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Dropleton – A New Particle ? Vol. XXIII No. 5 May 2015 Corporate Office : ccording to the authors who have discovered microscopic particle Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR). Tel : 0124-4951200 Aclusters in solids, which behave like a liquid have the properties of a e-mail : [email protected] website : www.mtg.in quasi particle. This particle has a very short life span. Stimulated by light, the Regd. Office 406, Taj apartment, Near Safdarjung Hospital, smaller particles briefly condense into a ‘droplet’ with the characteristics Ring Road, New Delhi - 110029. of liquid water. This can have ripples. The life time of this droplet is only Managing Editor : Mahabir Singh about 25 picoseconds (trillionth of a second). Editor : anil ahlawat (BE, MBa) This strangely behaves like a liquid. It is thought that five electrons contents are forming the new particle with five holes. The very short life-time of the particle, the changing positions of these particles (or electron-hole Physics Musing (Problem Set-22) 8 combinations) give it an appearance of a liquid drop. JEE Advanced 11 Practice Paper 2015 In order to evaluate and appreciate the new aspects in this experiment, I am quoting what is given in the Penguin Dictionary of Physics. Thought Provoking Problems 21 “Exciton : an electron in combination with a hole in a crystalline solid. JEE Workouts 24 The electron has gained sufficient energy to be in an excited state and Core Concept 26 is bound by electrostatic attraction to the positive hole. The exciton may You Ask We Answer 30 migrate through the solid and eventually the hole and electron recombine Target PMTs 31 with emission of a photon”. Practice Questions 2015 any new discovery should excite our students. Our advice to our research BITSAT 42 student is this: appreciate what others have done and learn what other Practice Paper 2015 scientists have performed earlier. Making that as a starting point, one has Brain Map 46 to go further devising your own methods and extensions. AIIMS 54 Practice Paper 2015 Anil Ahlawat JIPMER 64 Editor Practice Paper 2015 Physics Musing (Solution Set-21) 73
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Solved Paper 2015 individual subscription rates combined subscription rates 1 yr. 2 yrs. 3 yrs. 1 yr. 2 yrs. 3 yrs. Crossword 85 Mathematics Today 330 600 775 PCM 900 1500 1900 Chemistry Today 330 600 775 PCB 900 1500 1900 Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and Physics For You 330 600 775 PCMB 1000 1800 2400 printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are adviced to make appropriate thorough enquiries before acting upon any Biology Today 330 600 775 advertisements published in this magazine. Focus/Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers. All Send D.D/M.O in favour of MTG Learning Media (P) Ltd. disputes are subject to Delhi jurisdiction only. Payments should be made directly to : MTG Learning Media (P) Ltd, Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana) All rights reserved. Reproduction in any form is prohibited. We have not appointed any subscription agent.
Physics for you | May ‘15 7 PHYSICSPHYSICS MUSINGMUSING hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics PMusing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their solutions. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.
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1. A 50 kg woman is on a large swing of radius 9 m 5. A small ball of mass m moving with a velocity v that rotates in a vertical circle at 6 rev/min. What is (just before the impact) strikes a rough surface of the magnitude of her weight when she has moved frictional coefficient at an angle q with the vertical halfway up? as shown in the figure. If the vertical component (a) 522 N (b) 521 N of velocity of the ball just after impact is found (c) 523 N (d) 520 N to be (v/2) in vertically upward direction then 2. A car is moving uniformly along a circle of radius the horizontal velocity of the ball after impact in 250 3 meter with a time period of 20 seconds. horizontal direction is A stone is thrown from the car such that it again lands on the car when the car is at a diametrically opposite point in the orbit. The angle of projection of the stone as seen from the ground should be (Take g = 10 m s–2 ) (a) 45° (b) 60° (c) 30° (d) none of these 3. A hemispherical shell rests on a rough inclined plane whose angle of friction is l, then the maximum (a) vsinq inclination of the plane of base of rim from the (b) vsinq – m{(v/2) – vcosq} horizontal is (c) vsinq + m{(v/2) + vcosq} (d) vsinq – m{(v/2) + vcosq} 6. A heavy string of mass m hangs between two fixed points A and B at an angle q with the horizontal as shown in the figure. The tension at the lowest point in the string is (a) q = sin–1(sinl) (b) q = sin–1(2 sinl) (c) q = sin–1(3sinl) (d) q = sin–1(4sinl) 4. At NTP the density of a gas is 1.3 kg m–3 and the velocity of sound propagation in the gas is 330 m s–1. The degree of freedom of the gas molecule is (a) 3 (b) 5 (a) mg/(2sinq) (b) mg/(2cosq) (c) 6 (d) 7 (c) mg/(2tanq) (d) mg/(2cotq) By Akhil Tewari, author Foundation of Physics for JEE Main & advanced, Senior Professor Physics, RaO IIT aCaDEMy, Mumbai.
8 Physics for you | May ‘15
7. On a rough table, three blocks (including the first 9. Two balls of masses 1 kg each are connected by an block) are placed as shown in the figure. Mass of inextensible massless string. The system is resting each block is m and coefficient of friction for each on a smooth horizontal surface. An impulse of block is m. A force F is applied on the first block 10 N s is applied to one of the balls at an angle so as to move the system. The minimum value of F 30° with the line joining two balls in horizontal should be direction as shown in the figure. Assuming that the string remains taut after the impulse, the magnitude of impulse of tension is
1 kg 10 N s (a) 8 mmg (b) 9 mmg (c) 7 mmg (d) 5 mmg 30º 1 kg 8. A fixed wedge ABC in the A shape of an equilateral 5 triangle of side l. Initially, a (a) 6 N s (b) 3Ns chain of length 2l and and 2 mass m rests on the wedge 5 as shown. The chain is 60º (b) 5 N s (d) Ns 60º 3 slowly being pulled down B C l by the application of a 10. A skier travels with a constant speed of 6 m s–1 along force F as shown. Work F x2 done by gravity till the a parabolic path y = . Find the acceleration of time, the chain leaves the 20 wedge will be the skier when he is at (10, 5). Neglect the size of the skier. ()31+ mgl ()32+ mgl − (a) (b) 9 92 2 2 (a) units (b) units 10 2 10 ()32+ mgl ()34+ mgl (c) − (d) (c) 1 units (d) None of these 4 4
10 Physics for you | May ‘15 paper-1 Section-1 3. A voltmeter and an ammeter are joined in series to One or More Than One Options Correct Type an ideal cell, giving readings V and A respectively. This section contains 10 multiple choice questions. Each question If a resistance equal to resistance of ammeter is now has four choices (a), (b), (c) and (d) out of which ONE or MORE THAN joined parallel to ammeter then ONE are correct. (a) V will decrease slightly 1. A particle of mass m moves on the x-axis as follows: (b) V will increase slightly it starts from rest at t = 0 from the point x = 0, (c) A will become half of its initial value and comes to rest at t = 1 at the point x = 1. No (d) A will become slightly more than half of its other information is available about its motion initial value at intermediate times (0 < t < 1). If a denotes the 4. A charged particle goes undeflected in a region instantaneous acceleration of the particle, then containing electric and magnetic fields. It is possible (a) a cannot remain positive for all t in the interval that 0 ≤ t ≤ 1. (b) |a| cannot exceed 2 at any point in its path. (a) EB|| ,|vE| (c) |a| must be ≥ 4 at some point or points in its (b) E is not parallel to B path. (c) vB|| but E is not parallel to B (d) a must change sign during the motion, but no other assertion can be made with the (d) EB|| but v is not parallel to E information given. 5. A disc of mass 35 kg and radius of gyration 2. A highly rigid cubical block A of small mass M and 0.75 m is rotating 600 rev/min as shown in side L is fixed rigidly onto another cubical block figure. What force F must be applied to braking B of the same dimensions and of low modulus of mechanism to stop the disc in 25 seconds? rigidity h such that the lower face of A completely Coefficient of friction between the disc and rod covers the upper face of B. The lower face of B is is 0.3. rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces F 1000 mm 1250 mm of A. After the force is withdrawn, blockA executes small oscillations, the time period of which is given mm by 800 = Mh r 600 rpm (a) 2phML (b) 2p L ML M (c) 2p (d) 2p . (a) 91.34 N (b) 39.30 N h hL (c) 95.83 N (d) 85.00 N
Physics for you | May ‘15 11 6. A spherical conductor contains two spherical are the pressure amplitude and displacement cavities. The total charge on the conductor itself is amplitude respectively, then the intensity of zero. However, sound wave is there are the point 1 vP2 (a) wBkA2 (b) charges +q1, +q2 2 2B at the centre of 2 P2 two cavities, (c) P (d) 2rv which may be 2 rB unsymmetrical. 10. Water is flowing smoothly through a closed pipe At a considerable distance r away from the centre of system. At one point A, the speed of the water is the spherical conductor, there is another charge 3.0 m s–1 while at another point B, 1.0 m higher, the +q3. Forces acting on q1, q2 and q3 are F1, F2 and F3 speed is 4.0 m s–1. The pressure at A is 20 kPa when respectively. Choose the incorrect statement(s). the water is flowing and 18 kPa when the water flow
(a) F1 < F2 < F3 (b) F1 = F2 < F3 stops. Then (c) F1 = F2 > F3 (d) F1 > F2 > F3 (a) the pressure at B when water is flowing is 7. Two particles 1 and 2 are projected with same speed 6.7 kPa. u as shown in figure. Particle 2 is on the ground and (b) the pressure at B when water is flowing is moves without friction on the horizontal surface. 8.2 kPa. Particle 1 is initially at a (c) the pressure at B when water stops flowing is height h from the ground 10.2 kPa. and at a horizontal (d) the pressure at B when water stops flowing is distance S from particle 2. 8.2 kPa. If a graph is plotted Section-2 between u and S for the one integer Value correct Type condition of collision of the two then (u on y-axis This section contains 10 questions. Each question, when worked out and S on x-axis) will result in one integer from 0 to 9 (both inclusive). (a) It will be a parabola passing through origin 11. A pop gun consists of a tube 25 cm long closed (b) It will be a straight line passing through the at one end by a cork and at the other end by a g tightly fitted piston. The piston is pushed slowly in. origin and having a slope of . 8h When the pressure rises to one and half times the (c) It will be a straight line passing through the atmospheric pressure, the cork is violently blown out. The frequency of the pop caused by its ejection g origin and having a slope of . is 170n Hz. Find n. (v = 340 m s–1) 4h (d) It will be a parabola not passing through the 12. A hydrogen-like atom of atomic number Z is in an origin. excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a 8. In the circuit shown in the figure the battery is ideal. transition to quantum state n, a photon of energy A voltmeter of resistance 40.8 eV is emitted. Find Z. 600 W is connected in R1 = 600 Ω 13. Image of an object approaching a convex mirror turn across R1 and R2, 120 V of radius of curvature 20 m along its optical giving readings V and R2 = 300 Ω 1 25 50 V respectively. Then axis is observed to move from m to m 2 3 7 (a) V1 = 80 V (b) V1 = 60 V in 30 seconds. What is the speed of the object in (c) V2 = 30 V (d) V2 = 40 V km h–1? 9. A sound wave of angular frequency w travels with a 14. A force F = 2.0 N acts on a particle P in the speed v in a medium of density r and bulk modulus xz-plane. The force F is parallel to x-axis. The B. Let k be the propagation constant. If P and A particle P is at a distance 3 m and the line joining
12 Physics for you | May ‘15 P with the origin makes angle 30° with the x-axis. and 127° respectively). Find the peak current (in What is the magnitude of torque on P with respect amp) through the resistor. – + to origin O (in N m)? ~ + V3 7 15. A box is kept on a rough horizontal surface. A ~ – V2 3 horizontal force just strong enough to move the box V1 is applied. This force is maintained for 2 seconds, ~ and is then removed. The total distance moved + – by the box is S meter. Then find the value of3S. 18. Two metallic spheres S1 and S2 are made of the same (Take ms = 0.20 and mk = 0.15) material and have got identical surface finishing. The mass of S is thrice that of S . Both the spheres 16. For the arrangement of the potentiometer shown in 1 2 are heated to the same high temperature but are the figure, the balance thermally insulated from each other. The ratio of point is obtained at a V 1 distance 75 cm from A 1 3 the initial rate of cooling of S1 to that of S2 is . n when the key K is open. Find the value of n. The second balance point V 19. There are two radioactive substances A and B. is obtained at 60 cm from Decay constant of B is two times that of A. Initially A when the key K is closed. K both have equal number of nuclei. Aftern half lives Find the internal resistance of A rate of disintegration of both are equal. What (in W) of the battery e1. is the value of n? 17. Three alternating voltage sources V1 = 3 sin wt V, 20. A particle of mass m attached to a string of length V2 = 5 sin (wt + f1) V and V3 = 5 sin (wt – f2) V l is describing circular motion on a smooth plane a 7 inclined at an angle with the horizontal. For the connected across a resistance R = W as shown particle to reach the highest point its velocity at the 3 lowest point should exceed : ngl sina . Find the in the figure (where f1 and f2 corresponds to 30° value of n. paper-2
Section-1 a non conducting light rod of length A Qm, only one option correct Type L. This system is released in a This section contains 10 multiple choice questions. Each question has uniform electric field of strength E four choices (a), (b), (c) and (d) out of which ONLy ONE is correct. as shown. Just after the release L E (assume no other force acts on the 1. A particle of charge Magnetic field system) q and mass m is B Qm, 2 (a) rod has zero angular projected with a O acceleration velocity v towards a QE 0 (b) rod has angular acceleration in circular region having v0 θ qm, anticlockwise direction. 2mL uniform magnetic P 2QE (c) acceleration of point A is towards right field B perpendicular and into the plane of paper, 3m from point P as shown in figure.R is the radius and QE O is the centre of the circular region. If the line OP (d) acceleration of point A is towards right m makes an angle q with the direction of v0 then the 3. Two unequal masses are connected on two sides of value of v0 so that particle passes through O is a light string passing over a light qBR qBR (a) (b) and smooth pulley as shown in msinq 2sm inq figure. The system is released from 2qBR 3qBR rest. The larger mass is stopped for a (c) (d) msinq 2msinq moment, 1.0 s after the system is set into motion. The time elapsed before 2. Two small balls A and B of positive charge Q each the string is tight again is and masses m and 2m respectively are connected by (Take g = 10 m s–2)
Physics for you | May ‘15 13 (a) 1/4 s (b) 1/2 s 9. A reeled carpet of radius R is gently pushed. If it (c) 2/3 s (d) 1/3 s rolls without sliding 4. A block of mass 1 kg is attached to one end of a and unreels, the speed spring of force constant k = 20 N m–1. The other end of the centre of mass of of the spring is attached to a fixed rigid support. This the rolling portion spring block system is made to oscillate on a rough when its radius is R horizontal surface (m = 0.04). Initial displacement halved is (Neglect the of the block from the equilibrium position is vertical motion of the a = 30 cm. How many times the block passes centre of mass) from the mean position before coming to rest? 7 14 (Take g = 10 m s–2) (a) gR (b) gR 3 3 (a) 11 (b) 7 (c) 6 (d) 15 14R 7R (c) (d) 5. An object of mass 0.2 kg executes simple harmonic 3g 3g 25 oscillations along the x-axis with a frequency Hz. 10. A concave mirror has the form of a hemisphere p of radius R = 20 cm. A thin layer of an unknown At the position x = 0.04 m, the object has kinetic transparent liquid is poured into this mirror and energy 0.5 J and potential energy 0.4 J. The amplitude it was found that the given optical system with of oscillation is the source in a certain position produces two real (a) 6 cm (b) 4 cm images, one of which (formed by direct reflection) (c) 9 cm (d) 2 cm coincides with the source and the other is at a 6. In Young’s double slit experiment, on introducing distance 8 cm from it. The refractive index m of the a thin sheet of mica of thickness 12 × 10–5 cm in liquid is the path of one of the interfering beams, the central 7 6 (a) (b) fringe is shifted through a distance equal to spacing 6 7 between the successive bright fringes. The refractive 4 3 (c) (d) index of mica, if the wavelength of light used is 3 2 6 × 10–5 cm will be Section-2 (a) 1.6 (b) 1.4 comprehension Type (only one option correct) (c) 1.5 (d) 1.55 This section contains 3 paragraphs, each describing theory, 7. A galvanometer of resistance 50 W is connected experiments, data etc. Six questions relate to the three paragraphs to a battery of 3 V along with a resistance of with two questions on each paragraph. Each question has only one 2950 W in series. A full scale deflection of correct answer among the four given options (a), (b), (c) and (d). 30 divisions is obtained in the galvanometer. In Paragraph for questions 11 and 12 order to reduce this deflection to 20 divisions, the resistance to be added in series should be A metal block is placed in a room which is at 10°C (a) 6050 W (b) 4450 W for long time. Now it is heated by an electric heater (c) 5050 W (d) 5550 W of power 500 W till its temperature becomes 50°C. Its initial rate of rise of temperature is 2.5°C s–1. The heater 8. Consider a thin square is switched off and now a heater of 100 W is required sheet of side L and thickness to maintain the temperature of the block at 50°C. t, made of a material of (Assume Newton’s law of cooling to be valid) resistivity r. The resistance between two opposite faces, 11. What is the heat capacity of the block? (a) 50 J °C–1 (b) 100 J °C–1 shown by the shaded areas –1 –1 in the figure is (c) 150 J °C (d) 200 J °C (a) directly proportional to L 12. What is the rate of cooling of block at 50°C if the (b) directly proportional to t 100 W heater is also switched off? (c) independent of L (a) 5°C s–1 (b) 0.5°C s–1 (d) independent of t (c) 1°C s–1 (d) 0.1°C s–1
14 Physics for you | May ‘15 Paragraph for questions 13 and 14 (a) 2.76 × 10–46 kg m2 (b) 1.87 × 10–46 kg m2 –47 2 –47 2 In the given circuit the capacitor (C) may be charged (c) 4.67 × 10 kg m (d) 1.17 × 10 kg m through resistance R by battery V by closing switch S1. 16. In a CO molecule, the distance between Also when S1 is opened and S2 is closed the capacitor is C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.), where connected in series with inductor (L). 5 −27 1 a.m.u. =×10 kg, is close to 3 (a) 2.4 × 10–10 m (b) 1.9 × 10–10 m (c) 1.3 × 10–10 m (d) 4.4 × 10–11 m Section-3 Matching List Type (Only One Option Correct) This section contains four questions, each having two matching lists. 13. At the start, the capacitor was uncharged. When Choices for the correct combination of elements from List-I and List-II are given as options (a), (b), (c) and (d), out of which one is correct. switch S1 is closed and S2 is kept open, the time constant of this circuit is t. Which of the following 17. Figure gives the x-t plot of a particle executing one- is correct? dimensional simple harmonic motion. (a) After time interval t, charge on the capacitor CV is. 2 (b) After time interval 2t, charge on the capacitor of CV(1 – e–2) (c) The work done by the voltage source will be half of the heat dissipated when the capacitor is fully Match the List-I with List-II charge List-I List-II (d) After time interval 2t, charge on the capacitor is Time Signs of position (x), velocity CV(1 –e–1) (v) and acceleration (a) 14. When the capacitor gets charged completely, S1 is P. At t = –1.2 s 1. x < 0, v < 0, a > 0 opened and S2 is closed. Then, (a) at t = 0, energy stored in the circuit is purely in Q. At t = –0.3 s 2. x > 0, v > 0, a < 0 the form of magnetic energy R. At t = 0.3 s 3. x > 0, v < 0, a < 0 (b) at any time t > 0, current in the circuit is in the S. At t = 1.2 s 4. x < 0, v > 0, a > 0 same direction (c) at t > 0, there is no exchange of energy between Code: the inductor and capacitor (a) P - 4, Q - 3, R - 1, S - 2 (d) at any time t > 0, instantaneous current in the (b) P - 3, Q - 1, R - 2, S - 4 C (c) P - 4, Q - 3, R - 2, S - 1 circuit may be V L (d) P - 3, Q - 4, R - 1, S - 2 Paragraph for questions 15 and 16 18. In the following, List-I lists some physical The key feature of Bohr’s theory of spectrum of hydrogen quantities and the List-II gives approximate atom is the quantization of angular momentum when energy values associated with some of them. an electron is revolving around a proton. We will extend Choose the appropriate value of energy from this to a general rotational motion to find quantized List-II for each of the physical quantities in List-I. rotational energy of a diatomic molecule assuming it to List-I List-II be rigid. The rule to be applied is Bohr’s quantization P. Energy of thermal neutrons 1. 0.025 eV condition. Q. Energy of X-rays 2. 8 MeV 15. It is found that the excitation frequency from ground to the first excited state of rotation for the CO R. Binding energy per nucleon 3. 3 eV 4 molecule is close to ×1011Hz.Then the moment of p S. Photoelectric threshold of a 4. 10 keV inertia of CO moelcule about its center of mass is metal close to (Take h = 2p × 10–34 J s)
Physics for you | May ‘15 15 Code: R. Friction force 3. (a) P - 1, Q - 2, R - 3, S - 4 between the block (b) P - 3, Q - 1, R - 2, S - 4 and the incline (c) P - 1, Q - 4, R - 2, S - 3 versus q θ (d) P - 3, Q - 4, R - 1, S - 2 S. Net interaction 4. 19. The velocity of an aircraft as seen by the driver of force between –1 a car is 5 m s upwards. A passenger in a train the block and the simultaneously sees the car to move southwards incline versus q θ with 5 m s–1. The conductor of a bus feels that the train is moving north with a velocity of 10 m s–1. Code: A decoit running towards the bus feels it moving (a) P - 2, Q - 4, R - 3, S - 1 6 m s–1 eastwards. A police jeep chasing the decoit (b) P - 3, Q - 1, R - 2, S - 4 feels him to be moving westwards with 3 m s–1. A (c) P - 2, Q - 3, R - 4, S - 1 person standing on the ground sees the police jeep (d) P - 3, Q - 4, R - 1, S - 2 moving north-west with 15 2 m s–1. SolutionS List-I List-II Paper-1 P. Velocity of the aircraft as 1. 1. (a, c, d) : The body is at rest at t = 0 and t = 1. −−35ij− 5k a seen by the conductor Initially is positive so that the body acquires some velocity. Thena should be negative so that the body Q. Velocity of the conductor 2. 55jk+ comes to rest. Hence a cannot remain positive for as seen by the police all time in the interval 0 ≤ t ≤ 1 R. Velocity of the aircraft 3. 3i The journey is depicted in the following v-t graph. as seen by the person on Total time of journey = 1 sec ground Total displacement = 1 m v A B C vm S. Velocity of police as seen 4. = Area under (v-t) graph −+12ij20 + 5k by pilot of aircraft 2s 21× −1 vmax = ==2 ms t 1 D t Code: 0 1 (a) P - 1, Q - 2, R - 3, S - 4 For path OB, acceleration (a) Change in velocity 2 − (b) P - 2, Q - 3, R - 4, S - 1 = = = 4 ms 2 time 1/2 (c) P - 1, Q - 4, R - 2, S - 3 For path BD, retardation = – 4 m s–2 (d) P - 2, Q - 3, R - 1, S - 4 For path OA, a (acceleration) > 4 m s–2 20. A block of mass m is put on For path AD, a (retardation) < – 4 m s–2 a rough inclined plane of For path OC, acceleration a < 4 m s–2 –2 inclination q, and is tied with a µ For path CD, retardation a > – 4 m s light thread shown. Inclination Hence |a| ≥ 4 at some point or points in its path. q is increased gradually from θ Hence (a), (c) and (d) are correct options. q = 0° to q = 90°. 2. (d) : The cubical block A is placed above cubical Match the column according to corresponding block B. They have same dimensions. curves. The lower face of B is held rigidly on a horizontal List-I List-II surface. A force F is applied to the upper cube A P. Tension in the 1. at right angles to one of the side faces. The block thread versus q A executes SHM when the force is withdrawn. The lower block gets distorted. x θ \ Modulus of rigidity, F A L −F Q. Normal reaction 2. h = Aq between the block B F F and the incline = −−= 2 x Lx versus q θ ()L L
16 Physics for you | May ‘15 ⇒ Restoring force F = –hLx 6. (a, c, d) 7. (b) F hLx ⇒ Acceleration = =− 8. (b, c) : Equivalent resistance of R1 (= 600 W) and voltmeter (= 600 W) in parallel is 300 W. M M hL or Acceleration = –w2x where w2 = . Total resistance in the circuit = 300 W + 300 W M = 600 W Obviously the motion is simple harmonic. 2 120 V 2 hpL 2 hL M Since current, I ==02. A, \=w or ==or T 2p . 600 W MT M hL Potential difference (V1) across R1 and voltmeter ee× R = 0.2 A × 300 W = 60 V 3. (b, d): In first case, V = and A = Rr+ Rr+ Equivalent resistance of R2 (= 300 W) and voltmeter e × R (= 600 W) in parallel is 200 W. In second case, V′= r Total resistance in the circuit = 200 W + 600 W R + = 800 W 2 V A 120V R r Since current, I′= = 01. 5 A, 1 e ee 800 W A′ = = > 2 r 22Rr+ ()Rr+ Potential difference (V2) across R2 and voltmeter R + 2 = (0.15 A) (200 W) = 30 V Clearly V increases and A becomes slightly more 9. (a, b, c, d) : Intensity, by definition, is the energy than half of its initial value. flowing per unit area per unit time. 4. (a, b) : If electric field is parallel to magnetic field, The intensity is related to the displacement amplitude the charged particle moving parallel to E experience A of the sound wave by 1 a force in the direction of E; due to B there will not Iv= rw22A 2 be any force. Hence, no deflection. P The displacement amplitude is given by A = , Bk The particle may go undeflected in the case when w where k = is the propagation constant. forces due to electric field and magnetic field v balance each other. B 5. (a) : R The speed is given by v = . F r 1000 mm 1250 mm Use these relations to get the required expressions. fR= O mm 10. (a, d) : Let P1, h1, v1 and P2, h2, v2 represent the 800 r = pressures, heights and velocities of flow at the 600 rpm two points A and B respectively. According to the Bernoulli’s theorem Taking moment about O, 1 2 1 2 Pg++rrhv=+Pgrrhv+ ...(i) F × (1000 + 1250) – R × 1000 + mR × 0 = 0 112 1 222 2 –1 –1 2250 Putting v1 = 3.0 m s , v2 = 4.0 m s , (h2 – h1) = 1 m, \=RF= 22. 5 F 1000 P1 = 20 kPa \ Frictional force, f = mR = 0.3 × 2.25F = 0.675F we get, 800 103 Braking torque, t = f × r = 0.675F × Nm P =+20 103 ×−98.(1)[+−9161] × 0−3 1000 2 2 t = 0.54F = 20 – 9.8 – 3.5 = 6.7 kPa Also, t = Ia Also when the flow stops, v1 = v2 = 0 and then from (i), 05..40F 54F \=IFaa05. 4 or == P2 = 18 – 9.8 = 8.2 kPa I mk2 05. 4F a= = 0.0275F 11. (3) : 35 × (.075)2 600 ww− 2p × Now, t = 0 ⇒=25 60 a 0.0275F Solving, we get, F = 91.34 N
Physics for you | May ‘15 17 Let A be area of cross-section. 25 Speed of the object = ms−1 In isothermal process, 30 P1V1 = P2V2 25 18 3 50 =× km h−1 = 3 km h–1 PA××25 =×PL××A ⇒=L cm 30 5 2 3 Now, after the ejection of cork, for oscillating air 14. (3) : Torque, t=rF× node will be at piston while antinode will be at the In magnitude, t = rFsinq z open end and as minimum distance between node where q is the angle between and antinode is (l/4). P F l 50 2 r and F. So,c==L mm⇒=l Here, F = 2.0 N, r = 3 m, q = 30° 30° 4 3 3 x v 340 × 3 \ The magnitude of torque on O andhence,Hu == = 510 z=170 ×3Hz l 2 P with respect to origin O is Hence, n = 3 t = (3 m) (2.0 N) sin 30° 1 12. (4) : The energy of nth orbit of hydrogen-like atom = (3 m) (2.0 N) = 3 N m 2 is 13.6Z2 En =− 15. (4) : Applied force = ms N = msmg, when the block is n2 moving, friction acting on it is m mg Maximum energy corresponds to transition 2n → 1 k mmmg − mg 2 1 1 sk −2 \=204 13.6Z − \ a = =−()mmskgg==00..505 ms 1 ()2n 2 ...(i) m When the applied force is removed, retardation Also for transition 2n → n –2 = mk g = 1.5 m s 11 3 40..81=−36Z2 or 40..81= 36Z2 22 2 1 2 nn4 4n Distance covered in 2 s, d1 = ××05.;21= m 2 Z2 or 40..84==08 or 4nZ22 –1 2 velocity after 2 s = 0.5 × 2 = 1 m s 4n Distance covered during retardation; or 2n = Z ...(ii) 2 2 1 From (i) and (ii), we get 0 = 1 – 2(1.5) d2 ⇒ d2 = 3 2 1 2 204 =−13..61Z =−13 61Z 36. 1 4 Z2 \ Total distance S = 1 + ==; 34S m 3 3 13.6 Z2 = 204 + 13.6 = 217.6 16. (6) : Let l is resistance 2 217.6 ZZ==16 or = 4 per unit length of wire 13.6 AB. 13. (3) : Focal length of a convex mirror, R 20 When K is opened. f == m = 10 m 2 2 I(lx1) = e1 ... (i) 25 K is closed, Ilx2 = e1 – ir ... (ii) vf1 =+ mm, =+10 3 e i = 1 ... (iii) 111 Rr+ Using mirror formula += vu f From eqns (i), (ii) and (iii), we get 1 11 113 x 07. 5 \ += or =− r = 1 −1 Rr=−1246; = W (/25 3) u1 10 u1 10 25 x2 06. 0 or u1 = – 50 m 17. (3) : V1 = 3 sin wt; V2 = 5 sin (wt + f1); 50 vf=+ mm, =+10 V3 = 5 sin (wt – f2) 2 7 111 \+= vu22f 1 11 117 += or =− (/50 7) u2 10 u2 10 50
or u2 = – 25 m
18 Physics for you | May ‘15 = (gl sin a) + 2g (2l sin a) V =+(c35os30°−53sin)75°+22(sin30°−53cos)7° max \ vA = 5gl sina 2 Paper-2 53 2 = +=(.15)V21 1. (b) : Let r be the radius of circular path. If the particle 2 passes through O, then from the given figure, V 21 21× 3 R /2 R I ==max = = 3 A sin q = = ...(i) max R 7 73/ r 2r mv0 18. (3) : The rate at which energy leaves the object is As r = qB DQ 4 =seAT qBr Dt \ v = 0 m Since, DQ = mCDT, we get qB(/R 2 sin)q DT eAs T 4 = (Using (i)) = m Dt mC qBR = 4 3 2sm inq Also, since m = prr for a sphere, we get 3 23/ 2. (d) : Resultant force on arrangement is 2QE, thus 2 3m Ar==44pp acceleration of centre of mass is given by 4pr 2QE (towards right) aC = 4 23/ 13/ 3m DT eTs 3m 1 Hence, = 4p = K D pr As per conservation of angular momentum t mC 4 m 2L L 4L22L For the given two bodies QE − QE =×m +×2m a 33 9 9 13/ 13/ (/DDTt)1 m2 1 = = QE (/DDTt)2 m1 3 a= (clockwise) 2mL 19. (1): Let lA = l and lB = 2l From constraint Initially rate of disintegration of A is lN0 and that equation, we get of B is 2lN0. 2L aaAC=+a After one half-life of A, rate of disintegration of 3 lN 0 A will become and that of B would also be 2QE QE 2L lN 2 1 =+ 0 (half-life of B = half-life of A) 32m mL 3 2 2 3QE QE So, after one half-life of A or two half-lives of B. == (towards right) dN dN 3m m − =− dt dt 3. (d) : Net pulling force = 2 g – 1 g = 10 N AB Mass being pulled = 2 + 1 = 3 kg \ n = 1 10 \ Acceleration of the system is a = m s–2 20. (5) : h = 2 l sin a 3 A is the lowest point and B \ Velocity of both the blocks at t = 1 s will be 10 10 is the highest point. At B, in va==t ()1 = ms−1 0 critical case tension is zero. Let 3 3 velocity of particle at B at this Now, at this moment velocity of 2 kg block instant be vB. Then 10 –1 mv 2 becomes zero, while that of 1 kg block is m s mg sin a = B 3 2 l upwards. Hence, string becomes tight again when or vB = gl sin a displacement of 1 kg block equals displacement of 2 2 Now vA = vB + 2gh 2 kg block.
Physics for you | May ‘15 19 7. (b) : Total initial resistance 1 221 i.e., vt0 −=gt gt = G + R = 50 W + 2950 W = 3000 W 2 2 3 V −3 v 10 / 3 1 Current, I ==11× 0 A=1mA or t ==0 = s 3000 W g 10 3 If the deflection has to be reduced to 20 divisions, then current 4. (b) : Let the initial amplitude a decreases to a1 to the 1 mA 2 other side, i.e., after the first sweep ; I′ =×20 = mA 30 3 then decrease in elastic potential energy Let x be the effective resistance of the circuit, then = work done against friction 2 3 Vm=×3000 WW1 Am=×x A 1 2 1 2 3 or ka −=ka1 mmg()aa+ 1 3 2 2 or x =×3000 1 ×=4500 W 2 1 \ W W W or ka()+−aa11()am=+m ga()a1 Resistance to be added = (4500 – 50 ) = 4450 2 rl 8. (c) : As R = 2mmg or aa−= A 1 k Here, l = L, A = Lt 2mmg rrL Similarly, aa−= \=R = 12 k Lt t ...... Hence, the resistance between two opposite 2mmg faces, shown by the shaded areas in the figure is aann−1 −= k independent of L. Adding all the equations 9. (b) 10. (c) 2nmm g aa−= ...(i) 11. (d) : Pheater = Pgiven to block n k dT The block stops when \ 500 = C ; 500 = 2.5C mmg dt mmg = kan or a = . n k or C = 200 J °C–1 Substituting in eqn. (i), we get 12. (b) : Power loss to surroundings = 100 W; mmg dT ()21n + = a C =−100 k dt ka ()20 (.03) dT 100 −1 or (2n + 1) = ==15 =− =−05.C° s mmg (.0041)( )(10) dt 200 \ n = 7 13. (b) 14. (d) 15. (b) k 16. (c) : The moment of inertia of CO molecule is 5. (a) : As w = 2pu = m I = mr2 ...(i) \ k = (2pu)2 m where, m = reduced mass of the CO molecule Total energy of oscillation is r = distance between C and O or bond length E = 0.5 + 0.4 = 0.9 J The reduced mass m of the CO molecule is 1 2 mm ()12 ()16 5 −27 \ 0.9 = kA m= 12 = ××10 kg 2 mm 12+ 12 + 16 3 18..18 From equation (i), we get == or A 2 −46 k 2 I 18. 71× 0 ()2pu m r = = m 12 × 16 5 −27 1 18. 3 ××10 = = m = 6 cm 28 3 25 02. 50 –10 2p or r = 1.3 × 10 m p 17. (a) 18. (c) 19. (b) 20. (a) 6. (c)
20 Physics for you | May ‘15 By : Prof. Rajinder Singh Randhawa*
1. A small bead of mass m, charge –q is constrained to ensure that the electron can move away from the move along a frictionless wire. A positive charge Q axis of the cylinder to a distance greater than r. lies at a distance L from the wire. Show that if the bead is displaced a distance x, where x << L, and R0 released, it will exhibit simple harmonic motion. Obtain an expression for the time period of simple harmonic motion. L
2. Three charges –q, +2q and –q are arranged on a line r as shown in figure. Calculate the field at a distance r > a on the line.
6. A spherical shell of radius R1 with a uniform charge q has a point charge q0 at its centre. Find the work performed by the electric forces during the shell expansion from radius R to radius R . 3. A small ball of mass 2 × 10–3 kg having a charge of 1 1 2 mC is suspended by a string of length 0.8 m. Another SOLUTIONS identical ball having the same charge is kept at the 1. From figure, applying Newton’s nd2 law, ^ point of suspension. Determine the minimum −=Ficosq ma horizontal velocity which should be imparted to the Qq x ^ − im= a lower ball so that it can make complete revolution. 22 22 4pe0()xL+ xL+ 4. A circular ring of radius R with uniform positive Qq x ^ charge density l per unit length is located in the \=a − i 2232/ Y-Z plane with its centre at the origin O. A particle 4pe0m ()xL+ of mass m and positive charge +q is projected from the point PR(,30,)0 on the positive X-axis directly towards O, with initial velocity v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P.
5. A straight infinitely long cylinder of radius R0 is uniformly charged having surface charge density s. The cylinder serves as a source of electrons, with the velocity vector of emitted electrons perpendicular For small linear displacement, x << L, so x2 is to its surface. What must be electron velocity to neglected. *Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699
Physics for you | May ‘15 21 nd Qq ^ From Newton’s 2 law \=a − xi 3 2 2 4pe0mL q mv which is equation of SHM with Tm2 +−g = ...(i) 4pe l2 l Qq 0 w = 3 As T2 = 0, then 4pe mL 2 2 0 q mv 3 mg −= 2p 4pe mL 2 l ...(ii) As w =\T = 2p 0 4pe0l T Qq Using conservation of energy 2. Energy at topmost point = Energy at lowest point
1 1 mv22+=mg()2lmu 2 2 v2 = u2 – 4gl ...(iii) From figure, Put (iii) in (ii), we get q q2 E =− ug=−5 l 1 2 4pe ml 4pe0()ra− 0 –1 2q q Put the given values, we get u = 5.8 m s E ==and E − 4. 2 2 3 2 Y 44pe0r pe0()ra+ r \=EE12++EE3 R a q 121 X =− +− O v P 22 2 ( 3,R 0, 0) 4pe0 ()ra− rr()+ a −−Z 22 q a a E =−12− +−1+ From figure, 4pe r2 {}r {}r 0 Potential at the centre of ring, If r > a, we can use binomial approximation, lR V = n ()nn()−1 2 (1 + x) ≈ 1 + nx + x for x < 1 2e aR22+ 2! 0 2 2 lR l q 23a a 23a a V = = \=E −+1 + +−21 −+ 2 2 2 224e0 4pe0r r r r r 23e0 ()RR+ ql q 23a a2 23a a2 \=Potential energy at P =− 1−− +−21+− 4e pe 2 r 2 r 2 0 4 0r r r lq 1 2 6aq2 \=Total energy at P + mv =− 4e 2 4 0 4pe r lq 0 The potential energy at centre = 3. If the ball has to just complete the circle then the 2e0 tension must vanish at the topmost point, i.e. The particle will not return to P, when T2 = 0. lq 1 lq +=mv2 4e0 22e0 lq Solving, we get v = 2e0m 5. According to Gauss’s theorem, q sp()2 RL ∫ Ed⋅=s ⇒=Er()2p L 0 e0 e0
22 Physics for you | May ‘15 sR0 2eR s ln ()rR/ \=E = 00 e r ...(i) v0 0 e0 me Applying Newton’s 2nd law, 6. The electrical potential energy of the system is dr2 R s m =−e 0 ...(ii) 2 e 2 q qq dt e0r U =+0 Using energy conservation 84pe0R pe0R 1 2 Initial potential energy of the mve 0 −=eV0 − eV ...(iii) 2 system is where V0 is the potential of the cylinder and V is the 2 potential at a distance r from the cylinder’s axis. q qq0 Ui =+ dV R s dV 84pe01R pe01R Since E =− , \=0 − dr e0r dr Final potential energy of the system is Integrating, we get q2 qq =+0 R0s Uf V =− ln rC+ ...(iv) 84pe02R pe02R e0 From work-energy theorem, where C is constant of integration. W = – DU = – (Uf – Ui) R s Also,lV =− 0 n RC+ ...(v) 0 0 q e0 qq+ 0 2 11 WU=−ifU = − From eqns. (iii), (iv) and (v), we get 4pe01RR2
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Physics for you | May ‘15 23 JEEVibration and SHM 1. A uniform bar with mass m lies symmetrically stiffnessK . Find the time period of small horizontal across two rapidly rotating fixed rollers, A and B oscillation of the block along x-axis. with distance L = 2.0 cm between the bar’s centre of mass and each roller. The rollers , whose directions of rotation are shown in figures slip against the bar with coefficient of kinetic friction mk = 0.40. Suppose the bar is displaced horizontally by a distance x as shown in figure and then released. 6. A stepped pulley having mass m, radius R and radius What is the angular frequency w of the resulting of gyration k is connected with two ideal springs of horizontal simple harmonic motion of the bar? stiffness K1 and K2 as shown in figure. If the pulley rolls without sliding, find the angular frequency of L L its oscillation.
x AB mg
2. A particle of mass m is located in a unidimensional potential field where the potential energy of the particle depends on the coordinates as SOLUTIONS U(x) = U0(1 – cos ax), where U0 and a are constants. Find the period of small oscillations that the particle 1. Consider horizontal forces performs about the equilibrium position. mkNA – mkNB = ma 3. The speed v of a particle moving along x-axis is 2 2 2 mmkANN− kB …(i) given by v = 8bx – x – 12b where b is a constant. or a = Find amplitude of oscillation. m NA NB 4. A pendulum has a string of length 99.39 cm. How much length of the pendulum must be shortened f=KA KAN f=KB KBN to keep the current time of the pendulum if it loses x 4 s a day. mg 5. A block of mass m placed on a smooth horizontal Taking torque about an axis perpendicular to floor is connected with two light springs each of the plane and through the contact point between
Aditya Sharma, Prime Classes Pvt Ltd, Durgakund, Varanasi
24 Physics for you | May ‘15 bar and roller A. The bar experiences no angular 5. When the block is pushed by x along x-axis, right acceleration about that axis. spring will be compressed by x while left spring will NA(0) + NB(2L) – mg(L + x) + fKA(0) + fKB(0) = 0 …(ii) be stretched by xcosq. By balancing vertical forces, we have Net restoring force along x-axis, NA + NB – mg = 0 …(iii) 2 2 Fnet = Kx + K x cos q = xK(1 + cos q) Solving eqns. (i), (ii) and (iii) 2 ⇒ Keq = K(1 + cos q) mg()LX+ mg()LX− N = , N = m BA22L L \ T = 2p + 2 q m g K(c1 os ) a =− k X L 6. Torque about contact point P, Comparing it with a = –w2X tP = – K1x12r – K2x2 (R – r) = IPa Now, x1 = 2rq, x2 = (R – r)q mk g 04. 01× 0 −1 2 2 we get, w == 14 rads \ IPa = –4K1r q – K2(R – r) q ...(i) L −2 2 2 21× 0 Also, IP = mk + mr ...(ii) From eqns. (i) and (ii) 2. U = U0(1 – cosax) This is zero when cos ax = 1 or ax = 0 or x = 0 ((4Kr2 +−KRr))2 aq=− 1 2 \ x = 0 is mean position of system. mk()22+ r At extreme end, potential energy becomes Comparing it with a = –w2q,we get maximum, i.e., 2U0 \ 2U = U (1 – cos Aa) where A is amplitude. 4Kr2 +−KR()r 2 0 0 w= 1 2 2 = 1 – cos Aa mk()22+ r ()Aa 2 Since A is small, cos aA = 1− soluTioN of APril 2015 crossworD 2 2 ()Aa 12 3 4 5 6 \−11 − = 2 GB W A PIT CH 2 A 7P E R MEA N CE I N H M C 8T E N S I O N I 9M A 10 11 or A = 2/a M Q N D S H A R M O N I C 12 As maximum kinetic energy = 2U0 A CCU R A CY E V O S N N E U A T 13I S O T H E R M 1 1 4 14 15 \ 2Um==ww22Am2 G R EGE L A T ION R D L 0 2 16 2 2 S E A R E O E M a 17 18 T E S L A M O P F O S 2 19 Ua U 2p m R I N F R I N GES D O 2 0 0 20 or ww==or a ⇒=T O R BIT N S C C E N m m a U0 M 21F R A C T U R 22EP O I N T R O R A L A G 23 24 3. At extreme position v becomes zero, so T F R V E N T U R IME T E R 25 2 2 E R L C O A 8bx – x – 12b = 0 26 27 2 2 U S I P H O N B E T A D E C A Y R P or x – 8bx + 12b = 0 T G W R H (x – 6b)(x – 2b) = 0 E G 28A B S O L U T E Z E R O Y \ x = 2b and 6b C E M T 29T R U T H T A BLE 30Q U A LIT Y It means particle moves along x–axis from x = 2b I T 31PLU SE to 6b, 32C A R N O T E N GIN E A 33 \ 2A = 6b – 2b = 4b ⇒ A = 2b R E T A R D A T ION K 4. Time lost by pendulum per day = 4 s dn = –4 s WInnerS (April 2015) Vipul Ahuja Number of seconds in a day, n = 24 × 3600 s Sattwik Sadhu Let dl be the length by which the pendulum should Atriz roy be shortened to keep the correct time Solution Senders (April 2015) dn dl −4 −dl Anubhab Banerjee Asma Saifi =− ⇒ = n 2l 24 × 3600 29× 93. 9 Solution Senders (March 2015) Hemen Ved Poonam Sharma ⇒ dl = 0.00920 cm
Physics for you | May ‘15 25 Standing Waves
When two waves of identical frequency (of similar kind) to the particles which have a positive amplitude, as travelling from opposite directions meet, the resultant the particles located betweeen Q and R. But it should wave obtained is known as standing waves. be noted that all particles located between P and Q Standing waves can either be pure standing waves or oscillate in same phase, i.e. reach their extreme ends partial standing waves. together, cross their mean position together but they Pure Standing Waves have different amplitude of oscillation. Here the amplitude of the two superposing waves need Similar is the situation with particles located between to be identical too. Q and R. Let the two travelling waves be Let me clarify this with a more suitable pictorial y1(x, t) = A sin(wt – kx) representation. y2 (x, t) = Asin(wt + kx) Due to their superposition, the resultant wave is yR(x, t) = y1(x, t) + y2(x, t) = A[sin(wt – kx) + sin(wt + kx)] P Q R = (2Acoskx)sinwt where (2Acoskx) = Ax is clearly a position dependent function (cosine, hence periodic too) and is a constant for a particular location and is known as amplitude of Note that, when particles located between P and Q oscillation at that location. are at their lower extreme, the particles between Q While the second term, sin(wt) indicates SHM of the and R are at their upper extreme and when one region particle at position x with amplitude Ax. Therefore, particles are travelling up, the others are travelling sinwt term is common for all particles, hence each down but they would cross the mean position particle has same frequency of oscillation. But what together and twice in every oscillation. Hence 180° about the phase? Let us see. out of phase! The amplitude,A x = 2Acos(kx) as a function of x can be But why were they named standing waves? plotted as below. The answer again lies in the amplitude expression, Ax Ax = 2Acos(kx) Since, its a cosine function, it will have all values between +2A –1 to +1 including zero. P R What would zero amplitude of oscillation mean? Q Amplitude indicates the maximum displacement from mean position and when the maximum displacement itself is zero, it means that the particles located here do But, what would a negative amplitude indicate, as in not oscillate at all. Hence, we can say that these particles from location P to Q on x-axis? of zero amplitude are standing at their respective It first indicates that the instantaneous phase of these points. medium particles are 180° out of phase with respect These points of zero amplitude are known as nodes,
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
26 Physics for you | May ‘15 whereas the particles with maximum amplitude (±2A) No! form the location of antinodes. sintyR(,xt) = (2Akcos xt) sin coskx We always represent standing waves by drawing the costy(,xt) = (2Akcos xt) cos envelope of oscillation of the particles of the medium, R as below. [which clearly can be obtained from superposition of two cosine travelling wave functions, cos(wt – kx) and Nodes yxR() cos(wt + kx)]. sintyR(,xt) = (2Aksin xt) sin sinkx 2Ax|cos| costyR(,xt) = (2Aksin xt) cos 2Ax|cos| Each of the above four equations form the standing waves. (2Acoskx) or (2Asinkx) which can be used to indicate the magnitude at a location, depends on the boundary Antinodes condition at x = 0, i.e. if We can easily find out the location of (a) at x = 0, node exists we use 2Asinkx (i) Nodes, by using coskx = 0 (b) at x = 0, antinode exists we use 2Acoskx p ⇒=kx ()21n + cos(wt) or sin(wt) will be used seeing if the particles start 2 from mean position (sinwt) or extreme end (coswt) of 2p p oscillaton at t = 0. ⇒=xn()21+ l 2 So, how to find the amplitude of oscillation at any l lll3 5 l ⇒=xn()21+=,, arbitrary location, say at a distance of from node? 444 4 6 Onlinel Test SeriesWe can use both (2Acoskx) as well as (2AsinOnlinekx) to find Test Series i.e. at odd multiples of the amplitude, as below. Practice4 Part Syllabus/ Full Syllabus Practice Part Syllabus/ Full Syllabus (ii) Antinodes, by using coskx = ±1 Mock Test Papers for � = Mock Test Papers for ⇒ kx = np 4 6 12 6 2p ⇒ xn= p l Ax ll 3l ⇒ xn= = 0,,l, 2 Log2 JEE on2 to http://test.pcmbtoday Main.com Log onAIPMT to http://test.pcmbtoday l i.e. at integral multiples of 2 Hence the distance between two consecutive nodes or ll l AA= 2 cosskAor 2 in k antinodes is . x 2 12 6 Node-to-node is a loop of standing wave and hence is 2pl 2pl Attempt lfree online test= 2AAcoss or 2 in Interested in more tests! also said to be the loop length which is . ll12 6 Log on to http://test.pcmbtoday2 .com Log on to http://test.pcmbtoday It is also important to note here that unlike travelling pp = 2AAcoss or 2 in = 3A waves, energy is not transformed here from one end to 6 3 the other. The energy is confined between node to node i.e. within a loop. This can easily be understood since, one wave transports energy in positive x-direction while the other transports equal amount in negative x-direction. Attempt free online test and Hence overall, there is no transfer of energy. So, is yR(x, t) = (2Acoskx)sin(wt) the only standing analyse your performance wave equation? Log on to http://test.pcmbtoday.com
Physics for you | May ‘15 27 Since each particle executes simple harmonic motion of Since both the ends are fixed, they will necessarily form different amplitude, their energy of oscillation are also nodes. different, clearly maximum at antinodes. Therefore, to have largest wavelength we will have to Considering a standing wave in a stretched string as in insert one antinode between these two nodes. the strings of a guitar where T = tension in string m = mass per unit length
w = angular frequency of wave l = 0 2p 2 k = wave number of wave = l \ v = f0l0 = f0(2l) We can find the energy confined within a loop. v T / m ⇒=f = 0 22l l Overtones are the higher frequencies, with respect to the fundamental frequency with which standing waves can be set up. For example, 1st overtone is the immediate higher next frequency (hence immediate lower next wavelength). Therefore we insert one more antinode here. 1 22 \=dE ()dm w Ax 2 [This result is a standard result of energy of a particle in simple harmonic motion] l = 1 1 22 = (mwdx)(s2Akin x) \ v = f1l1 2 v l/2 ⇒=f 22= f ⇒=mw22 2 10 EA2 ∫ sin(kx)dx 2l 0 st nd \ 1 overtone = 2f0 = 2 harmonic l/2 22 (c12− os()kx ) (Note : All integral multiples of fundamental frequency = 2mw A ∫ dx 0 2 are said to be harmonics of f0). Similarly for finding 2nd overtone, mw22A l \=E 2 Now, let us try to find out, what are the frequencies with which we can set up standing waves in a string fixed at both ends. 3 l = 2 2
Velocity of wave \ v = f2l2 l (length of string) v = T 3v ⇒=f = 3 f 202l The smallest frequency with which standing waves ⇒ 2nd overtone = 3rd harmonic can be set up in any system is said to be fundamental \ Generalising, we have, nth overtone frequency (f0). Finding f0, clearly means having largest th fn = (n + 1) harmonic wavelength so that l v v = f = constant, =+()nf110 =+()n since velocity of wave is only medium dependent. 2l
28 Physics for you | May ‘15 If the string oscillates in m loops, the frequency of are of different amplitudes, we obtain a partial standing oscillation is wave. v T / m The discussion below will clarify the name. fmm ==fm0 = m 22l l Let, y1(x, t) = A1sin(wt – kx) We can easily change the frequency of oscillation y2(x, t) = A2sin(wt + kx) (A2 < A1) by changing T (by increasing/decreasing tension), \ Resultant wave, m (thin/thick wire) or l (long/short wire). This is the yR(x, t) = A1sin(wt – kx) + A2sin(wt + kx) basic of sound production of different frequency in ⇒ yR(x, t) = (A1 – A2) sin(wt – kx) + A2sin(wt – kx) guitar wires. By the movement of the fingers, on the + A2sin(wt + kx) string one controls the location of nodes whereas by In this expression, clearly plucking it with the other hand, we control the location A2sin(wt – kx) + A2sin(wt + kx) will form standing of antinodes. wave. If the string’s tension is increased/decreased slightly \=yx(,tA)(−−At)sin()w kx + (c2Akos xt)sinw (less than 5%), we can apply error formula to find the R 12 2 Travelling wave transporting Pure standinng wave percentage change in frequency of string energgy in +ve x-direction Df 1 DT i.e. ≈ f 2 T Standing Waves in Composite Strings
Tl,, 11 Tl,,22
A light string is attached to a heavy string, with a Clearly, a travelling wave of amplitude A1 – A2 is common tension T. superposed over a pure standing wave. Hence the T amplitude of each point is raised by A1 – A2. If m > m , then the velocity of wave v = 2 1 m , \ At nodes, A = A1 – A2 + 0 = A1 – A2 = Amin At antinodes, A = (A – A ) + 2A = A + A = A v > v 1 2 2 1 2 max 1 2 The resultant envelope of the wave in this case would and as v = f l, the strings are supposed to oscillate with look like same frequency, therefore, () l1 > l2 yxR i.e. thin string will have larger wavelength and hence + lesser loops. – AA12 AA12 If p and q be the number of loops in the thin and thick strings respectively, the junction obviously being a node.
f1 = f2 Nodes Antinodes v v ⇒=p 1 q 2 22l1 l2
T //mm1 T 2 An experiment is a ⇒=p q question which science l1 l2 poses to Nature, and a measurement is the Partial Standing Waves recording of Nature’s This is a very less talked about topic. If the two answer. superposing waves travelling from opposite directions -Max Planck
Physics for you | May ‘15 29 bridle point (the point where the main, long string branches into separate strings running to various points on the kite frame). The rotation changes the YUASK kite’s angle of attack and thus alters the lift and drag. As a result, the kite not only rotates but also moves WE ANSWER vertically. The vertical motion changes the angle at which the string pulls on the bridle point and thus Do you have a question that you just can’t get also the horizontal and vertical pulls of the string. answered? Stable flight occurs if three quantities vanish (i) Use the vast expertise of our mtg team to get to the the torques, (ii) the net vertical force, and (iii) the bottom of the question. From the serious to the silly, net horizontal force. For these to vanish, not only the controversial to the trivial, the team will tackle the must the kite have the proper orientation but also questions, easy and tough. the string must pull at the proper angle and with The best questions and their solutions will be printed in the proper force. The kite is then said to be in an this column each month. equilibrium state. For a given wind speed, there Q1. When you load up a plastic shopping bag with might be more than one equilibrium state. If the groceries and then carry the bag by the loops at wind changes, both the orientation of the kite and the top of the bag, why will the loops initially the angle of the string must change for the kite to withstand the load but then, several minutes later, find a new equilibrium state. begin to stretch, perhaps to the point of tearing? Q3. How is an aircraft protected from lightning while – Saumya Shah (W.B.) in the air? Ans. If you suspend a load from the lower end of a – Abhishek Thomas (Kerala) spring hanging from a ceiling, the spring will Ans. Thunderclouds can be identified, and the first stretch by a certain amount and then stay stretched. precaution is to ensure that the aircraft stays away Plastic, which consists of polymers, is different. from these highly charged clouds. Aircraft are fitted If you suspend a load from the lower end of a with conducting brushes at the tips of wings and the plastic strip, the strip will initially stretch like the rudder to ensure that they stay close to the electrical spring but thereafter it will gradually stretch more potential of the air mass in which they are flying. which is called viscoelastic creep. The mechanism The electrical potential in the atmosphere varies of this creep can vary from polymer to polymer. significantly as a function of altitude. In addition, The polymer consists of many long and entangled just the movement of the airplane through the air molecules. When the polymer is put under load, mass can be fast enough to produce electrostatic these molecules gradually disentangle somewhat charge. This charge must be continuously dissipated. because they are pulled in the direction of the load. In comparison with a large cloud or even the ground, The gradual reorientation of the molecules allows an aircraft does not appear an attractive target for the plastic to gradually stretch. If the plastic stretches a lightning bolt. It is, after all, fairly well insulated enough, it may also get narrowed perpendicular to from the ground. Even if there is flow of high the direction of the load called as necking. current through the skin of the aircraft,the highly Q2. What keeps a kite aloft and what determines a conducting metallic construction ensures that stable flight, as opposed to a chaotic one in which passengers are protected from the effect of electrical the kite constantly loops and flutters? activity outside. Communication system could be – Aman Gupta (U.P.) affected. The chance of an accident appears greater Ans. A triangular kite is a flexible surface that faces into than the actual observed incidence. This must be the wind while tilted up at an angle, said to be the a reflection of the precautions taken by pilots. In angle of attack. Four forces act on the kite are (i) The spite of the brushes at the tips of wings, an aircraft gravitational force pulls downward. (ii) Because the can accumulate significant electrical charge on its wind is deflected downward by the kite’s face, the body prior to landing. Sparks or discharges must kite experiences an upward lift. (iii) The wind also be prevented. In the earlier days, airplanes used to produces a drag force in the direction of the wind. deploy a metal chain that trailed and touched the (iv) The string produces a force that is downward ground before the wheels of the plane did. Modern and into the wind. aircraft use tyres that are made of conducting If the kite is not in stable flight, the torques due to rubber. the forces automatically rotate the kite around the
30 Physics for you | May ‘15 1. The major contribution of Sir C.V. Raman is (a) explanation of photoelectric effect (b) principle of buoyancy Q P (c) scattering of light by molecules of a medium (d) electromagnetic theory (a) varies unpredictably 2. The period of revolution of Jupiter around the Sun (b) increases from P to Q is 12 times the period of revolution of the Earth (c) decreases from P to Q around the Sun. The distance between the Jupiter (d) is constant throughout the conductor and Sun is n times the distance between the Earth and Sun. Then the value of n is 7. A 0.01 H inductor and 3p W resistance are (a) (144)3/2 (b) (144)2/3 connected in series with a 220 V, 50 Hz ac source. (c) 3 144 (d) 4 144 The phase difference between the current and the voltage is 3. A source of sound producing wavelength of 50 cm p p p p is moving away from a stationary observer with (a) (b) (c) (d) 2 6 3 4 (1/5)th speed of sound. Then, what is the wavelength 8. A man moves 20 m north, then 10 m east and then of sound heard by the observer? (a) 70 cm (b) 55 cm 10 2 m south-west. His displacement is (c) 40 cm (d) 60 cm (a) 20 m north (b) 10 2 m north-west 4. A long straight wire along the z-axis carries a (c) 10 m north (d) 10 2 m south-east current I in the negative z-direction. The magnetic 9. When a light of photons of energy 4.2 eV is field B at a point having coordinates (x, y) in the incident on a metallic sphere of radius 10 cm and z = 0 plane is work function 2.4 eV, photoelectrons are emitted. The number of photoelectrons liberated before the m0I()xi − yj m0Ix()jy− i (a) (b) emission is stopped, is 2p()xy22+ 2p()xy22+ –19 1 92−2 (e = 1.6 × 10 C and =×910 N mC ) m0Ix()iy+ j m Iy()ix− j 4pe (c) (d) 0 0 2p()xy22+ 2p()xy22+ (a) 6.25 × 108 (b) 1.25 × 1018 8 18 5. If the absolute errors in two physical quantities A (c) 1.25 × 10 (d) 6.25 × 10 and B are a and b respectively, then the absolute 10. At time t = 0, two bodies A and B are at the same error in the value of A – B is point. A moves with constant velocity v and B starts (a) a – b (b) b – a from rest and moves with constant acceleration. (c) a ± b (d) a + b The relative velocity of B with respect to A when 6. A conductor has a non-uniform section as shown the bodies meet each other is v v in the figure. A steady current is flowing through it. (a) 2v (b) (c) (d) v 2 3 Then the drift speed of the electrons
Physics for you | May ‘15 31 11. For most materials the Young's modulus is n times 17. A stone tied to a rope is rotated in a vertical circle the rigidity modulus, where n is with uniform speed. If the difference between the (a) 2 (b) 3 (c) 4 (d) 5 maximum and minimum tensions in the rope is –2 12. Choose the incorrect statement. 20 N, mass of the stone is (g = 10 m s ) (a) No work is done if the displacement is (a) 2 kg (b) 1 kg perpendicular to the direction of the applied (c) 1.5 kg (d) 0.5 kg ^^ force. 18. A charged particle with velocity v =+xi yj moves ^^ (b) If the angle between the force and displacement in a magnetic field By=+ixj. The magnitude of vectors is obtuse, then the work done is negative. the force acting on the particle is F. The correct (c) Frictional force is non-conservative. option for F is (d) All the central forces are non-conservative. (i) No force will act on the particle if x = y. 13. A pipe closed at one end and open at the other end (ii) Force will act along y axis if y < x. resonates with a sound of frequency 135 Hz and (iii) Force is proportional to (x2 – y2) if x > y. also with 165 Hz, but not at any other frequency (iv) Force is proportional to (x2 + y2) if y > x. intermediate between these two. Then, the frequency (a) (i) and (ii) are true (b) (i) and (iii) are true of the fundamental note of the pipe is (c) (ii) and (iv) are true (d) (iii) and (iv) are true (a) 15 Hz (b) 60 Hz 19. An astronomical telescope arranged for normal (c) 7.5 Hz (d) 30 Hz adjustment has a magnification of 6. If the length 14. Choose the correct statement. of the telescope is 35 cm, then the focal lengths of (a) Polar molecules have permanent electric dipole objective and eyepiece respectively are moment. (a) 30 cm, 6 cm (b) 30 cm, 5 cm (b) CO2 molecule is a polar molecule. (c) 5 cm, 30 cm (d) 40 cm, 5 cm (c) H2O is a non-polar molecule. 20. Two deuterium nuclei each of mass m, fuse together (d) The dipole field at large distances falls off as to form a helium nucleus, releasing an energy E. If 1 . c is the velocity of light, the mass of helium nucleus 2 r formed will be E E 15. The linear momentum of a particle varies with time (a) 2m + (b) t as p = a + bt + ct2 where a, b and c are constants. c2 mc2 E E Then which of the following is correct? (c) m + (d) 2m − (a) Force is dependent linearly on time. c2 c2 (b) Velocity of particle is inversely proportional to 21. How much heat energy must be supplied to time. 14 g of nitrogen at room temperature to raise its (c) Displacement of the particle is independent of temperature by 40°C at constant pressure? time. (Molecular mass of N2 = 28 g, (d) Force varies with time in a quadratic manner. R = universal gas constant) 16. A mass M is suspended from a light spring. An (a) 50R (b) 60R (c) 70R (d) 80R additional mass m is added, displaces the spring 22. An electrical device which offers a low resistance to further by a distance x. Now the combined mass the current in one direction but a high resistance to will oscillate with a period the current in opposite direction is mg (a) current amplifier (b) oscillator (a) T = 2p xM()+ m (c) power amplifier (d) rectifier ()Mm+ x 23. A copper wire and a steel wire of the same length and (b) T = 2p same cross-section are joined end to end to form a mg composite wire. The composite wire is hung from mgx a rigid support and a load is suspended from the (c) T = 2p ()Mm+ other end. If the increase in length of the composite wire is 2.4 mm, then the increase in lengths of steel + (d) ()Mm and copper wires are T = 2p 10 –2 11 –2 mgx (YCu = 10 × 10 N m , Ysteel = 2 × 10 N m )
32 Physics for you | May ‘15 (a) 0.4 mm, 2.0 mm (b) 1.2 mm, 1.2 mm 30. A plane electromagnetic wave travels in free space. (c) 0.6 mm, 1.8 mm (d) 0.8 mm, 1.6 mm Then the ratio of the magnitudes of electric and magnetic fields at a point is equal to 24. Consider a parallel A/4 A/2 (a) energy of electromagnetic wave. plate capacitor of A/4 (b) inverse of the velocity of the electromagnetic capacitance 10 mF wave. filled with air. When AirADielectric ir (c) inverse of the energy of electromagnetic the gap between the wave. plates is filled partly (d) velocity of electromagnetic wave. with a dielectric of dielectric constant 4, 31. The pressureP for a gas is plotted against its absolute as shown in figure, the new capacitance of the temperature T for two different volumes V1 and capacitor is (A is the area of each plate) V2 where V1 > V2. If P is plotted on y-axis and T on (a) 20 mF (b) 40 mF x-axis, then (c) 2.5 mF (d) 25 mF (a) the curve for V1 has greater slope than that for V2. 25. The objective lens of an optical instrument is an (b) the curve for V2 has greater slope than that achromatic combination with a focal length of for V1. 90 cm. The two lenses possess dispersive powers (c) both curves have same slope. 0.024 and 0.036 respectively and are in contact with (d) the curves intersect at some point other than each other. Then their focal lengths are T = 0. (a) –30 cm, 45 cm (b) 45 cm, 30 cm 32. A paramagnetic sample shows a net magnetization (c) 30 cm, –45 cm (d) 30 cm, –30 cm of 0.8 A m–1, when placed in an external magnetic 26. A 3 kg block is placed field of strength 0.8 T at a temperature 5 K. over a 10 kg block and When the same sample is placed in an external both are placed on a magnetic field of 0.4 T at a temperature of 20 K, the smooth horizontal magnetization is surface. The coefficient (a) 0.1 A m–1 (b) 0.2 A m–1 of friction between the blocks is 0.2. If a horizontal (c) 0.4 A m–1 (d) 0.8 A m–1 force of 20 N is applied to 3 kg block, accelerations 33. A mass is suspended from the end of a spring. –2 –2 of the two blocks in m s are (g = 10 m s ) When the system is oscillating the amplitude of 13 14 (a) ,.06 (b) , 3 oscillation is 4 cm and the maximum kinetic energy 4 3 of oscillation of the system is 1 J. Then the force 13 14 constant of the spring is (c) , 3 (d) ,.06 –1 –1 4 3 (a) 2500 N m (b) 1250 N m (c) 500 N m–1 (d) 250 N m–1 27. A flash light lamp is marked 3.5 V and 0.28 A. 34. A p-n-p transistor is used in common emitter The filament temperature is 425°C. The filament mode in an amplifier circuit. When base current resistance at 0°C is 4 W. Then, the temperature is changed by an amount DI , the collector current coefficient of resistance of the material of the B changes by 2 mA. If the current amplification factor filament is is 50, then the value of DI is (a) 8.5 × 10–3 °C–1 (b) 3.5 × 10–3 °C–1 B –3 –1 –3 –1 (a) 15 mA (b) 40 mA (c) 0.5 × 10 °C (d) 5 × 10 °C (c) 50 mA (d) 60 mA 28. If in an amplitude modulated wave, the maximum 35. A shell is fired from a cannon with a velocityv at an amplitude is 10 V and the modulation index is 2/3, angle q with the horizontal. At the highest point in then the minimum amplitude is its path it explodes into two pieces of equal masses. (a) 2 V (b) 7 V (c) 9 V (d) 6 V One of the pieces retraces its path and reaches 29. A body of mass 300 kg is moved through 10 m the cannon. Then the velocity of the other piece along a smooth inclined plane of angle 30°. The immediately after the explosion is work done in moving is (g = 9.8 m s–2) 3 (a) 2vcosq (b) v cosq (a) 4900 J (b) 9800 J 2 (c) 14700 J (d) 2450 J (c) 3vcosq (d) 2vsinq
Physics for you | May ‘15 33 36. A metal rod of length 2 m is rotating with an angular 42. A radioactive substance has density r, volume V velocity of 100 rad s–1 in a plane perpendicular to and decay constant l. If the molecular weight of the a uniform magnetic field of 0.3 T. The potential substance is M, and Avogadro number is NA, then difference between the ends of the rod is the activity of the substance after time t is (a) 30 V (b) 40 V (c) 60 V (d) 100 V lrVNA −lt (a) ()1−e 37. Displacement of a body is ()53ij+−4k m when a M force ()66ij++4k N acts for 5 s. The power is NVA −l (b) e t/2 r (a) 1.6 W (b) 9.6 W (c) 6.4 W (d) 3.2 W M lN lrVN 38. If the wavelength of light that is emitted from (c) A −lt (d) A e−lt e hydrogen atom when an electron falls from orbit VMr M n = 2 to orbit n = 1 is 122 nm, then minimum 43. Two polarizers have their axes inclined at 45° to wavelength of the series is each other. If unpolarized light of intensity I0 is (a) 405 Å (b) 9150 Å incident on the first polarizer, then the intensity of (c) 812 Å (d) 915 Å light transmitted through second polarizer is 39. Three uniform circular discs, I0 I0 (a) (b) (c) I0 (d) 0 each of mass M and radius R are A 4 2 kept in contact with each other 44. An infinitely long thin straight wire has uniform as shown in the figure. Moment 1 linear charge density of Cm−1. Then, the of inertia of the system about the 3 axis AB is magnitude of the electric intensity at a point 18 cm 7 11 2 (a) MR2 (b) MR B away is 4 4 (a) 0.33 × 1011 N C–1 (b) 3 × 1011 N C–1 11 –1 11 –1 11 MR2 (c) 0.66 × 10 N C (d) 1.32 × 10 N C (c) MR2 (d) 2 4 45. The thermal conductivity of a rod depends on 40. A glass slab of thickness 8 cm contains the same (a) length (b) area of cross-section number of waves as 10 cm long path of water when (c) mass (d) material of the rod both are traversed by the same monochromatic 4 soluTioNs light. If the refractive index of water is , the refractive index of glass is 3 1. (c) : The major contribution of Sir C.V. Raman is 5 5 16 3 scattering of light by molecules of a medium. (a) (b) (c) (d) 3 4 15 2 2. (c) : According to Kepler's third law, The period of revolution of a planet around the Sun is 41. A syringe of diameter 1 cm having a nozzle 32/ 2pr 4p23r of diameter 1 mm, is placed horizontally at a T = or T2 = . height 5 m from the ground as shown below. An GMS GMS incompressible non-viscous liquid is filled in the where M is the mass of the Sun and r is the radius syringe and the liquid is compressed by moving S –1 of orbit. the piston at a speed of 0.5 m s , the horizontal 2 3 TJ rJS distance travelled by the liquid jet is \=, where rJS is the distance between the 2 3 (g = 10 m s–2) TE rES Jupiter and Sun and rES is the distance between the Earth and Sun. Here, TJ = 12TE, rJS = nrES 2 3 12TE nrES \ = TE rES 144 = n3 or n = 3 144 3. (d) : Here, (a) 12.5 m (b) 25 m (c) 50 m (d) 75 m Wavelength of the source, l = 50 cm
34 Physics for you | May ‘15 Physics for you | May ‘15 35 Speed of sound = v 8. (c) : The situation is shown in figure. 1 Speed of the source, vvs = 5 As the source is moving away from the stationary 10 m observer, so wavelength of sound heard by the observer is 1 vv+ vv+ s 5 ll′ = = l v v 6 6 ==l (c50 m) = 60 cm 5 5 4. (d) : The wire carries the current I in the negative z-direction. OA represents 20 m north, represents 10 m The magnetic field B at point P is perpendicular to AB OP in the direction shown in figure. east and BC represents 10 2m south-west. ^^ \ BB=−00sincqqiBos j If ijand be the unit vectors along east and north respectively, then m0I Here, B0 = ^ ^ 2pr OA = 20 j m , AB = 10 i m y x ^^ sincqq==and os BC =−(c10 24os 51°−ij02sin)45° m r r m 1 ^^ 1 0I =− 10 2 ij−10 2 m \ B =−()yi xj 2pr2 2 2 ^^ m Iy()ix− j 2 2 2 =−()10 ij−10 m = 0 (as r = x + y ) 22 2p()xy+ \ Displacement OC =+OA AB + BC
5. (d) : The absolute error in the difference of two ^^ ^^ =+−− physical quantities is equal to the sum of the OC 20 jim+10 mm()10 ij10 ^ absolute errors in the individual quantities. =10 j m = 10 m due north \ The absolute error in A – B is = a + b 9. (c) : Here, I Radius of the sphere, r = 10 cm = 10 × 10–2 m 6. (c) : The drift speed is vd = neA Work function, f0 = 2.4 eV where n is the number density of electrons and A is Energy of a photon, E = hu = 4.2 eV the area of cross-section of the conductor. According to Einstein's photoelectric equation u f 1 Kmax = h – 0 For steady current, vd ∝ But K = eV A max 0 As A increases from P to Q, so v decreases from P where V0 is the stopping potential. d \ eV = hu – f to Q. 0 0 eV0 = 4.2 eV – 2.4 eV = 1.8 eV 7. (b) : Here, or V0 = 1.8 V L = 0.01 H, R = 3pW, u = 50 Hz The sphere will stop emitting photoelectrons, when The inductive reactance is the potential on its surface becomes 1.8 V. XL = wL = 2puL Let n be the number of photoelectrons emitted from = 2p(50 Hz)(0.01 H) = p W the sphere. Then charge on the sphere is q = ne The phase difference f between the current and Potential on the surface of the sphere is 1 q 1 ne the voltage is V == 4pe00r 4pe r −1 XL XL − f=tan as tanf= Vr (.18 V)(101× 0 2 m) R R \ n = = 1 ()91××0192Nm CC−−21(.610 9 ) p W e −−111 p 4pe0 = tant = an =°30 = 3p W 3 6 = 1.25 × 108
36 Physics for you | May ‘15 10. (d) : Let the two bodies A and B meet at time t. 17. (b) : Let m be the mass of the stone. Distance travelled by A in time t, As Tmax – Tmin = 2mg sA = vt TTmaxm− in 20 N \=m ==1 kg Distance travelled by B in time t, 2g 21()0 ms−2 1 2 saB = t ( u = 0) 18. (b) : The force acting on a charged particle of charge 2 At the time of meeting, q moving with velocity v in a magnetic field B is sA = sB Fq=×()v B 1 2 2v ^^ ^^ \ vt = at or a = ...(i) =+ =+ 2 t Here, v xi yj and Byixj ^^ ^^ Velocity of A after timet , vA = v \=Fq[(xi+×yj)(yi+ xj)] Velocity of B after time t, ^ =−22 2v qx()yk va==t tv= 2 (using (i)) B t The magnitude of the force is \ The relative velocity of B with respect to A when Fq=−[(xy22)]22=−qx()y2 the bodies meet each other is (i) If x = y, then F = 0 v = v – v = 2v – v = v BA B A (iii) If x > y, then F ∝ (x2 – y2) 11. (b) : For most materials, the modulus of rigidity h 19. (b) : Let f and f be focal lengths of objective and is one third of the Young's modulus Y. o e 1 eyepiece respectively. i.e. h= Y or Y = 3h \ n = 3 3 For normal adjustment, f 12. (d) : All the central forces are conservative. Magnification of the telescope, m = o f 13. (d) e and length of the telescope, L = fo + fe 14. (a) : CO2 is a non-polar molecule. As m = 6 and L = 35 cm (given) H2O is a polar molecule. 1 fo The dipole field at large distances falls off as . \ 6 = or fo = 6 fe ... (i) r3 fe 15. (a) : Here, 2 and 35 cm = fo + fe ... (ii) Linear momentum, p = a + bt + ct On solving eqns. (i) and (ii), we get dp As force, F = fo = 30 cm and fe = 5 cm dt d 20. (d) : The given reaction is \=F ()ab++tct2 2 2 4 dt 1 HH+→1 2He F = b + 2ct The energy released in the reaction is 2 16. (b) : Let k be spring constant of the spring. E = (m + m – mHe)c 2 When the mass M is suspended from the spring, let E = (2m – mHe)c it be stretched by distance y. E E =−22mmHe or mmHe =− \ Mg = ky …(i) c22c Now when the additional mass m is added to it, it 21. (c) : Here, stretches further by the distance x. Mass of N2,m = 14 g \ (M + m)g = k(x + y) …(ii) Molecular mass of N2, M = 28 g Subtracting eqn. (i) from eqn. (ii), we get m 14 g 1 (M + m)g – Mg = k(x + y) – ky Number of moles of N2, n == = mg M 28 g 2 mg = kx or k = …(iii) x Rise in temperature, DT = 40°C The period of the combined mass is The amount of heat supplied at constant pressure is ()Mm+ ()Mm+ DQ = nCPDT T = 2p = 2p (using (iii)) As nitrogen is diatomic, its molar specific heat at k ()mg/x constant pressure is ()Mm+ x 7 = 2p CRP = mg 2
Physics for you | May ‘15 37 1 7 ee00()A/4 A \=DQR ()40°=C 70R C3 == 2 2 d 4d 22. (d) Q C1, C2 and C3 are in parallel. 23. (d) : As same load (say W) is being applied on both \ The new capacitance of the capacitor is the wires, which have same area of cross-section A, Cnew = C1 + C2 + C3 so stress is same for both wires. eeA KAeeA A KAe =+00+=+00 0 As stress = Young's modulus × strain 42d d 42d d 2d W DDLCu Lsteel e A 10 mF \=YCu = Ysteel 0 A L L =+()1 K =+()14 (using (i)) Cu steel 2d 2 DLCu LCu Ysteel = 25 mF = DL L Y steel steel Cu 25. (c) : Let f1 and f2 be focal lengths of the two lenses 10 –2 Here, LCu = Lsteel, YCu = 10 × 10 N m respectively. If F be focal length of the combination, then Y = 2 × 1011 N m–2 steel 11 1 − =+ DL Y 21× 011 N m 2 \=Cu steel = = 2 Ff12f 10 −2 DLsteel YCu 10 ×10 Nm As F = 90 cm (given) DLCu = 2DLsteel ... (i) 1 11 \ =+ ... (i) The increase in length of the composite wire is 90 cm ff12 = DLCu + DLsteel For an achromatic combination, But DLCu + DLsteel = 2.4 mm (given) ... (ii) ww 1 +=2 0 \ 2DLsteel + DLsteel = 2.4 mm (using (i)) ff1 2 3DLsteel = 2.4 mm or DLsteel = 0.8 mm where w1 and w2 be their dispersive powers. Putting this value in eqn. (i), we get As w1 = 0.024 and w2 = 0.036 (given) DLCu = 2(0.8 mm) = 1.6 mm \ 0..024 +=0 036 24. (d) : Capacitance of a parallel plate capacitor filled 0 ff12 with air is 0..024 0 036 e A =− C ==0 10 mF (given) ... (i) ff d 12 3 where d is the distance between the plates. ff21=− ... (ii) 2 When the gap between the plates is filled partly with a dielectric of dielectric constant K(= 4) as shown in Putting this value of f2 in eqn. (i), we get 32− figure. Then, 1 =−12= = 1 90 cm ff113 3 ff113 90 f1 ==cm 30 cm 3 Putting this value of f1 in eqn. (ii), we get 3 f2 =− (c30 m) =−45 cm 2 26. (d) : Here, Capacitance of part I is Mass of upper block, m1 = 3 kg Mass of lower block, m2 = 10 kg ee00()A/4 A C1 == Horizontal force, F = 20 N d 4d Coefficient of friction between the blocks, Capacitance of part II is m = 0.2 ee KA00()/2 KA Let a1 and a2 be accelerations of upper and lower C2 == d 2d blocks respectively. Then Capacitance of part III is the equation of motion of upper block is
38 Physics for you | May ‘15 Here, m = 300 kg, s = 10 m, g = 9.8 m s–2, q = 30° \ W = (300 kg)(9.8 m s–2)(10 m)sin30° = 14700 J F – f = m a 1 1 30. (d) : The magnitude of electric and magnetic fields Ff− Fm−m 1g E a1 = = in an electromagnetic wave are related as B = 0 m m 0 1 1 c −2 20 Nk− (.02)(31gm)( 0 s ) E0 = Thus, = c 3 kg B0 − where c is the velocity of electromagnetic wave in 20 NN6 14 −2 = = ms free space. 33kg The equation of motion of lower block is 31. (b) : According to gas equation, PV = nRT f = m2a2 a 2 f nR m \=P T f mmg1 2 V a2 == m2 m2 Comparing this with y = mx, P-T graph is a straight nR −2 line passing through origin with slope = . (.02)(31kg)( 0 ms ) 6 N − = ==06. ms 2 1 V Slope ∝ 10 kg 10 kg P V V2 27. (d) : The filament resistance at 0°C is As V1 > V2 (given) V1 \ (Slope) < (Slope) R0 = 4 W (given) 1 2 and at 425°C is or (Slope)2 > (Slope)1 T 35. V 32. (a) : The magnetisation of a paramagnetic sample R425 ==12.5 W B 02. 8 A is MC= Let a be required temperature coefficient of T where B is the external magnetic field, T is the resistance. absolute temperature and C is the Curie's constant. As RT = R0(1 + aT) RR− RR− M1 CB11/T B1 T2 \=a T 0 = 425 0 \= = M2 CB22/T B2 T1 RT0 R0()425°C WW− 12.54 −−31 B2 T1 = =×510 °C MM21= 4W()425°C B1 T2 2 m= −1 04. T 5 K 28. (a) : Here, Amax = 10 V, = (.08 A m) 3 08. T 20 K AA− Modulation index, m= maxmin = 0.1 A m–1 AAmaxm+ in 10 V − A 33. (b) : Here, 2 = min Amplitude of oscillation, A = 4 cm = 4 × 10–2 m 3 10 V + Amin Maximum kinetic energy, Kmax = 1 J 2(10 V + Amin) = 3(10 V – Amin) 1 2 20 V + 2Amin = 30 V – 3Amin As Kkmax = A 2 5Amin = 10 V or Amin = 2 V \ The force constant of the spring is 29. (c) : The work done in moving a body along a 2K 21()J smooth inclined plane is k ==max = 1250 N m–1 22− 2 W = mgsinqs A ()41× 0 m where m is the mass of the body, q is the angle of 34. (b) : Here, inclination, s is the distance moved by the body Change in collector current, DIC = 2 mA along the inclined plane. Current amplification factor, b = 50
Physics for you | May ‘15 39 DI 3 4 As b= C =+()66ij+⋅4ki Nm+−jk s−1 DIB 5 5 18 16 DIC 2 mA −3 =+6 − W \ DIB == ==00..40mA 04 ×10 A b 50 5 5 –6 30 +−18 16 32 = 40 × 10 A = 40 mA = WW==64. W 5 5 35. (c) : The flight of the shell before explosion is shown 38. (d) : The given transitionn = 2 to n = 1 corresponds in figure where H is the highest point. to first line of Lyman series. For Lyman series, 11 1 =−R where,n = 23,,4 .... l 122n For first line, n = 2 11 1 3 ... (i) \=RR − = l 1222 4 Velocity of the shell at H = vcosq For minimum wavelength of the series, n = ∞ At H, it explodes into two pieces each of mass m/2 11 1 and denoted by 1 and 2 respectively. Let v1 and v2 \=RR− = ... (ii) l 22 be velocities of the pieces 1 and 2 immediately after min 1 ∞ the explosion. Dividing eqn. (i) by eqn. (ii), we get As the piece 1 retraces its path to the cannon, lmin 3 3 \ v = –vcosq ==or llmin 1 l 4 4 Applying the law of conservation of momentum, m m But l = 122 nm (given) mv cosq= v12+ v 22 3 \=lmin (122 nm) = 91.5 nm = 915 Å m m 4 mv cos(qq=−v cos)+ v2 22 39. (b) : Moment of inertia of each disc of mass M and v2 = 3vcosq radius R about its diameter is 36. (c) : Here, 1 2 IMdia = R Length of the rod, l = 2 m 4 A Moment of inertia of disc 1 Angular velocity, w = 100 rad s–1 1 2 Magnetic field, B = 0.3 T about the axis AB is 2 The potential difference between the ends of the I1 = Idia + MR 3 rod is (by theorem of parallel axes) 1 2 1 2 −1 B ew= Bl = (.03T)(2 mr)(100 ad s) 1 225 2 2 2 =+MR MR = MR 4 4 = 60 V 5 2 Similarly, IM2 = R 37. (c) : Here, 4 Moment of inertia of disc 3 about the axis AB is Displacement,(Dri=+53jk− 4) m 1 2 II3 ==dia MR Force),(Fi=+66jk+ 4 N 4 \ Moment of inertia of the system about the axis Dr ()53ij+−4k m Velocity, v == AB is I = I1 + I2 + I3 Dt 5 s 5 2225 1 3 4 =+MR MR + MR =+ij− k ms−1 4 4 4 5 5 11 = MR2 Power, PF=⋅v 4
40 Physics for you | May ‘15 40. (a) : Number of waves in glass slab is 42. (d) : Number of atoms present in a radioactive Thickness of glass slab 8cm m = N0 = NA N1 = substance is M Wavelength of light in glass lglass Number of waves in water is where m is the mass of the substance, M its molecular Length of path of water 10 cm weight and NA is Avogadro number. N2 ==As m = Vr Wavelengthof light in water lwater Vr As N1 = N2 (given) \ N0 = NA ...(i) M 81cm 0 cm \ = The activity of the substance after time t is ll –lt –lt glasswater R = lN = lN0e ( N = N0e ) r lglass 8 cm 4 V −lt or ...(i) R = l NeA == M (using (i)) lwater 10 cm 5 As refractive index of a medium, lrVNA −l R = e t Wavelengthof light in vacuum M m= Wavelengthof light in medium 43. (a) : If I0 is the intensity of unpolarised light, then intensity of light transmitted through first polariser m glass lwater \ = I0 m l is I1 = water glass 2 lwater Intensity of light transmitted through second or mglass = mwater lglass polariser is I = I cos2q 5 4 5 2 1 = = (using (i)) (where q is the angle between axes of two 4 3 3 polarisers) 2 41. (c) : II0 2 0 1 I0 =°cos 45 = = 2 2 2 4 44. (a) : The magnitude of electric field intensity due to an infinitely long straight wire of uniform linear charge density l is Here, l 1 2l –2 == Diameter of syringe, d1 = 1 cm = 10 m E 2pe00rr4pe –3 Diameter of nozzle, d2 = 1 mm = 10 m where r is the perpendicular distance of the point According to equation of continuity, A v = A v 1 1 2 2 from the wire. 2 2 dd1 2 1 −1 –2 pp vv1 = 2 Here, l= Cm , r = 18 cm = 18 × 10 m 22 3 1 2 2 =× 92−2 10−2 m 10−3 m 910 Nm C −1 4pe0 pp (.05ms ) = v2 2 2 92−−211 −42 ()91× 02NmCC() m 10 m −1 3 or v2 = (.05ms ) \ E = 10−62m ()18 ×10−2 m –1 = 50 m s 1 11 −−1111 Horizontal distance travelled by the liquid jet is =×10 NC =×03. 310 NC 3 2h 25( m) R ==v ()50 ms−1 = 50 m 45. (d) : The thermal conductivity of a rod depends on 2 −2 g 10 ms material of the rod. mm
Physics for you | May ‘15 41 EXAM ON 14th to 29th MAY BITSAT PRACTICE PAPER 1. A short pulse of white light is incident from air to 5. The velocity displacement a glass slab at normal incidence. After travelling graph of a particle moving through the slab, the first colour to emerge is along a straight line is (a) blue (b) green shown. The most suitable (c) violet (d) red acceleration-displacement 2. A rectangular loop graph will be carrying a current I is situated near a long straight wire (a) (b) such that the wire is parallel to one of the sides of the (c) (d) loop and is in the plane of the loop. If steady current I′ is established in the wire as shown in the figure, the loop will 6. A thin circular ring of area A is held perpendicular (a) rotate about an axis parallel to the wire to a uniform magnetic field of induction B. A (b) move away from the wire small cut is made in the ring and a galvanometer (c) move towards the wire is connected across the ends such that the total (d) remain stationary. resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing 3. Force between two identical charges placed at through the galvanometer is a distance of r in a vacuum is F. Now a slab of 2 BR AB BA dielectric constant 4 is inserted between these two (a) (b) (c) ABR (d) . charges. If the thickness of the slab is r/2, then the A R R2 force between the charges will become 7. What is the tension in a rod of length L and mass 3 (a) F (b) F M at a distance y from F1, when the rod is acted on 5 by two unequal forces F1 and F2 ( 42 Physics for you | May ‘15 (–50 cm, 0). The coordinates of the image will be 11e0A (c) e0A/6d (d) (a) (50 cm, –0.5 cm) (b) (–50 cm, –1 cm) 6d (c) (50 cm, –1 cm) (d) (–50 cm, –0.5 cm) 15. What is the power of an engine, which can just 9. Imagine a light planet revolving around a massive pull a train of mass 5000 quintals up an incline of –1 star in a circular orbit of radius r with a period of 1 in 50 at the rate of 54 km h . The resistance due –2 revolution T. If the gravitational force of attraction to friction is 0.8 N/quintal. Take g = 9.8 m s . between planet and the star is proportional to r–5/2, (a) 1530 kW (b) 1470 kW then the relation between T and r will be (c) 3060 kW (d) 5000 kW –2 2 7/2 (a) T ∝ r (b) T ∝ r 16. Consider a metal exposed to light of wavelength (c) T 3 ∝ r4 (d) T ∝ r3/2 600 nm. The maximum energy of the electron 10. A 4 mF capacitor and a resistance of 2.5 MW are in doubles when light of wavelength 400 nm is used. series with 12 V battery. The time after which the The work function in eV could be potential difference across the capacitor is 3 times (a) 1.04 (b) 2.08 the potential difference across the resistor is (c) 1.40 (d) 1.68 [Given, ln2 = 0.693] 17. If the shear modulus of a wire material is (a) 13.86 s (b) 6.93 s (c) 7 s (d) 14 s. 5.9 × 1011 dyne/cm2 then the potential energy of a wire of 4 × 10–3 cm in diameter and 5 cm long 11. A boat which has a speed of 5 km h–1 in still water twisted through an angle of 10′, is crosses a river of width 1 km along the shortest (a) 1.253 × 10–12 J (b) 2 × 10–12 J possible path in 15 minute. The velocity of the river (c) 1.00 × 10–12 J (d) 0.8 × 10–12 J water in km h–1 is 18. In given figure, the potential difference between 41 (a) 1 (b) 3 (c) 4 (d) the points C and D is 1F 2F 12. A steady current flows in a metallic conductor of C non-uniform cross-section. The quantity/quantities B A 3 6 constant along the length of the conductor is/are (a) current, electric field and drift speed D (b) drift speed only = 12 V 1 (c) current and drift speed (d) current only. (a) 7.2 V (b) 3.6 V (c) 10.8 V (d) 0 V nn21− 13. The number of particles given byn = – D 19. A point mass is subjected to two simultaneous xx21− sinusoidal displacements in X-direction : are crossing a unit area perpendicular to x-axis in 2p x1(t) = A sin wt and x2(t) = A sin wt + . unit time, n1, n2 are number of particles per unit 3 volume, for the values of x meant to be x1 and Adding a third sinusoidal displacement x2. What is the dimensional formula of diffusion x3(t) = Bsin(wt + f) brings the mass to a complete constant D? rest. The values of B and f are 2 –2 (a) [L] (b) [L T ] 3p 4p 2 –1 2 (a) 2A, (b) A, (c) [L T ] (d) [L T] 4 3 5p p 14. The expression for the A (c) 3A, (d) A, equivalent capacitance 6 3 d of the system shown in 20. A 100 V ac source of frequency 500 Hz is connected figure is (A is the corss- A to an LCR circuit with L = 8.1 mH, C = 12.5 mF sectional area of one of d 3d and R = 10 W, all connected in series. The potential the plates) difference across the resistance is (a) e A/3d A 0 d (a) 25 V (b) 50 V 3e A (b) 0 (c) 75 V (d) 100 V d 3A Physics for you | May ‘15 43 21. A particle is projected upward from the surface of 28. A small sphere of mass 0.2 g hangs by a thread the earth (radius R) with a kinetic energy equal to between two parallel vertical plates 5 cm apart. half the minimum value needed for it to escape. To The charge on the sphere is 6 × 10–9 C. What is the which height does it rise above the surface of the potential difference between the plates if the thread earth? assumes an angles of 30° with the vertical? (a) 2 R (b) R (c) 3R/2 (d) R/2 (a) 9.4 × 103 V (b) 9.4 × 104 V 3 4 22. If Ka-radiation of Mo (Z = 42) has a wavelength (c) 5.2 × 10 V (d) 5.2 × 10 V 0.71 Å, the wavelength of the corresponding 29. A lift is moving down with accelerationa . A man in radiation of Cu(Z = 29) is equal to the lift drops a ball inside the lift. The acceleration (a) 1.52 Å (b) 0.71 Å of the ball as observed by the man in the lift and (c) 0.36 Å (d) 2.5 Å a man standing stationary on the ground are 23. In the test experiment on a model aeroplane in respectively a wind tunnel, the flow speeds on the upper and (a) g, g (b) a, g – a lower surfaces of the wing are 70 m s–1 and 63 m s–1 (c) g – a, g (d) a, g respectively. What is the lift on the wing if its area is 30. A coil having N turns is wound tightly in the 2 –3 2.5 m ? Take the density of air to be 1.3 kg m . form of a spiral with inner and outer radii a and b 3 5 (a) 10 N (b) 10 N respectively. When a current I passes through the 3 3 (c) 1.5 × 10 N (d) 3 × 10 N coil, the magnetic field at the centre is –5 24. Two particles have charges + 4.0 × 10 C and m0NI 2m0NI –5 (a) (b) –4.0 × 10 C and each has a mass of 5 g. When b a they are 1.0 m apart, they are released from rest. m m0NI b 0IN a (c) ln (d) ln . The speed of each particle when their separation 2(ba − ) a 2(ba − ) b becomes 50 cm would be (a) 26 m s–1 (b) 120 m s–1 31. The opposite faces of a cubical block of iron of (c) 53.7 m s–1 (d) 82.7 m s–1 cross-section 4 cm2 are kept in contact with steam 25. A closed container of volume 0.02 m3 contains a and melting ice. The quantity of ice melted at the end of 10 min would be mixture of neon and argon gases at a temperature –1 –1 –1 of 27°C and pressure 1 × 105 N m–2. The total mass (Given, K for iron = 0.2 cal s cm C° ). of the mixture is 28 g. If the gram molecular weights (a) 0.3 kg (b) 0.5 kg of neon and argon are 20 and 40 respectively, find (c) 1.0 kg (d) 1.5 kg the mass of the individual gases in the container, 32. A sphere of mass m moving with a velocity u hits assuming them to be ideal. (Universal gas constant, another stationary sphere of same mass. If e is the R = 8.314 J mol–1 K–1) coefficient of restitution, what is the ratio of the (a) 4 g, 24 g (b) 20 g, 8 g velocities of two spheres after the collision? (c) 26 g, 2 g (d) 6 g, 22 g (a) 1 (b) e 1+ e 2 − e 26. For the given uniform square lamina ABCD, whose (c) (d) centre is O, 1− e 1+ e 33. The output of a step-down transformer is measured (a) I = 2I AC EF to be 24 V when connected to a 12 W light bulb. (b) 2IAC = IEF The value of the peak current is (a) (b) (c) IAD = 3IEF ()12/ A 2A (d) IAC = IEF (c) 2 A (d) ()22A 27. One gram of radium is reduced by 2 mg in 5 years 34. When a mass m is connected by a-decay. The half-life period of radium is individually to two springs (a) 300 years (b) 585 years S1 and S2, the oscillation (c) 1735 years (d) 2105 years frequencies are u1 and u2 respectively. If the same 44 Physics for you | May ‘15 mass is attached to the two springs as shown in the Out of the following diagrams, which represents figure, the oscillation frequency would be the T-P diagram? 2 2 (a) u1 + u2 (b) uu1 + 2 −1 11 2 2 (c) + (d) uu1 − 2 uu12 35. Mercury has an angle of contact equal to 140° (i) (ii) with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury (iii) (iv) at the temperature of the experiment is 0.465 –1 3 –3 (a) (iv) (b) (ii) N m . Density of mercury is 13.6 × 10 kg m (c) (iii) (d) (i) (cos140° = – 0.7660) 40. A solid cylinder rolls down an inclined plane. Its (a) 2.9 mm (b) 7.2 mm mass is 2 kg and radius 0.1 m. If the height of the (c) 3.5 mm (d) 5.3 mm inclined plane is 4 m, what is its rotational K.E., 36. A string of length l fixed at one when it reaches the foot of the plane? end carries a mass m at the (a) 26 J (b) 13 J (c) 52 J (d) 78 J other end. The string makes soluTioN 2 rev s –1 around the horizontal 2 p 1. (d) : Since lr > lb > lv and m = a + b/l axis through the fixed end as So, mr < mg < mb < mv . . shown in the figure. The tension in the string is or vr > vg > vb > vb ( . v = c/m), (a) 16 ml (b) 4 ml So, red colour (with largest velocity in glass slab) is (c) 8 ml (d) 2 ml the first to emerge. 37. A whistle producing sound waves of frequencies 2. (c) : Current I′ in wire and current I in AB are 9500 Hz and above is approaching a stationary parallel and in same direction. person with speed v m s–1. The velocity of sound in s B air is 300 m s–1. If the person can hear frequencies C upto a maximum of 10,000 Hz, the maximum value I I of vs upto which he can hear the whistle is (a) 15/ m s–1 (b) 15 m s–1 2 D –1 –1 A (c) 30 m s (d) 15 2 m s AB will be attracted towards wire. Wire and CD 38. A sinusoidal voltage is applied directly across an carry parallel currents in opposite directions. 8.00 mF capacitor. The frequency of the source is Repulsion occurs between them. The distance is 3.00 kHz and the voltage amplitude is 30.0 V. The more from wire and so the force of repulsion on CD displacement current between the plates of the will be less than the force of attraction on AB. capacitor is BC and DA are at right angles to wire and carry (a) 1.9 A (b) 2.5 A currents in opposite directions. Their net force will (c) 4.5 A (d) 5.2 A be zero. 39. Consider P-V diagram for an ideal gas shown in Due to greater attractive force on AB, the loop will figure. move towards the wire. P 1 2 constant q P = 3. (c) : In vacuum, F = V 2 4pe0r 2 With the dielectric in between the charges, V Physics for you | May ‘15 45 ELECTROSTATICS BRAIN In dry weather, a spark is produced by walking across certain types of carpet and then bringing one of the fingers near a metal doorknob, or even a person. Multiple sparks can be produced when a sweater or cloth is pulled from the body. It reveals that electric charge is present in our bodies, sweaters, carpets, doorknobs, computers etc. In fact, every object contains a vast MAP amount of electric charge. Coulomb's Law Electric Charge Electric Dipole lElectric force between two point charges : lA phy s i c a l qu ant it y re s p ons i bl e for lA pair of equal and opposite charges q and –q r r 1 q1q2 electromagnetism. There are only two kinds of separated by a small distance 2a. F12 = rˆ12 =−F21 2 l 4πε0K r12 charge, positive and negative. The direction of dipole is the direction from –q to q. 1 where K = 1 for free space, =9×109 N m2 C−2 lConservation of charge : The total charge of an lDipole moment, p = q × 2a r 4πε 1 pcosθ1 p⋅rˆ 0 isolated system is always conserved. lElectric Potential, V = = lCoulomb's law agrees with Newton's third law. 4πεr2 4πεr2 lQuantisation of charge : Q = ne 0 0 l Force on charge q1 due to remaining charges q2, q3, lDipole field at an axial point 1 2pr where, n = ±1, ±2, ±3, ... and e = 1.6 × 10–19 C E = q , ... q in the region. axial 2 2 2 4 n at distance r from the centre 4πε0 (r −a ) rq n q of the dipole F =1 i rˆ Electric Potential 1 2p 1 ∑2 1i E = 4πε0 i=2 r when r >> a axial 3 1i lWork done by an external force (equal and opposite 4πε0 r It is a vector sum of all the forces acting on point l 1 p to the electric force) in bringing a unit positive Dipole field at an E = charge q . equatorial 2 2 3/2 1 charge from infinity to a point = electrostatic equatorial point 4πε0 (r +a ) potential (V) at that point. at distance r from the centre of the dipole is 1 q 1 p lDue to a point charge : V(r) = E = Electric Field 4πεr when r >> a equatorial 3 r 0 4πε0 r rF 1 q lPotential difference between two points, lDue to a point charge, E == rˆ lTorque, τ = pE sin θ 2 l qtest 4πε0 r UP −UR WRP Potential energy of an electric dipole in a uniform VP −VR = = lDue to positive charge, electric field is away from q q electric field, U = –pE (cos q2 – cos q1) l ? the charge and due to negative charge, it is towards Potential at a point P, due to system of charges If initially the dipole is perpendicular to trher field E, q1 = 90° and q2 = q , then U =−pE cosθ=−p⋅E the charge. q1, q2, q3, ... 1 q q q ?If initially the dipole is parallel to the field E, q1 = 0° lDue to charge distribution, V =V +V +V +... =1 +2 +3 +... P 1 2 3 4πεr r r and q2 = q , then U = – pE (cos q – 1) = pE(1 – cos q) r1 λ∆L 0 1 2 3 Linear charge, E = rˆ ∑2 It is algebraic sum of the potentials due to the 4πε0 r individual charges. r1 σ∆S Dielectric and Polarisation Surface charge, E = rˆ ∑2 4πε0 r lA dielectric develops a net dipole moment in an r1 ρ∆V Relation between Field and Potential Volume charge, E = rˆ external electric field. 4 ∑2 dV πε0 r lE =− net dipole moment dr lPolarisation = volume rr lElectric field is in the direction in which the For a linear isotropic dielectric, P =χE Electric Flux and Gauss's Law e potential decreases steepest. lSurface charge density due to polarisation, r lElectric flux through a plane surface area DS held in lIts magnitude is given by the change in the σP =P ⋅nˆ =P cosθ a uniform electricr fieldr magnitude of potential per unit displacement ∆φ=E ⋅∆S =E∆Scosθ E normal to the equipotential surface at the point. Capacitors and Capacitance lAccording to Gauss's law, electric flux through a lA capacitor is a system of two conductors separated closed surface S is equal to (1/e) times charge Electrostatics of Conductors 0 by an insulator. enclosed. lAt the surface of a charged conductor, electrostatic q q rrq lCapacitance, C == φ=E ⋅dS =en E Ñ∫ field must be normal to the surface at every point. Vo V1 −V2 ε0 l l pe The interior of a conductor can have no excess For a spherical conductor, C = 4 0R charge in the static situation. lEnergy stored in a charged conductor or capacitor, Applications of Gauss's Law lElectrostatic potential is constant throughout the 1 1 q2 1 U =CV 2 ==qV volume of the conductor and has the same value (as 2 2 C 2 lElectric field due to a long straight wire λ E = inside) on its surface. of uniform linear charge density l, 2π εr 0 lElectric field at the surface of a charged conductor Different Types of Capacitors lElectric field due to an infinite plane sheet σ rσ E = E =nˆ where s is the surface charge density and s Kε0A of uniform surface charge density , 2ε0 ε0 lCapacity of parallel plate capacitor, C = lElectric field due to two positively charged parallel nˆ is a unit vector normal to the surface in the d ss outward direction. 1 1 plates with charge densities 1 and such that l 2 Induced charges, σi =σ1−or qi =q1− ss K K 1 > 2 > 0, Potential of Charge Distributions lCapacity of capacitor εA 1 1 C =0 E =(σ1 +σ2) E =(σ1 −σ2) Charge distribution Formula partially filled with a t 2ε0 2ε0 d −t + (Outside the plates) (Inside the plates) Kq σR Kq dielectric, K Uniformly charged V =V ==, V = 2 l i s o lForce of attraction between q Electric field due to two equally and oppositely spherical shell R ε0 r F = charged parallel plates, plates of capacitors, 2Aε0 σ Kq ab E = Uniformly V =(1.5R2 −0.5r2) lCapacity of spherical capacitor, C =4πε E = 0 (For outside points), (For inside points) i 3 0 ε0 charged solid sphere R b −a lElectric field due to a thin spherical shell of charge Kq Kq lCapacity of cylindrical capacitor 2πε0 V =, V = density s and radius R, s R o r per unit length, ln(b /a) 1 q 1 q Kq E = E = On the axis of 2 E = 0 2 V = Combination of Capacitors 4πε0 r 4πε0 R uniformly charged 2 2 R +x Kq For r > R For r < R For r = R ring At centre, x = 0 ∴V = 1 1 1 1 1 (Outside points) (Inside points) (At the surface) R lCapacitors in series : =+++... =∑ where, q = 4pR2 s C C1 C2 C3 i Ci . Infinitely long λr2 l S lElectric field of a solid sphere of uniform charge PD =ln Capacitors in parallel : C = C1 + C2 + C3 + ... = Ci line charge 2πεr i density r and radius R 0 1 1 q 1 qr 1 q E = E = E = Electric Potential Energy Van de Graaff Generator 4πεr2 4πε3 4πε2 0 0 R 0 R lElectric potential energy of a 1 q q lBy means of a moving belt and suitable brushes U = 1 2 For r > R For r < R For r = R charge is continuously transferred to the shell and system of two point charges, 4πε0 r12 (Outside points) (Inside points) (At the surface) potential difference of the order of several million l 1 q j qk 4 Electric potential energy of U = Σ q =πR3ρ volts is built up, which can be used for accelerating where a system of N point charges, 4πε0 all pairs rjk 3 (j > k) charged particles. 2 7. (d) : Here, as is clear in figure, net force on the rod q F ′ = 2 F = (F1 – F2) 4pe0Kr′ If F ′ = F, then Kr′2 = r2 or r = K r′. In the given situation, q2 4 q2 4 \ F ′ = ==F Acceleration of rod, along F1 2 2 rr 9 4pe r 9 F ()FF12− 4pe + 4 0 a = = …(i) 0 2 2 M M M 4. (a) : When glass is introduced in the path of one Mass of part AB of the rod = y of interfering beams then an extra path difference L is produced. Intensity of the position of central Let T be the tension in the rod at B. maxima occured previously remains unchanged if Equation of motion of part AB of the rod is extra path difference is equal to integral multiple of M FF12− F1 – T = ma = y ⋅ wavelength of light i.e., Dx = (m – 1)t = nl, n = 1, 2, 3 L M nl 1l for minimum thickness, n = 1, t = = y y m − − F1 – T = (F1 – F2) or T = F1 – (F1 – F2) t = 2l ()1 (.15 1) L L 5. (a) : The velocity-displacement graph is given. T = F1(1 – y/L) + F2(y/L) Corresponding acceleration-displacement graph is 8. (c) : In figureOO ′ be the principal axis of the convex required. lens before it is cut into two pieces. In this case, the From v-x graph, the corresponding equation is object is 0.5 cm above OO′. v v =−0 x + v 0 (Pattern, y = mx + c) Top part x0 v (– 50 cm, 0) dv v0 dx 0 \ = − or a = − v Object 0.5 cm (0, 0) 0.5 cm x0 dt x0 dt O 0.5 cm ImOage v v Principal (50 cm, –1 cm) 0 0 axis or a = − ×− xv+ 0 x0 x0 2 v v2 or a =0 x − 0 (Pattern, y = mx + c) 11 1 11111 1 x x As −= , =+= += 0 0 vu f vuf −50 25 50 2 v0 or v = 50 cm Slope of the line= x0 Further, magnification of the image, i.e., q v 50 Slope is positive. Therefore is acute. m = = = – 1 v2 u −50 Intercept = 0 . It is negative. Therefore, the image of the object is formed at 50 cm x0 from the pole and 0.5 cm below the principal axis. The slope and intercept reveal that the graph (a) is Thus, the co-ordinates of the image with respect to the most suitable representation. time the X-axis through the edge of the top part of 6. (b) : EMF induced in the ring = e the cut lens are (50 cm, –1 cm). − df \ e = , where f = flux = BA 9. (b) : Force of gravitation on the planet dt = Centripetal force −f d 2 or IR = or RI dt = −f d –5/2 2 2p dt kr = mrw = mr T or − f or Rq = | f | 2 2 RI∫ dt = ∫ d 44ppmr m or T2 = = ⋅r7/2 BA −52/ k or q = kr R \ T2 ∝ r7/2. 48 Physics for you | May ‘15 t e A e V − e0A 0 0A 10. (a) : Charging current = I = 0 e RC 14. (d) : Here, C1 = , C2 = , C3 = R d 2d 3d As the three condensers are in parallel in figure. t − RC ee00A A e0A \ Potential difference across RV, R = Ve0 \ Cp = C1 + C2 + C3 = ++ −tR/ C d 23d d \ Potential difference acrossCV , C = VV00− e ee00A 1 1 11 A Cp = 1++ = According to question, VC = 3VR d 2 3 6d −tR/ C or − −−tR//CtRC or 1 = 4 e 1 ee = 3 15. (a) : Here, power, P′ = ?, m = 5000 quintals t 5 or ln 1 = ln 4 + − = 5 × 10 kg. RC 1 –1 54 ×1000 –1 –1 or t = RC × 2ln2 sinq = , v = 54 km h = m s = 15 m s 50 60 × 60 t = 2.5 × 106 × 4 × 10–6 × 2 × 0.693 or t = 13.86 sec. Force of friction, F = 0.8 N/quintal = 0.8 × 5000 N = 4000 N 15 1 11. (b) : Time t = 15 min = = horu W S 60 4 P′ = = (mg sinq + F) × = (mg sinq + F) × v t distance t velocity = 1 time P′ = 51××095 .8 ×+4000 15 1 km 50 v cos q = = 4 b 1/4 h P′ = (98000 + 4000) × 15 = 1530000 W = 1530 kW or 5 cos q = 4 hc 16. (a) : As Kmax = hu – f0 = –f0, 4 3 l or cosq= \ sinq = 5 5 According to question, 2Kmax = K′max Consider the triangle of velocity ABC, hc hc or 2 − f =−f vr 3 vr 3 –1 0 0 = or = or vr = 3 km h . l1 l2 vb 5 5 5 2hc hc or −=2f − f 12. (d) : I = neAvd where 0 0 l1 l2 vd = drift speed, A = cross-sectional area. 21 As A is different at different sections, since the or f0 = hc − ll conductor is non-uniform, vd is different. 12 I 2 1 Again electric field E = . = (1242 eV nm) − sA 600 nm 400 nm E does not remain constant as A varies from section to section of a non-uniform conductor. (as hc = 1242 eV nm) 1 Hence current is the only quantity that remains = (1242 eV nm) = 1.035 eV constant along the length of the conductor. 1200 nm 17. (a) : To twist wire through the angle dq, it is nn21− 13. (c) : From, n = –D necessary to do the work xx− 21 dW = tdq nx()21− x 10 pp D = andrq =10′ =× = ad −−()nn21 60 180 1080 Dimensions of n = number/area/time 4 42 q q hprdqq hpr q –2 –1 ==tq = = [L T ] Wd∫0 ∫0 –3 24l l Dimensions of n1, n2 = number/volume = [L ] 11 −−5542 Dimensions of x – x = difference between two 59.(××10 10 ××pp210 ) 2 1 W = positions = [L] 10−−42××45××10 ()1080 2 []LT−−21[]L –12 \ D = = [L2T –1] W = 1.253 × 10 J [L−3] Physics for you | May ‘15 49 18. (b) : Since the capacitors are in series, charge on p =+Atsin wp+ each capacitor is the same. At steady state, the 3 charge on the capacitor is due to terminal potential 4p difference of cell e. xA=+sin wt 3 3 When steady state is reached, no current flows . . . x3 = Bsin(wt + f) through the capacitors. The current in the circuit is, 4p C1 =1 F C2 =2 F Hence, BA==, f C 3 B A –3 R1 = 3 R2 = 6 20. (d) : As XL = 2puL = 2 × 3.14 × 500 × 8.1 × 10 D 1 = 25.4 W, XC = = 12 V 1 2puC 1 = 25.4 W, 12 −6 I = = 1.2 A = 23××..14 500 ××12 510 13++6 Terminal potential difference across A and B, \ Z = RX22+−()X = 10 W V = e – Ir = 12 – 1.2 × 1 = 10.8 V LC Capacitors are in series, their effective capacitance e 100V I = rms = = 10 A 12× 2 rms Z 10 W C = = mF 12+ 3 VR = IrmsR = (10 A)(10 W) = 100 V \ Charge on each capacitor, 21. (b) : For the particle to escape, K.E. = P.E. 2 –6 –6 q = × 10 × 10.8 = 7.2 × 10 C 1 2 GMm 3 mve = ...(i) –6 2 R Charge on capacitor C2, q = 7.2 × 10 C. 1 1 2 GMm Potential difference between A and C = VA – VC = But supplied K.E. = ×=mve −6 2 22R q 72. ×10 Suppose the particle rises to a height h, then = −6 = 3.6 V C2 21× 0 1 1 2 GMm ×=mve Potential difference between A and D = VA – VD 2 2 Rh+ = 6 × 1.2 = 7.2 V. GMm GMm Potential difference between C and D = V – V = (Using (i)) C D 2R Rh+ = (VA – VD) – (VA – VC) = 7.2 – 3.6 = 3.6 V \ h = R 19. (b) : Here, x = Asinwt 1 1 22. (a) : From Moseley’s law, as l ∝ 2p ()Z −1 2 xA2 =+sin wt 3 2 l ()Z −1 ()29 −1 2 2p Mo = cu = = 0.4663 \ 2 2 xx12+=Atsinsww++Atin l − − 3 Cu ()ZMo 1 ()42 1 ° 2p 2p lMo 07. 1Α =+AtsinswwAtin cosc+ oswt sin or lCu = = = 1.52 Å 3 3 0.4663 0.4663 23. (c) : Here, v = 70 m s–1, v = 63 m s–1, A = 2.5 m2 1 3 1 2 =+ww− + w AtsinsAt in cos t 1 2 1 2 2 2 For horizontal flow, P1 + rv1 = P2 + rv2 2 2 A 3 1 2 2 =+sincwwtAos t or (P2 – P1) = r (v1 – v2 ) 2 2 2 p p p 1 –3 –1 2 –1 2 =+Atsincw os cosswt in =+Atsin w or (P2 – P1) = (1.3 kg m )[(70 m s ) – (63 m s ) ] 33 3 2 . . 1 –3 2 –2 –2 . xx12++x3 = 0 = (1.3 kg m )(931 m s ) = 605 N m p 2 ⇒ x = – (x + x ) =−Atsin w + Lift (force) on the wing, F = (P2 – P1) A 3 1 2 3 = (605 N m–2) (2.5 m2) = 1.5 × 103 N 50 Physics for you | May ‘15 Physics for you | May ‘15 51 28. (a) : Refer to figure. 1 2 qq12 qq12 24. (c) : As |DK| = |DU|, 2 mv =−k The sphere is at equilibrium, if 2 r r 2 1 T sinq = qE or (5 × 10–3)v2 T cosq = mg T sinq qE 9 –5 –5 1 1 \ = = (9 × 10 )(4 × 10 )(4 × 10 ) − = 14.4 05. 1 T cosq mg mg tanq or, E = 14.4 q or v = = 53.7 m s–1 mgd tanq −3 \ V = Ed = 51× 0 q −− 25. (a) : mN + mA = 28 g …(i) 21××0942.t85××10 ×°an30 Here, m and m represent masses of neon and = N A 3 61× 0−9 argon gases respectively. = 9.4 × 10 V m 29. (c) : When the ball dropped, its acceleration is g as mN A nN = , nA = is observed by a man standing stationary on the 20 40 ground. Now the man inside the lift has downward As PV = nRT = (nN + nA)RT acceleration a, the acceleration of the ball as observed by this man will be g – a. PV 105 × 00. 2 nN + nA = = = 0.8 30. (c) : Consider an element of thickness (dr) at a × RT 8.314 300 distance r from the centre of spiral - coil. m m \ or N + A = 0.8 Number of turns in spiral = N 20 40 N \ Turns per unit length = or 2mN + mA = 32 …(ii) (ba − ) From eqns. (i) and (ii), Magnetic field due to element = dB m = 4 g, m = 24 g N A m ()dN I \ dB = 0 26. (d) : Let I be the M.I. of the square lamina about 2r an axis through O and perpendicular to its plane. m I N dr or dB = 0 Then by perpendicular axes theorem, 2 (ba − )r F DC b m0NI dr I + I = I or ∫∫dB = AC BD 2( − ) r GH baa I O or IAC = [ IAC = IBD] m NI b 2 or B = 0 ln . AB Again by perpendicular axes theorem, E 2(ba − ) a IEF + IGH = I 31. (a) : Let x be each side of the cube. As area (A) of each face is 4 cm2, I 2 or IEF = [ IEF = IGH] A = x × x = 4 cm or x = 2 cm 2 Further, with usual notation, Hence IAC = IEF (T1 – T2) = (100°C – 0°C) = 100C° (T1 = temperature of steam, T2 = temperature of 27. (c) : Given that N0 = 1 g melting ice) –1 –1 –1 N = 1 g – 2 mg = 1 g – 0.002 g = 0.998 g, t = 5 y k = 0.2 cal s cm C° , t = 10 min = 10 × 60 s = 600 s kA()TT12− t –lt N –lt lt As Q = , As N = N0e , = e or e = N0/N x N0 02. ××4 100 × 600 Q = cal = 24,000 cal 5l 1g 2 or e = = 1.0020 0.g998 If m gram of ice be melted with this heat and L be the latent heat of fusion of ice, ln1.0020 or 5l = ln 1.0020 or l = y–1 Q = mL 5 or m = Q 24,000 (as L = 80 cal g–1) ln25ln2 = or T1/2 = = = 1735 y L 80 l ln1.002 or m = 300 g = 0.3 kg 52 Physics for you | May ‘15 –1 32. (c) : Here u1 = u, u2 = 0 37. (b) : u = 9500 Hz, us = ?, v = 300 m s , u′ = 10000 Hz vv− vv− As the source is approaching a stationary observer, \ e = 21= 21 uu− u − 0 v′ 300 21 u′ = u or 10000 = 9500 or v – v = eu …(i) 2 1 vv− s 300 − us By the law of conservation of momentum, –1 vs = 15 m s mu + m × 0 = mv1 + mv2 38. (c) : Here; C = 8.00 mF = 8.00 × 10–6 F, u = 3.00 kHz, or v1 + v2 = u …(ii) Adding (i) and (ii), V0 = 30.0 V w pu p 3 –1 2v = u + eu = u(1 + e) Clearly, = 2 = 2 × (3.00 × 10 s ) 2 = 6p × 103 s–1 ue()1+ or v2 = Voltage across the capacitor, V = V0 sin wt 2 = (30.0 V) sin(6p × 103 t) Again, from (ii), Displacement current, Id ue()1+ ue()1− v = u – v = u – = dQ d d dV 1 2 = ==()Q ()CV = C 2 2 dt dt dt dt v2 1+ e Hence, = . –6 d 3 − = (8.00 × 10 ) [30.0sin (6p × 10 t)] v1 1 e dt –6 3 3 12 W 1 P = (8.00 × 10 )(30.0)(6p × 10 )cos(6p × 10 t) 33. (a) : Irms = = A (as I = ) = (8.00 × 10–6)(30.0)(6p × 103)cos(6p × 103 t) 24 V 2 V = (4.52 A)cos(6p × 103 t) 1 1 I = I = = Hence, the displacement current varies sinusoidally 0 rms 2 A 2 A 2 2 with time and has a maximum value of 4.52 A. 39. (c) : P-V diagram represents, PV = constant (Boyle’s law) 1 k1 1 k2 34. (b) : As u1 = , u2 = , This implies that temperature (T) is constant. 2p m 2p m Further, from 1 to 2, pressure decreases. Thus, Frequency of the system, corresponding T-P diagram is represented by (iii). 40. (a) : Here m = 2 kg, R = 0.1 m 1 kk12+ u = Height of inclined plane, h = 4 m 2p m At the top of the inclined plane, the cylinder has 1 k k 1 u = 12+= ()22pu 2 + ()pu 2 P.E. = mgh 2ppm m 2 1 2 At the bottom of the inclined plane, the cylinder = 2 2 1 2 uu1 + 2 has translational K.E. = mv and rotational 2 –3 1 35. (d) : Here, r = 1.00 mm = 1.00 × 10 m K.E. = Iw2 T = 0.465 N m–1 2 3 –3 r = 13.6 × 10 kg m , q = 140° By conservation of energy, If h is the difference in the level of mercury in the 1 2 1 2 tube and the surface outside, mv + Iw = mgh 2 2 2T cosq h = 1 2 rgr But v = Rw and I = mR 2 20(.462 N m−1)(−0.)7660 or h = 1 2 1 1 2 2 −−333 −2 \ m(Rw) + × mR w = mgh (.1001××01mk)( 36.)10 g mm(.98 s ) 2 2 2 –3 = – 5.3 × 10 m = – 5.3 m 3 4gh or mR2w2 = mgh or w2 = 36. (a) : For equilibrium, 4 3R2 Tsinq = mrw2 1 221 1 4gh 2 \ Rotational K.E. = Imw= ××R r 2 2 2 2 3R2 T × =⋅mr 2p mgh 29××.84 l p = = = 26.13 J. 3 3 T = 16 ml mm Physics for you | May ‘15 53 Exam on : 1st June 1. A car travels equal distances in the same direction 6. The greatest length of steel wire that can hang with velocities 60 km h–1, 20 km h–1 and 10 km h–1 vertically without breaking is respectively. The average velocity of the car over the (Given, breaking stress of steel = 8.0 × 108 N m–2, whole journey of motion is Density of steel = 8.0 × 103 kg m–3) (a) 8 m s–1 (b) 7 m s–1 (a) 10 m (b) 102 m (c) 103 m (d) 104 m –1 –1 (c) 6 m s (d) 5 m s 7. A rectangular vessel when full of water, takes 2. Moment of inertia of a thin rod of mass M and 10 min to be emptied through an orifice in its length L about an axis passing through its centre is bottom. How much time will it take to be emptied ML2 when half filled with water? . Its moment of inertia about a parallel axis at 12 (a) 9 min (b) 7 min (c) 5 min (d) 3 min L a distance of from this axis is given by 8. If the frequency of sound produced by a siren 4 increases from 400 Hz to 1200 Hz while the wave ML2 ML3 ML2 7ML2 (a) (b) (c) (d) amplitude remains constant, the ratio of the 48 48 12 48 intensity of the 1200 Hz to that of the 400 Hz wave 3. Two linear SHMs of equal amplitude A and angular will be frequencies w and 2w are impressed on a particle (a) 1 : 1 (b) 1 : 3 (c) 3 : 1 (d) 9 : 1 along the axes x and y respectively. If the initial 9. A spherical charged conductor has surface charge phase difference between them is p/2, the resultant density s. The electric field on its surface is E path followed by the particle is and electric potential of conductor is V. Now the (a) y2 = x2(1 – x2/A2) (b) y2 = 2x2(1 – x2/A2) radius of the sphere is halved keeping the charge (c) y2 = 4x2(1 – x2/A2) (d) y2 = 8x2(1 –x2/A2) to be constant. The new values of electric field and 4. Equal charges, q each are placed at the vertices A potential would be and B of an equilateral triangle ABC of side a. The (a) 2E, 2V (b) 4E, 2V magnitude of electric intensity at the point C is (c) 4E, 4V (d) 2E, 4V q 2 q 10. In a series LCR circuit, the voltage across the (a) (b) 4pe a2 4pe a2 resistance, capacitance and inductance is 10 V each. 0 0 If the capacitance is short circuited, the voltage 3 q 2 q across the inductance will be (c) 2 (d) 2 4pe0a 4pe0a (a) 10 V (b) 10 2V 10 5. In the arrangement shown in figure, the current (c) V (d) 20 V through 5 W resistor is 2 11. Light of two different frequencies whose photons 2 2 have energies 1 eV and 2.5 eV respectively, 12 V 5 12 V successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the 12 (a) 2 A (b) 1 A (c) A (d) zero emitted electrons will be 7 (a) 1 : 5 (b) 1 : 4 (c) 1 : 2 (d) 1 : 1 54 Physics for you | May ‘15 12. If the series limit wavelength of the Lyman series 19. A potential difference of 2 V is applied between for hydrogen atom is 912 Å, then the series limit the opposite faces of a Ge crystal plate of area wavelength for the Balmer series for the hydrogen 1 cm2 and thickness 0.5 mm. If the concentration atom is of electrons in Ge is 2 × 1019 m–3 and mobilities (a) 912 Å (b) 912 × 2 Å of electrons and holes are 0.36 m2 V–1 s–1 and 912 2 –1 –1 (c) 912 × 4 Å (d) Å 0.14 m V s respectively, then the current 2 flowing through the plate will be 13. A freshly prepared radioactive source of half-life (a) 0.25 A (b) 0.45 A (c) 0.56 A (d) 0.64 A 2 hours emits radiation of intensity which is 20. At 1 atm pressure, 1 g of water having a volume 64 times the permissible safe level. The minimum of 1000 cm3 becomes 1671 cm3 of steam when time after which it would be possible to work safely boiled. The heat of vaporization of water at 1 atm is with the source is 539 cal g–1. What is the change in internal energy (a) 6 hours (b) 12 hours during the process? (c) 24 hours (d) 128 hours (a) 539 cal (b) 417 cal 14. A constant voltage is applied between the two ends (c) 498.5 cal (d) 835.5 cal of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if 21. A particle is projected from a point at a certain angle (a) both the length and radius of the wire are with the horizontal. At any instant t, if p is the linear halved momentum and E the kinetic energy, then which of (b) both the length and radius of the wire are the following graphs is correct? doubled (c) the radius of the wire is doubled (d) the length of the wire is doubled. (a) (b) 15. A bar magnet when placed at an angle of 30° to the direction of magnetic field induction of 5 × 10–2 T, experience a moment of couple 2.5 × 10–6 N m. If the length of the magnet is 5 cm, its pole strength is (c) (d) (a) 2 × 102 A m (b) 2 × 10–3 A m (c) 5 A m (d) 5 × 10–2 A m 16. A concave lens forms the image of an object such 22. A man measures the period of a simple pendulum that the distance between the object and image inside a stationary lift and finds it to beT second. If g is 10 cm and the magnification produced is 1/4. the lift accelerates upwards with an acceleration , 4 The focal length of the lens will be then the period of the pendulum will be (a) 8.6 cm (b) 6.2 cm (c) 10 cm (d) 4.4 cm T 2T (a) T (b) (c) (d) 25T 17. In Young’s double slit experiment, the spacing 4 5 between the slits is d and wavelength of light used is 23. A pendulum bob of mass 10–2 kg is raised to a height 6000 Å. If the angular width of a fringe formed on a 5 × 10–2 m and then released. At the bottom of its distant screen is 1°, then value of d is swing, it picks up a mass 10–3 kg. To what height (a) 1 mm (b) 0.05 mm will the combined mass rise? (c) 0.03 mm (d) 0.01 mm (a) 4.1 × 10–2 m (b) 2.5 × 10–2 m 18. An oil drop of mass 50 mg and of charge –5 µC is (c) 1.3 × 10–2 m (d) 0.7 × 10–2 m just balanced in air against the force of gravity. The 24. A rod of length 1.0 m and mass 0.5 kg fixed at one strength of the electric field required to balance, is end is initially hanging vertical. The other end is –2 (Take g = 9.8 m s ) now raised until it makes an angle 60° with the –1 (a) 98 N C upwards vertical. The work required to be done for this, is –1 (b) 98 N C downwards (a) 1.2 J (b) 1.8 J (c) 9.8 N C–1 towards north (c) 2.4 J (d) 3.0 J (d) 9.8 N C–1 towards south Physics for you | May ‘15 55 25. A common emitter transistor amplifier has a current see objects beyond 40 cm, are gain of 50. If the load resistance is 4 kW and input (a) –2.5 D and concave lens resistance is 500 W, the voltage gain of the amplifier (b) –2.5 D and convex lens is (c) –3.5 D and concave lens (a) 200 (b) 400 (c) 600 (d) 800 (d) –3.5 D and convex lens 26. When an electron jumps from the fourth orbit to 33. A reactor is developing energy at the rate of the second orbit, one gets the 3000 kW. How many atoms of U235 undergo fission (a) second line of Paschen series (b) second line of Balmer series per second, if 200 MeV energy is released per (c) first line of Pfund series fission? (d) second line of Lyman series (a) 6.5 × 1022 (b) 5.15 × 1021 (c) 3.384 × 1023 (d) 9.4 × 1016 –1 3 27. The critical angle of a certain medium is sin . 5 34. Power supplied to a particle of mass 2 kg varies with The polarising angle of the medium is 3t2 time as P = W, where t is in seconds. If velocity of −1 4 −1 5 2 (a) sin (b) tan 5 3 particle at t = 0 is zero, then the velocity of particle at −1 3 −1 4 t = 2 s will be (c) tan (d) tan 4 3 (a) 1 m s–1 (b) 4 m s–1 28. Which of the following types of electromagnetic (c) 2 m s–1 (d) 22ms−1 radiation travels at the greatest speed in vacuum? (a) Radio waves (b) Visible light 35. Three identical spherical ballsA , B and C are placed (c) X-rays on a table as shown in the figure along a straight (d) All of these travel at the same speed line. 29. A vector A is along the positive z-axis and its vector product with another vector B is zero, then vector B could be B and C are at rest initially. The ball A hits B head –1 ^^ ^ ^ ^ ^ on with a speed of 10 m s . Then after all collisions (a) ij+ (b) (c) jk+ (d) − 4 i 7k (assumed to be elastic) A and B come to rest and C 30. Two perfect gases having masses m1 and m2 at takes off with a velocity of temperatures T1 and T2 respectively are mixed (a) 5 m s–1 (b) 10 m s–1 without any loss of energy. If the molecular weights (c) 2.5 m s–1 (d) 7.5 m s–1 of the gases are M and M respectively, then the 1 2 36. The time period of an artificial satellite in a circular final temperature of the mixture is orbit of radius R is 2 days and its orbital velocity is ()mT + mT ()MT + MT (a) 11 22 (b) 11 22 v . If time period of another satellite in a circular + o ()mm12 ()MM12+ orbit is 16 days then mT11 mT22 MT11 MT22 (a) its radius of orbit is 4R and orbital velocity is + + M1 M2 m1 m2 vo. (c) (d) (b) its radius of orbit is 4R and orbital velocity is m1 m2 M1 M2 + + M M m m v 1 2 1 2 o . 2 31. A proton with energy of 2 MeV enters a region of (c) its radius of orbit is 2R and orbital velocity is a uniform magnetic field of 2.5 T normally. The vo. magnetic force on the proton is (d) its radius of orbit is 2R and orbital velocity is (Take, mass of proton to be 1.6 × 10–27 kg) vo –12 –8 . (a) 3 × 10 N (b) 3 × 10 N 2 (c) 8 × 10–12 N (d) 2 × 10–10 N 37. Force constant of a spring (k) is analogous to 32. The power and type of lens by which a person can YL AL (a) YA (b) (c) (d) ALY see clearly the distant objects, if the person cannot L A Y 56 Physics for you | May ‘15 38. Two rings of same radius r and mass m are placed reason: The peak emission wavelength of a black such that their centres are at a common point and body is proportional to the fourth power of absolute their planes are perpendicular to each other. The temperature. moment of inertia of the system about an axis 44. Assertion : Work done by a gas in isothermal passing through the centre and perpendicular to expansion is more than the work done by the gas in plane of one of the rings is 1 3 the same expansion, adiabatically. mr2 2 mr2 2 (a) 2 (b) mr (c) 2 (d) 2mr reason : Temperature remains constant in 39. Four equal and parallel forces are acting on a rod isothermal expansion, and not in adiabatic of length 100 cm, as shown in figure, at distances of expansion. 20 cm, 40 cm, 60 cm and 80 cm respectively from 45. Assertion : A current carrying conductor produces one end of the rod. Under the influence of these only an electric field. forces, the rod (neglecting its weight) reason : Electrons in motion give rise to only an electric field. 46. Assertion : The energy stored in the inductor of 2 H, when a current of 10 A flows through it is (a) experiences no torque 100 J. (b) experiences torque reason : Energy stored in an inductor is directly (c) experiences a linear motion proportional to its inductance. (d) experiences torque and also a linear motion 47. Assertion : The electric field and hence electric 40. The current gain of a transistor in a common base field lines are everywhere at right angle to an arrangement is 0.98. Find the change in collector equipotential surface. current corresponding to a change of 5.0 mA in reason : Equipotential surfaces are closer together emitter current. What would be the change in base where the electric field is stronger and farther apart current? where the field is weaker. (a) 4.9 mA, 0.1 mA (b) 4.9 mA, 0.2 mA (c) 5.9 mA , 0.3 mA (d) 5.9 mA, 0.8 mA 48. Assertion : When temperature difference across the two sides of a wall is increased, its thermal Directions : In the following questions (41-60), a conductivity remains constant. statement of assertion (A) is followed by a statement reason : Thermal conductivity depends on nature of reason (r). Mark the correct choice as of material of the wall. (a) If both assertion and reason are true and reason is the 49. Assertion : Young’s modulus for a perfectly plastic correct explanation of assertion. body is zero. (b) If both assertion and reason are true but reason is not reason : For a perfectly plastic body, restoring force the correct explanation of assertion. is zero. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 50. Assertion : Orbit of a satellite is within the gravitational field of earth whereas escaping is 41. Assertion : During rapid pumping of air in tyres, beyond the gravitational field of earth. air inside the tyre is hotter than atmosphere air. reason : Orbital velocity of a satellite is greater than reason : Adiabatic process occurs at very high its escape velocity. rate. 51. Assertion: A piece of ice, with a stone frozen inside 42. Assertion: When a cyclist doubles his speed of it, floats on water in a beaker. When the ice melts, turning, the chance of his skidding becomes nearly the level of water in the beaker decreases. four times. reason : Density of stone is more than that of water. reason : When the speed of the vehicle increases, the angle of bending also increases. 52. Assertion : Adding a scalar to a vector of the same dimensions is a meaningful algebraic operation. 43. Assertion: At higher temperatures, the peak reason : The displacement can be added with emission wavelength of a black body shifts to lower distance. wavelengths. Physics for you | May ‘15 57 53. Assertion: A p-n junction with reverse bias can be 2 2 ML L 7ML2 used as a photodiode to measure light intensity. =+M = reason : In a reverse bias condition the current 12 4 48 though small is more sensitive to change with 3. (c) : x = A sin(wt + p/2) = A cos wt incident light intensity. x x2 \ cos wt = andsinw=t 1 − 54. Assertion : Most amplifiers use common emitter 2 circuit configuration. A A reason : Its input resistance is comparatively y = A sin 2wt = 2A sin wt cos wt higher. or y2 = 4A2 sin2 wt cos2 wt 2 22 2 55. Assertion : Magnetic force between two short 2 x Ax− 2 x magnets, when they are co-axial follows inverse =×44A × =−x 1 A2 A2 A2 square law of distance. reason : The magnetic forces between two poles do q 4. (c) : EE== ,actingat60° not follow inverse square law of distance. 12 2 4pe0a 56. Assertion: A photon has no rest mass, yet it carries \ Resultant intensity definite momentum. EE=+2 EE2 + 2cE osq reason: Momentum of photon is due to its energy 1 2 12 and hence its equivalent mass. =+2 2 +°2 EE1 1 26E1 cos 0 57. Assertion : A capacitor of suitable capacitance can q 3 be used in an A.C. circuit in place of the choke ==E 3 1 2 coil. 4pe0a reason : A capacitor blocks D.C. and allows A.C. only. 5. (a) : The circuit may be redrawn as shown in the figure. 58. Assertion : Static crashes are heard on radio, when a lightning flash occurs. reason : Light and radio waves travel with the same 2 2 speed. 5 12 V 12 V 59. Assertion : Diffraction is common in sound but not common in light waves. reason : Wavelength of light is more than the 22× wavelength of sound. Here, e ==12V, r =1 W eq eq + 60. Assertion : Improper banking of roads causes wear 22 eeq 12 12 and tear of tyres. \=I = ==2A reason : The necessary centripetal force is provided Rr+ eq 51+ 6 by the force of friction between the tyres and the road. 6. (d) : Let l be the length of the wire that can hang vertically without breaking. Then the stretching soluTioNs force on it is equal to its own weight. If A is the area 1. (d) : Let the total distance be 3x km. of cross-section and r the density, then total distance weight ()Alr g Then, average velocity = maximumstress(sm) = or sm = total time taken A A s 3x 3x \=l m = = r xxx xx++36x g ++ Substituting the values, we get 60 20 10 60 8 80. ×10 4 36x × 0 −1 18 × 5 −−11 l = =10 m = =18 km h = ms = 5 ms (.80×103)(10) 10x 18 7. (b) : If A is the area of orifice at the bottom below 2. (d) : Applying parallel axis theorem 0 the free surface and A that of vessel, time t taken to I = I + Md2 CM be emptied the tank, 58 Physics for you | May ‘15 (... series limit of Lyman series = 1/R = 912 Å) A 2H t H H t = \=1 1 = 1 13. (b) : Let safe level activity be A. A0 g t2 H2 H1 / 2 Initial activity = 64A. t1 t1 10 N A A 1 ⇒=2 \ t2 ===52≈ 7min Hence, == = t2 2 2 N00A 64A 64 1 n 6 8. (d) : IA= rw22v or I ∝ w2 or I ∝ u2 1 1 1 2 or == or n = 6 2 2 2 64 2 I u 1200 \ 2 = 2 = = 9 t . . I1 u1 400 ⇒ = 6 \ t = 12 hours ( . T = 2 hours) T 1 q 2 9. (b) : E = V 2 14. (b) : Heat developed, H =×t 4pe0 R R 1 As q is constant, so E ∝ Heat developed will be doubled when R is halved. 2 R Further, R = (rl/pr2) Radius is halved. Therefore electric field will become Vr22××p t 4 times or 4E. \=H 1 q rl = Further,V So, heat produced will be doubled when both the 4pe0 R 1 length and radius of the wire are doubled. As q is constant, so V ∝ –2 R 15. (b) : Here, q = 30°, B = 5 × 10 T, –6 Radius is halved, so potential will become two times t = 2.5 × 10 N m, 2l = 5 cm = 0.05 m or 2V. t = MB sin q = m(2l)B sin q − t 25. ×10 6 10. (c) : Here, R = XL = XC \ m == (... voltage across them is same) Bl()2 sin q 51×°00−2 (.05)sin 30 Total voltage in the circuit, –3 2 2 1/2 m = 2 × 10 A m V = I[R + (XL – XC) ] = IR = 10 V 16. (d) : Concave lens forms the virtual image of a real object. When capacitor is short circuited, 1 Let u = –4x, then v = –x m = V 10 I′= = As per question, 3x = 10 cm 4 2212/ 2 R ()RX+ L 10 or x = cm \ Potential drop across inductance 3 10 O I =′IXL =′IR= V 40 2 \=u − cm and 3x x 3 11. (c) : If E is the energy of incident photon and f 0 10 the work function, then according to Einstein’s v =− cm 4x 3 photoelectric equation, 111 Substituting in lens formula, =− 1 2()E − f0 fvu Em−=f v2 or v = 0 m 2 13− 3 40 ⇒=+ =− v E − f 10− .5 05. 1 or f = –4.4 cm \=1 10= == f 10 40 9 v E − f 25..− 05 2 2 2 20 y 17. (c) : Here,sinqq = 12. (c) : For series limit of Balmer series, D n = ∞, n = 2 Dy i f So, Dq= 111 11 1 R D or =−R = =−R 22 For Dy = b, angular fringe width q = Dq l 22 l 2 ∞ 4 0 nnfi blD 1 l 4 \ q == ×= \ l= =×4 912 Å 0 D dDd R Physics for you | May ‘15 59 m1v1 + m2v2 = (m1 + m2)v p –7 As q0 = 1° = rad and l = 6 × 10 m 10–2 × 1 + 10–3 × 0 = (10–2 + 10–3)v 180 −2 l 180 10 10 −1 \ d == ××610−7 or v = = ms × −2 11 qp0 11. 10 v22(/10 11) = 3.44 × 10–5 m ≈ 0.03 mm Now, h == =×41.m10−2 2g 21× 0 18. (b) : As qE = mg −−62 mg ()50 ××10 kg 98. ms 24. (a) : or E == = 98 NC−1 q 51× 0−6 C Since the force due to electric field on charged particle should be opposite to the gravity pull and charge on the drop is negative, hence the electric W = Change in potential energy field must act vertically downwards. l =−mg (c1 os q) 2 19. (d) : Conductivity, s = ne (µe + µh) = 2 × 1019 × 1.6 × 10–19 × (0.36 + 0.14) Substituting the values, we have = 1.6 W–1 m–1 10. W = (.05)(98.) (c16−°os 0 ) = 1.2 J −3 2 l l 05. ×10 25 \ Resistance, R ==r = = W I A sA × −4 8 25. (b) : R = 4 kW, R = 500 W, b= C = 50 16. 10 L I I V 2 16 B Hence, current I == ==A.064 A VCE = ICRL, VBE = IBRI R 25/ 8 25 V I R 41× 03 20. (c) : Heat spent during vaporisation A ==CE C ×=L 50 × = 400 V V I R 500 Q = mL = 1 × 539 = 539 cal BE B I Work done W = P(Vv – Vl) 26. (b) : Electron jump, to second orbit leads to Balmer = 1.013 × 105 × (1671 – 1) × 10–6 J series. When an electron jumps from 4th orbit to 2nd 169.2 orbit, it gives rise to second line of Balmer series. == cal.40 5 cal 41. 8 3 15 \ Change in internal energy 27. (b) : Given, sin.C =\m == U = 539 cal – 40.5 cal = 498.5 cal 5 sinC 3 If ip is the polarising angle, then 21. (d) : As p2 = 2Em or p2 ∝ E according to Brewster’s law i.e., p2 versus E graph is a straight line passing –1 –1 5 through origin. tanip = m or ip = tan (m) = tan 3 l 22. (c) : In stationary lift, T = 2p 28. (d) : In vacuum, all electromagnetic waves travel g with the same speed c(= 3 × 108 m s–1). In upward moving lift, 29. (d) : The vector product of two non-zero vectors l is zero if they are in the same direction or in T′ = 2p where a =acceleration of lift ()ga+ the opposite direction. Hence vector B must be parallel or antiparallel to vector A , i.e. along T′ g g 4 2 ⇒ = = == ± z-axis. T ga+ g 5 5 g + m1 4 30. (c) : Number of moles of gas 1, n1 = 2T M1 \ ′ = m2 T Number of moles of gas 2, n2 = 5 M2 23. (a) : Velocity of pendulum bob in mean position As there is no loss of energy, the final temperature of the mixture is vg= 2 h =×2105××10−−21=1ms 11 m1 m2 –3 T1 + T2 When the bob picks up a mass 10 kg at the nT11+ nT22 M1 M2 Tmixture = = bottom, then by conservation of linear momentum nn12+ m1 m2 + the velocity of combined mass is given by M1 M2 60 Physics for you | May ‘15 Physics for you | May ‘15 61 31. (c) : Here, E = 2 MeV = 2 × 1.6 × 10–13 J 36. (b) : According to Kepler’s third law B = 2.5 T, m = 1.6 × 10–27 kg, T2 ∝ R3 –19 32/ q = 90°, q = 1.6 × 10 C T R \=1 1 1 2 2E As, Em= v or v = T2 R2 2 m 23/ 23/ T 16 Now F = qvBsinq or RR= () 2 = ()R 21 T 1 2 2E 2E 1 = q × × B × sin90° = q × × B = 4R = 4R ...(i) m m 1 −13 GM 22××16. ×10 Orbital velocity, vo = = (1.6 × 10–19) × × 2.5 R 16. ×10−27 v R R 1 1 1 –12 \=o2 1 ==or vv==v = 8 × 10 N oo212 2 o vo1 R2 4R 2 32. (a) : Here, in this case lens used by person should FL YA form the image of distant object at a distance of 37. (a) : Y ==or F DL D 40 cm in front of it. AL L YA \ u = –∞, v = –40 cm Comparing this with F = kDx, we get, k = L 111 11 1 2 and =− or = − 38. (c) : I1 = mr fvu f −40 −∞ I = moment of inertia about 2 11 2 or = or f = –40 cm the diameter f −40 1 1 2 100 100 = mr Power ===−25.D 2 Axis f ()cm −40 Therefore, the moment of inertia Negative sign shows that lens used is concave lens. of the system about the desired 3 2 33. (d) : Rate of development of energy by the reactor axis is II=+12Im= r = 3000 kW = 3 × 106 J s–1 2 Energy released per fission = 200 MeV 39. (b) : Since, upward forces on the rod are equal to downward forces, hence, the resultant force on the = 200 × 1.6 × 10–13 J = 32 × 10–12 J rod is zero and there is no linear motion. Moment Number of atoms undergoing fission per second of forces about one end of rod 31× 06 = =×94. 1016 = –F × 20 + F × 40 – F × 60 + F × 80 = F × 40 32 ×10−12 As it is not zero, hence a torque acts on the rod. 34. (c) : From work energy theorem, D K.E. = Wnet 40. (a) : Given, a = 0.98 and DIE = 5.0 mA \ DI KKfi−=∫ Pdt From the definition of, a= C 2 D 2 3 IE 1 22 3 1 2 3 t mv −=0 ∫ td t , ()2 v = = 4 Change in collector current 2 2 0 2 23 DIC = (a)(DIE) = 0.98 × 5.0 mA = 4.9 mA –1 0 \ v = 2 m s Change in base current 35. (b) : Since A, B and C are identical balls, if any of DIB = DIE – DIC = (5.0 – 4.9) mA = 0.1 mA the two balls undergo elastic collision, they will 41. (a) exchange their velocities. v2 42. (b) : For the banking of curve, tanq= . When rg Thus, when A and B collide, A comes to rest, and B velocity v is doubled, tanq becomes four times. starts moving ahead with 10 m s–1. With increase in q, the chance of overturning also increases. When v increases, q increases and Similarly, when B collides with C, B comes to consequently chance of skidding or overturning rest and C starts moving ahead with a speed of becomes greater. 10 m s–1. 62 Physics for you | May ‘15 43. (c) : According to Wien’s displacement law, own volume, m2/rs where rs is the density of stone. the maxima of wavelength emitted is inversely mm Hence, total volume of extra water isV ′ =+12. proportional to the absolute temperature rr 11 s 1 As rs > r, < . l ∝ . rr m T s Therefore, V′ < V and thus level of water in the 44. (b) : Adiabatic curve is steeper than isothermal beaker decreases. curve. Therefore, area under adiabatic curve is smaller than the area under isothermal curve i.e., 52. (d) : Scalar and vector are quantities of different work done by the gas in adiabatic expansion is nature. We can add only those quantities which smaller than the work done by the gas in isothermal have same nature and units. expansion. Assertion is true. Reason is also true but 53. (a) : The reason that the reverse bias current is reason does not explain assertion. more sensitive to change is true. The current is 45. (d) : A current carrying conductor has free due to minority carriers and is therefore small. The electrons in motion. They give rise to a magnetic reason is the explanation of the assertion. field. Electrons in motion in space produces both 54. (b) : Most amplifiers use the common emitter electric and magnetic fields. Thus, both assertion circuit configuration because the circuit offers both and reason are wrong. current and voltage gains resulting in much higher 46. (b) : The energy stored in the inductor is given by power gain that can be obtained by a common-base 1 1 amplifier. Input resistance of CE amplifier is higher UL==I22××21()0 =100 J 2 2 than that of CB amplifier. Energy stored is directly proportional to its 55. (d) inductance. 56. (a) : Equivalent mass of photon (m) is given from 47. (b) : Electric field and field lines are perpendicular equation to an equipotential surface because there is no hu E = mc2 = hu \ m = potential gradient along the equipotential surface c2 dV where E is energy, m is mass, c is speed of light, h is i.e., along the surface = 0. dr Planck’s constant, u is frequency. huuh 48. (a) : Thermal conductivity of the wall depends \ Momentum of photon = ×=c only on nature of material of the wall, and not on c2 c temperature difference across its two sides. 57. (b) : We can use a capacitor of suitable capacitance stress as a choke coil, because average power consumed 49. (a) : Young’s modulus of a material, Y = strain in an ideal capacitor is zero. Therefore, like a choke coil, a condenser can reduce A.C. without power restoringforce and, stress = dissipation. area As restoring force is zero for a perfectly plastic body, 58. (a) \ Y = 0 59. (c) : For diffraction of a wave, size of an obstacle 50. (c) : The orbital velocity, if a satellite close to earth or aperture should be comparable to the size of wavelength of the wave. As wavelength of light is of is vgoe= R , while the escape velocity for a body the order of 10–6 m and obstacle / aperture of this size are rare, therefore, diffraction is not common in thrown from the earth’s surface is vgee= 2 R light waves. On the contrary, wavelength of sound vo gRe 1 is of the order of 1 m and obstacle / aperture of this Thus ==or vveo= 2 ve 2gRe 2 size are readily available, therefore, diffraction is common in sound. 51. (a) : If m1 and m2 are the respective masses of ice and stone, the volume of water displaced is 60. (a) : When roads are not properly banked, force of friction between tyres and road provides partially V = (m1 + m2)/r. When ice melts, additional the necessary centripetal force. This cause wear and volume of water produced is m1/r. The stone sinks in water and displaces a volume of water equal to its tear of tyres. mm Physics for you | May ‘15 63 Exam on 7th June 1. A body of mass 0.5 kg travels in a straight line with diameter the change in the resistance of the wire velocity v = kx3/2 where k = 5 m–1/2 s–1. The work done will be by the net force during its displacement from x = 0 to (a) 300% (b) 200% (c) 100% (d) 50% x = 2 m is 8. Three long, straight and parallel wires carrying (a) 1.5 J (b) 50 J (c) 10 J (d) 100 J currents are arranged as shown in figure. The force 2. A ballet dancer spins with 2.8 rev s–1 with her experienced by 10 cm length of wire Q is arms out stretched. When the moment of inertia RQ P about the same axis becomes 0.7 times the earlier moment of inertia, the new rate of spin is 2 cm 10 cm (a) 3.2 rev s–1 (b) 4.0 rev s–1 (c) 4.8 rev s–1 (d) 5.6 rev s–1 20 A 10 A 30 A 3. A bag of sand of mass m is suspended by a rope. A (a) 1.4 × 10–4 N towards the right bullet of mass m/20 is fired at it with a velocity v (b) 1.4 × 10–4 N towards the left and gets embedded into it. The velocity of the bag (c) 2.6 × 10–4 N towards the right finally is (d) 2.6 × 10–4 N towards the left v v 20v 21v (a) (b) (c) (d) 9. A magnetic field given by B(t) = 0.2t – 0.05t2 tesla 21 20 21 20 (where t denotes the time), is directed 4. A car is moving on a road and rain is falling perpendicular to the plane of a circular coil vertically. Select the correct answer. containing 25 turns of radius 1.8 cm and whose (a) The rain will strike the hind screen only. total resistance is 1.5 W. The power dissipation at (b) The rain will strike the front screen only. t = 3 s is approximately (c) The rain will strike both the screens. (a) 1.37 mW (b) 7 mW (d) The rain will not strike any of the screens. (c) zero (d) 4 mW 5. Two equal masses m and m are hung from a balance 10. A prism, having refractive index 2 and refracting whose scale pan differs in vertical height by h/2. angle 30° has one of the refracting surfaces polished. The error in weighing in terms of density of the A beam of light incident on the other refracting earth r is surface will retrace its path, if the angle of incidence 1 is (a) prGmh (b) pGrmh (a) 0° (b) 30° (c) 45° (d) 60° 3 4 8 11. The wavelength ofK a line for an element of atomic (c) prGmh (d) prGmh number 43 is l. Then the wavelength ofK line for 3 3 a an element of atomic number 29 is 6. Electric charges of 1 mC, –1 mC and 2 mC are placed 49 42 9 4 in air at the corners A, B and C respectively of an (a) l (b) l (c) l (d) l 29 28 4 9 equilateral triangle ABC having length of each side 12. The reverse biasing in a p-n junction diode 10 cm. The resultant force on the charge at C is (a) decreases the potential barrier (a) 0.9 N (b) 1.8 N (c) 2.7 N (d) 3.6 N (b) increases the potential barrier 7. The length of a given cylindrical wire is increased (c) increases the number of minority charge carriers by 100%. Due to the consequent decrease in (d) increases the number of majority charge carriers 64 Physics for you | May ‘15 13. For a network shown in the figure, the value of the 20. A body constrained to move in the y direction is ^ ^ ^ current I is subjected to a force Fi=+()2156jk+ newton. 2 The work done by the force in moving the body through a distance of 10 m along y-axis is 4 4 3 (a) 100 J (b) 150 J (c) 60 J (d) 20 J 6 t x I 21. The equation yA=−sin2p , where the T l V symbols carry the usual meaning and A, T and l 9V 5V 5V 18V (a) (b) (c) (d) are positive, represents a wave of 35 18 9 5 T (a) amplitude 2A (b) period 14. Two spheres A and B of radius a and b respectively l are at same electric potential. The ratio of the l (c) speed surface charge densities of A and B is T 2 a b a2 b (d) velocity in negative x-direction (a) (b) (c) (d) 2 b a b2 a 22. A body starting from rest moves with constant 15. If the two waves represented by y1 = 4sinwt and acceleration. The ratio of distance covered by the th y2 = 3sin(wt + p/3) interfere at a point, the amplitude body during the 5 second to that covered in of the resulting wave will be approximately 5 seconds is (a) 7 (b) 5 (c) 6 (d) 3.5 9 3 25 1 (a) (b) (c) (d) 25 25 16. Figure shows the distance-time graph of the motion 25 9 of a car. It follows from the graph that the car is 23. A 40.0 kg boy is standing on a plank of mass (a) at rest x 160 kg. The plank originally at rest, is free to slide on a smooth frozen lake. The boy walks along the (b) in uniform motion 2 xt= 1.2 plank at a constant speed of 1.5 m s–1 relative to (c) in non-uniform motion the plank. The speed of the boy relative to the ice (d) uniformly accelerated t surface is 17. In figure, all the three rods are of equal length L (a) 1.8 m s–1 (b) 1.6 m s–1 and same mass M. The system is rotated such that (c) 1.2 m s–1 (d) 1.5 m s–1 rod B is the axis of rotation. What is the moment of inertia of the system? 24. A radioactive nucleus of mass M emits a photon of frequency u and the nucleus recoils. The recoil 2 2 ML ML A energy will be (a) (b) 22 3 6 B h u (a) Mc2 – hu (b) C 2 4 2 2 2Mc (c) ML (d) ML2 3 3 (c) zero (d) hu 18. The change in the gravitational potential energy 25. A drunkard walking in a narrow lane takes 8 steps when a body of mass m is raised to a height nR forward and 6 steps backward, followed again by above the surface of the earth is (here R is the 8 steps forward and 6 steps backward and so on. radius of the earth) Each step is 1 m long and requires 1 s. Determine how long the drunkard takes to fall in a pit n n (a) mgR (b) mgR 18 m away from the start. n +1 n −1 (a) 18 s (b) 126 s (c) 78 s (d) 62 s mgR 26. Which of the following statements is true? (c) nmgR (d) n (a) Sound waves cannot interfere. (b) Only light waves may interfere. 19. If Q, E and W denote respectively the heat added, (c) The de Broglie waves associated with moving change in internal energy and the work done in a particles can interfere. closed cyclic process, then (d) The Bragg formula for crystal structure is (a) W = 0 (b) Q = W = 0 an example of the corpuscular nature of (c) E = 0 (d) Q = 0 electromagnetic radiation. Physics for you | May ‘15 65 27. Which of the following physical quantity has a ratio 35. There is a mine of depth about 2.0 km. In this mine of 103 between its SI units and CGS units? the conditions as compared to those at the surface (a) Universal gravitational constant are (b) Boltzmann’s constant (a) lower air pressure, higher acceleration due to (c) Planck’s constant gravity (d) Young’s modulus of elasticity (b) higher air pressure, lower acceleration due to gravity 28. Which of the following letters do not suffer lateral (c) higher air pressure, higher acceleration due to inversion? gravity (a) HGA (b) HOX (d) lower air pressure, lower acceleration due to (c) VET (d) YUL gravity. ^ ^ 29. If unit vectors ABand are inclined at an angle q, 36. A ball is rolled off along the edge of table (horizontal) ^ ^ –1 then ||AB− is with velocity 4 m s . It hits the ground after time q q 0.4 s. Which one of the following is wrong? (a) 2 sin (b) 2 cos (a) The height of the table is 0.8 m. 2 2 q (b) It hits the ground at angle of 60° with the vertical. (c) 2 tan (d) tanq (c) It covers a horizontal distance 1.6 m from the 2 table. –1 30. Two liquid drops of equal radii are falling through (d) It hits the ground with vertical velocity 4 m s . air with the terminal velocity v. If these two drops 37. An archaeologist analyses the wood in a prehistoric coalesce to form a single drop, its terminal velocity structure and finds that C14 (half-life = 5700 years) will be to C12 is only one-fourth of that found in the cells (a) 2 v (b) 2v (c) 3 4 v (d) 3 2 v of buried plants. The age of the wood is about (a) 5700 years (b) 2580 years 31. The given figure shows the volume V versus (c) 11,400 years (d) 22,800 years temperature T graphs for a certain mass of a perfect 38. Which one of the following is the property of a gas at two constant pressures of P1 and P2. What monochromatic, plane electromagnetic wave in inference can you draw from the V graphs? P2 free space? (a) Electric and magnetic fields have a phase (a) P1 > P2 P1 difference of p/2. (b) P1 < P2 (b) The energy contribution of both electric and (c) P1 = P2 T (d) No inference can be drawn due to insufficient magnetic fields are equal. (c) The direction of propagation is in the direction information of BE× . 32. If R is universal gas constant, the amount of heat (d) The pressure exerted by the wave is the product needed to raise the temperature of 2 moles of an of its speed and energy density. ideal monoatomic gas from 273 K to 373 K when 39. What happens to the fringe pattern if in the path no work is done will be of one of the slits a glass plate which absorbs 50% (a) 100R (b) 150R (c) 300R (d) 500R energy is interposed? 33. Two pendulums of lengths 121 cm and 100 cm (a) The bright fringes become brighter and dark start vibrating. At some instant the two are in the fringes become darker mean position in the same phase. After how many (b) No fringes are observed vibrations of the shorter pendulum, the two will be (c) The fringe width decreases in phase in the mean position? (d) None of these (a) 10 (b) 11 (c) 20 (d) 21 40. If the electron velocity is ()23ij+ m s–1 and it is 34. An air-tight cage with a parrot sitting in it is subjected to a magnetic field, 4k T, then suspended from the spring balance. The parrot (a) speed of electron will change starts flying. The reading of the spring balance will (b) path of electron will change (a) increase (b) decrease (c) Both (a) and (b) (c) not change (d) be zero. (d) None of these 66 Physics for you | May ‘15 41. The magnetic field at a distance d from a short bar and q is the angle the string makes with the vertical. magnet in longitudinal and transverse positions What should be the frequency (u) of rotation of are in the ratio mass m so that mass M remains stationary? (a) 1 : 1 (b) 2 : 3 (c) 2 : 1 (d) 3 : 2 1 mg (a) u = 42. A resistor of 500 W, an inductance of 0.5 H are in 2p M series with an ac which is given by l 1 Mg m V = 100 2sin(1000t). The power factor of the (b) u = combination is 2p ml 1 1 (a) (b) (c) 0.5 (d) 0.6 1 Mg (c) u = 2 3 p m M 43. A ray of light is incident at an angle of 60° on one 1 M g face of a prism of angle 30°. The ray emerging out of (d) u = the prism makes an angle of 30° with the incident 2p m l ray. The emergent ray is 48. The moment of inertia of a sphere of mass M and (a) normal to the face through which it emerges radius R about an axis passing through its centre (b) inclined at 30° to the face through which it 2 is MR2. The radius of gyration of the sphere emerges 5 (c) inclined at 60° to the face through which it about a parallel axis to the above and tangent to the emerges sphere is 7 3 7 3 (d) None of these R (a) R (b) R (c) R (d) 44. The decreasing order of wavelength of infrared, 5 5 5 5 microwaves, ultraviolet and gamma rays is 49. A lift is tied with thick iron wires and its mass is (a) microwaves, infrared, ultraviolet, gamma rays 1000 kg. The minimum diameter of wire if the –2 (b) gamma rays, ultraviolet, infrared, microwaves maximum acceleration of lift is 1.2 m s and the 8 –2 (c) microwaves, gamma rays, infrared, ultraviolet maximum safe stress is 1.4 × 10 N m is –2 (d) infrared, microwaves, ultraviolet, gamma rays (g = 10 m s ) (a) 0.00141 m (b) 0.00282 m 45. Two blocks of masses 1 kg and 2 kg are connected (c) 0.005 m (d) 0.01 m by a metal wire going over a smooth pulley as shown in figure. The breaking stress of the metal is 50. When an AC voltage, of variable frequency is 2 × 109 N m–2. What should be the minimum applied to series L-C-R circuit, the current in the circuit is the same at 4 kHz and 9 kHz. The current radius of the wire used if it is not in the circuit is maximum at to break? (Take g = 10 m s–2) (a) 5 kHz (b) 6.5 kHz (a) 4.6 × 10–5 m –6 (c) 4.2 kHz (d) 6 kHz (b) 4.6 × 10 m 1 kg × –6 51. A concave mirror of focal length f (in air) is (c) 2.5 10 m 2 kg (d) 2.5 × 10–5 m immersed in water (m = 4/3). The focal length of the mirror in water will be 46. A uniform metal chain is placed on a rough table 4 3 7 such that one end of it hangs down over the edge (a) f (b) f (c) f (d) f 3 4 3 of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then the 52. In Young’s double slit experiment, when two light coefficient of static friction is waves form third minimum intensity, they have p (a) 3/4 (b) 1/4 (c) 2/3 (d) 1/2 (a) phase difference of 3 5p (b) phase difference of 47. A large mass M and a small mass m hang at the 2 two ends of a string that passes through a smooth (c) path difference of 3l tube as shown in figure. The massm moves around 5l a circular path in a horizontal plane. The length of (d) path difference of the string from mass m to the top of the tube is l, 2 Physics for you | May ‘15 67 53. An LC resonant circuit contains a 400 pF capacitor soluTioNs and a 100 mH inductor. It is set into oscillation 1. (b) : Given : m = 0.5 kg, v = kx3/2 coupled to an antenna. The wavelength of the where, k = 5 m–1/2 s–1 radiated electromagnetic waves is dv dv dx dv dx Acceleration, a == = v v = (a) 377 mm (b) 377 m dt dx dt dx dt (c) 377 cm (d) 3.77 cm As v2 = k2x3 54. If the wavelength of the first line of the Balmer Differentiating both sides with respect to x, we get dv 3 22 series of hydrogen atom is 6561 Å, the wavelength 23v = kx22 \ Acceleration, ak= x of the second line of the series should be dx 2 3 (a) 13122 Å (b) 3280 Å Force, Fm=×Mass Acceleration = kx22 2 (c) 4860 Å (d) 2187 Å 2 3 22 55. A p-type semiconductor has acceptor level 57 meV Work done, WF==∫∫dx mk xdx 0 2 above the valence band. The maximum wavelength 2 3 of light required to create hole is 3 2 x 3 23 –3 Wm= k =×05.[××52−=05]J0 (a) 57 Å (b) 57 × 10 Å 23 0 6 (c) 217100 Å (d) 11.61 × 10–33 Å 2. (b) : Here, u1 = 2.8 rps, u2 = ?, I2 = 0.7I1 56. A particle is executing simple harmonic motion Using conservation of angular momentum, with an amplitude A and time period T. The I1w1 = I2w2 displacement of the particle after 2T period from w I 1 u 10 2 ==1 \=2 its initial position is w1 I2 07. u1 7 (a) A (b) 4A (c) 8A (d) zero 10 10 uu==×=28..40rps 57. A thin disc having radius r and charge q distributed 217 7 uniformly over the disc is rotated with u rotations 3. (a) : Applying principle of conservation of linear per second about its axis. The magnetic field at the momentum, we get centre of the disc is m m 21 v m×+0 ×=vm+ vm′ = vv′ ⇒ ′ = muq muq muq 3muq (a) 0 (b) 0 (c) 0 (d) 0 20 20 20 21 r 2r 4r 4r 4. (b) : The relative velocity of rain with respect to car 58. The following configuration of gates is equivalent is inclined to the vertical in the backward direction. to G1 Therefore, it will strike the front screen. A (a) NAND 5. (c) : Error in weighing = difference in weight at two B (b) XOR Y different heights (c) OR G 3 2h12 2h (d) AND G =−mg 1 −−mg 1 2 R R 59. At a given temperature, velocity of sound in oxygen 2mg 2m GM h and in hydrogen has the ratio =−()hh21=× × R R R2 2 (a) 4 : 1 (b) 1 : 4 (c) 1 : 1 (d) 2 : 1 [where h2 – h1 = h/2] 24m h 4 60. In the Davisson and Germer experiment, the =×GR××pr3 = prGm h velocity of electrons emitted from the electron gun R3 323 can be increased by 6. (b) : FA = Force on charge at C due to charge at A (a) increasing the potential difference between the −−66 F 9 10 ××210 A anode and filament =×910 × C −22 +2 C 120° (b) increasing the filament current ()10×10 (c) decreasing the filament current =18.N FB (d) decreasing the potential difference between the FB = Force on charge at C +1 C –1 C 10 cm anode and filament due to charge at B A B 68 Physics for you | May ‘15 −−66 2 10 ××210 l − =×9109 × =18.N 2 Z1 1 2 So, = (.01) l1 Z2 −1 Net force on charge at C, 22 l 43 −1 42 ⇒=2 = 9 FF=+22FF+°2 F cos120 ⇒=ll AB AB l 29 −1 28 2 4 =+(.18)(2218.) +−21(.81)( .)81(/21).= 8 N 12. (b) : In reverse biasing of p-n junction diode, the applied reverse voltage establishes an electric field 7. (a) : Resistance of wire, which acts in the same direction as the electric field l rl2 in the potential barrier. Therefore, the height of R ==r ⇒ R ∝ l2 A V potential barrier is increased. 2 2 R1 l1 l 13. (b) : Given circuit is balanced Wheatstone bridge \= = or R2 = 4R1 and hence no current flows through 4 W resistance R2 l2 2l connected across diagonal. \ Change in resistance, So, equivalent resistance of circuit, 6 − − DR RR21 4RR11 69× 18 ×=100 ×=100 ×100 = = W R R R Req 1 1 69+ 5 9 = 300% V V 5V I \ Current, I == = A 8. (c) : Force on wire Q due to wire R Q R 18 / 5 18 V − 41p× 0 7 22××010 F F = × ×01. R 14. (b) 15. (c) R F 4p 00. 2 P 16. (d) : x = 1.2t2, = 20 × 10–5 N (towards right) dx d 2 Force on wire Q due to wire P Velocity,(v ==12.)tt= 24. dt dt −7 41p× 0 23××010 dv d FP = × ×01. Acceleration,aa ==(.24t).==24 constant 4p 01. dt dt –5 = 6 × 10 N (towards right) Thus the given motion is uniformly accelerated. Net force on Q –5 –5 17. (b) : As is clear from figure, rod B passes through F = FR + FP = 20 × 10 + 6 × 10 –5 centre of rods A and C. Therefore, moment of = 26 × 10 N inertia of the system about rod B is = 2.6 × 10–4 N (towards right) I = IA + IB + IC A 9. (d) : f = NAB = (25)(p)(1.8 × 10–2)2 [0.2t – 0.05t2] ML22ML ML2 B = 0.025(0.2t – 0.05t2) I =+0 += 12 12 6 C df e ==0.(025 02..− 01t) 18. (a) : Gravitational potential energy of mass m at dt any point at a distance r from the centre of earth is e At t = 3 s, = 0.0025 V GMm e U =− I ==0.A0016 r R At the surface of earth r = R, –6 \ P = eI = 4 × 10 W = 4 mW GMm GM \=U −=−=mgRg 10. (c) : Light should fall normally on the silvered s 2 R R face. At the height h = nR from the surface of earth r = 0° \ r = A = 30° 2 1 r = R + h = R + nR = R(1 + n) sini sini1 Now, m= 1 or 2 = GMm mgR sin30° \ U =− =− sinr1 h + + Rn()11()n This gives i1 = 45° Change in gravitational potential energy is 1 mgR 11. (b) : For Ka line, l ∝ Ka DU = Uh – Us =− −−()mgR ()Z −1 2 ()1 + n Physics for you | May ‘15 69 mgR 1 n huuh =− + mgR =−mgR 1 = mgR \=Mv or v = 1+ n 1 + n 1+ n c Mc The recoil energy of the nucleus 19. (c) : In a closed cyclic process, the system returns 2 1 1 huu h22 to its initial state. Therefore, the change in internal ==Mv2 M = 2 energy is zero, i.e., E = 0. 2 2 Mc 2 Mc ^^^ 25. (c) : When the drunkard walks 8 steps forward and 20. (b) : Here,(Fi=+2156jk+ ) N ^ 6 steps backward, the displacement in the first 14 and rj=10 m steps = 8 m – 6 m = 2 m ^^^^ Time taken for first 14 steps = 14 s \=WF⋅=ri()21++56jk⋅=10 j 150 J Time taken by drunkard to cover first 10 m of t x 14 21. (c) : yA=−sin2p journey =×10 = 70 s T l 2 l If the drunkard takes 8 steps more, he will fall into Speed of wave = T the pit, so the time taken by the last 8 steps = 8 s Amplitude = A Total time taken = 70 s + 8 s = 78 s Period = T, 26. (c) : Both sound and light waves exhibit the Velocity is along +x-direction. phenomenon of interference. The Bragg formula for crystal structure is an example 22. (a) : Distance covered in 5th second is of the wave nature of electromagnetic radiation. aa9 D5 =+0 ()25×−1 ==() u 0 27. (a) : The value of universal gravitational constant 2 2 –8 2 –2 Distance covered in 5 seconds is in CGS system is 6.67 × 10 dyne cm g and in SI system is 6.67 × 10–11 N m2 kg–2. 1 2 25a Sa5 =+0 ×=5 () u = 0 Their corresponding ratio is 2 2 G in CGSunit 66. 71× 0−8 D 9 = =103 \=5 − G in SI unit 66. 71× 0 11 S5 25 23. (c) : The system is not subjected to any external 28. (b) force and hence conservation of momentum can be ^^ ^^^^ 29. (a) : ||AB−=2 ()AB−⋅()AB− used. Let mb and mp represent the masses of the boy ^^ ^^ ^^ ^^ and the plank respectively. Let vbi, vpi and vbp be the =⋅AA−⋅AB−⋅BA+⋅BB velocity of the boy with respect to ice, that of the ^^ ^^ =−11AB⋅−AB⋅+=−22cosq =−21[cosq] plank with respect to ice and that of the boy with 22qq respect to the plank respectively. Then, =−21 12+ sins = 4 in 2 2 mbvbi + mpvpi = 0 ...(i) ^^ q vbi = vbp + vpi ...(ii) or ||AB−=2sin or vpi = vbi – vbp 2 Substituting the value of vpi in eq. (i), we get 30. (c) : Let R be the radius of big drop formed and r be mbvbi + mp(vbi – vbp) = 0 radius of each small drop. Then mvpbp 4 334 13/ or vbi(mb + mp) = mpvbp = ppRr=×2 or Rr= 2 or vbi 3 3 mmbp+ Terminal velocity ∝r2 Substituting the given values, we get 2 −1 2 13/ 160 kg ×15. ms 240 −−11 v′ R 2 r v = ==ms 12. ms \ == = 3 4 or vv′ = 3 4 bi ()40 +160 kg 200 2 v r r hu 24. (b) : Momentum of emitted photon, p = 1 V2 P1 photon c 31. (a) : For a given T, V ∝ \= Let v be the recoil speed of nucleus. P V1 P2 According to law of conservation of momentum, For a given T, V2 > V1, so P1 > P2. pnucleus = pphoton 70 Physics for you | May ‘15 32. (c) : As no work is done, 2 and IImin =−11I = 0 \ dQ = dU + dW = dU = nCV dT () R R When one of the slits is covered by a glass plate = n dT =×2 ×=100 300R g −1 5 I1 < I0, I2 = I0 \ Imax < 4I0 and Imin > 0 −1 3 40. (b) : Velocity is in x-y plane and magnetic field is along z-axis. Therefore, path of the electron will be 121 100 33. (b) : T ==2ppand T 2 a circle. Magnetic force cannot change the speed of 12g g a particle. So, T > T . Let the shorter pendulum makes n 1 2 2 M vibrations, then the longer pendulum will make 41. (c) : For longitudinal positions, B ∝ 1 3 one less than n vibrations to come in phase again. For transverse positions, d So, nT2 = (n – 1)T1 M B B ∝ \=1 21: 100 121 2 3 B or n×=2pp()n −×12 d 2 g g 42. (a) : Here, R = 500 W, L = 0.5 H or 10n = (n – 1)11 or n = 11 Compare Vt=100 2 sin(1000 ) with V = V0sinwt, 34. (c) : The system is closed. The weight of parrot is we get w = 1000 supported by the air pushed downwards which in The inductive reactance is turn presses the base of cage with a weight equal XL = wL= (1000)(0.5) = 500 W to weight of parrot. These are internal forces. So Impedance of the RL circuit is there will be no change in the reading of the spring ZR=+22X =+()500 WW22()500 = 500 2 W balance. L R 500 W 1 35. (b) : Below the sea level the pressure is increasing Power factor, cosf= = = with depth due to presence of atmospheric air there. Z 500 2 W 2 The acceleration due to gravity below the surface 43. (a) : Angle of deviation of ray in prism is given by of the earth decreases with the distance from the d = i + e – A d So, e = d + A – i = 30° + 30° – 60° = 0° surface of the earth, as gg′ =−1. So, emergent ray will be perpendicular to face. R 36. (b) : Taking vertical downward motion, Or emergent ray will make an angle of 90° with the 1 1 face through which it emerges. sg==t22××10 04..= 08m 2 2 44. (a) : The decreasing order of wavelength of the given electromagnetic waves is as follows: Vertical velocity on ground, –1 lMicrowave > lInfrared > lUltraviolet > lGamma rays vy = gt = 10 × 0.4 = 4 m s Horizontal range = v t = 4 × 0.4 = 1.6 m 45. (a) : Here, m1 = 1 kg, m2 = 2 kg x 9 –2 If q is the angle at which the ball hits the ground Breaking stress = 2 × 10 N m The tension in the string is vx 4 with the vertical, then tanq= ==1 −2 v 4 2mm12 2 ××121kg kg × 0 ms 40 or q = 45° y T = g = = N mm12+ ()12+ kg 3 t/5700 2 t/5700 N0 1 1 1 1 If r is the minimum radius, then 37. (c) : == \ = N 4 2 2 2 (/40 3) t Breaking stress = \=2 or t = 11,400 years 2 5700 pr 2 40 2 − 38. (b) : The energy in EM waves is divided equally r = or r28=×10 32××p × 109 3p between the electric and magnetic fields. 39. (d) : In normal Young’s double slit apparatus, r = 0.46 × 10–4 m = 4.6 × 10–5 m I1 = I2 = I0 46. (d) : The chain starts sliding, when 2 applied force = force of friction \=IImax1+ II1 = 4 0 () (due to hanging part) (between chain and table) Physics for you | May ‘15 71 1 2 1 11 1 mg ==fRmm= mg m= =−R ; 22 3 3 2 l 2 n 47. (b) : Tension in the string T = mg. For first line of Balmer series, n = 3 2 2 Centripetal force on the body = mrw = mr(2pu) . 11 1 5R \=R − = This is provided by the component of tension acting 22 l1 2 3 36 horizontally For second line of Balmer series, n = 4 i.e. Tsinq = Mgsinq 11 1 3R l 20 2 Mgr 1 Mg =−R = \=2 \ mr(2pu) = Mgsinq = or u = l 2216 l 2p ml 2 2 4 l1 27 2 = 2 20 48. (c) : Given, IMR \=l ×6561 = 4860 Å 5 2 27 Using the theorem of parallel axes, moment of inertia of the sphere about a parallel axis tangent to 12375 12375 55. (c) : l(in Å) = = Å ≈ 217100 Å the sphere is E (ineV) 57×10−3 2 7 II′ =+MR22=+MR MR22= MR 56. (d) : The particle completes one oscillation in 5 5 time T. Therefore, in time 2T, it will complete two 7 227 ⇒ = oscillations and will reach to its starting point, \ IM′ ==KMR KR 5 5 i.e., initial position. Therefore, the displacement is zero. 49. (d) : When the lift is accelerated upwards with acceleration a, the tension in the rope is 57. (b) T = m(g + a) = 1000(9.8 + 1.2) = 11000 N 58. (b) : Output of G1 = (A + B) F T T Output of GA=⋅B Now, stress == or r2 = 2 A pr2 p×stress Output of G3 is =+⋅+ 11000×7 1 YA=+()BA⋅⋅()B ()AB()AB = = 84 22××14. 10 41× 0 =⋅AA+⋅AB+⋅BA+⋅BB=⋅AB+⋅AB 1 1 or rD==;.so 2r ==001 m It is the Boolean expression of XOR gate. 200 100 Hence, the given configuration of gates is equivalent 50. (d) to XOR gate. 51. (a) : On immersing a mirror in water, focal length 59. (b) : Speed of sound in gas is of the mirror remains unchanged. gRT 52. (d) : For minima, path difference should be an odd v = multiple of half wavelength. M l where the symbols have their usual meanings. or D=()21n − Since both oxygen and hydrogen are diatomic gases 2 For third minima, and at same temperature, the ratio of velocities will ll5 be n ==32,(D ×−31) = 2 2 vO MH 2 1 2 ==2 = 53. (b) : Frequency of LC oscillations is vH MO 32 4 1 2 2 1 = u = 1 2p LC 23××.14 100 10−−61××400 10 2 60. (a) : As eV =∝mv2 or vV 7 2 = 0.0796 × 10 Hz Thus velocity of electron emitted from electron c 31× 08 gun can be increased by increasing the potential So, wavelength, l == = 377 m u 0.0796×107 difference between anode and filament in Davisson and Germer experiment. 54. (c) : Wavelength of spectral line in Balmer series is given by 72 Physics for you | May ‘15 Using momentum conservation principle Mv = m[u(cos45°) × cos45° – v] H G mu Solution Set-21 ⇒+()Mmv = l 2 D C v u 1. (b) : FF=+22F mu mgl 2 TR ⇒=v = E F 2()Mm+ 2()Mm+ R l 2 l p l A S Fm==awand Fm =×m2a B TR22 22 5. (d) : F = pt – qx P Q Acceleration of the particle, 2 2 []Using ww=+0 2aq F 1 da pq− v a ==()pt − qx ⇒= l mm dt m \=Fmap1+ 2 2 2 da q 2 q 2. (d) : Magnetic induction at the centre of an arc ⇒=−=aa−w ⇒=w dt2 m m carrying current I is Hence, the acceleration of particle varies sinusoidally m I B =∝0 qqor BI with time. 4pr F L The total current I is divided along two arcs such 6. (a) : Young’s modulus, Y = . A l that F YA D k == 1 I2 Force constant, I ∝ O B l L Resistance of arc where l is the extension in the rod of original length I1 1 I L and cross-sectional area A when a force F = Mg is or I ∝ A Length of arc I applied. Now, the time period of vertical oscillations 1 is given by or I ∝ Anglesubtended at centre ()q M ML T1 Y2 3 T ==22pp\= = \ Iq = constant k YA T2 Y1 2 or I q = I (2p – q) 1 2 7. (a) : Mass per unit area of the disc, Hence the magnetic fields at the centre due to arcs Mass M M ACB and ADB are equal and opposite. The net s= = = Area 22 2 magnetic induction at the centre O is zero. p()()43RR− () 7pR Consider a ring of radius x and thickness dx as 1 2 3. (a) : For first fragment of particle, ut −=gt −H shown in the figure. 112 1 gt2 −−ut H = 0 2 1 1 uu±+2222gH uu++gH t = = 1 g g For second fragment of particle, along the vertical direction, Mass of the ring, dM = s2pxdx 1 2 2pMxdx gt2 +−ut2 H = 0 = 2 7pR2 22 Potential at point P due to annular disc is −±uu+ 22gH −+uu+ gH t = = 4R 4R 2 g g GdM GM2p xdx VP =− =− ∫ 22 2 ∫ 22 \ Dt = t1 – t2 = 2u/g 3R ()4Rx+ () 7pR 3R 16Rx+ 4. (a) : Let v is the velocity of carriage with respect to Solving, we get ground and u is the velocity of shot with respect to GM2p 4R V =− 16Rx22+ carriage as shown in the figure. P 7pR2 3R u2 sin(24×°5 ) 2GM =⇒lu22= gl ...(i) =− ()42− 5 g 7R Physics for you | May ‘15 73 Work done in moving a unit mass from P to ∞ = V∞ – VP −2GM 2GM =−0 42− 5 =−()42 5 7R () 7R 8. (a) : DQ = –DU; nCdT = –nCVdT C = –CV = – R/(g – 1) dQ = dU + dW; 2dQ = dW RT 2CdT = dV V ⇒ TV(g–1)/2 = constant and TV1/5 = constant (given) g −1 1 7 So =⇒g = 2 5 5 9. (c) : Net magnetic field at point P, m00I 3p m I m0I B = +⊗+⊗ 4pR 24 pR 4pR m I m I B =−00()3p 8pR 2pR m I =−0 ()34p 8pR 10. (c) : f1l =ml m1 g = 0.4 × 2 × 10 = 8 N f2l = m2 m2 g = 0.6 × 4 × 10 = 24 N fl > Fapplied So, system will not move i.e. it will be in equilibrium For equilibrium of 2 kg block 10 = 8 + T ⇒ T = 2 N 10 N 2 kg T f1 For equilibrium of 4 kg block T + f2 = 20 ⇒ f2 = 18 N Solution Senders of Physics Musing sET-21 1. Swayangdipta Bera 2. Priyadeep Kaur (New Delhi) 3. Sattwik Sadhu (West Bengal) 4. Anubhab Banerjee (West Bengal) 5. Arun Nayan (UP) sET-20 1. Anupama Saha (Indore) 2. Divyanshu Kaushik (Pune) 3. Manpreet Sandhu (Punjab) 74 Physics for you | May ‘15 E 1. Distance of the centre of mass of a solid uniform KE E PE cone from its vertex is z . If the radius of its base is 0 (c) (d) R and its height is h then z0 is equal to PE KE d d 2 5h 3h h2 3h (a) (b) (c) (d) 5. A train is moving on a straight track with speed 8 8R 4R 4 20 m s–1. It is blowing its whistle at the frequency of 2. A red LED emits light at 0.1 watt uniformly around 1000 Hz. The percentage change in the frequency it. The amplitude of the electric field of the light at a heard by a person standing near the track as the distance of 1 m from the diode is train passes him is (speed of sound = 320 m s–1) (a) 5.48 V/m (b) 7.75 V/m close to (c) 1.73 V/m (d) 2.45 V/m (a) 18% (b) 24% 3. A pendulum made of a uniform wire of cross (c) 6% (d) 12% sectional area A has time period T. When an 6. When 5 V potential difference is applied across a additional mass M is added to its bob, the time wire of length 0.1 m, the drift speed of electrons is –4 –1 period changes to TM. If the Young’s modulus of the 2.5 × 10 m s . If the electron density in the wire is 1 28 –3 material of the wire is Y then is equal to 8 × 10 m , the resistivity of the material is close to Y (a) 1.6 × 10–6 W m (b) 1.6 × 10–5 W m (g = gravitational acceleration) (c) 1.6 × 10–8 W m (d) 1.6 × 10–7 W m 2 2 7. Two long current carrying thin wires, both with TM A T A (a) 1 − (b) 1 − current I, are held by insulating threads of length L T Mg T Mg M and are in equilibrium as shown in the figure, with threads making an angle q with the vertical. If wires 2 2 T Mg have mass l per unit length then the value of I is TM A M − 1 (c) − 1 (d) (g = gravitational acceleration) T Mg T A 4. For a simple pendulum, a graph is plotted between L its kinetic energy (KE) and potential energy (PE) θ against its displacement d. Which one of the following represents these correctly? I I (graphs are schematic and not drawn to scale) pgL plgL E KE (a) 2 tanq (b) tanq E m0 m0 PE (a) d (b) KE plgL plgL (c) sinq (d) 2sinq d mqcos 0 mq0 cos PE Physics for you | May ‘15 75 8. In the circuit shown, Ω 12. In the given circuit, charge Q 6 V P 2 2 the current in the 1 W on the 2 mF capacitor changes 1Fµ resistor is as C is varied from 1 mF to C 1 Ω 9 V µ (a) 0.13 A, from Q to P 3 mF. Q2 as a function of ‘C’ is 2F (b) 0.13 A, from P to Q given properly by E Q (c) 0.3 A, from P to Q 3 Ω 3 Ω (figures are drawn schematically and are not to scale) (d) 0 A Charge Charge 9. Assuming human pupil to have a radius of Q (a) Q2 (b) 2 0.25 cm and a comfortable viewing distance of C C 25 cm, the minimum separation between two 1Fµ 3Fµ 1Fµ 3Fµ objects that human eye can resolve at 500 nm Charge Charge wavelength is (c) Q2 (d) Q2 (a) 100 mm (b) 300 mm C C (c) 1 mm (d) 30 mm 1Fµ 3Fµ 1Fµ 3Fµ 10. An inductor (L = 0.03 H) and a resistor (R = 0.15 kW) 13. From a solid sphere of mass M and radius R a cube are connected in series to a battery of 15 V EMF of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its in a circuit shown below. The key K1 has been kept centre and perpendicular to one of its faces is closed for a long time. Then at t = 0, K1 is opened 2 2 and key K2 is closed simultaneously. At t = 1 ms, the 4MR 4MR 5 (a) (b) current in the circuit will be (e @ 150) 93p 33p 2 0.03 H 0.15 kΩ MR MR2 (c) (d) 32 2p 16 2p K2 14. The period of oscillation of a simple pendulum is L K T = 2p . Measured value of L is 20.0 cm known 1 g 15 V to 1 mm accuracy and time for 100 oscillations of (a) 6.7 mA (b) 0.67 mA the pendulum is found to be 90 s using a wrist watch (c) 100 mA (d) 67 mA of 1 s resolution. The accuracy in the determination 11. An LCR circuit is equivalent of g is R L to a damped pendulum. In (a) 1% (b) 5% (c) 2% (d) 3% an LCR circuit the capacitor 15. On a hot summer night, the refractive index of air is is charged to Q0 and then smallest near the ground and increases with height connected to the L and R as C from the ground. When a light beam is directed shown here. horizontally, the Huygens’ principle leads us to If a student plots graphs of the square of maximum conclude that as it travels, the light beam 2 charge (Q Max) on capacitor with time (t) for two (a) bends downwards different values L1 and L2 (L1 > L2) of L then which (b) bends upwards of the following represents this graph correctly ? (c) becomes narrower (plots are schematic and not drawn to scale) (d) goes horizontally without any deflection Q2 2 16. A signal of 5 kHz frequency is amplitude modulated Max Q L1 Max Q (For both (a) (b) 0 on a carrier wave of frequency 2 MHz. The L2 LL12and) frequencies of the resultant signal is/are t t (a) 2005 kHz, 2000 kHz and 1995 kHz Q2 Q2 (b) 2000 kHz and 1995 kHz Max L Max 1 L2 (c) 2 MHz only (c) L (d) 2 L1 (d) 2005 kHz and 1995 kHz t t 76 Physics for you | May ‘15 17. A solid body of constant heat capacity 1 J/°C is being (a) R1 = 0 and R2 < (R4 – R3) heated by keeping it in contact with reservoirs in (b) 2R < R4 two ways: (c) R1 = 0 and R2 > (R4 – R3) (i) Sequentially keeping in contact with 2 reservoirs (d) R1 ≠ 0 and (R2 – R1) > (R4 – R3) such that each reservoir supplies same amount 21. Monochromatic light is incident on a glass prism of of heat. angle A. If the refractive index of the material of the (ii) Sequentially keeping in contact with 8 reservoirs m q such that each reservoir supplies same amount prism is , a ray, incident at an angle , on the face of heat. AB would get transmitted through the face AC of the prism provided In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases A respectively is θ (a) ln2, 2ln2 (b) 2ln2, 8ln2 (c) ln2, 4ln2 (d) ln2, ln2 B C 18. Consider a spherical shell of radius R at temperature −−11 1 T. The black body radiation inside it can be (a) qm>+coss in A sin considered as an ideal gas of photons with internal m U 4 energy per unit volume u =∝T and pressure −−11 1 (b) qm<+coss in A sin V m 1 U p = . If the shell now undergoes an adiabatic 3 V −−11 1 expansion the relation between T and R is (c) qm>−sins in A sin m 1 ∝ 1 (a) T (b) T ∝ R R3 −−11 1 (d) qm<−sins in A sin (c) T ∝ e–R (d) T ∝ e–3R m 19. Two stones are thrown up simultaneously from 22. A rectangular loop of sides 10 cm and 5 cm carrying the edge of a cliff 240 m high with initial speed of a current I of 12 A is placed in different orientations 10 m/s and 40 m/s respectively. Which of the as shown in the figure below. following graph best represents the time variation z z of relative position of the second stone with respect I B B to the first? (1) I I (2) (Assume stones do not rebound after hitting the y I y I I ground and neglect air resistance, take g = 10 m/s2) I x x I (The figures are schematic and not drawn to scale) (–) m (–)21 m yy21 z z 240 yy 240 B (a) (b) I B (3) (4) (s) (s) I I 8 12 t 8 12 t I y y I I (–) m (–) m I 240 yy21 240 yy21 x x I (c) (d) If there is a uniform magnetic field of 0.3 T in the (s) (s) positive z direction, in which orientations the loop 8 12 t 12 t would be in (i) stable equilibrium and (ii) unstable 20. A uniformly charged solid sphere of radius R has equilibrium. potential V (measured with respect to ∞) on its 0 (a) (2) and (4), respectively surface. For this sphere the equipotential surfaces (b) (2) and (3), respectively 3VV5 3VV with potentials 00,,00and have radius (c) (1) and (2), respectively 2 44 4 (d) (1) and (3), respectively R1, R2, R3 and R4 respectively. Then Physics for you | May ‘15 77 23. Two coaxial solenoids of different radii carry current I in the same direction. Let F be the magnetic force F 1 A B on the inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the inner one. Then (a) F1 is radially inwards and F2 = 0 (a) 120 N (b) 150 N (c) 100 N (d) 80 N = (b) F1 is radially outwards and F2 0 28. A long cylindrical shell carries positive surface charge s in the upper half and negative surface (c) FF12==0 charge –s in the lower half. The electric field lines (d) F1 is radially inwards and F2 = 0 is radially around the cylinder will look like figure given in outwards (Figures are schematic and not drawn to scale) 24. A particle of mass m moving in the x direction + + with speed 2v is hit by another particle of mass ++ ++ + + (a) + + (b) + + 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to (a) 56% (b) 62% (c) 44% (d) 50% ++ ++ 25. Consider an ideal gas confined in an isolated (c) + + (d) closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V q, where V is the volume of the gas. The value of q is 29. As an electron makes a transition from an excited state to the ground state of a hydrogen -like atom/ion Cp (a) kinetic energy decreases, potential energy g= increases but total energy remains same Cv (b) kinetic energy and total energy decrease but g+1 g−1 potential energy increases (a) (b) 2 2 (c) its kinetic energy increases but potential energy and total energy decrease 35g+ 35g− (c) (d) (d) kinetic energy, potential energy and total energy 6 6 decrease 26. From a solid sphere of mass M and radius R, a 30. Match List-I (Fundamental Experiment) with R spherical portion of radius is removed, as shown List-II (its conclusion) and select the correct option 2 from the choices given below the list: in the figure. Taking gravitational potentialV = 0 at r = ∞, the potential at the centre of the cavity thus List-I List-II formed is P. Franck-Hertz (i) Particle nature of (G = gravitational constant) Experiment. light −2GM −2GM (a) (b) Q. Photo-electric (ii) Discrete energy 3R R Experiment. levels of atom −GM −GM (c) (d) R. Davison-Germer (iii) Wave nature of 2R R Experiment. electron 27. Given in the figure are two blocksA and B of weight (iv) Structure of atom 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the (a) P - (ii), Q - (i), R - (iii) coefficient of friction between the blocks is 0.1 and (b) P - (iv), Q - (iii), R - (ii) between block B and the wall is 0.15, the frictional (c) P - (i), Q - (iv), R - (iii) force applied by the wall on block B is (d) P - (ii), Q - (iv), R - (iii) 78 Physics for you | May ‘15 soluTioNs 3. (c) : Let length of the pendulum wire be l. 1. (d) : Let r be the density of solid cone. l \ Time period, T = 2p ...(i) Consider a disc of radius r, thickness dy at a distance g of y from its vertex. Then mass of this disc is When an additional mass M is added to bob, let Dl rp 2 dm = r dy be the extension produced in wire. h ll+ D ydm y ...(ii) ∫ Then TM = 2p 0 h dy g \=ycm r h stress Mg / A Dl Mg dm Now, Y == ⇒= ...(iii) ∫ R strain Dll/ l AY 0 From eqns. (i) and (ii), we get h 2 T ll+D ∫ rprydy M = T l = 0 ...(i) h 2 T ll+ D Dl Mg rprd2 y or M = =+1 =+1 ∫ T l l AY 0 r R Ry (Using (iii)) From figure, tanq= = or r = ...(ii) y h h 2 2 Mg T 1 A TM or = M −1 or = −1 Putting eqn. (ii) in (i), we get AY T Y Mg T h rpR2 yd3 y 4. (d) : For a simple pendulum, variation of kinetic ∫ 2 h energy and potential energy with displacement d is = 0 ycm h rpR2 1 22 2 1 22 yd2 y KE =−mAw ()d and PE = mdw ∫ 2 2 2 0 h where A is amplitude of oscillation. h h 4 y 1 22 yd3 y When d = 0, KE ==mAw ,PE 0 ∫ 4 4 h 33h 2 ==0 0 =×= h h 3 1 22 3 4 h 4 When d = ± A, KE = 0, PE = mAw 2 y 2 ydy ∫ 3 Therefore, graph (d) represents the variation 0 0 correctly. \ Distance of the centre of mass of a solid uniform 5. (d) : Frequency of sound emitted by train, 3h cone from its vertex, zy==. u = 1000 Hz 0 cm 4 Speed of train (source), v = 20 m s–1 2. (d) : Intensity of light, I = u c s av Speed of sound, v = 320 m s–1 P 1 Also,aI ==nd uEe 2 Observer is stationary. 2 av 2 00 4pr Frequency heard by person as train approaches him P 1 \=e Ec2 v 2 00 uu1 = 4pr 2 vv− s 2P 320 3200 or E0 = = ×=1000 Hz 4pe rc2 320 − 20 3 0 8 –1 Here, P = 0.1 W, r = 1 m, c = 3 × 10 m s Frequency heard by person as train moves away 1 =×9109 N C–2 m2 from him 4pe 0 v 320 32000 uu= = × = 9 2 + 1000 Hz 20××.191× 0 −1 vvs 320 + 20 34 \=E0 ==62.V45 m 1328××10 Physics for you | May ‘15 79 \ Percentage change in frequency Applying KVL in loop PQCDP uu21− –1I2 – 3I1 +9 – 2I1 = 0 ⇒ 5I1 + I2 = 9 ...(i) = ×100 Applying KVL in loop PQBAP u1 –1I + 3(I – I ) – 6 = 0 ⇒ 3I – 4I = 6 ...(ii) 32000 3200 2 1 2 1 2 − Solving eqns. (i) and (ii), we get = 34 3 ×100 ≈ – 12% 3200 I1 = 1.83 A, I2 = –0.13 A 3 \ The current in the 1 W resistor is 0.13 A, from Q to P. Negative sign implies that the frequency heard by 9. (d) : Given, wavelength of light, l = 500 nm person decreases as the train passes him. = 500 × 10–9 m 6. (b) : V = IR Least distance of distinct vision, D = 25 cm –2 rl = 25 × 10 m As I = neAv and R = d A Radius of pupil, r = 0.25 cm r V \ Diameter of pupil, d = 2r = 0.50 cm \ l r= Vn=×eAvd or = 0.50 × 10–2 m A nevld 12. 2 l Here, V = 5 V, n = 8 × 1028 m–3, v = 2.5 × 10 –4 m s–1, Resolving power of eye, Dq = d d l = 0.1 m, e = 1.6 × 10–19 C 12. 2 ××500 10−9 5 = =×−4 \=r −2 12.r210 ad 81××0128 ..61××02−−19 51××004 .1 05. 01× 0 = 0.156 × 10–4 W m 1.6 × 10–5 W m \ Minimum separation that eye can resolve, Dq 7. (d) : Let the length of right wire be l, then its mass is ll. x = D = 1.22 × 10–4 × 25 × 10–2 –6 = 30.5 × 10 m 30 µm T T cos 10. (b) : When key K1 is kept closed, a steady current T sin e F I = flows through the circuit. 2L sin 0 R lg When K1 is opened and K2 is closed, current at any Force acting on this wire are tension (T), weight time t in the circuit is (llg) and force of repulsion due to other wire (F). tR e − L From figure, T cosq = llg ...(i) II==e−t/t e L t = 0 R R T sinq = F ...(ii) m Il2 Here, e = 15 V, R = 0.15 kW = 150 W Here, F = 0 –3 22pq(sL in ) L = 0.03 H, t = 1 ms = 10 s −3 2 10 × 150 m0 Il − −5 or T sin q = (Using (ii)) 15 00. 3 e 1 1 22pq(sL in ) \=Ie == = 150 10 10e5 10 ×150 llg m Il2 or sinq = 0 (Using (i)) = 6.67 × 10–4 A cosq 22pq(sL in ) = 0.67 mA 4plLgsin2 q plgL 11. (c) : At any time t, the equation of the given circuit is ⇒=I = 2 sinq mqcos mqcos dq2 dq 1 00 L ++R q = 0 ...(i) 2 dt C 8. (a) : 6 V dt II12– 2 I1 P which is equivalent to that of a damped pendulum. A D I2 The solution to eqn. (i) is –Rt/2L 9 V q = Q0 e cos (w′t + f) 1 2 I1 1 R B C where w′ =− . 3 Q 3 LC 2L 80 Physics for you | May ‘15 The square of maximum charge on capacitor at any 01. 1 time t is =×100 +×2 ×100 20.0 90 QQ2 = 22et−Rt/L cos(wf′ + ) max 0 = 2.72% 3% \ It decays exponentially with time. \ Accuracy in the determination of g is approximately 3% For L2 (L2 < L1), the curve is more steep. 12. (d) : Equivalent capacitance of the circuit 15. (b) : Consider a plane wavefront 11 1 11 travelling horizontally. =+ =+ As refractive index of air CCeq ()12+ C 3 increases with height, so speed 3C of wavefront decreases with or C = eq C + 3 height. Hence, the light beam 3CE Total charge in the circuit, Q = C E = bends upwards. eq C + 3 Charge on the 2 µF capacitor, 16. (a) : Given, um = 5 kHz, uc = 2 MHz = 2000 kHz The frequencies of the resultant signal are 2 2 3CE 2CE QQ= =× = u + u = (2000 + 5) kHz = 2005 kHz 2 3 3 ()C + 3 C + 3 c m u = 2000 kHz and 2E c or Q = u – u = (2000 – 5) kHz = 1995 kHz 2 3 c m 1 + C 17. (*) : Since entropy is a state function and the entropy change is independent of the path followed, dQ2 6E and = therefore for both cases dC ()C + 3 2 T2 dQ dT T2 As C increases, Q2 increases and slope of Q2 – C DS = ∫∫==C C ln T T T curve decreases. Hence, graph (d) represents the T1 1 correct variation. Here, T1 = 100°C = 373 K 13. (a) : A cube of maximum possible volume is cut T2 = 200°C = 473 K 473 from a solid sphere of radius R, it implies that the \ DS = C ln diagonal of the cube is equal to the diameter of 373 sphere, ie..,3aR= 2 *None of the given options is correct. If unit of 2R temperatures in question paper were Kelvin, then or a = 200 3 DS = C ln = Cln2 = ln2 i.e., option (d) would M M Density of solid sphere, r′= = 100 V 4p have been correct. R3 3 18. (a) : According to first law of thermodynamics, 3 \ Mass of cube, M′ = rV′ = ra ( r = r′) dQ = dU + dW 3 M 2RM 2 Since the shell undergoes an adiabatic expansion = = 4p \ dQ = 0, i.e., dU = – dW = – pdV R3 3 3 p 3 dU 1 U 1 U or =−p =− Given:p = Moment of inertia of the cube about an axis passing dV 3 V 3 V through its center and perpendicular to one of its dU 1 dV faces is ⇒ =− 2 U 3 V Ma′ 2 1 2MR 2 4MR2 I = = = Integrating both sides 6 6 p p 3 3 93 1 lnU = − lnV + lnC L 44pp2L 22Ln t 3 14. (d) : T =⇒2p g == T = 1/3 g T 2 t2 n or UV = C ...(i) Maximum percentage error in g U Given, u = ∝ T 4 or U = KVT4 DDg L Dt ×=100 ×+100 2 ×100 V g L t Physics for you | May ‘15 81 Putting this in eqn. (i) 21. (c) : According to Snell’s law A KVT 4V1/3 = C sinq = msinr1 T 4V 4/3 = C/K sinq ⇒ sinr = 43/ 1 m 4p 4p r1 r or TR43 = C/K ( V = R3) q 2 –1 sin 3 3 or r1 = sin 1 m ⇒ T 4R4 = C′ ⇒ T ∝ BC R Now, A = r1 + r2 –1 sinq 1 2 \ r2 = A – r1 = A – sin …(i) 19. (a) : Using h = ut + at m 2 1 2 For the ray to get transmitted through the face AC, For stone 1, y1 = 10t – gt 2 r2 must be less than critical angle, 1 2 –1 1 For stone 2, y2 = 40t – gt i.e., r2 < sin 2 m Relative position of the second stone with respect to –1 sinq –1 1 or A – sin < sin (using (i)) 1 2 1 2 m the first, Dy = y2 – y1 = 40t – gt – 10t + gt m 2 2 = 30 t –1 sinq –1 1 ⇒ sin > A – sin After 8 seconds, stone 1 reaches ground, m m i.e., y1 = –240 m sinq 1 ⇒ −1 1 2 >−sinsA in \ Dy = y2 – y1 = 40t – gt + 240 mm 2 Therefore, it will be a parabolic curve till other stone –1 −1 1 ⇒ q > sin msinsA − in reaches ground. m 20. (a, b) : Potential on the surface of charged solid 22. (a) : I = 12 A, Bk= 03.T, A = 10 × 5 cm2 = 50 × 10–4 m2 sphere Kq MI==An12 ××50 10−42n Am V0 = R = 6 × 10–2 n A m2 Spherical surface of radius r inside this sphere will −22 −22 Here, Mi1 =×610 Am , Mk2 =×610 Am be equipotential surface with potential V(> V0) Kq V Mj=−61× 0−22 Am , Mk=−61× 0−22 Am V = (3R2 – r2) = 0 (3R2 – r2) 3 4 3 2 2R 2R M2 is parallel to B , it means potential energy is minimum, therefore in orientation (2) the loop is 3VV003 V0 2 2 \ For V = , = (3R – R1 ) ⇒ R1 = 0 in stable equilibrium. 2 2 2R2 M is antiparallel to B , it means potential energy 5VV5 V R 4 00, = 0 2 2 ⇒ For V = 2 (3R – R2 ) R2 = is maximum, therefore in orientation (4) the loop is 4 4 2R 2 in unstable equilibrium. Spherical surface of radius r′ outside this sphere will 23. (c) be equipotential surface with potential V′(< V0) 24. (a) : Applying the principle of momentum Kq VR0 conservation V′ = = r′ r′ mv()22im+=()(vj mm+ 2 )v′ y 3VV003 VR0 4R \ For V′ = ; = ⇒ R3 = 4 4 R3 3 VV00VR0 For V′ = ; = ⇒ R4 = 4R m 2v x x 44R4 v Here R1 = 0, R2 < (R4 – R3), 2R < R4 and 2m (R2 – R1) < (R4 – R3) So, options (a) and (b) are correct. y 82 Physics for you | May ‘15 Physics for you | May ‘15 83 +=′ MV ′ 22mvimvj 3mv Mass of spherical portion to be removed, M′ = 2v V ⇒ v′ =+ij () 3 3 4p R M 22 M or vv′ = ′ = v = 32 = 3 4p 3 8 R Initial energy of the system, 3 1 2 1 2 Potential at point P due to spherical portion to be Ei = m(2v) + (2m)v 2 2 removed 2 2 2 −3GM′ −38GM(/) −3GM = 2mv + mv = 3mv V = = = 2 ′ 1 2 2R 22(/R ) 8R Final energy of the system, Ef = (3m)v′ 2 \ Potential at the centre of cavity formed 2 VP = V1 – V2 3m 22 = v −11GM −3GM −GM 2 3 = − = 8R 8R R 4 = mv2 3 27. (a) : Various forces acting on the system are shown in the figure. EEif− \ Percentage loss in the energy = × 100 Ei 4 3mv22− mv = 3 × 100 3mv2 5 = × 100 ≈ 56% 9 For vertical equilibrium of the system, 25. (a) : Average time of collision between molecules, fB = 100 N + 20 N = 120 N Mean free path()l 1 t = = 28. (c) : The electric field lines around the cylinder Mean speed ()v 2 N 8kT must resemble that due to a dipole. 2pd B V mp 29. (c) : For an electron in nth excited state of hydrogen 2 V V atom, ⇒ t ∝ or T ∝ …(i) 2 t2 e T kinetic energy = g–1 2 For adiabatic expansion, TV = constant 8p∈0 na0 2 V −e2 or V g–1 = constant t2 potential energy = 2 4p ∈0 na0 ()g+1 −e2 ⇒ t ∝ V 2 and total energy = 2 8p ∈0 na0 g+1 where a0 is Bohr radius. Comparing it with t ∝ V q, we get q = 2 As electron makes a transition from an excited state to the ground state, n decreases. Therefore kinetic 26. (d) : Potential at point P (centre of cavity) before energy increases but potential energy and total removing the spherical portion, energy decrease. 2 −GM 2 R 30. (a) : Franck-Hertz Experiment - Discrete energy V1 = 3R − 2R3 2 levels of atom. R/2 2 O P −GM 2 R Photo-electric experiment - Particle nature of light. = 3R − R 3 Davison-Germer Experiment - Wave nature of 2R 4 −11GM electron. = 8R 84 Physics for you | May ‘15 Readers can send their answer with complete address before 15th of every month to win exciting prizes. Winners' name with their valuable feedback will be published in next issue. ACroSS 3. A substance that can sustain Cut Here an electric field and act as an 123 4 5 6 insulator. (10) 7 5. Atomic unit of energy. (7) 819 10 1 7. A collective name for both hadrons 12 and bosons. (5) 8. A star that explodes and leaves a 13 neutron star remnant. (9) 14 13. Substance which does not allow 15 the passage of X-rays or other 16 17 radiations. (10) 14. The study of universe and its contents. (9) 18 19 20 16. An instrument used for measuring 21 the moisture content in the 22 23 24 atmosphere. (10) 25 26 17. Scale which measures power of an earthquake. (7) 19. The emission of light as a result of 27 the excitation of atoms by a source of energy other than heat. (12) 28 29 26. A cataract operation in which the diseased lens is reduced to a liquid 30 by ultrasonic vibrations and drained out of the eye. (19) 31 27. Process of bonding one metal over 32 33 another to protect the inner metal 34 from corrosion. (8) 35 28. Academic study of musical instrument. (10) 30. The hottest planet. (5) 9. Two electrons spinning in opposite directions. (8, 4) 31. A hypothetical form of matter invisible to electromagnetic 10. An instrument for measuring high temperature. (9) radiation. (4, 6) 11. Sound producing organ in humans. (6) 32. A device used to maintain constant pressure in a closed chamber. 12. A region in a magnetic material in which magnetization is in a (8) uniform direction. (8, 6) 33. Smallest unit of matter. (4) 15. A class of electron orbits in an atom in which the electrons have 34. An optical instrument for viewing objects that are in an the same principal quantum number. (5) obstructed field of vision. (9) 18. Discoverer of the moons of mars. (5, 4) 35. A wave in which each point on the axis of the wave has an 20. The flow of current in a path other than that intended, due to associated constant amplitude. (8, 4) imperfect insulation. (7) 21. Category under which pluto comes. (5, 6) DoWn 22. The shadow cast by the moon onto the earth during a solar 1. A device used to detect and find range of an obstacle by using eclipse. (5) the echo method. (5) 23. An instrument for the detection of charge. (12) 2. SI unit of thermodynamic temperature. (6) 24. A reaction in which a particle and its antiparticle collide and 4. A combination of electric or magnetic fields used to capture disappear, releasing energy. (12) charged particles in vacuum. (3, 4) 25. A fluid used as the working fluid of a refrigerator. (11) 6. An optical phenomena that is caused by reflection, refraction 29. One of the universal logic gates. (4) and dispersion of light in water droplets. (7) Physics for you | May ‘15 85 86 Physics for you | May ‘15