonding electrons in the case of C lie in the, second orbit, whereas in the case of Si they lie in the, third., (d) The four bonding electrons in the case of C lie in the, third orbit, whereas for Si they lie in the fourth orbit., 32. Which one of the following represents forward bias diode ?, R, –3V, (a) –4V, +2V, , (c), , 3V, , R, , 5V, , (d), , 0V, , R, , –2V, , An oscillator is nothing but an amplifer with, (a) positive feedback (b) negative feedback, (c) large gain, (d) no feedback, 34. The current gain in the common emitter mode of a transistor, is 10. The input impedance is 20kW and load of resistance is, 100kW. The power gain is, (a) 300, (b) 500, (c) 200, (d) 100, 35. The input signal given to a CE amplifier having a voltage, pö, æ, gain of 150 is Vi = 2 cos çè15t + ÷ø . The corresponding, 3, output signal will be :, 5p ö, 2p, æ, (a) 75cos æç15t + ö÷, (b) 2cos ç15t + ÷, è, è, 6ø, 3ø, , (d), –10V, , 22., 27., 32., , R, , 4p ö, æ, p, (d) 300cos æç15t + ö÷, 300cos ç15t + ÷, è, ø, è, 3, 3ø, To use a transistor as an amplifier, (a) the emitter base junction is forward biased and the, base collector junction is reverse biased, (b) no bias voltage is required, (c) both junctions are forward biased, (d) both junctions are reverse biased., (c), , 29. When n-type semiconductor is heated, (a) number of electrons increases while that of holes, decreases, (b) number of holes increases while that of electrons, decreases, , RESPONSE, GRID, , –2V, , 33., , 5V, , (c), , (b), , 23., 28., 33., , 36., , 24., 29., 34., Space for Rough Work, , 25., 30., 35., , 26., 31., 36.
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t.me/Magazines4all, , DPP/ CP28, , P-112, , I (mA), , 37., , A piece of copper and another of germanium are cooled, from room temperature to 77K. The resistance of, (a) copper increases and germanium decreases, (b) each of them decreases, (c) each of them increases, (d) copper decreases and germanium increases, 38. A d.c. battery of V volt is connected to a series combination, of a resistor R and an ideal diode D as shown in the figure, below. The potential difference across R will be, D, , R, , 800, 400, , 42., , 2 2.1 V (volt), (a) 1 W, (b) 0.25 W (c) 0.5 W (d) 5 W, The circuit diagram shows a logic combination with the, states of outputs X, Y and Z given for inputs P, Q, R and S, all at state 1. When inputs P and R change to state 0 with, inputs Q and S still at 1, the states of outputs X, Y and Z, change to, P(1), , V, , (a) 2V when diode is forward biased, (b) Zero when diode is forward biased, (c) 5V when diode is reverse biased, (d) 6V when diode is forward biased, 39. The current gain for a transistor working as common-base, amplifier is 0.96. If the emitter current is 7.2 mA, then the, base current is, (a) 0.29 mA (b) 0.35 mA (c) 0.39 mA (d) 0.43 mA, 40. In the circuit given below, A and B represent two inputs and, C represents the output., , X(1), , Q(1), , Z(0), R(1), S(1), , 43., , Y(1), , (a) 1, 0, 0 (b) 1, 1, 1 (c) 0, 1, 0 (d) 0, 0, 1, The following configuration of gate is equivalent to, A, OR, B, Y, , AND, , A, C, B, , 44., , The circuit represents, (a) NOR gate, (b) AND gate, (c) NAND gate, (d) OR gate, 41. The I-V characteristic of a P-N junction diode is shown, below. The approximate dynamic resistance of the p-n, junction when a forward bias voltage of 2 volt is applied is, , 37., 42., , RESPONSE, GRID, , 38., 43., , 45., , NAND, (a) NAND gate, (b) XOR gate, (c) OR gate, (d) NOR gate, A p-n photodiode is made of a material with a band gap of, 2.0 eV. The minimum frequency of the radiation that can be, absorbed by the material is nearly, (a) 10 × 1014 Hz, (b) 5 ×1014 Hz, 14, (c) 1 × 10 Hz, (d) 20 × 1014 Hz, The average value of output direct current in a full wave, rectifier is, (a) I0/p, (b) I0/2, (c) p I0/2 (d) 2 I0/p, , 39., 44., , 40., 45., , 41., , DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP28 - PHYSICS, Total Questions, Attempted, Incorrect, Cut-off Score, , 45, , Total Marks, Correct, Net Score, 50, Qualifying Score, Success Gap = Net Score – Qualifying Score, Net Score = (Correct × 4) – (Incorrect × 1), Space for Rough Work, , 180, , 70, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ CP01, , S-2, , 20., , 21., , b+ c= 1, a + b + 2c = 0; –a – 2b – 2c = 0, which give a = –2, b = 0 and c = 1., Therefore (M) = (V–2 F0 E)., (d) Number of significant figures in multiplication is three,, corresponding to the minimum number, 107.88 × 0.610 = 65.8068 = 65.8, (d) A quantity which has dimensions and a constant value, is called dimensional constant. Therefore, gravitational, constant (G) is a dimensional constant., [ML2 T -2 ][ML2 T -1 ]2, , (a), , 23., , (a) The mean value of refractive index,, m=, , and, , 1.34 + 1.38 + 1.32 + 1.36, = 1.35, 4, , | (1.35 - 1.34) | + | (1.35 - 1.38) | + | (1.35 - 1.32) | + | (1.35 - 1.36) |, Dm =, 4, , =, , 24., , 25., , (c), , 0.02, , Dm, ´ 100 =, m, , Thus, , e, = ML3T–2, 4pe0, , L = [LT–1]x [M–1L3T–2]y[ML3T–2]z, [L] = [Lx + 3y + 3z M –y + z T–x – 2y – 2z], Comparing both sides, –y + z = 0 Þ y = z, ...(i), x + 3y + 3z = 1, ...(ii), –x – 4z = 0, , 0.02, ´ 100 = 1.48, 1.35, , drift velocity Vd (ms -1) m 2 s -3, =, =, electric field E (Vm -1 ), V, , 30., 31., 32., , =, , = kg–1 s2 A = M–1 T2 A, 26., 27., , (a), (b), , a, , b, , \ n = 1; - n + m = -2, \ m = -2 + n = -2 + 1 = -1, 33., , v = kl r g, , [M 0 LT -1 ] = La (ML-3 ) b (LT -2 ) c, b, , a - 3b + c, , =M L, , T, , 28., , 34., , 37., , -2c, , (d) Let dimensions of length is related as,, é e2 ù, L = [c]x [G]y ê, ú, ëê 4pe 0 ûú, , z, , T, , 2, , (kT)2 = I k 2, , (a) The unit of l, x and A are the same, (c) L + B = 2.331 + 2.1 @ 4.4 cm, Since minimum significant figure is 2., (c) Given, x = cos(wt + kx), (wt + kx) is an angle and hence it is a dimension less, quantity., [(wt + kx)] = [M0L0T0], , 1, \ a=, 2, , E [M][LT-1]2, = ML2T -1, =, -1, n, T, , I, , p(r – x 2 ) [ML–1T –2 ][L2 ], =, = [ML–1T –1 ], (a) h =, 4vl, [LT –1[L], , [wt] = [M0L0T0], , or, , v µ l1/ 2 r0 g1 / 2 or v 2 µ l g, (b) [momentum] = [M][L][T–1] = [MLT–1], , Planck’s constant =, 29., , (c) I, Dimensions of A = I /T2;, Dimensions of B = kT, (Q power of exponential is dimensionless), , 2, , \ b = 0; a - 3b + c = 1, , -2c = -1 Þ c = 1 / 2, , \ m = -n, , = AT2 e–B/kT, , AB2 =, , 35., 36., , c, , Q2, is energy of capacitor so it represent, 2C, the dimension of energy = [ML2T–2]., (b) Let M = pnvm, , (c) We know that, , = M n L- n + m T -2n - m, , 2 -1, , m s C m s As, =, [Coulomb,c = As], J, kg m 2 s -2, , é, e2 ù, Hence, L = c-2 êG ×, ú, ëê 4pe 0 ûú, (c) Impulse = change in momentum, , ML-2 T -1 = ( ML-1 T -2 ) n (LT -1 ) m, , æ, joule(J) ö, çèQ Volt = V = coulomb(C) ÷ø, 2 -1, , ...(iii), , 1/2, , eV W PV, =, =, =R, T, T, T, R, and = Boltzmann constant., N, , (b) Mobility m =, , (Q y = z), , From (i), (ii) & (iii), 1, z = y = , x = –2, 2, , = [M 0 L0 T 0 ] = angle., , 22., , [M 5 ][M -1L3 T -2 ]2, , 2, , [w] =, , 38., , [M0 L0T0 ], = [M0L0T -1 ], [T], , 9, MD, 10, Vernier constant = 1 MD – 1 VD, , (c) 10 VD = 9MD, 1VD =, , æ, è, , = ç1-, , 9ö, 1, 1 1, ÷ MD = MD = ´ = 0.05 mm, 10ø, 10, 10 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ CP02, , S-4, , DAILY PRACTICE, PROBLEMS, 1., , 2., , PHYSICS, SOLUTIONS, , (a), , Acceleration of the particle a = 2t – 1, The particle retards when acceleration is opposite to, velocity., Þ a . v < 0 Þ (2t – 1) (t2 – t) < 0 Þ t (2t – 1) (t – 1) < 0, Now t is always positive, \ (2t – 1) (t – 1) < 0, 1, or 2t – 1 < 0 and t – 1 > 0 Þ t < and t > 1., 2, This is not possible, or 2t – 1 > 0 & t – 1 < 0 Þ 1/2 < t < 1, (b) x = at3 and y = bt3, vx =, , 8., , Total distance travelled, Total time taken, 5v1v2, x, =, =, 2 x / 5 3 x / 5 3v1 + 2v2, +, v1, v2, , (a) Instantaneous speed is the distance being covered by, the particle per unit time at the given instant. It is equal, to the magnitude of the instantaneous velocity at the, given instant., dx, dx, =a xÞ, (a) v = a x ,, = a dt, dt, x, , 5., , Average speed=, , x, , ò, , 0, , dx, x, , (d), , 10., , t, , 1 2, 1, at = 0 ´ 5 + ´ 6 ´ 52 = 75 m, 2, 2, IIIrd part: s = 45m, u = 30m/s, v = 0, s = ut +, , 0, , x, , é2 x ù, t, ê, ú = a[t ]0, 1, ë, û0, , -30 ´ 30, v2 - u 2, = -10m / s 2, =, 2s, 2 ´ 45, v = u + at Þ 0 = 30 – 10 × t Þ t = 3s, IInd part :, s = s1 + s2 + s3, 395 = 75 + s2 + 45 Þ s2 = 275 m, , a=, , Þ 2 x = at Þ x =, , 6., 7., , - b´, a, b ) = a (1 - e -1 ) = a (1 - 1 ), (1 - e, b, b, b, e, , a (e - 1), a ( 2 .718 - 1) a (1 .718 ), a 2, =, =, = 0 .637 ~, - a/b, b, 2, ., 718, b, 2, ., 718, b 3, b e, dx, velocity v =, = ae - bt , v0 = a, dt, dv, accleration a =, = - abe - bt & a 0 = - ab, dt, a, At t = 0, x = (1 - 1) = 0 and, b, 1, a, a, 1, 2, At t = , x = (1 - e -1 ) = (1 - ) = a / b, b, b, b, e, 3, a, At t = ¥, x =, b, a, It cannot go beyond this, so point x > is not reached, b, by the particle., a, At t = 0, x = 0, at t = ¥, x = , therefore the particle, b, does not come back to its starting point at t = ¥., (d) Ist part: u = 0, t = 5s, v = 108 km/hr = 30 m/s, v = u + at Þ 30 = 0 + a × 5 Þ a = 6 m/s2, , = a ò dt, , a2 2, t, 4, 1, 1, 1, (1 + 4) ´ 4 – ´ 1´ 2 - ´ 3 ´ 4 = 3 m, (c), 2, 2, 2, (b) The distance travel in n th second is, Sn = u + ½ (2n–1)a, ....(1), so distance travel in tth & (t+1)th second are, St = u +½ (2t–1)a, ....(2), St+1= u+½ (2t+1)a, ....(3), As per question,, St+St+1 = 100 = 2(u + at) ....(4), Now from first equation of motion the velocity, of, particle after time t, if it moves with an accleration a is, v=u+at, ....(5), where u is initial velocity, So from eq(4) and (5), we get v = 50 cm/sec., , x=, , =, , = 3t 2 a 2 + b2, , 4., , D istance 100, =, = 100sec., Velocity, 1, 1, , 9., , \ v = v x2 + v2y = 9a 2t 4 + 9b2t 4, , (d), , (d) Relative speed of police with respect to thief, = 10 – 9 = 1 m/s, Instantaneous separation = 100 m, , Time =, , dx, dy, = 3at 2 and v y =, = 3bt 2, dt, dt, , 3., , DPP/CP02, , 275, = 9.16 = 9.2s., 30, Total time taken = (5 + 9.2 + 3) sec = 17.2 sec, t=, , 11., , (a), , dv, dv, = - kv 3 or 3 = - k dt, dt, v, Integrating we get, At t = 0, v = v0 \ -, , 1, 2v 2, , 1, 2v2o, , = - kt + c, , =c, , ...(1), , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP02, , S-5, , Putting in (1), , -, , 1, 2v2, , = - kt -, , 1, 2v02, , or, , 1, 2v20, , -, , 1, 2v2, , = - kt, , [, , v 20, 1 + 2 v 20, , or v =, , kt, , 12. (c) We know that, v =, , ], , v, ; the slope is positive, a, These characteristics are represented by graph (b)., (d) The stone reaches its maximum height after time t1, given by, , For t >, , v0, , 15., , 1 + 2v 02 kt, , u = 10 m/s, , dx, Þ dx = v dt, dt, , x, , t, , 0, , 0, , u, t1 = (Q v = u - gt), g, , 40 m, 10, =, = 1 sec, 10, Again it reaches to its initial position in 1 sec and falls, with same initial speed of 10 m/s., Let t2 be the time taken to reach the ground, then, vground = u + gt 2, , Integrating, ò dx = ò v dt, t, , or x = ò (v0 + gt + ft 2 ) dt, 0, , t, , é, gt 2 ft 3 ù, = êv0 t +, +, ú, 2, 3 úû, ëê, 0, , 2, But vground, = u 2 + 2gh, , 2, = (10) + 2 ´ 10 ´ 40 = 900, , gt 2 ft 3, or, x = v0 t +, +, 2, 3, , g f, + ., 2 3, 13. (c) Let man will catch the bus after ‘t’ sec. So he will cover, distance ut., 1 2, Similarly, distance travelled by the bus will be at, 2, For the given condition, 1, ut = 45 + at 2 = 45 + 1.25 t 2 [As a = 2.5 m/s2], 2, 45, Þ u = +1.25 t, t, du, To find the minimum value of u, =0, dt, so we get t = 6 sec then,, , 900 = 30 m / s, , vground - u, , 30 - 10, = 2 sec., g, 10, \ Total required time = (1 + 1 + 2) sec = 4 sec, , \ t2 =, , 16., , (b), Þ, , 17., , (b), , 45, + 1.25 ´ 6 = 7.5 + 7.5 = 15 m/s, 6, 14. (b) For the body starting from rest, , 18., , 1 2 1, gt - g (t - T )2, 2, 2, , t =, , T L, + ., 2 gt, , 2S, g, 4S, , t2 =, g, , t1 + t 2 + t 3 =, , t, , L =, , 1, 1, g t 12 Þ 2S = AC = g (t1 + t 2 )2, 2, 2, 1, 2, and 3S = AD = g ( t 1 + t 2 + t 3 ), 2, , t1 + t 2 =, , t3 =, , =, , S = AB =, , t1 =, , u=, , 1, \ x1 - x2 = at 2 - vt, 2, at t = 0, x1–x2 = 0, , vground =, , Þ, , At t = 1, x = v0 +, , x1 – x2, 1, x1 = 0 + at2, 2, 1 2, Þ x1 = at, 2, For the body moving, with constant speed, x2 = vt, v/a, , v, ; the slope is zero, a, , For t =, , 2, é 1, ù, 1, or ê, or 1 + 2v 02 kt = v 0, + kt ú =, êë 2 v 02, úû 2 v 2, v2, , or v 2 =, , v, ; the slope is negative, a, , For t <, , 6S, g, , 6S, g, , 4S, g, , A, , 2S, g, , S, B, 2S S, , 4S, g, , t1 : t2 : t3 : :1: ( 2 - 1) : ( 3 - 2), (c) Height of tap = 5m and (g) = 10 m/sec2., For the first drop,, 5 = ut +, , 3S, C, , S, D, , 1 2, 1, gt = (0 ´ t ) + ´10t 2 = 5t2 or t2 = 1 or t = 1., 2, 2
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t.me/Magazines4all, , DPP/ CP02, , S-6, , It means that the third drop leaves after one second of, the first drop. Or, each drop leaves after every 0.5 sec., Distance covered by the second drop in 0.5 sec, 1, 1, = ut + gt 2 = (0 ´ 0.5) + ´ 10 = (0.5) 2 = 1.25m ., 2, 2, Therefore, distance of the second drop above the, ground = 5 – 1.25 = 3.75 m., , 19., , (c) Q t =, , Now, s1 = 0 ´ 2 +, , 23., , Þ x = t – 3 Þ x = (t – 3)2, dx, = 2(t – 3) = 0, dt, Þt=3, \ x = (3 – 3)2, Þ x = 0., , (c) We have, Sn = u +, or 65 = u +, , \, , 25., ..... (1), , a, (2 ´ 9 - 1), 2, , 17, a, 2, Equation (2) – (1) gives,, , or 105 = u +, , ..... (2), , 9, ´ 10 = 65 - 45 = 20 m / s, 2, \ The distance travelled by the body in 20 s is,, u = 65 -, , 1, 1, s = ut + at 2 = 20 ´ 20 + ´ 10 ´ (20)2, 2, 2, = 400 + 2000 = 2400 m., , 26., , 5, 50, m/s =, m/s, 18, 3, , 5 100, =, m/s, 18, 3, Let declaration be a then (0)2 – u2 = –2ad, or u2 = 2ad, … (1), and (0)2 – u'2 = –2ad', or u¢2 = 2ad¢, …(2), (2) divided by (1) gives,, d = 20m, u' = 120 ´, , d', Þ d ' = 4 ´ 20 = 80m, d, (b) 8 = a t1 and 0 = 8 – a (4 – t1), 4=, , 22., , 8, 8ö, æ, \ 8 = a ç4 - ÷, a, aø, è, 8 = 4 a – 8 or a = 4 and t1 = 8/4 = 2 sec, , or, , t1 =, , 1, (t + 5)3, , 3, µ v2, , µa, , 4 sec, , B (v = 0), 4 sec, , (d), , C, , (t = 0), , 17, 9, a - a = 4a or a = 10 m/s2., 2, 2, Substitute this value in (1) we get,, , Speed, u = 60 ´, , 1, µ v2, (t + 5), , (2 sec) A, , 40 =, , (d), , 1, t +5, , Now, , a, (2n - 1), 2, , 9, or 65 = u + a, 2, , 21., , (a) x =, , s2 = 8 m, , 1, , a, (2 ´ 5 - 1), 2, , Also, 105 = u +, , (d), , s1 = 8 m, , -1, dx, \ v = dt = (t + 5) 2, 2, d2x, = 2x3, \ a= 2 =, (t + 5)3, dt, , v=, , 20., , 1, ´ 4 ´ ( 2 ) 2 or, 2, \ s1 + s 2 = 16 m, s2 = 8 ´ 2 -, , 24., , x +3, , 1, ´ 4 ( 2 ) 2 or, 2, , 27., , D, As the time taken from D to A = 2 sec., and D ® A ® B ® C = 10 sec (given)., As ball goes from B ® C (u = 0, t = 4 sec), vc = 0 + 4g., 1, As it moves from C to D, s = ut + gt 2, 2, 1, s = 4g ´ 2 + g ´ 4 = 10 g., 2, 1, 1, (d) y = g (n + 1) 2 - gn 2, 2, 2, g, g, 2, 2, = [(n + 1) - n ] = (2n + 1) ......(i), 2, 2, g, Also, h = (2n - 1), ......(ii), 2, From (i) and (ii), y= h + g, (b) The stone rises up till its vertical velocity is zero and, again reached the top of the tower with a speed u, (downward). The speed of the stone at the base is 3u., u, , –, +, v, g, h, Hence (3u)2 = (-u)2 + 2gh or h =, , 4u 2, g, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ CP02, , S-8, , Þ v = 98 = 9.9, Also for upward motion, , 42., , 02 - u 2 = 2 ´ ( -9.8) ´1.8, , 38., 39., , 9, (2t - 1), =, 25, t2, 2, 9t = 50t – 25, 9t2 – 50t + 25 = 0, Solving, we get, 5, t = 5s or t = s, 9, Substituting t = 5s in (ii), we get, 1, h = ´ 9.8 ´ (5) 2 = 122.5 m, 2, (b) y µ t 2 ; v- µ t'; a µ t°, (b) Average velocity for the second half of the distance is, v + v2 4 + 8, = 1, =, = 6 m s -1, 2, 2, Given that first half distance is covered with a velocity, , 43., , 41., , 100, = 0.1 sec to reach target., 1000, , During this period vertical distance (downward), travelled by the bullet, 1 2 1, 2, = gt = ´ 10 ´ (0.1) = 0.0 5 m = 5cm, 2, 2, So the gun should be aimed 5 cm above the target., (c) The distance covered in n th second is, 1, S n = u + ( 2 n - 1)a, 2, where u is initial velocity & a is acceleration, 19 a, then 26 = u +, ....(1), 2, 21a, 28 = u +, ....(2), 2, 23 a, 30 = u +, ....(3), 2, 25 a, 32 = u +, ....(4), 2, From eqs. (1) and (2) we get u = 7m/sec, a=2m/sec2, \ The body starts with initial velocity u =7m/sec, and moves with uniform acceleration a = 2m/sec2, , A2, , 10, , A1, , 0, , A3 A, 4, , = 10 + 20 + 15 + 10 = 55 m, 44., , (a) Q h =, , 1 2, gt, 2, , 1, g(5)2 = 125, 2, , \ h1 =, , h1 + h2 =, , 1, g(10)2 = 500, 2, , Þ h2 = 375, h1 + h2 + h3 =, , 1, g(15)2 = 1125, 2, , Þ h3 = 625, h2 = 3h1 , h3 = 5h 1, , whole time of motion is 6 m s -1, (b) Bullet will take, , 20, , 1 2 3 4, Time (in sec), 1, 1, = ´1 ´ 20 + (20 ´ 1) + (20 + 10) ´1 + (10 ´ 1), 2, 2, , of 6 m s -1 . Therefore, the average velocity for the, , 40., , 30, , Velocity (m/s), , Þ u = 3528 = 5.94, Fractional loss = 9.9 - 5.94 = 0.4, 9.9, 37. (c) Distance travelled by the stone in the last second is, 9h g, = (2t - 1) (Q u = 0), ...(i), 25 2, Distance travelled by the stone in t s is, 1, 1, h = gt 2 (using s = ut + at 2 ), ...(ii), 2, 2, Divide (i) by (ii), we get, , x, x, (a) 8 = , 12 =, t1, t2, 2x, 2x, 2 ´ 8 ´ 12, v=, =, =, = 9.6 ms -1, x, x, t1 + t 2, 12, +, 8, +, 8 12, (b) Distance = Area under v – t graph = A1 + A2 + A3 + A4, , h3, h2, =, 3, 5, , or h1 =, 45., , 1 2, ft1, 2, Distance from B to C = ( ft1 ) t, , (d) Distance from A to B = S =, , Distance from C to D =, A f B, t1, , ( ft1 )2, u2, =, = ft12 = 2 S, 2a 2( f / 2), C f /2 D, 2t 1, , t, 15 S, , Þ S + f t1t + 2 S = 15 S, Þ, , f t1t = 12 S, , ............. (i), , 1 2, f t1 = S, 2, , ............ (ii), , Dividing (i) by (ii), we get t1 =, , Þ S=, , 2, , 1 ætö, f t2, fç ÷ =, 2 è 6ø, 72, , t, 6, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ CP03, , S-10, , 11., , (c) Yes, the person can catch the ball when horizontal, velocity is equal to the horizontal component of ball’s, velocity, the motion of ball will be only in vertical, direction with respect to person for that,, , 15., , (d) s = t3 + 5, , Tangential acceleration at =, , vo, = vo cos q or q = 60°, 2, , 12., , So,, , wt p, =, 2 2, , \ t=, , =, , v22, , + v12, , 2, , 9 ´ 16, = 7.2 m/s2, 20, \ Resultant acceleration, ac =, , A, B, B, B2, Hence, from (i), = A2 + B2 – 2A2 Þ A= 3, 2, A, A, 3, Þ cos q = – = –, \ q = 150°, B, 2, (b) Suppose velocity of rain, r, v R = v x ˆi - v y ˆj, \, , 17., , = 50 + 50 = 70.7 km/h, , 10, , 8, , q, , uur 6, Comparing with v = vx ˆi + v yˆj , we get, vx =, , 6ms–1 and, , R=, , v 2 sin 2q 2v 2 sin q cos q, =, g, g, , vy = 8, , ms–1, , 2, 2, Also, v 2 = v x + v y = 36 + 64 = 100, or v = 10 ms–1, 8, 6, sin q =, and cos q =, 10, 10, , 8 6 1, R = 2 ´ 10 ´ 10 ´ ´ ´, = 9.6 m, 10 10 10, , (12) 2 + (7.2) 2 = 144 + 51.84, , = 195.84 = 14 m/s2, B, = A 2 + B2 + 2AB cos q ...... (i), (b), 2, B sin q, Þ A + B cos q = 0, \ tan 90° =, A + B cos q, , 2, , The direction of this change in velocity is in South-West., (b) v = 6 î + 8ˆj, , at2 + ac2 =, , =, , p, w, , \ Change in velocity, uur ur, uur, ur, = | v2 - v1 | = | v2 + (-v1 ) |, , 14., , v 2 9t 4, =, R, R, At t = 2s, at = 6 ´ 2 = 12 m/s2, , 16., , ur, 13. (a) v1 = 50 km h –1 due North;, ur, uur, v2 = 50 km h –1 due West. Angle between v1 and, uur, v2 = 90º, ur, - v1 = 50 km h -1 due South, , dv, = 6t, dt, , Radial acceleration ac =, , (b) Two vectors are, r, ˆ, ˆ, A = cos wti + sin wtj, r, wt, wt, B = cos ˆi + sin ˆj, 2, 2, r, r, For two vectors A and B to be orthogonal A.B = 0, wt, wt, r r, A.B = 0 = cos wt.cos 2 + sin wt.sin 2, wt ö, æ, æ wt ö, = cos ç wt - ÷ = cos ç ÷, 2 ø, è, è 2 ø, , ds, = 3t 2, dt, , Þ velocity, v =, , \, , cos q = –, , and the velocity of the man, r, v m = u ˆi, Velocity of rain relative to man, r, r, r, v Rm = v R - v m = ( v x - u ) ˆi - v y ˆj, , According to given condition that rain appears to fall, vertically, so (vx – u) must be zero., \, vx – u = 0 or vx = u, When he doubles his speed,, uur, v'm = 2u ˆi, r, r, uur, Now v Rm = v R - v' m, , (, , ), , = v x ˆi - v y ˆj - (2uˆi ), = ( v x - 2u ) ˆi – v y ˆj, r, The v Rm makes an angle q with the vertical, r, x - componend of v Rm, r, tan q =, y - componend of v Rm, =, , ( v x - 2u ), -v y, , u - 2u, = -v, y, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP03, , S-11, , which gives, , v2 = u2 + 2gh \ v = u 2 + 2 gh, , u, vy =, tan q, , Another particle is thrown horizontally with same, velocity then velocity of particle at the surface of earth., u, h, , vx = u, , Thus the velocity of rain, r, v = v ˆi - v ˆi, x, , R, , v y = 2 gh, , y, , \ Resultant velocity, v = u 2 + 2 gh, , u ˆ, j., tanq, 18. (c) For projectile A, , = u iˆ -, , For both the particles, final velocities when they reach, the earth's surface are equal., u 2A sin 2 45°, 2g, , Maximum height, HA =, , 21. (b) rˆ = 0.5iˆ + 0.8 ˆj + ckˆ, | rˆ |= 1 = (0.5)2 + (0.8) 2 + c 2, , For projectile B, , u 2B sin 2 q, Maximum height, HB =, 2g, As we know, HA = HB, u 2A sin 2 45° u 2B sin 2 q, =, 2g, 2g, sin 2 q, sin 2 45°, , =, , u 2A, , (0.5)2 + (0.8)2 + c 2 = 1, , 22., , 23., , u B2, 2, , æu ö, sin q = ç A ÷ sin 2 45°, è uB ø, 2, , 2, , v, , Horizontal component of velocity vx = u, , c 2 = 0.11 Þ c = 0.11, (c) The pilot will see the ball falling in straight line because, the reference frame is moving with the same horizontal, velocity but the observer at rest will see the ball falling, in parabolic path., (a) R = 2H (given), 1, We know, R = 4 H cot q Þ cot q =, 2, , 2, , æ 1 ö æ 1 ö, 1, sin 2 q = ç, ÷ ç, ÷ =, 2, 2, 4, è, ø è, ø, 1, -1 æ 1 ö, sin q = Þ q = sin ç ÷ = 30°, 2, è2ø, 19. (a) The angle for which the ranges are same is, complementary., Let one angle be q, then other is 90° – q, 2u sin q, 2u cos q, T1 =, , T2 =, g, g, , 4u 2 sin q cos q, u 2 sin 2 q, =2R, (Q R =, ), g, g, Hence it is proportional to R., 20. (c) When particle thrown in vertically downward direction, with velocity u then final velocity at the ground level, , Ö5, q, 1, , From triangle we can say that sin q =, \ Range of projectile R =, , T1T2 =, , u, h, v = u 2 + 2 gh, , 2, , 2, 5, , , cos q =, , 1, 5, , 2, , 2v sin q cos q, g, , 2v 2, 2, 1, 4v 2, ´, ´, =, g, 5, 5 5g, (a) Note that the given angles of projection add upto 90°., For complementary angles of projection (45° + a) and, (45° – a) with same initial velocity u, range R is same., , =, , 24., , 25., , q1 + q2 = (45° + a) + (45° – a) = 90°, So, the ratio of horizontal ranges is 1 : 1., (a) Th e components of 1 N and 2N forces, along + x axis = 1 cos 60° + 2 sin 30°, = 1´, , 1, 1 1, 3, + 2 ´ = + 1 = = 1.5N, 2, 2 2, 2
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t.me/Magazines4all, , DPP/ CP03, , S-12, , Y, , For B, going down with velocity u, , 4 cos 30° + 1 sin 60°, , 4N, , Þ v B = u 2 + 2gh, , 1N, , For C, horizontal velocity remains same, i.e. u. Vertical, velocity = 0 + 2gh = 2gh, , 30°, 4 sin 30°, 30°, , 1cos 60° + 2 sin 30°, 30., , 2cos30°, The component of 4 N force along –x-axis, 1, = 2N ., 2, Therefore, if a force of 0.5N is applied along + x-axis,, the resultant force along x-axis will become zero and, the resultant force will be obtained only along y-axis., , =, 31., , = 4 sin 30° = 4 ´, , (d), , Fx =, , d px, = - 2 sin q., dt, , Similarly, Fy =, , 32., , d py, , (13 - 2)iˆ + (14 - 3)ˆj 11 ˆ ˆ, = (i + j), 5, 5-0, , (c) Position vector, r = cos wt + sin wt ŷ, r, x̂, \ Velocity, vr = –wsin wt x̂ + wcos wt ŷ, and acceleration,, r, r, a = –w2 cos wt x̂ + w sin wt ŷ = –w2 r, r, r, r × r = 0 hence r ^ v and, r v, r is directed towards the origin., a, Y, (d), , = 2 cos q., dx, Angle q between two vectors, , Fx p x + Fy p y, cos q =, r r, |F || p |, , 27., , 28., , 29., , a, , ®, v, , q, , ®, , ®, , ®, , ®, , ®, , .... (1), , v A B = v A - vB = v - v1, , By taking x-components of equation (1), we get, v1, .... (2), v, By taking Y-components of equation (1), we get, v y = v cos q, .....(3), 0 = v sin q - v1 Þ sin q =, , Time taken by boy at A to catch the boy at B is given by, Relative displacement along Y - axis, t=, Relative velocity along Y - axis, a, a, a, =, =, =, 2, v cos q v . 1 - sin 2 q, æv ö, v. 1- ç 1 ÷, è vø, , u, , [From equation (1)], , u = vx, , a, , =, , u, , v., , h, , vC, , X, , Velocity of A relative to B is given by, , Using v 2 = u 2 + 2as Þ v A = u 2 + 2gh, , vB vA, , ®, vA/B, , A, , Þ cos q = 0 Þ q = 90°, (a) The motion of the train will affect only the horizontal, component of the velocity of the ball. Since, vertical, component is same for both observers, the ym will be, same, but R will be different., (d) As body covers equal angle in equal time intervals. Its, angular velocity and hence magnitude of linear velocity, is constant., (a) For A: It goes up with velocity u will it reaches its, maximum height (i.e. velocity becomes zero) and comes, back to O and attains velocity u., , O, , ®, v1, , B, O, , ( -2sin q) (2cos q) + (2cos q) (2sin q), r r, =, |F||p|, , u 2 + 2gh ., , Hence v A = v B = v C, r, D r (displacement), r, (d) vav =, Dt (time taken), , 2N, , 26., , v 2x + v 2y =, , The resultant v C =, , 60°, , u = vX vc = v 2x + v2y, , 33., , v 2 - v12, , (b) H =, , v2, , =, , a, v 2 - v12, , u 2 sin 2 45° u 2, =, 2g, 4g, , =, , a2, v 2 - v12, , ...(1), , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP03, , S-13, , u 2 sin 90° u 2, =, R=, g, g, \, , R=, , R u2, =, 2 2g, , =, , u2, 4g, u2, 2g, , =, , 8 6 1, ´ ´, = 9.6 m, 10 10 10, (a) Range of a projectile is maximum when it is projected at, an angle of 45° and is given by, R = 2 ´10 ´ 10 ´, , ...(2), 37., , H, \ tan a =, R/2, , Rmax =, , æ 1ö, \ a = tan -1 ç ÷, è 2ø, , 1, 2, , 45°, R/2, , H, R/2, , 34. (b) Here, x = 4sin(2pt), ...(i), y = 4cos(2pt), ...(ii), Squaring and adding equation (i) and (ii), x2 + y2 = 42 Þ R = 4, Motion of the particle is circular motion, acceleration, , ur, v2, vector is along – R and its magnitude =, R, Velocity of particle, v = wR = (2p) (4) = 8p, ur uur, ur uur, 35. (d) | A + B |2 = | A - B |2, ur ur, ur uur, r, r, | A + B |2 = | A |2 + | B |2 + 2 A . B = A2 + B 2 + 2 AB cos q, uur, ur uur, r r, r, | A - B |2 = | A |2 + | B |2 - 2 A . B, = A2 + B 2 - 2 AB cos q, So, A2 + B2 + 2AB cos q, = A2 + B2 – 2AB cos q, 4 AB cos q = 0 Þ cos q = 0, , R u 2 sin 2q, =, , where q is the angle of projection., 2, g, R Rg sin 2q, Þ =, ; from (i), g, 2, 1, Þ sin 2q = Þ sin 2q = sin 30°, 2, Þ 2q = 30° \ q = 15°, 38. (a) Distance covered in one circular loop = 2pr, = 2 × 3.14 × 100 = 628 m, 628, = 10 m / sec, 62.8, Displacement in one circular loop = 0, , Speed =, , 0, =0, time, uuur uuuur uuur, b', (a) PQ + QR = PR, q, uuur uur r, P, \ QR = b ' - b, b, uur ur 2 uur ur uur ur, Now | b ' - b | = (b ' - b ).(b ' - b ), , Velocity =, , 39., , \ q = 90º, So, angle between A & B is 90º., 36. (b) v = 6 î + 8ˆj, , 10, , 8, , sin q =, , 8, 6, and cos q =, 10, 10, , u 2 sin 2 q, 2g, , H1 H 2 =, , -1, , Also, v 2 = v x 2 + v y 2 = 36 + 64 = 100, or v = 10 m s -1, , (a), , H1 =, , and H 2 =, , uur 6, Comparing with v = vx ˆi + v yˆj , we get, and v y = 8 m s, , Q, , [Q b ' = b], = 2b2 (1 - cos q), uur r, b ' - b = 2b 1 - cos q, qö, æ, q, = 2b ç 2 sin ÷ = 2bsin, 2ø, è, 2, , q, , vx = 6ms, , R, , = b '2 - 2bb ' cos q + b 2, , 40., , -1, , u2, , where u is the velocity of projection, g, , 2, ÞR= u, \ u 2 = Rg … (i), g, Now, to hit a target at a distance (R/2) from the gun, we, must have, , u, a, , v 2 sin 2q 2v 2 sin q cos q, =, g, g, , 41., , u 2 sin 2 (90° - q) u 2 cos 2 q, =, 2g, 2g, , u 2 sin 2 q u 2 cos2 q (u 2 sin 2q ) 2 R 2, ´, =, =, 2g, 2g, 16, 16 g 2, , \ R = 4 H1 H 2, r, r r, (c) P = vector sum = A + B, r, r r, Q = vector differences = A - B
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t.me/Ebooks_Encyclopedia27., , DAILY PRACTICE, PROBLEMS, 1., , 2., , 3., , 4., , PHYSICS, SOLUTIONS, , (d) Here m = 0.5 kg ; u = – 10 m/s;, t = 1/50 s ; v = + 15 ms–1, Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N, (b), T cos q, T1, q 1, , T1 sin q, , 2m, , t.me/Magazines4all, , W = Fapplied × d = 10.5 × 0.10 = 1.05 J, Hence this statement is incorrect., 5., , (b) Momentum P = mv = m 2 gh, (Q v2 = u2 + 2gh; Here u = 0), When stone hits the ground momentum, P = m 2 gh, when same stone dropped from 2h (100% of initial), then momentum, , T2, T2, , W = 60N, In eqbm T1cos q = T2 = 60N., T1sin q = 60 N, \ tan q = 1, q = 45°., (d) Mass of rocket (m) = 5000 Kg, Exhaust speed (v) = 800 m/s, Acceleration of rocket (a) = 20 m/s2, Gravitational acceleration (g) = 10 m/s2, We know that upward force, F = m (g + a) = 5000 (10 +20), = 5000 × 30 = 150000 N., We also know that amount of gas ejected, , DPP/CP04, , …(1), …(2), , æ dm ö F 150000, = 187.5 kg / s, ÷= =, ç, 800, è dt ø v, (b) (i) If a body is moved up an inclined plane, then the, work done against friction will be zero as there is no, friction. But work must be done against gravity. So, this statement is incorrect., (ii) This statement is correct, because moving vehicles, are stopped by air friction only., (iii) The normal reaction acting on a body on an, inclined plane is given by,, R = mg cosq, Where q is the angle of inclination., As q increases, cos q decreases and hence R, decreases. So this statement is incorrect., (iv) The applied force needed to rub the duster upward,, , 6., , =, , 7., , 2 ´ 3 ´ 10 ´ 3, = 50 ´ 3 3, 2 ´ 20, , = 150 3 newton, (b), , q, , q, , S/2, ugh, Ro, , S/2, h, oot, Sm, , S/2 sin q, , 2 sin q, = 2 tan q, cos q, (c) Forces acting on the block are as shown in the fig., Normal reaction N is provided by the force ma due to, acceleration a, , Þ m=, , Ff, , Fs = mR, mg, Fapplied = mg + mR = 0.5 × 10 + 0.5 ×11, = 5 + 5.5 = 10.5 N, \ The work done in rubbing it upward through a, distance of 10 cm,, , S/2 sin q, , For upper half of inclined plane, v2 = u2 + 2a S/2 = 2 (g sin q) S/2 = gS sin q, For lower half of inclined plane, 0 = u2 + 2 g (sin q – m cos q) S/2, Þ – gS sin q = gS ( sinq – m cos q), Þ 2 sin q = m cos q, , 8., 11 N, , P ¢= m 2 g (2h) = 2 P, Which is changed by 41% of initial., (c) Change in momentum along the wall, = mv cos60º – mv cos 60º = 0, Change in momentum perpendicular to the wall, = mv sin60º – (– mv sin60º) = 2mv sin60º, Change in momentum, \ Applied force =, Time, 2 mv sin 60º, =, 0.20, , N = ma, , mg
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t.me/Magazines4all, , DPP/ CP04, , S-16, , \ N= m a, For the block not to fall, frictional force,, Ff > mg, Þ m N > mg, Þ m m a > mg, Þ a > g/ m, , 9., , (b), , 17., , v = gr = 10 ´ 40 = 20 m s -1, 12 m/s, , 10., , m, , (a), , 4 m/s, , 1 kg, , 1, , F, , 2, , 3, 2 kg, , 8 m/s, According to conservation of linear momentum, P3 = p12 + p 22, , 11., , Þ m ´ 4 = (1´12)2 + (2 ´ 8) 2 = 20 Þ m = 5 kg., (c) Let T be the tension in the branch of a tree when, monkey is descending with acceleration a. Then mg –, T = ma; and T = 75% of weight of monkey, , 13., , æ 75 ö, æ1ö, g, =ç, ÷mg = ç ÷mg or a = ., 100, 4, è, ø, è ø, 4, u, (d) v = u - at Þ t = [ As v = 0], a, u ´ m 30 ´ 1000, t=, =, = 6sec, F, 5000, (c) Applying law of conservation of linear momentum, , 14., , v, m, m1, v, = - 2 or 1 = - 2, v2, m1, m2, v1, (a) The frictional force acting on M is µmg, , 12., , a = gsinq – mk(g) cos q, 0.9, = 0.5, Þ mk =, 3, (c) All blocks will move with the same aceleration, Let it be a . Then, F, F = 4Ma Þ a =, 4M, From the figures it is clear that, T1 = 3 Ma, T2 = 2 Ma and T3 = Ma, T, F, M, M 1, M, M, , 18., , M, , M, , 16., , 1, ms = tan 30° =, = 0.577 @ 0.6, 3, 1 2, S = ut + at, 2, 1, 1, 4 = a(4)2 Þ a = = 0.5, 2, 2, [Q s = 4m and t = 4s given], , M, , F = f 2 + R 2 = (µR) 2 + R 2 = R µ 2 + 1, Minimum force = R when there is no friction, , 19., 20., , Hence ranging from R to R µ2 + 1 [where, R = mg], (c) Motion with constant momentum along a straight line., According to Newton's second law rate of change of, momentum is directly proportional to force applied., (b) For the motion of both the blocks, m1a = T – mkm1g, m2g – T = m2a, a, mk m1g, , m1, , T, , mk, m2, , mmg, \ Acceleration =, M, , 15., , M, , T3, F, M, M, M, M, Putting the value of a, we get, 3, F, F, T1 = F , T2 =, and T3 =, 4, 2, 4, (c) Maximum force by surface when friction works, , m1v1 + m2v2 = 0,, , dM, = 0.1 kg/s, vgas = 50 m/s ,, (c), dt, Mass of the rocket = 2 kg. Mv = constant, dM, dv, dv 1 dM, –v, +M, =0. \, =, v, dt, dt, dt M dt, 1, 2, Þ Acceleration = ´ 50 ´ 0.1 = 2.5 m/s, 2, (a) Coefficient of static friction,, , T2, , a=, , m 2g – m k m1g, m1 + m 2, , a, , m2g, , æ m 2 g – m k m1g ö, ÷, m2g – T = (m2) ç m + m, è, 1, 2, ø, solving we get tension in the string, , T=, 21., , m1m 2 (1 + m k ) g, m1 + m 2, , (d) Acceleration of block while sliding down upper half =, g sin f;, retardation of block while sliding down lower half = –, (g sin f - mg cos f), For the block to come to rest at the bottom, acceleration, in I half = retardation in II half., g sin f = -(g sin f - mg cos f), Þ m = 2 tan f, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP04, , S-17, , 22. (d) The particle is moving in circular path, R, , Rsinq, q, , Rcosq, , mg, q, , h, , From the figure, mg = R sin q, , … (i), , mv 2, … (ii), = R cos q, r, From equations (i) and (ii) we get, r, rg, tan q = 2 but tan q =, h, v, , v 2 (0.5) 2, =, = 0.025m = 2.5cm, g, 10, 23. (b) By spitting or sneezing we get a momentum in opposite, direction which will help us in getting off the plane. In, all other cases we will slip on ice as there is no friction., 24. (a) Force required to just move a body, (F) = force due to static friction = µs mg, When body moves with a constant acceleration (a), then, F – fk = ma, where fk is the force of kinetic friction = µk, mg, m mg - m k mg, F - fk F - fk, = s, =, \ a=, m, m, m, g, = (µs – µk) g = (0.75 – 0.5) g = ., 4, 25. (c) Considering the two masses and the rope a system,, then, , 27., , \h =, , 10g, , mg sin q, q, , f1, , 28., 29., , mg cos q, , f2, mg sin q, q, , mg, , mg cos q, , For the upward motion of the body, mg sin q + f1 = F1, or, F1 = mg sin q + mmg cos q, , B, mBg, , 0 .2 3, × 10 ×g, 2, Þ mB = 3.268 » 3.3 kg, (a) When tension in the cable is equal to the weight of, cable, the system is in equilibrium. It means the system, is at rest or moving with uniform velocity., (c) Tension at the highest point, mv2, – mg = 2mg (\ vtop = 3gr ), r, Tension at the lowest point, Tbottom = 2mg + 6mg = 8mg, Ttop, 2mg 1, =, = ., \, Tbottom 8mg 4, (d) When brakes are on, the wheels of the cycle will slide, on the road instead of rolling there. It means the sliding, friction will come into play instead of rolling friction., The value of sliding friction is more than that of rolling, friction., (a) When car moves towards right with acceleration a then, due to pseudo force the plumb line will tilt in backward, direction making an angle q with vertical, a, From the figure, q, a, tan q = a / g, q, \q = tan -1 (a / g), g, Ttop =, , 2, , N2, , 10g cos30º, , 0 = 5g - m B g -, , 30., mg, , T a, , mN, , 10, g - T - m ´ 10 g cos 30 º, 2, but a = 0, T = mBg, , F1, , N1, , A, , \ 10 a =, , F, , (c), , si, , 0º, n3, , T, , 10g, , Þ (acceleration)final = 3 (acceleration)initial, 26., , a, , N, , Initial net force = [25 - (15 + 5)] g = 5g, Final net force = éë( 25 + 5 ) - 15 ùû g = 15 g, , For the downward motion of the body,, mg sin q – f 2 = F2, or F2 = mg sin q – mmg cos q, F1, sin q + m cos q, =, \, sin q - m cos q, F2, tan q + m 2m + m 3m, Þ, =, =, =3, tan q - m 2m - m m, (b) Considering the equilibrium of B, –mBg + T = mBa, Since the block A slides down with constant speed., a = 0., Therefore T = mBg, Considering the equilibrium of A, we get, 10a = 10g sin 30º – T – mN, where N = 10g cos 30°, , 31.
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t.me/Magazines4all, , S-18, , 32., , Q Coefficient of friction m = 0.5, , (b) See fig., , x2, 2, Þ x=+1, , mN, , \, , ma, , N, mg, , 33., , f sin 60°, , f = µR, , 34., , 35., , x3 1, = m, 6 6, (a) As the ball, m = 10 g = 0.01 kg rebounds after striking, the wall, \ Change in momentum = mv – (–mv) = 2 mv, Inpulse = Change in momentum = 2mv, Impulse, 0.54 N s, \n=, =, = 27 m s-1, 2m, 2 × 0.01 kg, (b) From the F.B.D., N = mg cos q, F = ma = mg sin q – mN, Þ a = g(sin q – m cos q), , 38., , W = 10Ö 3, , Fcos 60° = µ(W + Fsin 60°), 1, and W = 10 3 we get F = 20 N, Substituing µ =, 2 3, (d) When the block slides down the plane with a constant, speed, then the inclination of the plane is equal to angle, of repose (q)., Coeff. of friction = tan of the angle of repose = tan q., (d) Writing free body-diagrams for m & M,, m, , m, mg, , mg sin q, Now using,, , a, T T, , M, , F, , dy x 2, =, (from question), dx, 2, , m, q y, , – u2 = 2as, , 2gl (sin q - m cos q), , or, v =, , (a) During collision of ball with the wall horizontal, momentum changes (vertical momentum remains, constant), \, , Mg, , m g cos q, , or, v 2 = 2 ´ g (sin q - m cos q)l, (l = length of incline), 39., , N, , v2, , mN, , xmg, , q, , F, , we get T = ma and F – T = Ma, where T is force due to spring, Þ F – ma = Ma or,, F = Ma + ma, F, ., \ a=, M +m, Now, force acting on the block of mass m is, æ F ö = mF, ma = m ç, ., è M + m ÷ø m + M, (a) At limiting equilibrium, m = tan q, tanq = m =, , N, , M, K, N, , 36., , 37., , f cos 60°, , 0.5 =, , Now, y =, , If a = acceleration of the cart, then N = ma, \ mN = mg or m ma = mg or a = g/m, R, (a), F, 60°, , DPP/ CP04, , F=, , =, , Change in horizontal momentum, Time of contact, , 2P cos q 2mv cos q, =, 0.1, 0.1, , P = mv, , 60°, 30°, , 2 ´ 0.1 ´ 10 ´ cos 60°, = 10N, 0.1, 40. (d) Given F = 600 – (2 × 105 t), , =, , The force is zero at time t, given by, 0 = 600 – 2 × 105 t, 600, Þ, t=, = 3 ´ 10 –3 seconds, 2 ´ 105, t, , ò, , \ Impulse = Fdt =, 0, , 3´10 –3, , ò, 0, , (600 – 2 ´ 105 t ) dt, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , DPP/ CP04, 3´10, , é, 2 ´ 105 t 2 ù, = ê600t –, ú, 2, ëê, ûú 0, , –3, , = 600 ´ 3 ´ 10 –3 – 105 (3 ´10 –3 ) 2, , 44., , = 1.8 – 0.9 = 0.9Ns, 41. (d) According to question, two stones experience same, centripetal force, i.e. FC1 = FC2, , or,, , mv12 2mv 22, =, r, (r / 2), , or, V12 = 4V22, , So, V1 = 2V2 i.e., n = 2, 42. (b), , T1 = m(g + a ) = 0.1(10 + 5) = 1.5N, T2 = m(g - a ) = 0.1(10 - 5) = 0.5N, , Þ T1 - T2 = (1.5 - 0.5) N = 1N, 43. (d) As shown in the figure, the three forces are represented, by the sides of a triangle taken in the same order., , 45., , t.me/Magazines4all, , S-19, r, r, Therefore the resultant force is zero. Fnet = ma., Therefore acceleration is also zero i.e., velocity remains, unchanged., (b) Rate of flow of water will depend on the net acceleration, due to gravity., When the lift is moving upward with acceleration a ,, g 'u = g + a, When the lift is moving downward with acceleration, on a, g 'd = g – a, , \ g 'u > g > g 'd \ Ru > R0 > Rd, (d) According to law of conservation of momentum the, third piece has momentum, ˆ kg ms–1, = 1´ –(3iˆ + 4j), Impulse = Average force × time, Impulse, Þ Average force =, time, –(3iˆ + 4ˆj)kg ms –1, Change in momentum, =, =, time, 10 –4 s
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP05, 8., , S-21, , dw, (d) As we know power P =, dt, 1, 2, Þ w = Pt = mv, 2, , So, v =, , 12., , 2Pt, m, , 13., , (b) If the particle is released at the origin, it will try to go in, dU, the direction of force. Here, is positive and hence, dx, force is negative, as a result it will move towards –ve, x-axis., (c) The potential energy of a spring is given by,, 1 2, kx Þ 10 J = 1 ks2, .... (i), 2, 2, The potential energy stored when stretched, U=, , dv, 2P 1, =, ., dt, m 2 t, Therefore, force on the particle at time ‘t’, , Hence, acceleration a =, , = ma =, 9., , through(2s) =, , 2Km 2, 1, Km, mK –1/2, ., t, =, =, m, 2t, 2, 2 t, , Substituting from (i), P.E. = 40 J., But to increase ‘s’ to ‘2s’, the work done, = 40 – 10 = 30 J., , t3, dx 3t 2, (b) x =, =, = t 2 Þ v = t2, Þ, 3, dt, 3, when, t = 2 sec, v = t2 = (2)2 = 4 m/s, , Work done = K.E. acquired =, =, , 1, mv 2, 2, , 14., , 1, ´ (2) ´ (4)2 = 16 J, 2, , (m1 - m 2 )u1, 2m 2 u 2, +, (m1 + m 2 ), m1 + m 2, , 15., , As mass 2m, is at rest, So u 2 = 0, , Þ v1 =, , (b), , F(N), A B, 10, , 3, (8m - 2m )u, = u, 5, 8m + 2 m, , 2, , 1, 1, æ 3u ö, æ3ö, (8m )ç ÷ = (8m)u 2 ´ ç ÷, 2, 2, è 5 ø, è5ø, , 5, 0, –5, , 2, , 16., , 1 2, kx, 2, where k = spring constant, x = extension, Case (a) If extension (x) is same,, w=, , So, WQ > WP, , K, F I, l 2 3 4 5 6, D C, G H, , x(m), , (c), 10m, , (Q KP > KQ), , Case (b) If spring force (F) is same W =, , E, , = (2–1)×(10 – 0) + (3 –2) (5 –0) (4 –3) (–5 –0), 1, + (5 - 4)(10 - 0) = 15 J, 2, , 9, E = 0.36 E, 25, (b) As we know work done in stretching spring, , 1, W = K x2, 2, So, WP > WQ, , L, , J, , –10, , =, , 11., , ), , 2, 6 é, 2 ù, 1 2.05 ´ 10 ´ ë( 25) - 5 û, P= ´, 2, 5 ´ 60, P = 2.05 × 106 W = 2.05 MW, (b) Work done = Area under F-x graph, = area of rectangle ABCD + area of rectangle LCFE, + area of rectangle GFIH + area of triangle IJK, , Final energy of sphere = (K.E.) f, =, , (, , 1, 2, 2, Work done 2 m v - u, =, Power =, Time, t, , ( ), , 10. (b) For elastic collision in one dimension, v1 =, , 1, 1, k (2s 2 ) = k s 2 ´ 4, 2, 2, , F2, 2K, , h1, , Just before impact, energy, E = mgh = 10mg, ............. (1), Just after impact, 25, mgh = 0.75 mgh, 100, Hence, mgh 1 = E1, (from given figure), E1 = mgh -
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP06, , S-27, , Also xcm =, , m1Dx1 + m2 Dx2 m1v0t + 0, =, m1 + m2, m1 + m2, , mv t, m2l, \ final position =, + 10, m1 + m2 m1 + m2, 10. (a) Here, L = 1.8 kg m2 s–1, M = 1.5 kg,, w = 0.3 rad s–1, Angular momentum, L = Iw, L = k2 Mw, (QI = MK2), 2, or 1.8 = k × 1.5 × (0.3), 1.8, =4, Þ k2 =, 1.5 ´ 0.3, Þ k = 2 m., r r, r r r, r r, 11. (b) t = r ´ F Þ r . t = 0, F.t = 0, , 14., , Reference, level for P.E., , 1 2, The moment of inertia of the rod about O is ml ., 3, The maximum angular speed of the rod is when the rod, is instantaneously vertical. The energy of the rod in, 1, this condition is I w 2 where I is the moment of inertia, 2, of the rod about O. When the rod is in its extreme, portion, its angular velocity is zero momentarily. In this, case, the energy of the rod is mgh where h is the, maximum height to which the centre of mass (C.M), rises, \ mgh =, Þ h=, , 12 2 2, wr d, 25, , 1 2, Iw, 2, , 2 2ö, æ, çèQ ISolid sphere = mr ÷ø, 5, , K .Etranslational =, , Hence option (b) is correct, 1 2, Iw, 2, 1, 2, Kinetic energy (translational) KT = Mv, 2, , 15. (c) Kinetic energy (rotational) KR =, , M.I.(initial) Iring = MR2; winitial = w, , 16., 17., , 18., , R, (ii), (i), Moment of inertia of a ring about a given axis is, I = MR2, , l 2 w2, 6g, , 3M, 4, Moment of inertia of the remaining portion of the ring, about a given axis is, , Mass of the remaining portion of the ring =, v, qv, q, , 13. (b), , 3, I¢ = MR 2, 4, Given I¢ = kMR2, v R = v 2 + v 2 + 2 v 2 cos q = 2 v 2 (1 + cos q), , q, 2, , (v = Rw), , M.I.(new) I¢(system) = MR 2 + 2mR 2, Mw, w¢(system) =, M + 2m, Solving we get loss in K.E., Mm, w2 R 2, =, (M + 2m), (b) For no angular acceleration tnet = 0, Þ F1 × 5 = F2 × 30 (given F2 = 4N) Þ F1 = 24 N, L, (c) For toppling Mg = F1 × h, 2, For sliding, µMg = F2, For sliding to occur first, F1 > F2, mgL, > mMg or L > 2mh, or, 2, (a), R, , 1 2 1 æ 1 2ö 2, I w = ç ml ÷ w, ø, 2è3, 2, , = 2v cos, , 1 2, mv, 2, , K .Erotational, 2, =, \ K .E, 5, translational, , C. M, C. M, , K .Erotational =, , =, , r, r, r, Since, t is perpendicular to the plane of r and F ,, r, r, r, hence the dot product of t with r and F is zero., O, 12. (c), , h, , (b), , \, , k = 3/4.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP08, 39. (a), , S-39, , F/A F l, = ., Dl / l A Dl, 20 ´ 1, =, = 2 ´ 1011 Nm -2, -6, 10 ´ 10-4, , y=, , 40. (d) K =, , 43., , 44., , DP, 200 ´103 ´ 10, hrg, =, =, = 2 ´109, DV / V DV / V, 0.1/100, , 41. (d) Bulk Modulus =, , (d), , F, Fl, A, (b) As Y = Dl Þ Dl =, AY, l, V, But V = Al so A =, l, , dp, dv, v, , Therefore Dl =, , dv 0.1, =, v 100, , 42. (c), , 45., 200 ´103 ´ 9.8, = 19.6´108 N/ m2, 0.1/100, , x, , Fcom Fext, From the figure, it is clear that, Fcom < Fext., , Fl 2, µ l2, VY, , Hence graph of Dl versus l2 will give a straight line., , dp = h r g = 200 × 103 × 9.8, , Bulk modulus =, , Dr / r, 1 Dr 1 Dl, = 0.5 = ,, =, Dl / l, 2 r, 2 l, , F, , (b), , K=, , 4 ´ 9.8, F, =, = 19.6 ´ 102, x 2 ´ 10 -2, , Work done =, , 1, ´19.6 ´102 ´ (0.05)2 = 2.45 J, 2
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP09, , S-41, , v2 = 2 g ( H 0 - h), , Time of reaching the ground from hole B, t1 = 2( H 0 - h) / g, , Time of reaching the ground from hole A, , 20. (a) dv = 8 cm/s and dx = 0.1 cm, dv, 8, =, = 80 / s., Velocity gradient =, dx 0.1, 21. (a) Terminal velocities of rain drops are proportional to, square of their radii., Terminal velocity of a body is given by, , t2 = 2h / g, , 12. (a) Fluid resistance is given by R =, , vT =, , 8hL, , pr 4, When two capillary tubes of same size are joined in, parallel, then equivalent fluid resistance is, , R S = R1 + R 2 =, , 8 hL, pR 4, , +, , 8 h ´ 2L, p (2R) 4, , 2R 2, ( d - s ) g. or, V µ R 2, 9h, , 22. (b), vf, , æ 8hL ö 9, ÷÷ ´, è pR 4 ø 8, , 2, 2, v = W, 3 f 3 eff, , = çç, , é, 8, P pPR 4 8, pPR 4 ù, =, ´ = X ê as X =, Rate of flow =, ú, 9, RS 8hL 9, 8hL úû, êë, , 13. (b) The candle floats on the water with half its length above, and below water level. Let its length be 10 cm with 5 cm, below the surface and 5 cm above it. If its length is, reduced to 8 cm, it will have 4 cm above water surface., So we see tip going down by 1 cm., \ rate of fall of tip = 1 cm/hour., 4T, 14. (a) Inside pressure must be, greater than outside presr, sure in bubble., , Weff =, , Weff, When the, ball is released, , When the ball attains, terminal velocity, , vf, , Weff, When the ball attains, 2/3 of terminal velocity, , When the ball is just released, the net force on ball is, Weff (= mg - buoyant force), The terminal velocity vf of the ball is attained when net, force on the ball is zero., \ Viscous force 6phr vf = Weff, , pa, , When the ball acquires, , pa, , 2, rd of its maximum velocity vf, 3, , 2, Weff, 3, 2, 1, Hence net force is Weff - Weff = Weff, 3, 3, \ required acceleration is a/3, , the viscous force is =, This excess pressure is provided by charge on bubble., 4T, Q2, 4T s 2, Q ù, é, =, s=, ; r =, 2 4, ú, r, 2e 0, 16p r ´ 2e 0 êë, 4pr 2 û, Q = 8pr 2rTe 0, , 15. (a) Because film tries to cover minimum surface area., 16. (d), 17. (a) The condition for terminal speed (vt) is, Weight = Buoyant force + Viscous force, B=Vr2 g, , Fv, , W=V r 1g, , \ V r1 g = V r2 g + kvt2, , \ vt =, , Vg ( r1 - r 2 ), , k, 18. (d) According to Bernoulli's theorem, when velocity of, liquid flow increases, the pressure decreases., 19. (c) Wetability of a surface by a liquid primarily depends, on angle of contact between the surface and liquid., If angle of contact is acute liquids wet the solid and, vice-versa., , 23., , (a), , (2pr1 + 2pr2 )s = mg, 8.7, 8. 5 ù, é, ê2 p ´ 2 + 2p ´ 2 ús = 3.97 ´ 980, ë, û, , Þ s = 72 dyne cm-1, , 24., , (b) Over a small temperature ranges, S.T. of water decreases, linearly with rise of temperature., 4 3, 25. (c) Volume of air bubble V = pr, 3, 3, We get, V µ r, If r is 2 times, V becomes 8 times at the surface of lake., Pressure at the surface of lake is given by, P1 = 1 atmosphere, V1 = 8V, where, d = density of water, Þ P1 = Hdg, Pressure at the bottom of lake,, P2= Pressure of atmosphere + Pressure of water, P2 = Hdg + hdg = (H + h) dg where, h = depth of lake, Let final volume, V2 = V., Because temperature is constant, hence from Boyle's, law, P1V1 = P2V2 Þ Hdg × 8V = (H + h)dg × V Þ h = 7H., 26. (c)
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t.me/Magazines4all, , DPP/ CP09, , S-42, , 27., , 28., , 32, = 4 cm3, (b) Volume of first piece of metal =, 8, Upthrust = 4 gf, , Effective weight = (32 – 4) gf = 28 gf, If m be the mass of second body, volume of second, m, body is, 5, m, Now, 28 = m - Þ m = 35 g, 5, rm, 4, =, (b) geff. = 12 m/s2, =, rw 10, , 35., , T –T, \ – cos q = 2 1, T, T1 – T2, \ cos q =, T, (a) Let L = PQ = length of rod, , L, 2, Weight of rod, W = Alrg., Acting at point S, , \ SP = SQ =, , Q, R, , V, , FB, , S, , L, h = L/2, , a=, , 29., 30., , 31., , V rw ´ 12 - V rm ´ 12, = 18 m / s2, V rm, , d, , pL, , Here, L A d g = (pL) A (nr)g + (1 – p)L A r g, Þ d = (1 – p)r + pn r = [1 + (n – 1)p]r, (d) At equilibrium, weight of the given block is balanced, by force due to surface tension, i.e.,, 2L. S = W, or S =, , W, 1.5 ´ 10 -2 N, =, = 0.025 Nm -1, 2L, 2 ´ 0.3 m, , (a), (d) T1 + T cos (p – q) = T2, T, p–q, cos (p – q) =, , Þ, , l2, 2, , L, , =, , r, l, r, Þ =, r0, L, r0, , From figure, sin q =, 36., , h L 1 r0, = =, l 2l 2 r, , (b) Terminal velocity, v 0 =, , 2 r 2 (r - r0 )g, 9h, , 2 ´ (2 ´10 -3 ) 2 ´ (8 - 1.3) ´103 ´ 9.8, = 0.07 ms–1, 9 ´ 0.83, 37. (c) Work done = Surface tension × increase in area of the, film, W = S × DA, Increase in area = Final area – initial area, = 10 × (0.5 + 0.1) – 10 × 0.5 = 1 cm2, \ W = 72 × 2 × 1 = 144 erg, [Q There are 2 free surfaces; \ DA = 2 × 1]., 38. (b) Waterproofing agents are used so that the material, does not get wet. This means angle of contact is, obtuse., 39. (c), Fb, q, , q, , T1, \, , l, L, Alr0 g ´ cos q = ALrg ´ cos q, 2, 2, , =, , nr, , 33., 34., , (1 – p)L, , q, , And force of buoyancy,, FB = Alr0 g . [l = PR], Which acts at mid-point of PR., For rotational equilibrium., , 1, 1, 1=, ´ 18 t2, t = s., 2, 3, (b) Due to increase in velocity, pressure will be low above, the surface of water., (c) If r is the density of the ball and r¢ that of the another, ball, m for the balls are the same, but r¢= 2r, \ mg = 6prhv (by Stoke's law), v, or, 6prhv = 6p2rhv¢ So, v¢ =, 2, (d) As we know,, Pressure P = Vdg, , r, , 32., , P, , W, , T2, T2 – T1, T, , T, W¢, For floating disc, Fnet = 0, Fb + 2prT cos q = W ¢, or, or, W + 2 prT cos q = W ', T, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP10, , S-47, , 37. (b) According to Wein's displacement law,, lmT = 2.88 × 10–3, When T = 2000 K,, lm (2000) = 2.88 × 10–3 ....(1), When T = 3000 K,, l'm (3000) = 2.88 × 10–3 ....(2), Dividing (1) by (2),, l, 2 lm, 3, 2, = 1 Þ m = Þ l 'm = l m, 3 l 'm, 3, l 'm 2, , 38. (c) AB represents latent heat of fusion, Q1 = mLF, Here, LF µ length of line AB, CD represents latent heat of vaporization Q2 = mLV, Here, LV µ length of line CD, \ Q2 = 2Q1, [Q As CD = 2AB], 39. (b) Let the final temperature be T, Heat gained by ice = mL + m × s × (T – 0), = 10 × 80 + 10 × 1 × T, Heat lost by water = 55 × 1× (40 – T), By using law of calorimetery,, 800 + 10 T = 55 × (40 – T), Þ T = 21.54°C = 22°C, 40. (a) According to Newton's law of cooling,, q1 - q 2, éq + q, ù, = K ê 1 2 - q0 ú, t, 2, ë, û, , where q0 is the surrounding temperature., , 60 - 40, æ 60 + 40, ö, = Kç, - 10÷, \, è 2, ø, 7, , Þ, \, , 20, 1, = 40K Þ K =, 7, 14, , 40 - 28, é 40 + 28, ù, 12, = Kê, - 10ú Þ, = 24K, t, 2, ë, û, t, , or t =, , 12, 12 ´ 14, =, = 7 min, 24K, 24, , Q, dθ, dθ, = K1 A1, = K 2A2, t, dx, dx, 42. (c) Since specific heat = 0.6 kcal/g × °C = 0.6 cal/g × °C, From graph it is clear that in a minute, the temperature is, raised from 0°C to 50°C., Þ Heat required for a minute = 50 × 0.6 × 50 = 1500 cal., Also from graph, Boiling point of wax is 200°C., 43. (b) Temperature of B will be higher because, due to expansion centre of mass B will come down same heat is supplied but in B, potential energy is decreased therefore, internal energy gain will be more., 44. (d) According to the principle of calorimetry., Heat lost = Heat gained, mLv + mswDq = mwswDq, , 41., , (d), , Þ m × 540 + m × 1 × (100 – 80), = 20 × 1 × (80 – 10), Þ m = 2.5 g, Therefore total mass of water at 80°C, = (20 + 2.5) g = 22.5 g, 45. (a) Initial rate of loss of heat =, , s T 4 ´ A1 ´ e, s T4 ´ A2 ´ e, , =, , R 12, R 22
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t.me/Magazines4all, , DPP/ CP13, , S-58, , 8., , (c) Time to complete 1/4th oscillation is, , T, s. Time to, 4, , 13., , 1, th vibration from extreme position is, 8, obtained from, , complete, , T, a, 2p, = a cos w t = a cos, t or t = s, 6, 2, T, So time to complete 3/8th oscillation, , y=, , =, 9., , 10., , T = 2p, , l, g, , and T ' = 2p, , 1.21l, g, , (Q l ' = l + 21% of l), % increase =, =, , T '- T, ´ 100, T, , 1.21l - l, l, , ´ 100 =, , (, , ), , 1.21 - 1 ´ 100, , m, k, When a spring is cut into n parts, Spring constant for each part = nk, Here, n = 4, , m T, =, 4k, 2, , 12. (d) x = A cos (wt + d ), , When d = a +, , (c), , T = 2p, , M, k, , T ' = 2p, , M + m 5T, =, k, 3, , M +m 5, M, = ´ 2p, 3, k, k, 25, M +m =, ´M, 9, m 25, m 25, 16, 1+, =, Þ, =, -1 =, M, 9, M, 9, 9, , \ 2p, , Acceleration, a = 1.0 m/s2, We have, a = -w 2 y or | a |= w 2 y, , Þ 1.0 = w 2 ´ 0.01, 1.0, \ w2 =, = 100 Þ w = 10, 0.01, 2 p 2p p, =, = sec., Now, time period, T =, w, 10 5, 16. (c) We have, U + K = E, where, U = potential energy, K = Kinetic energy, E =, Total energy., Also, we know that, in S.H.M., when potential energy is, maximum, K.E. is zero and vice-versa., , 11. (d) T = 2p, , y = A cos(wt + a ), , 14., , 15. (d) A = 0.05 m, y = 0.01 m, , = (1.1 - 1) ´ 100 = 10%, , T1 = 2p, , é, pö ù, æ, = x0 w 2 cos ê p + ç wt - ÷ ú, è, 4øû, ë, 3p ö, æ, 2, = x0 w cos ç wt + ÷, ...(1), è, 4ø, Acceleration, a = A cos (wt + d) ...(2), Comparing the two equations, we get, 3p, A = x0w2 and d =, ., 4, , T T 5T, + =, 4 6 12, , 1, (a) As we know, kinetic energy = mw2 (A 2 – x 2 ), 2, 1, 2 2, Potential energy = mw x, 2, 1, mw2 (A 2 – x 2 ), 1, A2 – x 2 1, 2, = Þ, =, \, 1, 4, 4, x2, mw2 x 2, 2, 2, 4, A., 4A2 – 4x2 = x2 Þ x 2 = A 2, \x=, 5, 5, , (d), , (a) Here,, x = x0 cos (wt – p / 4 ), dx, pö, æ, \ Velocity, v =, = - x0 w sin ç wt - ÷, è, dt, 4ø, Acceleration,, dv, pö, æ, 2, a=, = - x0 w cos ç wt - ÷, è, dt, 4ø, , ....(1), , p, 2, , æp, ö, x = A cos ç + wt + a ÷, è2, ø, , x = - A sin (wt + a ), ....(2), Squaring (1) and (2) and then adding, x2 + y2 = A2 [cos2 (wt + a) + sin 2 (wt + a)], or x2 + y2 = A2, which is the equation of a circle. The, present motion is anticlockwise., , \U max + 0 = E Þ U max = E, Further,, 1, K .E . = mw 2 a 2 cos 2 w t, 2, But by question, K .E. = K0 cos 2 wt, 1, \ K 0 = mw 2 a 2, 2, 1, 2 2, Hence, total energy, E = mw a = K 0, 2, \U max = K 0 & E = K0 ., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP13, , S-59, , 17. (b) Distance covered by lift is given by, y = t2, \ Acceleration of lift upwards, , =, , d2 y, dt, , 2, , =, , d, g, (2t) = 2 m / s 2 =, dt, 5, , 18., , l, , (c) Here all the three springs are connected in parallel to, mass m. Hence equivalent spring constant, k = K + K + 2 K = 4 K., , 23., , (a), , æ 2p t, ö, y = a sin(wt + f ) = a sin ç, +f ÷, è T, ø, , = 2p, , a = 5m and, , ..... (i), , ......(ii), ....(iii), ...(iv), , (b), , a 1, =, A 2, Now, A – a = A coswt, , dx, = 2 × 10–2 p sin p t, dt, For the first time, the speed to be maximum,, p, sin p t = 1 or, sin p t = sin, 2, 1, p, Þ pt =, or,, t = = 0.5 sec., 2, 2, , 27., , (d), , Þ, Þ, , cos wt =, , Þ T= 6t, 19. (b), , 1, 2, , or, , 2p, p, t=, T, 3, , p p 2 p - 3p, p, - =, =–, 3 2, 6, 6, , g, f, 1, g, and f B = A =, LA, 2, 2p L B, , fA, 1, g, L, =, ´ 2p B Þ 2 =, f A / 2 2p L A, g, , regardless of mass, 21. (d), , Þ vmax = aw = 5 ´ 2 = 10, , dy2, pö, æ, = - 0.1p sin pt = 0.1p cos ç pt + ÷, è, 2ø, dt, , \ Phase diff. = f1 - f 2 =, , p, 2, , \ a = a12 + a22 = 32 + 42 = 5, , v2 =, , \, , 2 ´ 22 ´ 8 ´ 1000, min = 84.6 min, 49 ´ 60, , a1 = 3, a2 = 4, and f =, , dy1, pö, æ, = 0.1 ´ 100p cos ç100pt + ÷, è, dt, 3ø, , 1, 2p, , 22 8 ´ 10 3, R, 64 ´ 106, ´, = 2p, = 2´, 7 7´ 2, g, 9.8, , 28. (a) x = 3sin 2t + 4cos2t . From given equation, , v1 =, , 20. (c) f A =, , T = 2p, , =, , Þ, , A- a, cos wt =, A, , T =2π, , v=, , 2, , A - 3a 2 A2 + 2a 2 - 4 Aa - A2, =, A, A2, 2, 2, 2, A – 3aA = A + 2a – 4Aa, 2a2 = aA, A = 2a, , 2p t, = p t Þ T = 2 sec ., T, , x, displacement, = 2p / z, = 2p, zx, acceleration, 26. (b) Here, x = 2 × 10–2 cos p t, Speed is given by, 25., , A - 3a, æ A-a ö, = 2ç, ÷ -1, A, è A ø, , Þ, Þ, Þ, , 1, 2 2, 2, mw (a - x ), 2, , 1, 2 2, mw a ., 2, 24. (d) y = 5sin(p t + 4p ), comparing it with standard equation, , 5, l, =, T., g, 6, 6, g+, g, 5, 5, (d) In simple harmonic motion, starting from rest,, At t = 0 , x = A, x = Acoswt, When t = t , x = A – a, When t = 2 t , x = A –3a, From equation (i), A – a = Acosw t, A – 3a = A cos2w t, As cos2w t = 2 cos2 w t – 1, From equation (ii), (iii) and (iv), , Þ, , K.E. =, , When x = 0, K.E is maximum and is equal to, , l, Now, T = 2p, g, T ' = 2p, , 22., , L, LB, Þ 4= B ,, LA, LA, , 29. (d) Slope of F - x curve = – k = -, , 80, Þ k = 400 N/m,, 0.2, , 30., , m, = 0.0314 sec., k, (b) Phase change p in 50 oscillations., Phase change 2p in 100 oscillations., So frequency different ~ 1 in 100., , 31., , (a), , Time period, T = 2p, , 32., , l, l¢, ; 2 = 2p l = 2p, g, g, (g / 6 ), Time period will remain constant if on moon,, l' = l/6 = 1/6 m, (b) Let k be the force constant of spring of length l2. Since, l1 = n l2, where n is an integer, so the spring is made of, (n + 1) equal parts in length each of length l2., T = 2p
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t.me/Magazines4all, , DPP/ CP14, , S-62, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , l1, , (a), , l2, , l3, , 6., , (d) Third overtone has a frequency 7 n, which means, , 7., , 7l, = three full loops + one half loop, which would, 4, make four nodes and four antinodes., (c) Comparing it with y (x, t) = A cos (wt + p/2) cos kx., If kx = p/2, a node occurs ; \ 10 px = p/2 Þ x = 0.05 m, If kx = p, an antinode occurs Þ 10px = p, Þ x = 0.1 m, L=, , 110 cm, , n1 : n2 : n3 = 3 : 2 : 1, nµ, , 1, l, , l1 : l 2 : l 3 =, , 1 1 1, : : = 2:3:6, 3 2 1, , 50p, = 5m / s and, 10p, l = 2p / k = 2p / 10p = 0.2 m, , Also speed of wave w / k =, , l1 + l 2 + l 3 = 110, Þ 2x + 3x + 6x = 110 Þ x = 10, , 2., , \ The two bridges should be set at 2x i.e, 20 cm from, one end and 6x i.e, 60 cm from the other end., (a) Equation of the harmonic progressive wave given by :, y = a sin 2p (bt – cx)., Here u = b, k=, , 8., , (a), , 2p, 1, = 2pc take, = c, l, l, , b, 2, 1, given this is 2 ´, i.e. 2pa = or c =, c, c, pa, (c) y = 0.25 sin (10 px – 2pt), Comparing this equation with the standard wave, equation, y = asin (kx – wt), We get, k = 10p, , l, = 22.7, 4, , equation (1), , l2 + x =, , 3l, = 70.2, 4, , equation (2), , 5λ, equation (3), 4, From equation (1) and (2), , x=, , 9., , 5., , It would travel along negative direction of x-axis, and, on reflection at a rigid support, there occurs a phase, change of p., (c) Velocity of source = 18 km h–1 = 5 m s–1, (i) S moves towards listener (vS), (ii) listener moves towards source (vL), v + vL, n = 280 Hz , Beats = n' – n = 8., v - vS, , ENGINE, , (d), , A, , 0.9 km, C, , H, B I, L, L, , Let after 5 sec engine at point C, t=, , AB BC, +, 330 330, , 0.9 ´ 1000 BC, +, 330, 330, \ BC = 750 m, Distance travelled by engine in 5 sec, = 900 m – 750 m = 150 m, Therefore velocity of engine, , 5=, , 2, ´ 0 .9 = 0 .6, 3, , (b) Amplitude of reflected wave =, , l3 + x, =5, l1 + x, , l 3 = 5 l 1 + 4x = 5 × 22.7 + 4 × 1.05 =117.7 cm, , 2p, = 10p Þ l = 0.2 m, l, And w = 2p or, 2pv = 2p Þ v = 1Hz., The sign inside the bracket is negative, hence the wave, travels in + ve x- direction., , 4., , l 2 - 3 l 1 70 .2 - 68 .1, 2 .1, = 1.05 cm, =, =, 2, 2, 2, , From equation (2) and (3), , Þ, , n' =, , l1 + x =, , l3 + x =, , 1 b, \ Velocity of the wave = ul = b =, c c, dy, = a 2pb cos 2p (bt – cx) = aw cos (wt – kx), dt, Maximum particle velocity = aw= a2pb = 2p ab, , 3., , DPP/CP14, , 150 m, = 30 m/s, 5sec, (a) For fundamental mode,, =, , 10., , f=, , 1 T, 2l m, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP14, Taking logarithm on both sides, we get, , S-63, , 15., , æ Tö, æ 1ö, log f = log ç ÷ + log ç, ÷, è 2l ø, è mø, æTö, æ1 ö 1, = log çè ÷ø + log ç ÷, 2l, 2, èmø, , æ1ö 1, or log f = log ç ÷ + [log T - log m], è 2l ø 2, Differentiating both sides, we get, df 1 dT, =, (as l and m are constants), f, 2 T, dT, df, Þ, = 2´, T, f, Here df = 6, f = 600 Hz, dT 2 ´ 6, \, =, = 0.02, T, 600, 11. (b) Frequency received by listener from the rear source,, v -u, v -u v v-u, n¢ =, ´n =, ´ =, v, v, l, l, Frequency received by listener from the front source,, v+u v v+u, ´ =, v, l, l, No. of beats = n'' – n', v+u v -u, v + u - v + u 2u, =, =, =, l, l, l, l, vù, é, êv + 5 ú, é v + v0 ù, é6ù, n, ', n, n, =, =, 12. (c), ú = nê ú, ê, ê v ú, v, ë5û, ë, û, ú, ê, úû, êë, n' 6 n' - n 6 - 5, = ;, =, ´ 100 = 20%, n 5, n, 5, 13. (c) In a closed organ pipe the fundamental frequency is, u, u=, 4L, , uµ T, 17. (b) In fundamental mode,, l, = l Þ l = 2l, 2, v v, \ f = =, ....... (1), l 2l, 16., , l, , l \4, N, , l/2, , N, , A, Fundamental mode, , 18., , Half length dipped, in water, , In half length dipped in water mode,, l l, =, Þ l = 2l, 2 4, v v, \ f '= =, = f, l 2l, (a) Given wave equation is y(x,t), , (- ax +bt +2, 2, , =e, , 2, , ab xt, , ), , -[(, =e, , ax )2 + ( b t ) 2 + 2 a x . b t ], , -(, =e, , ax + bt ) 2, , æ, ö, b ÷, - çè x +, tø, a, e, , 2, , =, It is a function of type y = f (x + vt), Þ Speed of wave =, 19., , b, a, , (c) Particle velocity, d é, x öù, æ, v = dt ê x 0 sin 2pç nt - l ÷ú, øû, è, ë, , 14. (b), f' f", f', –1, –1, S 30 ms, O, O 30 ms, S, f ¢ is the apparent frequency received by an observer at, the hill. f ¢¢ is the frequency of the reflected sound as, heard by driver., v, f '=, f,, v - 30, v + 30, v + 30, 360, f"=, f'=, f=, ´ 600, v, v - 30, 300, = 720 Hz, , A, l \2, , u=, , f, , (a), , A, , n¢¢ =, , 320 ms -1, = 80 Hz, 4 ´1 m, In a closed organ pipe only odd harmonics are present., So, it can resonate with 80 Hz, 240 Hz, 400 Hz, 560 Hz., , (d) Figure(a) represents a harmonic wave of frequency, 7.0 Hz, figure (b) represents a harmonic wave of, frequency 5.0 Hz. Therefore beat frequency, vs = 7 – 5 = 2.0 Hz., , xö, æ, = 2 pnx 0 cos 2pç nt - ÷, l, ø, è, , \ Maximum particle velocity = 2pnx 0, Wave velocity =, , l, = nl, T, , 2pnx 0 px 0, =, 4n, 2, (d) As number of beats/sec = diff. in frequencies has to be, less than 10, therefore 0 < (n1 – n2) < 10, , Given, 2pnx 0 = 4nl Þ l =, , 20.
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t.me/Ebooks_Encyclopedia27., , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (a) Electric field intensity at the centre of the disc., s, E=, (given), 2 Î0, Electric field along the axis at any distance x from the, centre of the disc, æ, ö, x, ç1 ÷, ç, ÷, x2 + R 2 ø, è, From question, x = R (radius of disc), E' =, , s, 2 Î0, , æ, R, \ E ' = s ç1 2, 2 Î0 çè, R + R2, , =, , 4., 5., , ö, ÷, ÷, ø, , 6., , s æ 2R - R ö, ç, ÷, 2 Î0 çè, 2R ÷ø, , 7., , So sinner =, , 3., , –2Q, a, , Q, , +2Q, , 2, , ® ®, , W = qE . S = (qE) ´ S ´ cos q, , – Q + 2Q = Q, , 8., , b, , 9., , c, , 4pc, (a) Initial force between the two spheres carrying charge, (say q) is, 1 q2, (r is the distance between them), 4pe 0 r 2, Further when an uncharged sphere is kept in touch with, the sphere of charge q, the net charge on both become, F=, , B, , C, , A, , q, , q/2, , q/2, , F3 =, , 1, 4pe 0 æ r ö 2, ç ÷, è2ø, , 2, , 1, 4pe 0 æ r ö 2, ç ÷, è2ø, , \ 10 J = (0.5 C) × E × 2 cos 60°, E = 10 × 2 = 20 NC–1 = 20 Vm–1, (c) The charged sphere is a conductor. Therefore the field, inside is zero and outside it is proportional to 1/r2, q, (d) Since f total = f A + f B + fC =, ,, e0, where q is the total charge., As shown in the figure, flux associated with the curved, surface B is f = fB, Let us assume flux linked with the plane surfaces A and, C be, fA = fC = f', Therefore,, q, = 2f '+ f B = 2f '+ f, e0, , q+0 q, = . Force on the 3rd charge, when placed in, 2, 2, center of the 1st two, r/2, r/2, , æqö, ç ÷, è2ø, , S, , E, 60°, work done in moving through distance S,, , Charge, Surface area, , 4pb2, , æqö, qç ÷, è2ø -, , (d) Unit positive charge at O will be repelled equally by, three charges at the three corners of triangle. By, r, symmetry, resultant E at O would be zero., (c) When a dipole is placed in a uniform electric field, two, equal and opposite forces act on it. Therefore, a torque, acts which rotates the dipole., (d) Force acting on the charged particle due to electric, ®, , -2Q, , and sOuter =, , 1 q2, [2 - 1] = F, 4pe 0 r 2, r, (d) Since electric field E decreases inside water, therefore, r r, flux f = Ε.A also decreases., , field = qE, , 4 ö, æ, çè E - E÷ø ´ 100, 1000, 14, =, =, % ; 70.7%, E, 14, , (a) Surface charge density (s) =, , DPP/CP15, , =, , 4, = E, 14, \ % reduction in the value of electric field, , 2., , t.me/Magazines4all, , Þ f' =, , 10., 11., , ö, 1æ q, - f÷, ç, 2 è e0, ø, , kp, and Ee = 3 ; \ Ea = 2Ee, r, r, (c) Electric lines of force due to a positive charge is, spherically symmetric., All the charges are positive and equal in magnitude., So repulsion takes place. Due to which no lines of, force are present inside the equilateral triangle and, the resulting lines of force obtained as shown:, (b) We have Ea =, , 2kp, 3
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t.me/Magazines4all, , DPP/ CP15, , S-68, , 16., , +q, , +q, , +q, , 17., 18., , 12. (a), , a, Q1, , a, Q3, , Q2, , 19., 20., , Q 2 = -Q3 = Q, , (b) Net flux emmited from a spherical surface of radius a, according to Gauss’s theorem, q, fnet = in, e0, qin, or, (Aa) (4pa2) = e, 0, So, qin = 4pe0 A a3, (a) Since lines of force starts from A and ends at B, so A is, +ve and B is –ve. Lines of forces are more crowded near, A, so A > B., (c) Net force on each of the charge due to the other charges, is zero. However, disturbance in any direction other than, along the line on which the charges lie, will not make the, charges return., (c), (d) For distances far away from centre of dipole, , Force on Q3 due to Q 2 + Force on Q3 due to Q1 = 0., , 13., , E axis = E a =, , 1 æ - Q2 ö, 1 Q1 Q, = 0 Þ Q = 4Q, ç, ÷+, 1, 3, 4p Î0 çè a 2 ÷ø 4p Î0 4 a 2, (b) Charge per cm length of the wire = qC, \ Charge per metre of the wire = 100q C, According to Gauss’s law,, Total electric flux passing through the cylindrical, surface, q, 100q, f = enclosed =, e0, e0, , Eequa = E e =, , L, g, , = -6 ×, , g' = g +, \, , qE ö, æ, çè a =, ÷, mø, , qE, m, , T = 2p, , L, qE, g+, m, , = -3, , ... (i), , 1 p, 4pe0 r 4, , ... (ii), , From equation (i) and (ii) the magnitude of change in, electric field w.r.t. distance is more in case of axis of, dipole as compared to equatorial plane., (b), , q, A, , 2q, P, , –3q, B, , d, l, Let a charge 2q be placed at P, at a distance I from, A where charge q is placed, as shown in figure., The charge 2q will not experience any force, when, force, when force of repulsion on it due to q is, balanced by force of attraction on it due to –3q at B, where AB = d, , or, , 1/ 2, , T æ g ö, =, \, T0 çç g + qE ÷÷, m ø, è, 15. (c) Char ges (q) = 2 × 10 –6 C, Distan ce (d), = 3 cm = 3 × 10 –2 m and electric field (E), = 2 × 105 N/C. Torque (t) = q.d., E =(2 × 10–6) × (3 × 10–2) × (2 × 105), = 12 × 10–3 N–m ., , 1 p, 4pe 0 r 4, , d, 1, d, p ( r -3 ), ( Ee ) =, dr, 4pe 0 dr, , 21., , When the plates are charged, the net acceleration is,, g' = g +a, , 1, p, 4pe 0 r 3, , d, 1, d, 2p ( r -3 ), ( Ea ) =, dr, 4pe0, dr, , +, +, +, +, +, +, +, +, +, +, 1m, +, +, +, +, +, +, + 50 cm, +, +, +, , 14. (c) T0 = 2p, , 1 2p, 4pe0 r 3, , (2q)(q ), 4pe0 l, , 2, , =, , (2q )(-3q), 4pe 0 (l + d )2, , or, , (l + d)2 = 3l2, 2l2 – 2ld – d2 = 0, , \, , l=, , 2d ± 4d 2 + 2d 2 d, 3d, = ±, 4, 2, 2, , l=, , d + 3d, 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP15, , S-69, , 22. (d) Let F be the force between Q and Q. The force between, q and Q should be attractive for net force on Q to be, zero. Let F ¢ be the force between Q and q . The resultant, of F ¢ and F ¢ is R. For equilibrium, q, Q, R, , l, , F¢, , Now, force F¢ =, , 1 (2q) (- q), F, Þ F¢ =, 2, 4pe0, 8, r, , 30., , (a) – eE = mg, , 31., , uur, 9.1 ´ 10 -31 ´ 10, -11, E == -5.6 ´ 10 N / C, 1.6 ´ 10 -19, r r r, (d) Torque, t = p ´ E = pE sin q, , 4 = p × 2 × 105 × sin 30°, q, , F¢, , r r, R+ F =0, , Q, , Qq, Q2, Q, 2 ´ k 2 = -k, = -2 2, 2 Þ q, l, ( 2 l), 23. (b) Nuclear force binds the protons and neutrons in the, nucleus of an atom., 4 3, pr (r - r0 ) g, 3, For equilibrium, electric force must be upwards i.e., charge on the drop is positive., , 24. (c) Net downward force on the drop =, , neE =, , or, p =, , F, 2 F ' = -F, , 4 3, 4pr 3 (r - r0 ) g, pr (r - r0 ) g i.e. n =, 3, 3eE, , 25. (d) Electric flux, f = EA cos q ,, where q = angle between E and normal to the surface., , p 4 ´10-5, =, = 2 ´ 10-3 C = 2mC, 0.02, l, 32. (c) K.E. = Force × distance = qE.y, 33. (d) Charge (q) = 0.2 C; Distance (d) = 2 m; Angle q = 60º and, Work done (W) = 4J., Work done in moving the charge (W), = F.d cos q = qEd cos q, q=, , 1 q2, ×, 4pe 0 d 2, , Þ, , q = 4pe 0 d 2 F = ne, , \, , n=, , 4pe 0 Fd 2, e2, , 1 (4q) (-4q), 29. (b) F = 4pe, r2, 0, when C is touched with A, then charge on A & C each =, 2q after that C is touched with B, charge on, , B=, , 2q + (-4q), = -q, 2, , W, 4, 4, = 20 N/C., =, =, qd cos q 0.2 ´ 2 ´ cos 60º 0.4 ´ 0.5, r r, f = E. A = 4iˆ.(2iˆ + 3 ˆj ) = 8 V-m, , or, E =, 34., 35., , p, 2, Þ f=0, 26. (a) Potential energy of an electric dipole in an electric field, , F=, , = 4 ´ 10-5 Cm, , 2 ´ 10 ´ sin 30°, Dipole moment, p = q × l, , Here q =, , is, U = – p .E i.e. U = –pE cosq, For minimum U, q = 0º, Þ Umin = –pE cos 0 = –pE, 27. (c) Milikan demonstrated the quantisation of charge, experimentally. Charge on electron = –e = –1.6×10–19 C., Addition of charge can occur in integral multiples, of e., 28. (c) Let n be the number of electrons missing., , 4, 5, , (a), (d) The dipole is placed in a non-uniform field, therefore a, force as well as a couple acts on it. The force on the, negative charge is more (F µ E) and is directed along, negative x-axis. Thus the dipole moves along negative, x-axis and rotates in an anticlockwise direction., E 1q, , –q, a, , +q, , E, , E2q, , uur ur, Flux = E . A., ur, uur, E is electric field vector & A is area vector.., uur ur, Here, angle between E & A is 90º., uur ur, So, E . A = 0 ; Flux = 0, 37. (c) The charge on disc A is 10–6 mC. The charge on disc, B is 10 × 10–6 mC. The total charge on both = 11 mC., When touched, this charge will be distributed equally, i.e. 5.5 mC on each disc., 38. (a) According to Gauss’s law total electric flux through a, 1, closed surface is, times the total charge inside that, e0, surface., 36., , (b), , Electric flux, fE =, , q, e0
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t.me/Magazines4all, , S-70, , l, Electric field, E = 2pe r, 0, , Charge on a-particle = 2e, fE =, 39., , 40., 41., , 2e, e0, , -1.75 ´10-7, , =, , (b) It is possible to create or destroy charged particles but, it is not possible to create or destroy net charge. The, charge of an isolated system is conserved., (a) The flux is zero according to Gauss’ Law because it is a, open surface which enclosed a charge q., (d) By Gauss law, we know that, , 44., , 2 ´ 3.14 ´ 8.854 ´10 -12 ´ 4.6 ´ 10 -3, = –6.7 × 105 N C–1, (a) Charge resides on the outer surface of a conducting, hollow sphere of radius R. We consider a spherical, surface of radius r < R., By Gauss theorem, ++ +++, ++, +, +, +, +, +, +, R, +, +, +, +, O, S, +, +, r, +, +, E, +, +, +, +, +, +, +, ++, + + +, , q, f=, Here, Net electric flux, f = f2 – f1, e0, q, Þ q = 3 × 106 × e0., e0, r, r, (d) They will not experience any force if | FG |=| Fe |, , = 9 × 106 – 6 × 106 =, , 42., , Þ G, 43., , m2, -2 2, , (16 ´ 10 ), , =, , q2, q, 1, Þ =, ., 2, 2, 4 pe0 (16 ´ 10 ), m, , (c) Here, l = 2.4 m, r = 4.6 mm = 4.6 × 10–3 m, q = – 4.2 × 10–7 C, q, Linear charge density, l =, l, =, , -4.2 ´ 10-7, = –1.75 ×10–7 C m–1, 2.4, , DPP/ CP15, , rr, , 1, , òsE.ds = e0 ´ charge enclosed or, , 4pe 0 G, , E ´ 4 pr 2 =, , 1, ´0, e0, , ÞE=0, i.e electric field inside a hollow sphere is zero., , 45., , Q in, Î0, Thus, the net flux depends only on the charge enclosed, by the surface. Hence, there will be no effect on the net, flux if the radius of the surface is doubled., , (c) By Gauss’s theorem, f =, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (d) As volume remains constant, therefore,, 4, 4, \ R = n 1 / 3 r., p R 3 = n ´ p r3, 3, 3, nq, New potential = V´= n q =, 4 p eoR, , 4 p e o (n, , = n2/3, , 2., , 3., 4., , (a), , Ca =, , Î0 A, d, , and, , 6., , q, = n 2 / 3 V., 4 p eor, , X, , A, , C3, , C5, , 6mF, , 6mF, , 20mF, , C2, , mv 2, kq 2, kq 2, =, ; mv 2 =, 2, r, (2r), 4r, Kinetic energy of each particle, , 7., , +A, , 6mF, , C2, 6mF, , C3, 6mF, , C4, , 6mF, , Y ÞX, C3, , C4, , 9., , C', , C2, , X, , +, +, –, –, –, –, +, +, , –B, , Q2, , where K is the dielectric, 2KC, , constant., Again, when the dielectric slab is removed slowly its, energy increases to initial potential energy. Thus, work, done is zero., , Y, , C1 C2, Here, C = C, 3, 4, Hence, no charge will flow through 20mF, C1, , +, +, –, –, –, –, +, +, , (The plates of B, having negative charge do not, constitute a capacitor)., (a) The potential energy of a charged capacitor is given, , energy is given by, , 6mF, , 20mF, 6mF, , X, , +, +, –, –, –, –, +, +, , by U =, , Equivalent circuit, C1, , +, +, –, –, –, –, +, +, , Q2, ., 2C, If a dielectric slab is inserted between the plates, the, , C4, , 6mF, , 1, kq 2, mv 2 =, 2, 8r, (b) It consists of two capacitors in parallel, therefore, the, 2 Î0 A, total capacitance is =, d, , =, , 8., , Y, , +q, , O, , r), , A, A, Î0, Î0 K Î A, 2, 2 +, and C c =, = 0 (1 + K), 2d, d, d, Î0 A, or C b =, 2(1 + K ) > C a, d, Î0 A 1 + K, > C a \ C b and C c > C a, or C c =, 2, d, (c) Charges reside only on the outer surface of a conductor, with cavity., (b) In oil, C becomes twice, V becomes half. Therefore,, E = V/d becomes half., , (d), , (c), , –q, 1/ 3, , Î0 A, 2 Î0 A (1 + K ), =, d d, d, +, 2 2K, , Cb =, , DPP/CP16, r, , C1 6mF, , 5., , t.me/Magazines4all, , Y, , 10., C'', , C1 and C2 are in series, also C3 and C4 are in series., Hence, C' = 3 mF, C'' = 3 mF, C' and C'' are in parallel., Hence net capacitance = C' + C'' = 3 + 3 = 6 mF, , 1ö, æ, (a) As x = t ç1 - ÷ , where x is the addition distance of, K, ø, è, plate, to restore the capacity of original value., 1ö, æ, \ 3.5 ´ 10 -5 = 4 ´ 10 -5 ç1 - ÷ ., K, ø, è, Solving, we get, K = 8., (b) At. equipotential surface, the potential is same at any, point i.e., VA = VB as shown in figure. Hence no work is, required to move unit change from one point to another, i.e.,, W, VA - VB =, =0Þ W =0, unit ch arg e
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t.me/Magazines4all, , DPP/ CP16, , S-72, , equipotential, surface, VA, , 16., , (d) In equilibrium, F = q E = ( n e), , VB, , n=, , 11., , 12., , 17., , VAVB, , (b), (i) Electrostatic field is zero inside a charged conductor or, neutral conductor., (ii) Electrostatic field at the surface of a charged conductor, must be normal to the surface at every point., (iii) There is no net charge at any point inside the conductor, and any excess charge must reside at the surface., (iv) Electrostatic potential is constant throughout the, volume of the conductor and has the same value (as, insde) on its surface., (v) Electric field at the surface of a charged conductor is, r s, E = nˆ, e0, (b) In shell, q charge is uniformly distributed over its, surface, it behaves as a conductor., +, , +, , +, R, , +, +, , 18., , q, 4pe 0 R, , E=, , V2 =, , V, V – V2, V1, or k = 1 – 1 = 1, V2, V2, V2, , 19., , (b) Potential difference across the branch de is 6 V. Net, capacitance of de branch is 2.1 µF, So, q = CV, Þ q = 2.1 × 6 µC, Þ q = 12.6 µ C, Potential across 3 µF capacitance is, 12.6, V=, = 4.2 volt, 3, Potential across 2 and 5 combination in parallel is 6 –, 4.2 = 1.8 V, So, q' = (1.8) (5) = 9 µC, , 20., , (c), , V=, , 13., , 14., , V ' 1000r 1000, =, =, = 100, Þ, V, R, 10, \ V ' = 100V, (c) In a round trip, displacement is zero. Hence, work done, is zero., , +Q, , 15., , Q2 Q1, R1, r, , 21., 22., , +q, r, , 23., , R2, , Vr =, , Q2, Q1, +, 4 pe 0 r 4pe 0 R1, , Vr =, , 1 æ Q 2 Q1 ö, +, ç, ÷, 4 pe0 è r, R1 ø, , 1, U = QV = Area of triangle OAB, 2, (b) Charge on a particle, q = 2 e., K.E. = work done = q × V = 2e × 106 V = 2 MeV., (a) Let the side length of square be 'a' then potential at, centre O is, , (d), , eo A, (k1 + k2), t´2, , –q, , –Q, O, , (b) The two capacitors are in parallel so, C=, , Total charge, Total capacity, , C0 V1 + 0, V, = 1, C0 + kC0 1 + k, , 1+ k =, , and inside, , q, 4pe 0 R, Because of this it behaves as an equipotential surface., (c) Volume of big drop = 1000 × volume of each small drop, 4 3, 4, pR = 1000 ´ pr 3 Þ R = 10r, 3, 3, Q V = kq and V ' = kq ´1000, r, R, Total charge on one small droplet is q and on the big, drop is 1000q., , 1, 1, CV 2 = ´ 1 ´ 10-6 ´ (4000) 2 = 8 J., 2, 2, (d) As we know,, , (a), , Q1 = C0V1, Q2 = 0, therefore, , +, +, , V= potential at surface =, , mg d 1.96 ´ 10 -15 ´ 9.8 ´ 0.02, =, =3, eV, 1.6 ´ 10 -19 ´ 800, , Common potential =, , q, , +, , V, = mg, d, , 2Q, , 2q, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP16, V=, , S-73, , k (-Q) k (-q) k (2q ) k (2Q), +, +, +, =0, a, a, a, æ a ö, çè, ÷, 2, 2, 2, 2ø, , (Given), = – Q – q + 2q + 2Q = 0 = Q + q = 0, =Q=–q, 24. (b), , C0 =, C=, , k Î0 A, d, , 30., 31., , outer one. So capacitance that of outer one is 4p Î0 b ., 32., , 4 k Î0 A, C, 4, 3 d, \ C = kÎ A = 3, 0, 0, d, , Ceq.= C' = 2C., 1, U1 2, Now,, =, U2 1, 2, , 25. (b) Work done = Change in energy, C ö 2 1 æ 3C ö 2 3CV 2, 1æ, C, +, ç, ÷ V = çè ÷ø V =, 2è, 2ø, 2 2, 4, 26. (a) Potential at B, VB is maximum, =, , VB > VC > VA, As in the direction of electric field potential decreases., 27. (a) The equivalent circuit diagram as shown in the figure., Q2 2µF C Q2 3µF, , Q1, , 2µF, , B, Q, , 5V, , The equivalent capacitance between A and B is, , 2µF ´ 3µF, 16, + 2µF = µF, 2µF + 3µF, 5, Total charge of the given circuit is, Ceq =, , 16, µF ´ 5V = 16µC, 5, Q1 = (2µF) ´ 5V = 10µC, Q2 = Q – Q1 = 16 µC – 10 µC = 6 µC, Voltage between B and C is, Q=, , \, \, , VBC =, , (a) In Ist case when capacitor C attached with battery, charged with the energy., U1= U (stored energy on capacitor)., In IInd case after disconnect of battery similar capacitor, is attached in parallel with Ist capacitor then, , k Î0 2 2k Î0 A 4 k Î0 A, +, =, 3d, 3d, 3 d, , A, Q, , (ii) charge q remains unchanged, (iii) potential V decreases, (iv) energy E decreases, (b) Electric lines of force are always perpendicular to an, equipotential surface., (c) All the charge given to inner sphere will pass on to the, , Q 2 6µC, =, = 2V, 3µF 3µF, , 28. (d) Electric field, s, Q, E= =, e Ae, e of kerosine oil is more than that of air.., As e increases, E decreases., 29. (c) When a battery across the plates of capacitor is, disconnected and dielectric slab is placed in between, the plates, then, (i) capacity C increases, , q2, C = C', 2C, =, C, q2, C, C', , (Q C' = 2C), , U, 2, 33. (a) Here we have to findout the shape of equipotential, surface, these surface are perpendicular to the field, lines, so there must be electric field which can not be, without charge., So, the collection of charges, whose total sum is not, zero, with regard to great distance can be considered, as a point charge. The equipotentials due to point charge, are spherical in shape as electric potential due to point, charge q is given by, q, V = Ke, r, This suggest that electric potentials due to point charge, is same for all equidistant points. The locus of these, equidistant points which are at same potential, form, spherical surface., The lines of field from point charge are radial. So the, equipotential surface perpendicular to field lines from, a sphere., 34. (c) Equipotential surfaces are normal to the electric field, lines. The following figure shows the equipotential, surfaces along with electric field lines for a system of, two positive charges., U2 =, , 35., , (c) Let plate A plate B be carrying charges Q1 and Q2 respectively. When they are brought closer, they induce, equal and opposite charges on each other i.e. – Q2 on
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP17, , S-75, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (b), , V = IR = (neAvd )r, , l, A, , R1 : R2 : R3 =, , V, r = V lne, d, , \, , Here V = potential difference, l = length of wire, n = no. of electrons per unit volume of conductor., e = no. of electrons, Placing the value of above parameters we get resistivity, , r=, , 5, 8 ´ 10, , 28, , ´ 1.6 ´ 10, , -19, , ´ 2.5 ´ 10, , -4, , ´ 0.1, , 3., , (d) From the curve it is clear that slopes at points A, B, C,, D have following order A > B > C > D., And also resistance at any point equals to slope of the, V-i curve., So order of resistance at three points will be, RA > RB > RC > RD, (d) From the principle of potentiometer, V µ l, V l, Þ = ; where, E L, V = emf of battery, E = emf of standard cell., L = length of potentiometer wire, , El 30E, =, L 100, , V=, , 25 9 1, : : = 125 :15 :1, 1 3 5, (c) In series, Rs = nR, , 5., , In parallel,, , (a) Efficiency is given by h =, , 7., , 5 ´ 15 ´ 14, = 0.875 or 87.5 %, 10 ´ 8 ´ 15, (b) According to the condition of balancing, =, , 55 R, =, Þ R = 220W, 20 80, , 8., 9., , 10., , J = sE Þ J r = E, J is current density, E is electric field, so B = r = resistivity., (d) Kirchhoff's first law is based on conservation of charge, and Kirchhoff's second law is based on conservation, of energy., (a), , (c), , NOTE In this arrangement, the internal resistance, , 12., , (d) Resistance of a conductor, R =, , pr, , \R =, , 13., , (b), , R3 =, , rl32 d, m3, , (n - 1)I g, , R, , l, A, ne t, As the temperature increases, the relaxation time t, decreases because the number of collisions of electrons, per second increases due to increase in thermal energy, of electrons., , m, ld, , 2, 2, rl 2 d R = rl1 d R = rl2 d, , 1, , 2, m2, m1, m, , nIg - I g, , ÞS=, , Ig, , (b) S =, , of the battery E does not play any role as current is not, passing through the battery., 2, . But m = pr2 ld \ pr =, , Ig R, , V2 V2, V2, V2, :, :, :, R R / 4 R /16 R / 2, , 11., , E', , 2, , R, Þ PC = 16P, 16, , PA : PB : PC : PD =, , i, , rl, , rl, A, When wire is cut into 4 pieces and connected in parallel., R=, , Reff. =, , r, , output, input, , 6., , i, , (d) R =, , 1, 1 1, = + + ...n terms, Rp R R, , \ Rs/Rp = n2 /1 = n2, , E, , 4., , l12 l2 2 l32, :, :, m1 m2 m3, , R1 : R2 : R3 =, , = 1.6 × 10–5Wm, 2., , DPP/CP17, , R1 i1, R2 i2, V, , m, , 2
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t.me/Magazines4all, , DPP/ CP17, , S-78, , 33. (a), , In steady state, flow fo current through capacitor will, be zero., Current through the circuit,, E, r + r2, Potential difference through capacitor, i=, , Vc =, , \, 34., , æ E ö, Q, = E - ir = E - ç, r, C, è r + r2 ÷ø, , r, Q = CE 2, r + r2, , or, 10 + (10 × 0.002 × t) = 11 + 330 × 0.002, or, 0.02t = 1 + 0.66 = 1.066 or t =, 39., , 40., , (b) As P = Ι 2 R, so P1 = (1.01 Ι)2 R = 1.02 I 2 R = 1.02 P., It means % increase in power, æP, ö, = ç 1 - 1÷ ´100 = 2%., P, è, ø, (b) Let I1 be the current throug 5 W resistance, I2 through, (6 + 9) W resistance. Then as per question,, I12 ´ 5 = 20 or, I1 = 2A., Potential difference across C and D = 2 × 5 = 10V, Current I 2 = 10 = 2 A., 6+9 3, Heat produced per second in 2 W, , (c) i = neAVd and Vd µ E (Given), or, i µ E, i2 µ E, i2 µ V, Hence graph (c) correctly dipicts the V-I graph for a, wire made of such type of material., , 35., 36., , 37., , 38., , (b) Current, I = (2.9 ´ 1018 + 1.2 ´1018 ) × 1.6 × 10–19, = 0.66A towards right., (a) Copper rod and iron rod are joined in series., l, \ R = RCu + RFe = (r1 + r2), A, lö, æ, çèQ R = r ÷ø, A, From ohm’s law V = RI, = (1.7 × 10–6 × 10–2 + 10–5 × 10–2) ¸, 0.01 × 10–4 volt, = 0.117 volt (Q I = 1A), E, (d) I =, , Internal resistance (r) is, R+r, E, zero, I =, = constant., R, (b) Rt = R0 (1 + at), Initially, R0 (1 + 30a) = 10 W, Finally, R0 (1 + at) = 11 W, \, , 11 1 + at, =, 10 1 + 30a, , 1.66, = 83°C., 0.02, , 2, , 41., , (b), , æ8ö, = I2 R ç ÷ ´ 2 = 14.2cal / s., è 3ø, P R, SS, =, where S = 1 2, Q S, S +S, 1, , rl, 42. (c) R = 1 , now l2 = 2l1, A1, , 2, , A2 = p(r2)2 = p (2r1)2 = 4p r12 = 4A1, , r(2 l1 ) r l1 R, =, =, 4 A1, 2 A1 2, \ Resistance is halved, but specific resistance remains, the same., (d) E = V + Ir, V = 12 – 3 = 9 volt, (c) I = neAVd, \, , 43., 44., , R2 =, , I, = 5 × 10–3 m/sec, neA, (d) Since due to wrong connection of each cell the total, emf reduced to 2e then for wrong connection of three, cells the total emf will reduced to (ne – 6e) whereas, the total or equivalent resistance of cell combination, will be nr., , Vd =, , 45., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ CP18, , S-80, , Here i g = Full scale deflection current, =, , \ S¢ =, , 150, = 15 mA, 10, , 2, , G, ., G +S, , I, , I, , G, , G, , V = voltage to be measured = 150 volts, (such that each division reads 1 volt), Þ R=, , 12., , 150, , - 5 = 9995W, 15 ´ 10 -3, (d) Magnetic field at the centre of the current loop is, , S, , 18., , (d) Current in a small element, dI =, , dq, I, p, , Magnetic field due to the element, , µ 2 pI, B= 0, 4 pR, , dB =, , µ0 2 p q u, µ 2 pqu, , R= 0, 4 pR, 4 pB, Substituting the given values, we get, , m 0 2dI, 4p R, , The component dB cos q, of the field is cancelled by, another opposite component., Therefore,, , or, B =, , 4p ´ 10-7 ´ 2p ´ 2 ´ 10-6 ´ 6.25 ´ 1012, = 1.25 m, 4p ´ 6.28, r, r, (b) Here, E and B are perpendicular to each other and, r, the velocity v does not change; therefore, E, qE = qvB Þ v =, B, Also,, r r, E B sin 90° E, r, E´B, E B sin q, =, = = |v| =v, =, 2, 2, 2, B, B, B, B, , S¢, , R=, , 13., , 14., , (b) The force on the two arms parallel to the field is zero., , dB, , Bnet =, , ò, , dB sin q =, , m0 I, , p, , ò, , sin qd q =, , 2p 2 R 0, I, , 19. (a), , A, , C, , <, <, , <, , O, , F, , B, , B, –F, <, , 15. (d) Magnetic field at a point on the axis of a current carrying, wire is always zero., , 20., , m0I, , p2 R, , I, D, , Net magnetic field on AB is zero because magnetic field, due to both current carrying wires is equal in magnitude, but opposite in direction., (b) According to the figure the magnitude of force on the, segment QM is F3 –F1 and PM is F2., , Y, P, , Q, , i, x=-, , a, 2, , x=, x, , M, , a, 2, a, , 16. (b) Current carrying conductors will attract each other, while, electron beams will repel each other., 17. (c) To keep the main current in the circuit unchanged, the, resistance of the galvanometer should be equal to the, net resistance., æ GS ö, \G = ç, + S¢, è G + S ÷ø, ÞG-, , GS, = S¢, G+S, , Therefore, the magnitude of the force on, segment PQ is, , (F3 – F1 )2 + F22, , 21. (c) B = m 0 ni, , æ nö, B1 = (m0 ) ç ÷ (2 i ) = m 0 ni = B, è 2ø, , Þ B1 = B, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP18, , S-81, , 22. (c) The angular momentum L of the particle is given by, L = mr2w where w = 2pn., w, wq, ; Further i = q × n =, \ Frequency n =, 2p, 2p, wq, ´ pr 2 ;, Magnetic moment, M = iA =, 2p, wqr 2, M, wqr 2, q, \M=, So,, =, =, 2, L 2 mr 2 w 2 m, 23. (c) A current loop in a magnetic field is in equilibrium in, two orientations one is stable and another unstable., r uur ur, Q t = M ´ B = M B sin q, If q = 0° Þ t = 0 (stable), If q = p Þ t = 0 (unstable), , 1 2, mv and v2 is the square of the, 2, magnitude of velocity which does not change., , it is given by, , 29., , (a) I = 50 k; Ig = 20k, where k is the figure of merit of, galvanometer; S = Ig Rg(I – Ig); so 12 =, , For Ist case : B =, , m0 In m0 I ´ p, where 2pr = l, =, 2r, l, , and n = 1, , B' =, , 24. (c), , B=, , m0, , 2 i2, , 4 p (r / 2), , -, , m0, , 2 i1, , 4 p (r / 2), , =, , 27. (a) We know that the magnetic field produced by a current, carrying circular coil of radius r at its centre is, m I, B = 0 ´ 2p, 4p r, Here B A =, , m0 I, ´ 2p, 4p R, , and BB =, , m0 2 I, ´ 2p, 4p 2 R, , 31., , m0 4, (i - i ), 4p r 2 1, , m 4, m, = 0 (5 - 2.5) = 0 ., 4p 5, 2p, r, 25. (a) The direction of B is along ( - kˆ ), \ The magnetic force, ur, r ur, F = Q (v ´ B) = Q (viˆ) ´ B( - kˆ) = QvBjˆ, r, Þ F is along OY., 26. (a) According to Ampere's circuit law, r r, Ñò B.dI = µ0Ienclosed = µ0 (2A - 1A) = µ0, , (c), , l, 2np, , m0 nI m0 nI n 2 m 0 pI, =, =, = n 2B, l, l, 2r ', 2, 2np, , B=, , m 0i a 2, 2(x 2 + a 2 )3/2, , æ (x 2 + a 2 )3/2 ö, m 0i a 2, m 0i =, ç, ÷÷, B' =, 2a(x 2 + a 2 )3/2 çè, a2, 2a, ø, B' =, , B.(x 2 + a 2 )3/2, a3, , Put x = 4 & a = 3 Þ B' =, , 54(53 ), = 250mT, 3´ 3´ 3, , 32. (a) The force acting on a charged particle in magnetic field, is given by, r r, F = q(v ´ B) or F = qvB sin q, , 33., , when angle between v and B is 180°,, F= 0, (b) The force acting on electron will be perpendicular to, the direction of velocity till the electron remains in the, magnetic field. So the electron will follow the path as, given., y, , BA, Þ, =1, BB, , 28. (b) When a charged particle enters a magnetic field at a, direction perpendicular to the direction of motion, the, path of the motion is circular. In circular motion the, direction of velocity changes at every point (the, magnitude remains constant)., Therefore, the tangential momentum will change at, every point. But kinetic energy will remain constant as, , (50 k - 20 k ), , On solving we get Rg = 18 W., 30. (d) Let I be current and l be the length of the wire., , For IInd case : l = n(2pr ') Þ r ' =, , Do not experience a torque in some orientations, Hence option (c) is correct., , 20 k . R g, , e, , 34., , u, , X B, x, , (c) A voltmeter is a high resistance galvanometer and is, connected in parallel to circuit and ammeter is a low, resitance galvanometer so if we connect high resistance, in series with ammeter its resistance will be much high.
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t.me/Magazines4all, , DPP/ CP18, , S-82, , 35., , (d) Here, the wire does not produce any magnetic field at O, because the conductor lies on the line of O. Also, the, loop does not produce magnetic field at O., 36. (d) Magnetic field between the plates in this case is zero., 37. (d) For a given perimeter the area of circle is maximum. So, magnetic moment of (S) is greatest., r, ur r ur, 38. (a) Lorentz force, F = q {E + (v ´ B)}, iˆ ˆj kˆ, r ur, v ´ B = 1 2 0 = 8iˆ - 4jˆ - 7kˆ, , 1, , and F3 and F4 are equal and opposite., d, Hence, the net attraction force will be towards the, conductor., r r, (c) F1 = F2 = 0, , F1 > F2 as F µ, , 41., , because of action and reaction pair, 42., , 5 3 4, , (c) As electron move with constant velocity without, deflection. Hence, force due to magnetic field is equal, and opposite to force due to electric field., , r, ˆ = (10iˆ - 7ˆj - 7k), ˆ, F = 1 (2iˆ - 3jˆ + 8iˆ - 4jˆ - 7k), 39., , (b) Here, Rg = 100 W; Ig = 10–5 A; I =1A; S = ?, S=, , 40., , Ιg R g, I - Ig, , =, , 10, , -5, , ´ 100, , 1 - 10 -5, , = 10 -3 W in parallel, , I1, , (d), I, , (c), , 44., , (b) To measure AC voltage across a resistance a moving, coil galvanometer is used., r, urur, (c) As F = qVBsin q, F is zero for sin 0° or sin 180° and is non-zero for angle, ur, ur, between V and B any value other than zero and 180°., , F4, F2, , E 20, =, = 40 m / s, B 0.5, , 43., , 45., , F1, , F3, , qvB = qE Þ v =, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , DPP/ CP19, , S-83, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (b) As the axes are perpendicular, mid point lies on axial, line of one magnet and on equatorial line of other, magnet., \ B1 =, , t.me/Magazines4all, , m 0 2 M 10-7 ´ 2 ´ 1, =, = 2 ´ 10-7, 4 p d3, 13, , 7., 8., 9., , (a) The temperature above which a ferromagnetic substance, becomes paramagnetic is called Curie’s temperature., (a) Iron is ferromagnetic., (b) t = MB sin q, A, t = iAB sin 90º, t, iB, Also, A = 1/2 (BC) (AD), , \ A=, , m M, and B2 = 0 3 = 10-7, 4p d, 2., , \ Resultant field = B12 + B 22 = 5 ´ 10 -7 T, (c) Initial magnetic moment of each magnet = m × l., As is clear from Fig., S1 and N2 neutralize each other., Effective distance between, N1 and S2 =, , 3., , But, , 1, 1, (BC)(AD) = (l ) l 2, 2, 2, , Þ, , 3 2 t, (l ) =, 4, Bi, , 5., , (b) For a diamagnetic material, the value of µr is less than, one. For any material, the value of Îr is always greater, than 1., (a) The time period of oscillation of a freely suspended, magnet is given by, T = 2p, , \, , 6., , 2, , =, , 3 2, l, 4, , 1.2 ´ 10-3, -6, , =, , 1, = sin 30°, 2, , 11., , 60 ´ 40 ´ 10, Þ q = 30°, (b) Electro magnet should be amenable to magnetisation, and demagnetization., \ retentivity and coercivity should be low., , 12., , (b), , I, MH, , I, I, 1, = 2p, where I = ml 2, M ´B, MB, 12, When the magnet is cut into three pieces the pole, strength will remain the same and, T = 2p, , 2, , 1 æ mö æ lö, I, çè ÷ø çè ÷ø ´ 3 =, 12 3 3, 9, We have, Magnetic moment (M), = Pole strength (m) × l, \ New magnetic moment,, æ lö, M ' = m ´ ç ÷ ´ 3 = ml = M, è 3ø, , M.I. (I¢) =, , H', H, , H', =2, or, H, or, H' = 4H, (b) Diamagnetic materials are repelled in an external, magnetic field., Bar B represents diamagnetic materials., , ælö, -ç ÷, è2ø, , C, , 10. (a) M = 60 Am2, r, t = 1.2 × 10–3 Nm, BH = 40 × 10–6 Wb/m2, r r r, t = M ´ BH Þ t = MBH sin q, Þ 1.2 × 10–3 = 60 × 40 × 10–6 sin q, , I, T, MH, =, Thus,, T', I, 2p, MH ', Given, T = 4 sec, T ' = 2sec ,, , 4, =, 2, , l/2, , 1, , 2p, , So,, , D, , æ t ö2, ÷, l = 2ç, ç 3 B.i ÷, è, ø, , Þ sin q =, , 4., , l/2, , l, , \ M¢ = ml 2 ., (d) As shown in the figure, the magnetic lines of force are, directed from south to north inside a bar magnet., , S, , l, , l, , B, , l 2 + l2 = l 2, , N, , DPP/CP19, , 2, s., 9 3, (b) Graph [A] is for material used for making permanent, magnets (high coercivity), Graph [B] is for making electromagnets and, transformers., \T'=, , 13., , T, , =
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t.me/Magazines4all, , DPP/ CP19, , S-84, , 14., , (a) The earth’s core is hot and molten. Hence, convective, current in earth’s core is responsible for it’s magnetic, field., , 15. (b) T µ, , 1, H, , T1, =, T2, , Þ, , Þ F = 3H or, , H2, 2, Þ =, 1, H1, , H+F, H, , B=, , µ0, M, 3/2 ; M = ml., 4p é, l2 ù, 2, êr + ú, 4û, ë, , After substituting the values and simplifying we get, B = 6 × 1–5 A -m, 24. (c) Initially magnetic moment of system, , H 1, =, F 3, , M 1 = M 2 + M 2 = 2M and moment of inertia, , 16. (a) H = B cos q, V = B sin q, Here B = earth’s magnetic field, q = angle of dip = 90º at north pole, Þ H = B cos 90° = 0, V = B sin 90° = B, Þ V >> H, 17. (d) Initially for circular coil L = 2pr and M = i × pr2, , I1 = I + I = 2I., Finally when one of the magnet is removed then, M2 = M and I2 = I, So, T = 2p, , 2, , iL2, æ L ö, = i ´ pç ÷ =, è 2p ø, 4p, , ...(i), , Finally for square coil side a =, 2, , T1, =, T2, , L, and, 4, , iL, æ Lö, M '=i´ç ÷ =, è4ø, 16, , ...(ii), , 25., , 26., L/4, pM, Solving equation (i) and (ii) M ¢ =, 4, MB, (b) FL = MB (= Torque) Þ L =, F, (a) cd < cp < cf, For diamagnetic substance cd is small and negative, (10–5), For paramagnetic substances cp is small and positive, (10–3 to 10–5), For ferromagnetic substanes c f is very large, (103 to 105), i, , 18., 19., , 20., , 21., , 22., , B, = slope of B-H curve, H, According to the given graph, slope of the graph is, highest at point Q., (d) On increasing the temperature by 700ºC, the magnetic, needle is demagnetised. Therefore, the needle stops, vibrating., , (b), , (b), , t = MB Þ, , 23., , 27., , B = m0 mr H Þ m r µ, , t = MB sin q, , I1 M 2, ´, =, I 2 M1, , Þ T2 =, , 2, , (q = 90°), , B1 t1, =, (since magnetic moment is same), B2 t2, , (a) Magnetic field due to a bar magnet in the broad-side, on position is given by, , 28., , I, M BH, , 25 / 4, 21/ 4, , 2I, M, ´, I, 2M, , = 2sec, , (d) A magnetic needle kept in non uniform magnetic field, experience a force and torque due to unequal forces, acting on poles., 3ù, V 3 é, (d) tan d = =, êëQ tan 37º = 4 úû, H 4, 3, \ V= H, 4, V = 6 × 10–5 T, 4, –5, –5, H = ´ 6 ´ 10 T = 8 ´ 10 T, 3, \ Btotal = V2 + H2 = (36 + 64) ´ 10 –5, = 10 × 10–5 = 10–4T., (b) Ferromagnetic substance has magnetic domains, whereas paramagnetic substances have magnetic, dipoles which get attracted to a magnetic field., Diamagnetic substances do not have magnetic dipole, but in the presence of external magnetic field due to, their orbital motion of electrons these substances are, repelled., (d) PQ6 corresponds to the lowest potential energy among, all the configurations shown., tan q =, , tan q¢, 1, V, V, =, , tan q¢ =, ;, H, H cos x tan q cos x, , 29., , (a), , 30., , (d) In series, same current flows through two tangent, galvanometers., q, (c) Net magnetic dipole moment = 2 Mcos, 2, q, As value of cos is maximum in case (c) hence net, 2, magnetic dipole moment is maximum for option (c)., , 31., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP19, , S-85, , 32. (b) Since magnetic field is in vertical direction and needle, is free to totate in horizontal plane only so magnetic, force cannot rotate the needle in horizontal plane so, needle can stay in any position., 33. (b) Work done in rotating the magnetic dipole from position, q1= 0° to q2= 180°, Q W = MB (cosq1– cosq2), \ W = MB (cosq° – cos 180°) = 2MB, 34. (b) The time period of a bar magnet in a magnetic field is, given by., T = 2p, , (, , \ cm2 = cm1 / 2 = 0.5cm = 0.5c. Q cm1 = c, 1, , 40., , (a) We know that, \, , 41., , I, ;, MB, , (a) We know that, , =, =, µ =, =, , µr, , T1, , \, , = 2p, , B =, , T µ I µ m ; so, T becomes twice as mass becomes, four times, Given, B = 4 × 10–5 T, RE = 6.4 × 106 m, Dipole moment of the earth M = ?, m M, B= 0 3, 4p d, 4 ´10-5 =, , 1+ x, 1 + 5500 = 5501, µr µ0 = (5501) ´ (4p ´ 10 -7 ), 6.9 × 10–3, T, MBH1, , … (i), , BH1 = 24 ´ 10-6 T, Where, The magnetic field produced by, wire, , Here, I = moment of inertia µ m , M = moment of magnet,, B = magnetic field., , 35. (a), , ), , 4p´10-7 ´ M, , (, , 4p´ 6.4 ´ 106, , ), , 3, , M @ 1023 Am2, , 36. (b) From m r = 1 + c m ;, Magnetic suscaptibility, c m = m r - 1, cm = 0.075 - 1 = - 0.925., 37. (d) d1 = 40°, d2 = 30°, d = ?, , T2, , =, , = 2p, , I, MBH 2, , … (ii), , 4p´10-7 2 ´ 8 ´ 1022, ´, 4p, (6.4 ´ 106 )3, , @ 0.6 Gauss, 44. (c) Magnetic field in solenoid B = m0n i, Þ, , m0 C, T, where C is Curie constant, T = temperature, , 39. (b) According to Curie's law, cm =, , 1, \ cma, T, cm1 T2 273 + 333 606, =, =, =, =2, c m2 T1, 273 + 30 303, , (18), , –7, = (2 ´ 10 ) ´, 0.20, = 1.8 ´ 10-6 T, = BH1 + B = 42 × 10–6 T, , Using equations (i) and (ii) , and substituting the values,, we get, T2 = 0.076 s, 42. (a) As BI = m0MIM = m0(I + IM), Here, I = 0, Then m0MI = m0(IM), Þ IM = MI = 105 A, 43. (a) Given M = 8 × 1022 Am2, d = Re = 6.4 × 106m, m0 2M, ., Earth’s magnetic field, B =, 4p d3, , cot d = cot 2 d1 + cot 2 d2 = cot 2 40º + cot 2 30º, , cot d = 1.192 + 3 = 2.1, \ d = 25º i.e. d < 40º., 38. (d) In magnetic dipole, 1, Force µ, r4, In the given question,, Force µ x– n, Hence, n = 4, , BH 2, , Now, , µ0 i, ., 2p r, , B, = ni, m0, (Where n = number of turns per unit length), , Þ, , B, Ni, =, m0, L, , Þ, , 3 ´ 103 =, , 100i, 10 ´ 10-2, , Þ i = 3A, 45., , (a) As length of each part also becomes half, therefore, magnetic moment M = pole strength × length, Þ, , 1, 1, 1, ´ = th i.e. M/4., 2, 2, 4
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t.me/Magazines4all, , DPP/ CP20, , S-86, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (b) Induced emf produced between the centre and a point, on the disc is given by, , DPP/CP20, , Emf induced in side 2 of frame e2 = B2 Vl, moI, B2 =, 2p (x + a/ 2), , 1, wBR 2, 2, Putting the values,, e=, , x, I, , w = 60 rad/s, B=0.05 Wb/m 2, and R = 100 cm = 1m, , 1, x–, , 1, ´ 60 ´ 0.05 ´ (1) 2 = 1.5V, 2, (a) According to Faraday's law of electromagnetic, df, induction, e =, dt, Also, e = iR, df, \ iR =, Þ ò d f = R ò idt, dt, Magnitude of change in flux (df) = R × area under current vs, time graph, , v, , a, 2, , We get e =, , 2., , or,, 3., , (a), , 4., , (d), , 5., , (a), , \, , 1 1, df = 100 ´ ´ ´ 10 = 250 Wb, 2 2, If a wire , l meter in length, moves perpendicular to a, magnetic field of B weber/meter2 with a velocity of v, meter/second, then the e.m.f. induced in the wire is, given by, V = B vl volt., Here, B = 0.30 × 10-4 weber/meter 2,, v = 5.0 meter/second and l = 10 meter., \ B = 0.30 × 10–4 × 5.0 × 10 = 0.0015 volt., The magnetic field is increasing in the downward, direction. Therefore, according to Lenz’s law, the current, I1 will flow in the direction ab and I2 in the direction dc., Self inductance of a solenoid,, m N2 A m0 N2 pr 2, L= 0, =, l, l, 2, L1 æ r1 ö æ l2 ö, [Q N1 = N2], =, L 2 çè r2 ÷ø çè l1 ÷ø, l, 1 r, 1, Here, 1 = , 1 =, l2 2 r2 2, , a, a, x+, 2, , 9., , Emf induced in square frame, e = B1Vl – B2Vl, m0 I, m0 I, lv –, lv, =, 2p (x – a / 2), 2p (x + a/ 2), 1, or, e µ, (2x – a)(2 x + a), (a) When a north pole of a bar magnet moves towards the, coil, the induced current in the coil flows in a direction, such that the coil presents its north pole to the bar, magnet as shown in figure (a). Therefore, the induced, current flows in the coil in the anticlockwise direction., When a north pole of a bar magnet moves away from, the coil, the induced current in the coil flows in a, direction such that the coil presents its south pole to, the bar magnet as shown in figure (b)., , N, , 6., , 10., , 7., 8., , N, , (b), , (d) Given : f = 4t 2 + 2t + 1 wb, \, , df d, = (4t 2 + 2t + 1) = 8t + 2 =| e |, dt dt, , Induced current, I =, , L1 æ 1 ö æ 2 ö 1, =ç ÷ ç ÷=, L2 è 2 ø è 1 ø 2, , | e | 8t + 2 8t + 2, =, =, A, R, 10W, 10, , At t = 1 s,, I=, , (b) l = 1m, w = 5 rad/s, B = 0.2 ´ 10 T, -4, , e=, , (a), , Therefore induced current flows in the coil in the, clockwise direction., , 2, , \, , 2, , Bwl 0.2 ´ 10 -4 ´ 5 ´ 1, =, = 50mV, 2, 2, , (c), (c) Emf induced in side 1 of frame e1 = B1Vl, mo I, B1 =, 2p (x – a/ 2), , 11., , (d), , 8 ´1+ 2, A = 1A, 10, , M=, =, , m0 N1 N2 A, l, , 4p ´ 10-7 ´ 300 ´ 400 ´ 100 ´ 10 -4, 0.2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP20, , S-87, , m0 N1 N2 A, l, = 2.4p ´ 10 -4 H, , M=, , 19., , \ The current induced will be i =, , Df -D ( LI ), DI, =, = -L, 12. (d) e = Dt, Dt, Dt, 4, DI, Þ 8= L ´, \ | e |= L, 0.05, Dt, 8 ´ 0.05, Þ L=, = 0.1H, 4, , 500 ´ 4 ´ 10-3, henry = 1 H., 2, 14. (d) Electric field will be induced, as ABCD moves, in both, AD and BC. The metallic square loop moves in its own, plane with velocity v. A uniform magnetic field is, imposed perpendicular to the plane of the square loop., AD and BC are perpendicular to the velocity as well as, perpendicular to applied field so an emf is induced in, both, this will cause electric fields in both., =, , 20., , 21., , (b), , 22., , e=, , = 0.3 ´ 10-4 ´ 5 ´ 20, 23., , B, , = 3 × 10–3 V = 3 mV., (d) The self inductance of a long solenoid is given by, L = m r m 0 n 2 Al, Self inductance of a long solenoid is independent of, the current flowing through it., w, (d) Here, induced e.m.f., l, 2l, , e=, , C, , ur ur, df, d(NB.A), =15. (d) E.M.F. generated, e = dt, dt, d, = - N (BA cos wt ) = NBAw sin wt, dt, Þ e max = NBAw, 16. (c) L = 2mH, i = t2e–t, , di, = - L[ -t 2e -t + 2te - t ], dt, , E=, when E = 0,, –e–t t2 + 2te–t = 0, or, 2t e–t = e–t t2, Þ t = 2 sec., , df (W2 - W1 ), R tot = (R + 4R)W = 5R W, =, dt, t, - n(W2 - W1 ), ndj, =, i=, ., R tot dt, 5Rt, , (Q W2 & W1 are magnetic flux), , 3l, , = Bw, , 25., 26., 27., 28., 29., , 30., , 31., , x, , ò (wx) Bdx, , 2l, , 17. (b), 18. (b), , df, dB, =nA, dt, dt, , \ e = 10 ´ (10 ´10 -4 )(10 4 ) [108 Gauss/sec=10 4 T/s], = 100 V., Ι = (e/R) = (100/20) = 5amp., (a), W, E, eind = Bvl, , v, , -L, , |e|, 1 df, Þi=, R, R dt, , dq, dq 1 df, 1, BA, Þ, =, Þ ò dq = ò df Þ q =, dt R dt, R, R, dt, –5, (c) Induced emf = vBH l = 1.5 × 5 × 10 × 2, = 15 × 10–5, = 0.15 mV, , 24., , D, , - df, dt, , But i =, , 13. (c) Total number of turns in the solenoid, N = 500, Current, I = 2A., Magnetic flux linked with each turn, = 4 × 10–3 Wb, Nf, As, f = LI or N f = LI Þ L =, 1, , A, , (b) The individual emf produced in the coil e =, , dx, , [(3l) 2 – (2l)2 ] 5Bl 2 w, =, 2, 2, , (b), (b) Induced e.m.f. in the ring opposes the motion of the, magnet., (a), (b), (d) Magnetic flux, fB = BA cos q, Induced emf, e = BA sinq, Here, q = 0º, \ Magnetic flux is maximum and induced emf is zero., 1, 1, (c) e.m.f. induced = B R 2 w = BR 2 (2 p n), 2, 2, 1, 2, = ´ (0.1) ´ (0.1) ´ 2 p ´ 10 = (0.1)2 p volts, 2, NM I, d, dI, (a) E = ( NMI ) Þ E = NM Þ E =, t, dt, dt, emf induced per unit turn =, , E MI, =, N, t
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t.me/Magazines4all, , DPP/ CP20, , S-88, , 32., , (d) According to Lenz's law, when switch is closed, the, flux in the loop increases out of plane of paper, so, induced current will be clockwise., , Ndf df, 33. (a) Since e = –, if, is fast, so e is large., dt, dt, 34., , 35., , (d) The e.m.f. is induced when there is change of flux. As, in this case there is no change of flux, hence no e.m.f., will be induced in the wire., (b) Given, B = 0.01 T, A = pR2 = p × (1 m)2 = pm2, w = 100 rads–1, \ The maximum induced emf emax= BAw, = 0.01×p×100 V = pV, , 36., , (b), , 37., , (c), , 38., , (d), , e=, , - (f 2 - f1 ) -( 0 - NBA ) NBA, =, =, t, t, t, , t=, , NBA 50 ´ 2 ´10 –2 ´10 –2, =, = 0.1 s, e, 0. 1, , Df, = e = iR Þ Df = (iDt )R = QR, Dt, Df, ÞQ=, R, f = BA cos q = 2.0 ´ 0.5 ´ cos 60º, , =, , 2.0 ´ 0.5, = 0.5 weber., 2, , x=, , W, W, Þ V=, Þ W = QV, Q, Q, , 39., , (a), , 40., , (b) Mutual inductance depends on the relative position, and orientation of the two coils., (c), (a) As the magnetic field increases, its flux also increases, into the page and so induced current in bigger loop, will be anticlockwise. i.e., from D to C in bigger loop, and then from B to A in smaller loop., (c) As I increases, f increases, \ Ii is such that it opposes the increases in f., Hence, f decreases (By Right Hand Rule). The induced, current will be counterclockwise., (d) According to Faraday’s law of electromagnetic, induction,, , 41., 42., , 43., , 44., , Induced emf, e =, , 45., , Ldi, dt, , æ 5–2 ö, 50 = L ç, ÷, è 0.1sec ø, 50 ´ 0.1 5, = = 1.67 H, Þ L=, 3, 3, (d) Mutual inductance between two coil in the same plane, with their centers coinciding is given by, , M=, , m 0 æ 2p2 R22 N1 N 2 ö, ÷ henry., 4p çè, R1, ø, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (c) Across resistor, I =, At resonance,, , 2., 3., , 4., , 5., , 100, V, =, = 0.1 A, R 1000, , 1, 1, X L = XC =, =, = 2500, wC 200 ´ 2 ´ 10-6, Voltage across L is, I X L = 0.1 ´ 2500 = 250 V, (a) The phase angle between voltage V and current I is, p/2. Therefore, power factor cos f = cos (p/2) = 0. Hence, the power consumed is zero., (c) The current drawn by inductor and capacitor will be in, opposite phase. Hence net current drawn from generator, = IL – IC = 0.9 – 0.4 = 0.5 amp., (c) Capacitive reactance, X = 1 = 1, wC 2 pfC, 1, Þ Xµ, fC, X' f C, f, C, 1, X, \, = ´ =, ´, =, Þ X' =, X f ' C' 2f 2C 4, 4, (a) The charging of inductance given by,, Rt, æ, - ö, i = i0 ç1 - e L ÷, çè, ÷ø, Rt, , Rt, , i0, 1, = i0 (1 - e L ) Þ e L =, 2, 2, Taking log on both the sides,, , -, , 8., , Also, , L, From the rating of the bulb, the resistance of the bulb, can be calculated., 2, Vrms, = 100W, P, For the bulb to be operated at its rated value the rms, current through it should be 1A, , R=, , Vmax, 2 æ, R +, , ö, è w Cø, , i rms, 2, , 1, , .......(i), , 2, , Vrms, , =, , 2, , Vmax, , =, , 9, 2, æ 1 ö, R +, R +, 2 2, w C, çw ÷, çè C ÷ø, 3, From equation (i) and (ii) we get, 2, , .....(ii), , 1, , 3R, , 9., 10., , 2, , 5, , =, , 2 2, , Þ, , wC, , =, , 3, 5, , Þ, , XC, R, , =, , 3, 5, , R, w C, (c), (c) Growth in current in LR2 branch when switch is closed, is given by, E, i=, [1 - e - R2t / L ], R2, , di, E R2 - R2t / L E =, = e, . .e, dt R2 L, L, Hence, potential drop across L, Þ, , R2t, L, , æ E -R t / L ö, -R t / L, = ç e 2 ÷ L = Ee 2, èL, ø, , = 12e, , 200V,50Hz, , ~, , 100 + (2 p50.L) 2, , 3, H, p, (a) At angular frequency w, the current in RC circuit is given, by, imax =, , Rt, = log1 - log 2, L, , 1A, , 200, 2, , L=, , wL X L, =, (c) tan f =, R, R, Given f = 45°. Hence XL = R., , (d), , Vrms, Z, , 1=, , \, , 300 ´ 10 -3, L, log 2 =, ´ 0.69, R, 2, Þ t = 0.1 sec., , 7., , DPP/CP21, , Also, Irms =, , Þ t=, , 6., , t.me/Magazines4all, , 11., , (a) I = I o, , -, , 2t, 400´10 -3, , = 12e–5tV, , R, æ, - tö, ç1 - e L ÷, , çè, , ÷ø, (When current is in growth in LR circuit), , 5, R, æ, - ´2 ö, - tö, 5, Eæ, 10, L, ç, ÷, = ç1 - e, ÷ = 1- e, ÷ø, R çè, ÷ø 5 çè, = (1 – e–1), 12. (d) Power, P = Ι r.m.s ´ Vr.m.s ´ cos f, In the given problem, the phase difference between, voltage and current is p/2. Hence, , P = Ι r.m.s ´ Vr.m.s ´ cos( p / 2) = 0.
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t.me/Magazines4all, , DPP/ CP21, , S-90, , 13. (c) When the capacitor is completely charged, the total, energy in the LC circuit is with the capacitor and that, , 4V0, V, t, 20. (d) V = 0 t Þ V =, T, T/4, , 1 Q2, 2 C, When half energy is with the capacitor in the form of, electric field between the plates of the capacitor we get, , energy is E =, , E 1 Q '2, =, where Q ' is the charge on one plate of the, 2 2 C, capacitor, , Þ Vrms = < V2 > =, , (a) Energy stored in magnetic field =, , f=, 1 2, Li, 2, , Also q = q0 cos wt and w =, On solving t =, , p, LC, 4, , 1 æ, 1 ö, çè wL =, ÷ the circuit will have resistance, w, Cø, LC, nature. Hence (b) is false., Power factor, 1, R, cos f =, = 1 if wL =, wC, 2, 1 ö, æ, R 2 + ç wL ÷, è, wC ø, V, (T / 2)V0 2 + 0, (b) Vrms =, = 0 ., T, 2, (d) Option (d) is false because the reason why the voltage, 1, > Lw and if the, leads the current is because, Cw, voltage lags, the inductive reactance is greater than, the capacitive reactance., , 18. (b) P =, Þ, , =, , 23., , 1, , If w =, , 17., , 1, 2, , 4p LC, 1, 4p 2 f 2 L, , Þ wL =, , LC, , 1 ö, æ, çè wL >, ÷, LC ø, LC, Hence (a) is false. Also if circuit has inductive nature, the current will lag behind voltage. Hence (d) is also, false., , 16., , 2p LC, , ÞC=, , 1, , 15. (c) The circuit will have inductive nature if, , w>, , 3, , =, , 1, 4 ´ 10 ´ 10 -2 ´ 1012, , =, , 10 -12, = 2.5 pF, 0.4, , 22. (d) Current is maximum when XL = XC, , 1 2 1q, Li =, 2, 2 C, , \, , V0, , 1, , f2 =, , 1 q2, Energy stored in electric field =, 2 C, 2, , =, , 21. (a) L = 10 mHz = 10–2 Hz, f = 1MHz = 106 Hz, , Q, 1 1 Q 2 1 Q '2, Þ Q' =, \ ´, =, 2, 2 2 C, 2 C, 14., , 4V0, T, , 1/ 2, , ìT/4 2 ü, ï ò t dt ï, 4V0 ï 0, ï, 2, <t > =, í, ý, T ï T /4 ï, dt, ï ò, ï, î 0, þ, , 1, V0i 0 cos f Þ P = Ppeak .cos f, 2, , p, 1, 1, (Ppeak ) = Ppeak cos f Þ cos f = Þ f =, 2, 2, 3, , E I, 110 ´ 9, = 0.9 ´ 100% = 90%, 19. (c) h = s s \ h =, Ep I p, 220 ´ 5, , 1, Þ w=, wC, , 1, LC, , 1, , =, , 0.5 ´ 8 ´ 10-6, , 1, , = 500 rad/s., 2 ´10-3, (a) If w = 50 × 2p then wL = 20W, If w¢ = 100 × 2p then w¢L = 40W, Current flowing in the coil is, I=, , 200, 200, 200, =, =, 2, 2, Z, R + (w¢L), (30)2 + (40) 2, , I = 4A., 24. (a) At resonance impedance is minimum (\ XL = XC ), current is maximum, because VLand VC are equal in, magnitude, \ VLC = VL – VC = 0, Hence, voltmeter V2 read 0 volt., 25. (b), 26., , (d) As E p Ι p = Pi, , \ Ιp =, , Pi, 4000, =, = 40 A., Ep, 100, , 27. (c) The phase angle is given by, tan f =, , w L 2 p ´ 50 ´ 0.21, =, = 5. 5, R, 12, , f = tan -1 5.5 = 80º, , 28., , (c), , 29. (a) Since Vs = N s, Vp N p, Where, Ns = No. of turns across primary coil = 50, Np = No. of turns across secondary coil, = 1500, and V p =, Þ Vs =, , df d, = (f 0 + 4t ) = 4, dt dt, , 1500, ´ 4 = 120 V, 50, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP21, 30. (a), , S-91, , æn, Es, n, = s or E s = E p ´ ç s, ç np, Ep np, è, , ö, ÷, ÷, ø, , æ 200 ö, \ E s = 120 ´ ç, ÷ = 240 V, è 100 ø, , Ιp, Ιs, , =, , æ np ö, æ 100 ö, ns, = 5 amp, or Ι s = Ι p ç ÷ \ Is = 10 ç, è 200 ÷ø, np, è ns ø, dI, ® (- ve) ® e = ( + ve), dt, , 31. (b) R Þ I ¯ Þ, , 32., , dI ù, é, ê As e = - L dt ú, ë, û, Supporting ® I net , (b) We know that, i = i0 (1 - e-t / t ), , or, , 3, i0 = i0 (1 - e -4 / t ), 4, , or, , e -4 / t =, 4/ t, , 38. (b) V= V0 sin w t, Voltage in r.m.s. value, , V0 = 2 ´ 234 V = 331 volt, and w t = 2 p n t = 2 p ´ 50 ´ t = 100 p t, Thus, the equation of line voltage is given by, V = 331 sin (100 p t), 39. (a) The resistance in the middle plays no part in the charging, process of C, as it does not alter either the potential, difference across the RC combination or the current, through it., 40. (a) Here, C = 100 mF = 100 × 10–6 F, R = 40 W,, Vrms = 110 V, f = 60 Hz, Peak voltage,, , V0 = 2 . Vrms = 100 2 = 155.54 V, Circuit impedance,, , 1, 4, , =4, , or, , e, , \, , 4, = ln 4, t, , or, , t=, , 2, s, ln 2, , 34. (d), , 35. (c), 36. (b), 37. (d), , =, , 40 2 +, , 1, 2 2, , w C, , 1, (2 ´ p ´ 60 ´ 100 ´ 10 -6 ) 2, , V0 155.54, =, = 3.24 A, Z, 48, (a) Laminated core provide less area of cross-section for, the current to flow. Because of this, resistance of the, core increases and current decreases thereby, decreasing the eddy current losses., (b) V = 200V; r = 10W, R¢ = 10 + 100W = 110W, I0 =, , 1, . The circuit behaves as if it, wC, contains R only. So, phase difference = 0, At resonance, impedance is minimum Zmin = R and, current is maximum, given by, , E, , R2 +, , = 1600 + 703.60 = 2303.60 = 48 W, hence, maximum current in coil,, , 41., , 33. (d) At resonance, wL =, , E, Z min R, It is interesting to note that before resonance the current, leads the applied emf, at resonance it is in phase, and, after resonance it lags behind the emf. LCR series circuit, is also called as acceptor circuit and parallel LCR circuit, is called rejector circuit., Condition for which the current is maximum in a series, LCR circuit is,, 1, w=, LC, 1, 1000 =, L(10 ´ 10-6 ), Þ = L = 100 mH, When a circuit is broken, the induced e.m.f. is largest., So the answer is (c)., V= 50 × 2 sin 100 p cos 100 pt = 50 sin 200 pt, Þ V0 = 50Volts and n = 100 Hz, di, V = -L, dt, T, di, Here, time and, is + ve for, 2, dt, I max =, , Z=, , 42., , =, , di, T, is – ve for next, time, dt, 2, , V 220, =, = 2A, R¢ 100, P = I2R = 4 × 100 = 400 W, (b) For step-down transformer,, I=, , 43., , 44., , VP IS, =, Q VP > VS \ IS > IP, VS IP, (b) At t = 0 , no current will flow through L and R1, V, \ Current through battery = R, 2, At t = ¥ ,, RR, effective resistance, Reff = 1 2, R1 + R2, V, \ Current through battery = R, eff, V ( R1 + R2 ), R1R2, (d) These three inductors are connected in parallel. The, equivalent inductance Lp is given by, , =, , 45., , 1, 1, 1, 1 1 1 1 3, =, +, +, = + + = =1, L p L1 L2 L3 3 3 3 3, , \ Lp = 1
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t.me/Magazines4all, , DPP/ CP22, , S-92, , DAILY PRACTICE, PROBLEMS, 1., , 2., , PHYSICS, SOLUTIONS, , (b) Q The E.M. wave are transverse in nature i.e.,, r r, k ´E r, =H, …(i), =, mw, r, r, where H = B, m, r r, r, k ´H, and, … (ii), = -E, we, r, r, r, r, k is ^ H and k is also ^ to E, r r r, r, r, or In other words X || E and k || E ´ B, (a) Erms = 720, The average total energy density, 1, 1, 2, = Î0 E 20 = Î0 [ 2E rms ]2 =Î0 E rms, 2, 2, , 7., , 3., , dV I D, 10 -3, =, =, = 500 Vs–1, dt, C, 2 ´ 10-6, Therefore, applying a varying potential difference of, 500 V s–1 would produce a displacement current of, desired value., 1, (c) E0 = CB0 and C =, , (b) For electromagnetic waves we know that,, E, =c, B, , 9 ´10-4, = 3 ´108 ms -1, B, B = 3 × 10–12 T., , \, , 8., , (b) Here, k =, , 2p, , w = 2pu, l, , 1 1, k 2p / l, =, =, =, (Q c = u l), w 2pu pu c, where c is the speed of electromagnetic wave in, vacuum. It is a constant whose value is 3 × 106 m s–1, (a) E.M. wave always propagates in a direction, perpendicular to both electric and magnetic fields. So,, electric and magnetic fields should be along + X– and, + Y–directions respectively. Therefore, option (a) is, the correct option., 1, e 0 E 02 is electric energy density.., (d), 2, , \, , 9., , = 8.85 ´ 10-12 ´ (720)2, = 4.58 ´ 10-6 J / m3, (d) Id = 1 mA = 10–3 A, C = 2mF = 2 × 10–6 F, dV, d, ID = IC =, (CV) = C, dt, dt, , DPP/CP22, , 10., , B2, is magnetic energy density.., 2m 0, , Therefore,, , 4., , So, total energy =, 11., , m0 e 0, , 1, e 0 E0 2 = m E, 2, 1 Bo 2, = mB, Magnetic energy density =, 2 m0, Thus, mE = mB, Energy is equally divided between electric and magnetic, field, (c) In an electromagnetic wave electric field and magnetic, field are perpendicular to the direction of propagation, of wave. The vector equation for the electric field is, r, 2p ö, æ, E = E0 cos ç wt y zˆ, è, l ÷ø, (a) Frequency remains constant during refraction, 1, c, vmed =, =, µ0 Î0 ´4 2, , 5., , 6., , \, , l med vmed c / 2 1, =, =, =, l air, vair, c, 2, wavelength is halved and frequency remains, unchanged., , E, c, For perfectly reflecting surface with normal incidence, , (c) Incident momentum, p =, , Dp = 2p =, , Electric energy density =, , 1, B2, e 0 E 20 + 0, 2, 2m 0, , F=, , 2E, c, , Dp 2E, =, Dt, ct, , F 2E, =, A ctA, (a) E x and By would generate a plane EM wave travelling, r, r r, in z-direction, E , B and k from a right handed system, r, k is along z-axis. As ˆi ´ ˆj = kˆ, Þ E ˆi ´ B ˆj = Ckˆ i.e., E is along x-axis and B is along, P=, , 12., , x, , 13., , y, , y-axis., (b) From question,, B0 = 20 nT = 20 × 10–9T, (Q velocity of light in vacuum C = 3 × 108 ms–1), r, r, r, E0 = B0 ´ C, r, r, r, | E 0 |=| B | . | C |= 20 ´ 10 -9 ´ 3 ´ 108, = 6 V/m., , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP22, , S-93, , 14. (c) Microwave oven acts on the principle of giving, vibrational energy to water molecules., 15. (a) Displacement current is set up in a region where the, electric field is changing with time., 16. (a) On comparing the given equation to, r, E = a0iˆ cos (wt – kz), , 24., , (b) The average energy stored in the electric field, 1, e0 E 2, 2, The average energy stored in the magnetic field = UB =, UE =, , 1 B2, ,, 2 m0, , w = 6 × 108z,, 2p w, =, k=, r, c, , According to conservation of energy UE = UB, , w 6 ´ 108, =, = 2 m -1, k=, c 3 ´ 108, 17. (b) Displacement current, ID = conduction current, IC, , \, 18., , dq d, =, [q0 cos2put] = – q0 2pu sin2put, dt dt, , E, (a) Momentum of light falling on reflecting surface p =, C, As surface is perfectly reflecting so momentum reflect, E, E, E, p1 = –, =P, C, C, So, momentum transferred C, , E æ E ö 2E, = P – P1 = – ç – ÷ =, C è Cø, C, 19. (d) (1) Infrared rays are used to treat muscular strain, because these are heat rays., (2) Radio waves are used for broadcasting because, these waves have very long wavelength ranging, from few centimeters to few hundred kilometers, (3) X-rays are used to detect fracture of bones, because they have high penetrating power but, they can't penetrate through denser medium like, dones., (4) Ultraviolet rays are absorbed by ozone of the, atmosphere., r, ˆ, 20. (d) Given : B = 1.2 ´ 10-8 kT, r, E=?, From formula,, , E = Bc = (1.2 ´ 10-8 T)(3 ´ 108 ms -1 ) = 3.6 Vm -1, r, B is along Z-direction and the wave propagates along, r, X-direction. Therefore E should be along Y-direction., r, Thus, E = 3.6ˆj Vm-1, 21. (d), 22. (a), , E=, , hc, hc, Þ l=, l, E, , Þ l=, , 6.6 ´ 10-34 ´ 3 ´108, , = 12.4 Å, 11´1000 ´ 1.6 ´10 -19, Increasing order of frequency, , 23., , x-rays u-v rays visible Infrared, (a) Velocity of light, , C=, , E, E, 9.3, ÞB= =, = 3.1 ´ 10-8 T, B, C 3 ´ 108, , e 0m 0 =, , B2, E2, , 28., , B, 1, = e0m0 =, E, c, (b) EM waves carry momentum and hence can exert, pressure on surfaces. They also transfer energy to the, surface so p ¹ 0 and E ¹ 0., (b), (a) The decreasing order of the wavelengths is as given, below :, microwave, infrared, ultraviolet, gamma rays., (b) Infrared causes heating effect., , 29., , (c) Speed of EM waves in vacuum =, , 30., 31., , (b), (a), , 32., 33., 34., 35., 36., 37., , (d) n g- rays > n visible radiation > ninf rared > nRadio waves, (b), (d) Wave is uv rays., (b), (c) Electromagnetic waves are the combination of mutually, perpendicular electric and magnetic fields., (b) Audible waves are not electromagnetic wave., , 38., , (b) Wave impedance = Z =, , 39., 40., , (b), (b) Depends on the magnitude of frequency, , 25., 26., 27., , 41., 42., 43., 44., 45., , 1, m 0 Î0, , = constant, , Ι D = e o d fE / dt ., , m0, = 376.6 W, e0, , E 0 9 ´ 10 3, =, = 3 ´ 10 -5 T., 8, c, 3 ´ 10, (d) To generate electromagnetic waves we need, accelerating charge particle., (d), (d) For an E.M. wave power is transmitted in a direction, perpendicular to both the fields., , (d), , B0 =, , (c) Speed of light of vacuum c =, medium v =, \, , c, =, v, , 1, m 0 e0, , and in another, , 1, me, , me, = mr K Þ v =, m 0 e0, , c, mr K, , .
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP23, , S-95, , or i = tan -1 (m) i.e., i = tan -1 (1.62), , 1 -1 1, -2 + 1, =, +, v1 10 20 = 20, v1 = – 20 cm, For when u2 = – 30 cm, 1, 1, 1, =, f v2 30, , 9., , 13., , 1 -1 1, - 30 + 10 -20, =, +, =, v2 10 30 = 300, 300, v2 = – 15 cm, L = v2 – v1 = – 15 – (– 20), L = 5 cm, (a) Magnification, , 14., 15., , Angle subtended by, f0, final image on the eye, =, =, f e Angle subtended by, the object on eye (or objective), , (b) f0 = 100 cm, fe = 5 cm, When final image is formed at least distance of distinct, vision (d), then, f æ f ö 100 æ, 5ö, M = 0 ç1 + e ÷ =, ç 1 + ÷ [Q D = 25 cm], fe è, dø, 5 è 25 ø, 6, M = 20 ´ = 24, 5, (b) Secondary rainbow is formed by rays undergoing, internal reflection twice inside the drop., h, (b) tan 45° =, Þ h = 60 m, 60, Tower, h, , 0.3m, b, 30 cm, b, =, =, Þ, 3cm 0.5°, 3cm, 0.5°, Þ b = 5°, 10. (b) Due to difference in refractive indices images obtained, will be two. Two media will form images at two different, points due to difference in focal lengths., 11. (c) For reading purposes :, u = – 25 cm, v = – 50 cm, f = ?, , 45°, 45° 60 m, , Þ, , 1 1 1, 1, 1, 1, = - =+, =, ;, f, v u, 50 25 50, , 16., , (c) Using,, , m 1, v u, , =, , m -1, R, , or, , 2 1, v ¥, , =, , 2 -1, R, , \, 17., , 100, = +2 D, f, For distant vision, f' = distance of far point = –3 m, P=, , P=, , Image, , (a), , f = 20 cm, We have,, , æ 1, 1, 1 ö, = ( a nl - 1) ç, ÷, f, è R1 R 2 ø, æ 1, 1, 1 ö, = (1.6 - 1) ç, 20, è R1 R 2 ÷ø, , 1, 1, = - D = -0.33 D, f¢, 3, , 12. (a) Clearly,, i + r + 90° = 180°, Þ, , … (i), , i + r = 90°, , A, i i, B, , Now,, Þ, , sin i, =m, sin r, , sin i, = m Þ tan i = m, cos i, , ..... (1), , æ 1, 1, 1 ö, = ( w n l - 1) ç, ÷, f', è R1 R 2 ø, , æ n, öæ 1, 1 ö, = ç a l - 1÷ç, ÷, è a nw, ø è R1 R 2 ø, , r, , sin i, = m, from (1), sin (90° - i ), , or, , Also,, , C, , m = 1.62, , v = 2R, , anl = 1.6, anw = 1.33, , 1 æ 1.6, 1 ö, öæ 1, =ç, - 1÷ ç, f ' è 1.33 ø è R1 R 2 ÷ø, , D, , ..... (2), , Dividing equation (1) by (2), Þ, , f', 0.6, =, 20 (1.2 - 1), , 0.6 ´ 20, = 60 cm., 0.2, Hence it's focal length is three times longer than in air., f'=
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t.me/Magazines4all, , DPP/ CP23, , S-96, , 18., , 19., , (a), , m=, , v0 æ, dö, 1 + ÷ = 20 æç1 + 20 ö÷, ç, | u 0 | è fe ø, 5 è 10 ø, , æ 10 + 20 ö 4 ´ 30, = 4ç, = 12, ÷=, 10, è 10 ø, (a) Given, i = 60°, A=d=e, d = i + e – A Þ d = i (Q e = A), , 23., , (d), , 24., , (d), , 25., , (a), , 26., , (b), , æ A + dm ö, sin ç, è 2 ÷ø, m=, A, sin, 2, , Here angle of deviation is min. (Q i = e), , 20., , æ 60° + 60° ö, sin ç, ÷ø, è, 2, m=, = 1.73, 60°, sin, 2, (b) u = –50 cm = –0.5 m, , v = –30 cm = –0.3 m, , 21., , 1 1 1 -1 1, -0.2, +, P= = - =, =, = -1.33 D ., f v u 0.3 0.5, 0.15, (b) Object distance u = – 40 cm, , x æ m2 + m1 ö, x (m1 + m 2 ), ç, ÷ =, 2 è m1m 2 ø, 2m1m2, As r1 < i1 i.e., the incident ray bends towards the, normal Þ medium 2 is denser than medium 1., Or r2 < i1 Þ medium 3 is denser than medium 1., Also, r2 > r1 Þ medium 2 is denser than medium 3., Here, vA = 1.8 × 108 m s–1, vB = 2.4 × 108 m s–1, Light travels slower in denser medium. Hence medium, A is a denser medium and medium B is a rarer, medium. Here, Light travels from medium A to, medium B. Let C be the critical angle between them., 1, \ sinC = AmB = B, mA, Refractive index of medium B w.r.t. to medium A is, v, Velocity of light in medium A, A, mB =, = A, Velocity of light in medium B, vB, vA, 1.8 ´ 108 3, = or C = sin–1 æç 3 ö÷, \ sin C =, =, vB, 2.4 ´ 108 4, è 4ø, For a thin prism, D = (m – 1) A, Since lb < lr Þ mr < mb Þ D1 < D2, Difference between apparent and real depth of a pond, is due to the refraction of light, not due to the total, internal reflection. Other three phenomena are due to, the total internal reflection., , =, , Focal length f = – 20 cm, According to mirror formula, , 27. (b) Using the lens formula, , 1 1 1, 1 1 1, + = or = u v f, v f u, 1, 1, 1, 1, 1, =, +, or +, v -20 ( -40 ) -20 40, 1 -2 + 1, 1, =, =or v = -40 cm., v, 40, 40, Negative sign shows that image is infront of concave, mirror. The image is real., , ( -40 ) = -1, -v, Magnification, m =, =u, ( -40 ), The image is of the same size and inverted., 22., , (a), , Oil, , m1, , Water, , m2, , Given v = d, for equal size image v = u = d, By sign convention u = –d, , \, 28., 29., , or f =, , d, 2, , (a) Due to covering the reflection from lower part is not, there so it makes the image less bright., (b) From the fig., Angle of deviation,, , 3, and e = A, 4, 3, 3, A, A+ A– A =, 4, 4, 2, For equilateral prism, A = 60°, , \d=, , Real depth, Apparent depth, \ Apparent depth of the vessel when viewed from, above is, x, x, 1 ö, xæ 1, +, = ç +, ÷, 2m1 2m 2, 2 è m1 m 2 ø, , 1 1 1, = +, f d d, , d = i+e- A, Here, e = i, , As refractive index, m =, , dapparent =, , 1 1 1, = f v u, , 60°, = 30°, 2, (a) Power of lens, P (in dioptre), , \ d=, , 30., , A, i, , d, , e, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , DPP/ CP24, , S-99, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (b) For path difference l, phase, , \, , 2p, 2p, æ, ö, x = .l = 2p ÷, difference = 2p ç Q =, l, l, è, ø, Þ I = I0 + I0 + 2I0 cos 2p, Þ I = 4I0, For x =, , 3., , (c), , (b), \, \, , 4., , (a), , 5., , (b), , I I, If I1 = I2 = I0 then I ' = 2I0 = 2. =, 4 2, Here Angle of incidence, i = 57, tan 57° = 1.54, uglass = tan i, It means, Here Brewster’s law is followed and the, reflected ray is completely polarised., Now, when reflected ray is analysed through a polaroid, then intensity of light is given by malus law., i.e. I = I0 cos2q, on rotating polaroid ‘q’ changes. Due to which intensity, first decreases and then increases., Let nth fringe of 2500 Å coincide with (n – 2)th fringe, of 3500Å., 3500 (n – 2) = 2500 × n, 1000 n = 7000, n = 7, 7th order fringe of 1st source will coincide with 5th, order fringe of 2nd source., When incident wavefronts passes through a prism,, then lower portion of wavefront (B) is delayed, resulting in a tilt. So, time taken by light to reach A’, from A is equal to the time taken to reach B’ from B., aµ = tan q where q = polarising angle., g, P, P, or, aµg = tan 60°, , or,, , c, = 3, vg, , or, vg =, 6., , c, 3, , 3 ´ 108, 3, , = 3 ´ 108 ms -1, , 8., , and Dx2 =, 9., , =, 10., , 11., , 2lD 2 ´ 600 ´ 10 - 6 ´ 2, =, m = 2.4 mm, d, 1 ´ 10 - 3, , (d) Conditions for diffraction minima are, Path diff. Dx = nl and Phase diff. df = 2np, Path diff. = nl = 2 l, Phase diff. = 2np = 4p (Q n = 2), (c) When the wavelength of light used is comparable, with the separation between two points, the image of, the object will be a f diffraction pattern whose size, will be, 1.22l, D, where l = wavelength of light used, D = diameter of the objective, Two objects whose images are closer than this, distance, will not be resolved., (d) The waves reflected from the top layer of oil interfere, with the wave train reflected from the lower surface of, thin oil film producing light and dark coloured pattern., , q=, , 13., , (d) Phase difference, f =, , l, = 10 -3 (given), d, \ No. of fringes within 0.12° will be, , f=, , I = I0cos2, 14., , (b), , 2p, × Path difference, l, , 2p l p, ´ = = 60°, l 6 3, , As, I = Imax cos 2, , 0.12 ´ 2p, , (c) Here A 2 = a12 + a 22 + 2a1a 2 cos d, Q a1 = a2 = a, , 2, æ4 ö, (m2 - 1) ´ 2t = ç - 1÷ ´ 2t = t ., 3, 3, è, ø, Dx1 , so shift will be along –ve y-axis., , As Dx2 >, (d) Given: D = 2m; d = 1 mm = 1 × 10– 3 m, l = 600 nm = 600 × 10– 6 m, Width of central bright fringe (= 2b), , = [2.09], 360 ´10-3, \ The number of bright spots will be two., 7., , d, 2, , d, ., 2, Dx1 = (m1 - 1)t = (1.5 - 1)t = 0.5t, , (b), , (c) Angular width =, , n=, , \ I µ A 2 µ cos 2, , \ I µ cos 2, , 12., , =, , d, 2, , 2, Now, I µ A, , l, p, , phase difference =, 4, 2, p, 2, , DPP/CP24, , d ö, æ, A 2 = 2a 2 (1 + cos d) = 2a 2 ç1 + 2cos 2 - 1÷, è, 2 ø, , Þ A 2 µ cos 2, , (\ cos 2p = 1), , \ I' = I1 + I 2 + 2 I1 I 2 cos, , 2., , t.me/Magazines4all, , f, 2, , 2, I 3, æ 3ö, 60°, 3, = I0 ´ ç ÷ = I0 I = 4, 2, 0, 4, è 2 ø
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t.me/Magazines4all, , DPP/ CP24, , S-100, , 15., , (a) We know that for maxima, , Dq 10, =, q 100, So, from eq. (1),, Dl Dq 10, =, =, = 0.1, l, q 100, , Give,, , l, b sin q = (2n + 1), 2, , or sin q =, , 16., , 2n + 1 æ l ö, ç ÷, 2 èbø, , R=, =, , 17. (c), , 23., , 1.22l 1.22 ´ 5 ´10 -7, rad, =, a, 2 ´10 -3, 1.22 ´ 5 ´ 10 -7, 2 ´ 10 -3, , ´, , 24., , l, 2, For first minimum, m = 1, , Dx = m, , m = tan i, , l, DX C = l, DX A =, 2, l, DX C - DX A = = 300nm, 2, (a) For a circularly polarised light electric field remains, constant with time., , l, 2, (d) Optical path difference, Dx = (m 2 – m1)t ., (b) Separation between slits are (r1=) 16 cm and (r2=) 9 cm., Actual distance of separation, , \ Dx =, , 25., 26., , = r1r2 = 16 ´ 9 = 12cm, , 27. (b), , t, , b 0.133, =, = 0.1 cm, m 1.33, , 29., , 20., , (b), , 21., 22., , (b), (d) Let l be wavelength of monochromatic light incident, on slit S, then angular distance between two, consecutive fringes, that is the angular fringe width, is, l, q=, d, where d is distance between coherent sources., , f=, , x, , a12 + a 22 + 2a1a 2 cos f » 6, , Dl, , where D is the distance between the slits &, d, screen and d is the separation between the slits., 2Dl 4 Dl, b' =, =, = 4b, d, d/2, (b) For first minima at P, AP – BP = l, b=, , AP – MP =, , l, 2, , P, , A, , S1, , p, , a = 4, a2 = 3, 3 1, , So, A =, 28. (c), , |E|, , b¢ =, , 589 ´ 10-9, 1, l, =, = 10-3 =, = 0.001, (c), 1000, d 0.589 ´ 10-3, (c) In Fraunhoffer diffraction, for minimum intensity,, sin q =, , 180, ´ 60 mi nute = 1 minute, p, , Þ i = tan -1 (m) = tan -1 ( 3) = 60°., 18. (d) Order of the fringe can be counted on either side of the, central maximum. For example, no. 3 is first order bright, fringe., , 19., , Þ Dl = 0.1l = 0.1 ´ 5890Å = 589Å (increases), Note : Since, q µ l , as q increases, l increases., , So on decreasing the slit width, ‘b’, keeping l same, sin, q and hence q increases., (b) If the angular limit of resolution of human eye is R then, , f, , M f, , O, , B, , S, , d, , q, So phase difference, f =, , S2, D, , 2p l, ´ = p radian, l 2, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP24, 30. (a) Shift =, , t=, , S-101, , 37., , D, b, (m - 1) t = (m - 1) t = 4b, l, d, , A=, , 4l, 4 ´ 6000 ´ 10-10, =, = 4.8 µm, m -1, 1.5 - 1, , 31. (c) The fringe width is given by, b = lD, d, The angular width of fringe is given by, d l, 6 ´ 10-7, = =, = 5 × 10–3 rad., D b 0.12 ´ 10-3, , 32. (c) Distance of nth maxima, x = nl, As l b < l g, , D, µl, d, , \, , A=, =, , 38., 39., 40., , (c), (c), (c), , y, , P, O, , Plane WF, µ increases, Vel decreases, , 42., , (a), , 43., 44., , (a), (a) Where n is equivalent number of fringe by which the, centre fringe is shifted due to mica sheet, , 2, , p, æ sin f ö, ÷÷ and f = ( b sin q ), I = I 0 çç, l, è f ø, When the slit width is doubled, the amplitude of the, wave at the centre of the screen is doubled, so the, intensity at the centre is increased by a factor 4., , l=, , d/2, , 2 + 2 ´ (-1) = 0, , (b) (Light bends, upwards), Refracted, WF, , 2d d d, - =, 3 2 6, , 2d/3, , 12 + 12 + 2 ´1´ 1´ cos 3p, , 41., , \ x blue < x green, , lD, and l b < l y ,, d, \ fringe width b will decrease, 34. (c) At Brewster’s angle, only the reflected light is plane, polarised, but transmitted light is partially polarised., Dxmax = 2 l., 35. (b), So there are five maxima., These are for Dx = 0, ± l , ± 2l., 36. (c) The nearest white spot will be at P, the central maxima., , ( A1 )2 + ( A2 )2 + 2 A1 A2 cos q, , Here, A1 = A2 = 1 cm, f = 3p rad, , 33. (a) As b =, , \ y=, , (d) : Resultant amplitude,, , 45., , (m - 1) t (1.5 - 1) 6 ´10 -6, =, n, 5, , = 6 ´ 10-7 m = 6000 Å, (c) Suppose intensity of unpolarised light = 100., \ Intensity of polarised light from first nicol prism, , Ι0 1, = ´ 100 = 50, 2 2, According to law of Malus,, =, , 2, , æ1ö, Ι = Ι0 cos 2 q = 50 (cos 60º )2 = 50 ´ ç ÷ = 12.5, è2ø
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP25, 8., , 9., , (a) Emission of electron from a substance under the action, of light is photoelectric effect. Light must be at a, sufficiently high frequency. It may be visible light, U.V,, X-rays. So U.V. cause electron emission., (b), , 10. (c), , 11., , S-103, , l0 =, l=, , 15. (c) Applying Einstein's formula for photo-electricity, hn = f +, , c, 3 ´ 108, =, = 6 ´ 10-7 m = 6000Å, v0 5 ´ 1014, m0, , h, ,v=, mv, , æ vö, 1- ç ÷, è cø, , 2, , If we use 2n frequency then let the kinetic energy, becomes K', So,, h . 2n = f + K', 2hn = hn – K + K', K' = hn + K, 16., , hc, (a) Q l0 = f, , 25, æ hc ö, çè ÷ø ´ N = 200 ´, l, 100, , \ (l0)sodium=, , Where N is the No. of photons emitted per second, h, is planck’s constant and c is speed of light., , Q l0 µ, , N=, , 200 ´ 25 l, ´, 100, hc, , 100 ´ 6.2 ´ 10-34 ´ 3 ´ 108, 12. (d) For photon E = hn, E=, , hc, Þ, l, , l2 =, , hc, E, , ...(i), , 17., 18., , l=, , Þ p=, , 2mE, , h, h, Þ l1 = =, l1, p, , h, , ...(ii), , 2mE, , V = 5 volt, , =, , hc, f + eV0, , 6.6 ´ 10 1.6 ´ 10, , 34, , ´ 3 ´ 10, , 8, , -19, , (6.2 + 5), , » 10, , -7, , m, , This range lies in ultra violet range., , K. E, , l, 2, , 1, m n12 = 2 W0 - W0 = W0 and, 2, , (b), , 1, m n 22 = 10 W0 - W0 = 9W0, 2, , \, , hc, - f = eV0, l, , 1, , If K.E is doubled, wavelength becomes, 19., , f = 6.2 eV = 6.2 ´ 1.6 ´10 -19 J, , Þl=, , (f) copper, (l ), 1, Þ 0 sodium =, (l 0 )copper (f)sodium, f, , h, h, =, p, 2. m.(K.E), , \ lµ, , λ2, hc, =, ¥ E –1/2, h, λ1 E×, 2mE, 13. (a) hn = W0 + Ek = 3.5 + 1.2 = 4.7 eV, 14. (a), , = 6188 Å, , Hence for light of wavelength 4000 Å, sodium is, suitable., 1, hc, 2( hc - lf ), mv 2 =, -f Þ v =, (c), 2, l, lm, (d) de-Broglie wavelength,, , From De Broglie Eqn., p=, , 2 ´ 1.6 ´ 10-19, , 2, × 6188 = 3094 Å, 4, To eject photo-electrons from sodium the longest, wavelength is 6188 Å and that for copper is 3094 Å., , = 1.5 × 1020, , 1, 2, for proton E = m p vp, 2, 2 2, 1 mp vp, E=, 2 m, , 6.6 ´ 10-34 ´ 3 ´ 108, , Þ (l0)copper =, , 200 ´ 25 ´ 0.6 ´ 10-6, , =, , hn = f + K, , f = hn - K, , , v ® c, m ® ¥, , hence, l ® 0 ., (a) Give that, only 25% of 200W converter electrical energy, into light of yellow colour, , 1 2, mv ;, 2, , n1, W0, 1, =, =, n2, 9 W0 3, , 20. (a) The work function has no effect on photoelectric current, so long as hn > W0. The photoelectric current is, proportional to the intensity of incident light. Since there, is no change in the intensity of light, hence I1 = I2., 21. (c) n ® 2 – 1, E = 10.2 eV, kE = E – f, Q = 10.20 – 3.57, h u0 = 6.63 eV
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP25, , S-105, , 38. (d) Photoelectrons are emitted if the frequency of incident, light is greater than the threshold frequency., 39. (a) K.E. = hn – hnth = eV0 (V0 = cut off voltage), , Þ V0 =, =, 40. (d), , According to de Broglie's concept, l=, , h, (8.2 ´ 1014 - 3.3 ´ 1014 ), e, 6.6 ´ 10, , -34, , 14, , ´ 4.9 ´ 10, , 1.6 ´ 10, , -19, , m, , » 2V., , =, , hc f, el e, , h, , Þl=, , hc, - f = eV0, l, v0 =, , h, mv, , 2e(100), m, h, , 2me(100), , = 1.2 ´ 10 -10 = 1.2Å, , 43. (d) Since p = nhn, , For metal A, , For metal B, , fA 1, =, hc l, , fB 1, =, hc l, , Þn=, 44., , 1, (increasing and decreasing) is not, l, specified hence we cannot say that which metal has, comparatively greater or lesser work function (f)., , As the value of, , 1 2, mv = e(100) Þ v =, 2, , 2e(100), m, , (a) From formula, h, l=, 2 mKT, =, , 41. (c), 42. (d) Potential difference = 100 V, K.E. acquired by electron = e (100), , 6.63 ´10 -34, , m, 2 ´ 1.67 ´10 -27 ´1.38 ´10 -23 T, [By placing value of h, m and k), , 30.8, Å, T, (c) The photoelectric equation, Kmax = hn – f0, Explains that the intensity of incident radiation will, increase photocurrent only beyond the threshold, frequency., =, , 45., , p, 2 ´ 10-3, = 5 ´ 1015, =, 34, 14, hn 6.6 ´ 10, ´ 6 ´ 10
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t.me/Magazines4all, , DPP/ CP27, , S-110, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , (b) B.E. = 0.042 × 931 ; 42 MeV, , 8., , (a) Mass defect = ZMp + (A –Z)Mn–M(A,Z), , Number of nucleons in 37 Li is 7., , 2., , B.E./ nucleon =, , (d), , A, A, A, ® Z+1Y : b, Z + 1Y ¾¾, ®, Z X ¾¾, , A –4 *, Z -1 B, , 3., , or,, , 42, = 6 MeV ; 5.6 MeV, 7, , \, , A–4 *, Z -1B, , :a, , 1/ 3, , 1/ 3, , where R 1 = the radius of 27 Al, and, A1 = Atomic mass number of Al, R2 = the radius of 64Cu and A2 = Atomic mass number, of C4, , 11., , 12., , =, , 4., , (c), , 4, = 4.8m, 3, Nuclear forces are short range attractive forces which, balance the repulsive forces between the protons inside, the nucleus., , 5., , (a), , A, 1, 5000, 1, l = log e o = log e, 5, 1250, t, A, , 2, log e 2 = 0.4 log e 2, 5, , (d) Radioactivity at T1 , R1 = l N1, Radioactivity at T2, R2 = l N2, \ Number of atoms decayed in time, (T1 – T2) = (N1 –N2), , (c), , = ZMp + (A–Z) Mn–M(A,Z), , (d), , B.E., c2, , BE/A, , A, , 10., , æ 64 ö, çè ÷ø, 27, , R2 = 3.6 ´, , 7., , 9., , 4, 3, , =, , c2, , -4, ¾¾, ®A, Z -1 B : g, , R2 æ A2 ö, =, R1 çè A1 ÷ø, , 6., , B.E., , \ M (A, Z) = ZMp + (A–Z)Mn–, , (b, a, g) ( Q b = -01 e, a = 42 He, mass number and, charge number of a nucleus remains unchanged during, g decay), (c) The radius of the nuclears is directly proportional to, cube root of atomic number i.e. R µ A1/3, Þ R = R 0 A 1/3 , where R 0 is a constant of, proportionality, , =, , DPP/CP27, , (R 1 - R 2 ) (R1 - R 2 ) T, µ (R1 - R 2 ) T, =, l, 0.693, , 2, 1H, , and 13 H requires a and b amount of energies for, their nucleons to be separated., 4, releases c amount of energy in its formation i.e.,, 2 He, , in assembling the nucleons as nucleus., Hence, Energy released =c – (a + b) = c – a – b, , From the graph of BE/A versus mass number A it is, clear that, BE/A first increases and then decreases with, increase in mass number., (c) The range of energy of b-particles is from zero to some, maximum value., (a), , 180, 72 A, , a, , b, , ¾¾, ® 70 A1176 ¾¾, ® 71 A 2176, a, , g, , ¾¾, ®69 A3172 ¾¾®69 A 4172, (c), , dN, = KN, dt, 9750 = KN0, 975 = KN, Dividing (1) by (2), N, 1, =, N0 10, N, 2.303, 2.303, log 0 =, log10, K=, t, N, 5, = 0.4606 = 0.461 per minute, , .... (1), ..... (2), , 13. (d), 14. (d) Extremely high temps needed for fusion make K.E. large, enough to overcome repulsion between nuclei., 15. (c) Binding energy, = [ZMP + (A – Z)MN – M]c2, = [8MP + (17 – 8)MN – M]c2, = [8MP + 9MN – M]c2, = [8MP + 9MN – Mo]c2, 16. (c) In this reaction mass is not conserved., 17., , (a), , T1/ 2 =, , ln 2, ln 2, \l =, l, T1/ 2, , Þ lA =, , In2, ln 2, l, T, ,lB =, Þ A = B., TA, TB, l B TA, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Magazines4all, , DPP/ CP27, , S-112, , m2, Hence, m =, 1, , 36., , r23, r13, , 40., 41., , (b) The order of density of uranium nucleus is 1017 kg/m2., (b), , 42., , (a), , 1, , v, r3, r2 æ 1 ö 3, \ 1 = 23, Þ, =ç ÷, v2 r1, r1 è 2 ø, (d) In an explosion a body breaks up into two pieces of, unequal masses both part will have numerically equal, momentum and lighter part will have more velocity., U ® Th + He, P2, P2, , KEHe =, KETh =, 2m He, 2m Th, , since mHe is less so KEHe will be more., 37., , (a) As we know, R = R0 (A)1/3, where A = mass number, RAI = R0 (27)1/3 = 3R0, , 5, R, 3 AI, (a) Given : Mass of neutron = Mn, Mass of proton = Mp; Atomic mass of the element = M;, Number of neutrons in the element = N and number of, protons in the element = Z. We know that the atomic, mass (M) of any stable nucleus is always less than the, sum of the masses of the constituent particles., Therefore, M < [NMn + ZMp]., X is a neutrino, when b-particle is emitted., (a) Activity decreases, 5000 dps to 2500 dps in 150 days, \ Half life period T1/2 = 150 days, \ 300 days = 2T1/2, Therefore, initial activity = 5000 × 2T1/2 = 5000 × 2 × 2, = 20000 dps, RTe = R0 (125)1/3 = 5R0 =, , 38., , 39., , B.E H =, , 2.22, = 1.11, 2, , B.E He =, , 28.3, = 7.08, 4, , B.E Fe =, , 492, = 8.78 = maximum, 56, , B.E U =, , 1786, = 7.6, 235, , 56, 26 Fe, , is most stable as it has maximum binding energy, , per nucleon., 43. (d) Neutrons can’t be deflected by a magnetic field., 44. (b) –1 e 0 is known as b-particle & n is known as, , 45., , antineutrino. Since in this reaction n is emitted with, 0, –1e (b-particle or electron), so it is known as b-decay., (a) Given, l A = 8l, l B = l, NB =, , NA, e, , Þ N o e -l B t = No, , e -l A t, e, , e -lt = e -8lt e-1, e -lt = e -8lt -1, Comparing both side powers, , -lt = -8lt - 1, –1 = 7lt, , t=–, , 1, 7l, , The best possible answer is t =, , 1, 7l, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , DPP/ CP28, , S-113, , PHYSICS, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, 1., , t.me/Magazines4all, , (d) DIE = 8.0 mA, DIC = 7.9 mA, DI, 7.9, a= C =, = 0.9875 ; 0.99, DIE 8.0, , a, 0.9875, =, = 79, 1 - a (1 - 0.9875), (d) Here, ni = 1016 m–3, nh = 5 × 1022 m–3, As nenh = n2i, , 8., , (a) Current gain (a) = 0.96, Ie = 7.2 mA, Ic, = a = 0.96, Ie, I c = 0.96 ´ 7.2 mA = 6.91 mA, , Also, b =, , 2., , 3., , (d), , n 2 (1016 m -3 )2, \ ne = i =, = 2 ´ 109 m -3, n h 5 ´ 1022 m -3, Energy band gap range is given by,, hc, Eg =, l, For visible region l = (4 × 10–7 ~ 7 × 10–7)m, Eg =, , =, , Ie = Ic + Ib, 9., , Þ Ib = Ie – Ic = 7.2 – 6.91 = 0.29 mA, (c) No. of electrons reaching the collector,, , \, , 7 ´ 10-7, 19.8 ´10-26, , 0.96 ´ 1010, IC n C, = 0.96, =, =, IE n E, 1010, (c) Here diode is forward biased with, , a=, , 7 ´ 10-7, , 10., , 2.8 ´ 10-19, , voltage = 2 – 0 = 2 V., , 1.6 ´ 10-19, Eg = 1.75 eV, , 4., , 5., , 6., 7., , VB = Vknee+ IR, , (b) Voltage gain = b × Impedance gain, 200, 50 = b ×, = 2b Þ b = 25, 100, 200, = 1250., and power gain = b2 ×, 100, (b) When either of A or B is 1 i.e. closed then lamp will, glow., In this case, Truth table, Inputs, , 96, ´ 1010 = 0.96 ´ 1010, 100, n ´e, Emitter current, IE = E, t, nC ´ e, Collector current, IC =, t, Current transfer ratio,, nC =, , 6.6 ´10-34 ´ 3 ´108, , =, , DPP/CP28, , Output, , A, , B, , Y, , 0, , 0, , 0, , 0, , 1, , 1, , 1, , 0, , 1, , 1, , 1, , 1, , This represents OR gate., (c) In p-region of p-n junction, holes concentration > electrons concentration and in nregion electrons concentration > holes concentration., (c) Peak value of rectified output voltage, = peak value of input voltage – barrier voltage, = 2 – 0.7 = 1.3 V., , 2 = 0.7 + I× 200, (\ Total resistance = 180 + 20 = 200W), 1.3, = 6.5mA, 200, 11. (b) I ® ON, II ® OFF, In IInd state it is used as a amplifier it is active region., 12. (b) In half wave rectifier only half of the wave is rectified., \ I=, , 13. (c), , V' = V + IR = 0.5 + 0.1 × 20 = 2.5 V, 0.5V, , 20W, , 0.1A, , V, , 14., , (b), , Vo, R, 5 ´103 ´ 62, = o ´b =, = 10 ´ 62 = 620, Vin R in, 500, Vo = 620 × Vin= 620 × 0.01 = 6.2 V, \ Vo = 6.2 volt., , 15. (b) Conductivity s = n i em e = 1017 ´ (1.6 ´10 -19 ) ´ 3800, = 60.8 mho/cm
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t.me/Magazines4all, , DPP/ CP28, , S-114, , 16. (d) Negative feedback is applied to reduce the output, voltage of an amplifier. If there is no negative feedback,, the value of output voltage could be very high. In the, options given, the maximum value of voltage gain is 100., Hence it is the correct option., 17. (a) In the given system all four gate is NOR gate, Truth Table, A B (y ' = A + B) y '' = (A + y ') y ''' = (A + y '') y = y ''+ y ''', 0 0, 1, 0, 0, 1, 0 1, 1 0, 1 1, , i.e.,, , 18., , A, , B y, , 0, 0, , 0, 1, , 1, 0, , 1, 1, , 0, 1, , 0, 1, , 1, 0, 0, , 0, 1, 0, , 0, 0, 1, , 1, = e(n eμ e + n h μ h ), ρ, 2.13 = 1.6 × 10–19(0.38 + 0.18) ni, (Since in intrinsic semi-conductor, ne = nh= ni), \ density of charge carriers, n i, 2.13, 1.6 ´ 10, , -19, , 25., , Ai =, , 26., 27., 28., , 0, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , ´ 0.56, , = 2.37 ´ 1019 m -3, , 31., , 32., , 33., , 20. (d) ni2 = nenh, , 34., , (1.5 × 1016)2 = ne (4.5 × 1022), Þ, , ne = 0.5 × 1010, , or, , ne = 5 × 109, , Given, , nh = 4.5 × 1022, , 35., , Þ nh >> ne, \ Semiconductor is p-type and, 21., , (c), (b) It is a p-n-p transistor with R as base., (c) Here P-N junction diode rectifies half of the ac wave, i.e., acts as half wave rectifier. During + ve half cycle, Diode ® forward biased output across will be, , \, , A B Y, 0, , -hf e, -50, =, 1 + hoe RL 1 + 25 ´ 10-6 ´ 1 ´ 103 = –48.78, , During –ve half cycle Diode ® reverse biased output will, not obtained., 29. (d) Due to heating, when a free electron is produced then, simultaneously a hole is also produced., 30. (b) I = nA evd or I µ nvd, , (d) Here Y = ( A + B ) = A.B = A × B . Thus, it is an AND, gate for which truth table is, 0, , (b) D2 is forward biased whereas D1 is reversed biased., So effective resistance of the circuit, R = 4 + 2 = 6W, 12, \i =, = 2 A., 6, (d) In common emitter configuration current gain, , 5V, , (a) Conductivity, σ =, , =, , 19., , 0, 0, 0, , 24., , ne = 5 × 109 m–3., (c) In n-type semiconductors, electrons are the majority, charge carriers., 3/ 2, , -, , Eg, , 22. (d) For semiconductor, n = AT e, ;, so n µ T3/2, 23. (d) When PN junction diode is forward biased both, depletion layer width W and barrier height V0 decrease, and current due to molarity carrier increases., 2KT, , Ie, n v, n e Ie v h 7 4 7, = e e, = ´, = ´ =, I h n h v h or n h Ih v e 4 5 5, , (c) Electronic configuration of 6C, 6C = 1s2, 2s2 2p2, The electronic configuration of 14Si, 14Si = 1s2, 2s2 2p6, 3s2 3p2, As they are away from Nucleus, so effect of nucleus is low, for Si even for Sn and Pb are almost mettalic., V2, V, (d) 1, In forward bias, V1 > V2 i.e., in figure (d) p-type semiconductor is at higher potential w.r.t. n-type semiconductor., (a) A positive feed back from output to input in an amplifier, provides oscillations of constant amplitude., (b) The power gain in case of CE amplifier,, Power gain = b2 × Resistance gain, R, = b2 ´ o, Ri, = (10)2 × 5 = 500., (c) Given : Voltage gain AV = 150, , p, Vi = 2cos æç 15t + ö÷ ; V0 = ?, 3ø, è, For CE transistor phase difference between input and output, signal is p = 180°, V0, Using formula, AV = V, i, , Þ V0 = AV × Vi, , pö, æ, = 150 × 2cos ç 15t + ÷ , 3ø, è, , EBD_7156, , t.me/Ebooks_Encyclopedia27.
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t.me/Ebooks_Encyclopedia27., , t.me/Magazines4all, , DPP/ CP28, , S-115, , p, or V0 = 300 cos æç 15t + + p ö÷, 3, è, ø, , 36., , 37., , 38., , 39., , 4 ö, æ, V0 = 300 cos ç 15t + p ÷, 3 ø, è, (a) To use a transistor as an amplifier the emitter base, junction is forward biased while the collector base junction, is reverse biased., (d) Copper is a conductor, so its resistance decreases on, decreasing temperature as thermal agitation decreases;, whereas germanium is semiconductor therefore on, decreasing temperature resistance increases., (b) In forward biasing, the diode conducts. For ideal, junction diode, the forward resistance is zero; therefore,, entire applied voltage occurs across external resistance R, i.e., there occurs no potential drop, so potential across R is, V in forward biased., (a) Current gain (a) = 0.96, Ie = 7.2 mA, , A, 1, 1, 0, 0, , B, 1, 0, 1, 0, , C, 1, 1, 1, 0, , This truth table follows the boolean algebra C = A + B, which is for OR gate, 41., , (b), , R=, , 42., , (c), , P(0), Q(1), , DV, 2.1 - 2, 1, =, = = 0.25 W, 3, DI (800 - 400) ´ 10, 4, 0, , X(0), Z(0), , R(0), S(1), , 43., , (b), , 0, , 1 Y(1), , Y1, , A, B, , Ic, = a = 0.96, Ie, , Y, Y2, , I c = 0.96 ´ 7.2 mA = 6.91 mA, , Y1 = A + B, Y2 = A . B, , Ie = Ic + Ib, , Y = (A + B)gAB = A g A + A g B + Bg A + Bg B, = 0 + AgB + BgA + 0 = AgB + BgA (XOR gate), 44. (b) E g = 2.0 eV = 2 × 1.6 × 10–19 J, Eg = hv, , Þ Ib = Ie – Ic = 7.2 – 6.91 = 0.29 mA, 40. (d), , A, , \ v=, , C, , =, , 2 ´ 1.6 ´ 10-19 J, , h, 6.62 ´ 10-34 Js, 15, = 0. 4833 × 10 s–1 = 4.833 × 1014 Hz, , B, The truth table for the above logic gate is :, , Eg, , ; 5 ´ 1014 Hz, 45., , (d) The average value of output direct current in a full, wave rectifier = (average value of current over a cycle), 2I 0, = (2I0/p) =, π
