Volume 24 Managing Editor Mahabir Singh February 2016 Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR) Tel : 0124-4951200 e-mail : info@mtg.in website : www.mtg.in Regd Office: 406, Taj Apartment, Near Safdarjung Hospital, New Delhi - 110029 CONTENTS Editor Anil Ahlawat (BE, MBA) No Physics Musing Problem Set 31 JEE Main Practice Paper 10 Core Concept 22 JEE Workouts 26 JEE Accelerated Learning Series 31 Brain Map 46 Ace Your Way CBSE XII 54 AIPMT Practice Paper 63 Physics Musing Solution Set 30 73 Exam Prep 2016 75 You Ask We Answer 82 Live Physics 83 At a Glance 2015 84 Crossword 85 subscribe online at www.mtg.in individual subscription rates combined subscription rates yr yrs yrs yr yrs yrs Mathematics Today 330 600 775 PCM 900 1500 1900 Chemistry Today 330 600 775 PCB 900 1500 1900 Physics For You 330 600 775 PCMB 1000 1800 2300 Biology Today 330 600 775 Send D.D/M.O in favour of MTG Learning Media (P) Ltd Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No 99, Sector 44, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and 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 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 disputes are subject to Delhi jurisdiction only Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd All rights reserved Reproduction in any form is prohibited Physics for you | FEBRUARY ‘16 P PHYSICS MUSING hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh The aim of Physics Musing 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 detailed solutions with their names and complete address 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 31 single oPtion correct tyPe A particle starts from rest at A and moves with uniform acceleration a m s–2 in a straight line After 1/a seconds, a second particle starts from A and moves with uniform velocity u in the same line and same direction If u > m s–1 then during the entire motion, the second particle remains ahead of first particle for a duration a (a) u(u − 2) (b) u(u − 2) a (c) (d) u(u − 2) u(u − 2) a a A block of mass 100 g moves with a speed of m s–1 at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane The cross-section of the tube is such that the block just fits in it The block makes several oscillations inside the tube and finally stops at the lowest point The work done by the tube on the block during the process is (a) 1.45 J (b) – 1.45 J (c) 0.2 J (d) zero Two identical balls A and B are released from the positions shown in figure They collide elastically on horizontal portion MN All surfaces are smooth The ratio of heights attained by A and B after collision will be (neglect energy loss at M and N) (a) : B A (b) : 4h (c) : h 45° M N 60° (d) : A stone of mass m, tied to the end of a string, is whirled around in a horizontal circle (Neglect the force due to gravity.) The length of the string is reduced gradually, keeping the angular momentum of the stone about the centre of the circle constant Then, the tension in the string is given by T = Arn, where A is a constant, r is the instantaneous radius of the circle, and n is (a) (b) – (c) – (d) – Physics for you | february ‘16 Consider two hollow glass spheres, one containing water and the other containing mercury Each liquid fills about one-tenth of the volume of the sphere In zero gravity environment (a) water and mercury float freely inside the spheres (b) water forms a layer on the glass while mercury floats (c) mercury forms a layer on the glass while water floats (d) water and mercury both form a layer on the glass subjective tyPe A body starts from rest and moving with uniform acceleration of m s–2 covers half of its total path during the last second of its motion Find the time taken and the total distance covered A cone of height h and base radius r is fixed base to base on a hemisphere of equal radius Find h so that the centre of gravity of the composite solid lies on the common base (Assume same density for both objects.) A rod PQ of length l is pivoted at an end P and freely rotated in a horizontal plane at an angular speed w about a vertical axis passing through P If coefficient of linear expansion of material of rod is a, find the percentage change in its angular velocity , if temperature of system is increased by DT Distance between the centres of two stars is 10a The masses of these stars are M and 16 M and their radii a and 2a, respectively A body of mass m is fired straight from the surface of the large star towards the smaller star What should be its minimum initial speed to reach the surface of the smaller star? 10 The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz The fundamental frequency of the closed organ pipe is 110 Hz Find the lengths of the pipes (Take, speed of sound in air = 330 m s–1) nn Motion in a Plane PaPer-i (single oPtion correct tyPe) From a certain height, two bodies are projected horizontally with velocities m s–1 and 15 m s–1 They reach the ground in time t1 and t2 respectively Then (b) 3t1 = t2 (a) t1 = t2 (c) t1= 3t2 (d) none of these Two balls A and B are thrown with speeds u and u/2 respectively Both the balls cover the same horizontal distance before returning to the plane of projection If the angle of projection of ball B is 15° with the horizontal, then the angle of projection of A is (a) (1/2) sin–1 (1/8) (b) (1/4) sin–1 (1/8) (c) (1/3) sin–1 (1/8) (d) sin–1 (1/8) A projectile is fired at an angle 30° to the horizontal such that the vertical component of its initial velocity is 80 m s–1 Find approximate velocity of the projectile at time T/4 where T is time of flight (a) 155 m s–1 (b) 130 m s–1 (c) 145 m s–1 (d) 180 m s–1 A particle is projected with a velocity u so that the horizontal range is twice the greatest height attained, then the greatest height attained is (a) 2u/5g (c) 2u2/5g (b) 4u2/2g (d) u2/2g A bomb is dropped from a plane flying horizontally with velocity 720 km h–1 at an altitude of 980 m The bomb will hit the ground after (Take g = 9.8 m s–2) (a) 14.14 s (b) 1.414 s (c) 7.2 s (d) 141.4 s The trajectory equation of a particle is given as y = 4x – 2x2 where x and y are the horizontal and vertical displacements of the particle in m from origin(point of projection) Find the maximum distance of the projectile from x-axis (a) 1.5 m (b) m (c) m (d) m A car is moving horizontally along a straight line with a uniform velocity of 25 m s–1 A projectile is, to be fired from this car in such a way that it will return to it after it has moved 100 m The speed of the projectile with respect to car should be (Take g = 9.8 m s–2) (a) 19.6 m s–1 (b) 15.6 m s–1 –1 (c) 9.8 m s (d) 24.6 m s–1 The ceiling of a hall is 40 m high For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 m s–1 without hitting the ceiling of the hall is (Take g = 9.8 m s–2) (a) 45° (b) 60° (c) 30° (d) 25° Two paper screens A and B are separated by 150 m A bullet pierces A and then B The hole in B is 15 cm below the hole in A If the bullet is travelling horizontally at the time of hitting A, then the velocity of the bullet at A is (Take g = 10 m s–2) (a) 500 m s −1 (b) 200 m s −1 (c) 100 m s −1 (d) 300 m s −1 10 A flag is mounted on a car moving due north with velocity of 20 km h–1 Strong winds are blowing due west with velocity of 20 km h–1 The flag will point in direction (a) east (b) north-east (c) south-east (d) south-west 11 The graph shows position as a function of time for two trains running on parallel tracks Which statement is true? Contributed by : Shiv R Goel, Intelli Quest, 9878359179 10 Physics For you | FEBRUARY ‘16 (a) At time tB, both trains have the same velocity (b) Both trains have the same velocity at some time after tB (c) Both trains have the same velocity at some time before tB (d) Nowhere the trains have some velocity 12 A wedge is placed on a smooth horizontal plain and a rat runs on its sloping side The velocity of wedge is v = m s–1 towards right What should be the velocity of rat with respect to wedge (u), so that the rat appear to move in vertical direction to an observer standing on ground? (a) m s–1 (b) m s–1 (c) m s–1 (d) m s −1 13 A plank is moving on ground with a velocity v and a block is moving on the plank with respect to it with a velocity u as shown in figure What is the velocity of block with respect to ground? (a) v–u towards right (b) v–u towards left (c) u towards right (d) none of these 14 A man is crossing a river flowing with velocity of m s–1 He reaches a point directly across at a distance of 60 m in s His velocity in still water should be (a) 12 m s–1 (c) m s–1 12 (b) 13 m s–1 (d) 10 m s–1 Physics For you | FEBRUARY ‘16 15 Two particles P1 and P2 are moving with velocities v1 and v2 respectively Which of the statements about their relative velocity vr is true? (a) vr cannot be greater than v1 + v2 (b) vr cannot be greater than v1 – v2 (c) vr > (v1 + v2 ) (d) vr < (v1 – v2 ) 16 A boat having a speed of km h–1 in still water, crosses a river of width km along the shortest possible path in 15 minutes The speed of the river in km h–1 is (a) (b) (c) (d) 41 17 A man can swim in still water with a speed of m s–1 If he wants to cross a river of water current of speed m s–1 along shortest possible path, then in which direction should he swim? (a) At an angle 120° to the water current (b) At an angle 150° to the water current (c) At an angle 90° to the water current (d) None of these 18 A ship is travelling due east at 10 km h–1 What must be the speed of a second ship heading 30° east of north if it is always due north of the first ship? (b) 25 km h–1 (a) 30 km h–1 (c) 15 km h–1 (d) 20 km h–1 19 Width of a river is 100 m Water flows in the river with a velocity of 0.5 m s–1 A boat starts travelling in water from one bank If the direction of boat with respect to water makes an angle 60° with upstream, find the velocity of boat with which it should travel in water to reach the other bank by shortest route (a) 0.5 m s–1 (b) m s–1 –1 (c) m s (d) 1.5 m s–1 20 A bird flies to and fro between two cars which move with velocities v1 = 20 m s–1 and v2 = 30 m s–1 If the speed of the bird is v3 = 10 m s–1 and the initial distance of separation between them is d = km, find the total distance covered by the bird till the cars meet (a) 2000 m (c) 400 m (b) 1000 m (d) 200 m 21 A car is moving towards a wall with a fixed velocity of 20 m s–1 When its distance from the wall is 100 m, a bee starts to move towards the jeep with a constant velocity m s–1 The time taken by bee to reach the jeep is (a) s (b) s (c) s (d) none of these 22 A stone is allowed to fall from the top of a tower and covers half of the height of tower in the last second of the journey The time taken by the stone to reach the foot of the tower is (b) s (a) (2 − ) s (c) (2 + ) s (d) (2 ± ) s       23 If a, b, c are unit vectors such that a + b − c = 0,   then the angle between a and b is π π (b) (a) 2π π (c) (d) 24 The sum of magnitudes of two forces acting at a point is 16 N and magnitude of their resultant is N If the resultant is at 90° with the force of smaller magnitude, then their magnitudes(in N) are (a) 3, 13 (b) 2, 14 (c) 5, 11 (d) 4, 12 25 The resultant of two forces acting at an angle of 150° is 10 N and is perpendicular to one of the forces The other force is (a) 20/ N (b) 10 N (c) 20 N (d) 20 N PaPer-ii (one or More oPtions correct tyPe) 26 The velocity time graph of two bodies A and B is given here Choose correct statements (a) Acceleration of B > acceleration of A (b) Acceleration of A > acceleration of B 14 Physics For you | FEBRUARY ‘16 (c) Both are starting from same point (d) A covers greater distance than B in the same time 27 A man wishes to throw two darts one by one at the target at T so that they arrive at T at the same time as shown in figure Mark the correct statements about the two projections (a) Projectile that travels along trajectory A was projected earlier (b) Projectile that travels along trajectory B was projected earlier (c) Both were projected at same time (d) Darts must be projected such that qA + qB = 90° 28 In the figure shown, two boats start simultaneously with different speeds relative to water Water flow speed is same for both the boats Mark the correct statements qA and qB are angles from y-axis at which boats are heading at initial moment (a) If vA > vB then for reaching the other bank simultaneously qA > qB (b) In option (a), drift of boat A greater than boat B (c) If vB > vA and qA > qB, boat B reaches other bank earlier than boat A (d) If vB = vA and qA > qB, drift of A is greater 29 A particle has an initial velocity of 4i + j m s −1 and an acceleration of −0.4i m s–2, at what time will its speed be m s–1? (a) 2.5 s (b) 17.5 s (c) s (d) 8.5 s 30 A boat is traveling due east at 12 m s–1 A flag on the boat flaps at 53° north of west Another flag on the shore flaps due north (a) Speed of wind with respect to ground is 16 m s–1 (b) Speed of wind with respect to ground is 20 m s–1 (c) Speed of wind with respect to boat is 20 m s–1 (d) Speed of wind with respect to boat is 16 m s–1 laws oF Motion; work, energy and Power PaPer-i (single oPtion correct tyPe) 31 A child is sliding down a slide in a playground with a constant speed (a) Statement-1 : His kinetic energy is constant Statement-2 : His mechanical energy is constant (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1 (b) Statement-1 is true, statement-2 is true and statement-2 is not the correct explanation for statement (c) Statement-1 is true, statement-2 is false (d) Statement-1 is false, statement-2 is true 32 Statement - : Force F1 required to just lift block A of mass m in case (i) is more than that in case (ii) (b) (c) Statement -2 : Less work has to be done in case (ii) to lift the block from rest to rest by a distance h (d) F2 F1 m A (i) h B m (ii) (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1 (b) Statement-1 is true, statement-2 is true and statement-2 is not the correct explanation for statement (c) Statement-1 is true, statement-2 is false (d) Statement-1 is false, statement-2 is true 33 A particle initially at rest is displaced from x = –10 m to x = +10 m under the influence of force F as shown in the figure Now the kinetic energy vs position graph of the particle is 34 The potential energy of an object is given by U(x) = 3x2 – 2x3, where U is in joules and x is in meters (a) x = is stable and x = is unstable (b) x = is unstable and x = is stable (c) x = is stable and x = is stable (d) x = is unstable and x = is unstable 35 A boy blowing a whistle sends out air at one gram per second with a speed of 200 m s–1 Find his lung power (a) 20 W (b) 0.2 W (c) W (d) 200 W 36 A stone tied to a string of length m is whirled in a vertical circle with the other end of the string at the centre At a certain instant of time, the stone is at its lowest position and has a speed 10 m s–1 The magnitude of the change in its velocity as it reaches a position where the string is horizontal, is (a) 60 m s–1 (b) 40 m s–1 (c) 80 m s–1 (d) 160 m s–1 37 A ball of mass m is on a thread The thread is held taut and horizontal, and the ball is released as shown At what angle between the thread and vertical will the tension in thread be equal to weight in magnitude? Physics For you | FEBRUARY ‘16 15 37 (b) : Here, x = Asinwt At t = t1, x = A A \ = A sin wt1 or sin wt1 = 2 p p or wt1 = or t1 = (i) 6w At t = t1 + t2, x = A \ A = Asinw(t1 + t2) or sinw(t1 + t2) = p p or w(t1 + t2 ) = or t1 + t2 = 2w p p p 2p −t = − or t2 = = (using (i)) 2w 2w 6w 6w t Thus, = t2 38 (c) : The radiation pressure on a perfectly reflecting surface is 2I P= c Here, I = W m–2, c = × 108 m s–1 2(5 W m −2 ) \ P= = 3.33 × 10–8 N m–2 −1 × 10 m s 39 (a) : The Boolean expression for output Y is Y = A⋅ B (by de Morgan’s theorem) = A+B =A+B which is Boolean expression for OR gate Thus the combination represents OR gate 40 (a) : For first ball, u = 0, a = g = 10 m s–2, t = 18 s As s = ut + at 2 \ s1 = + (10 m s −2 )(18 s)2 = 1620 m For second ball, u = v, a = g = 10 m s–2, t = 18 s – s = 12 s \ s2 = ut + at = v(12 s) + (10 m s −2 )(12 s)2 = v(12 s) + 720 m At the time of meeting, s2 = s1 \ v(12 s) + 720 m = 1620 m 1620 m − 720 m = 75 m s −1 or v = 12 s 72 Physics for you | february ‘16 41 (c) : Let the stone be thrown with velocity u at an angle q to the horizontal Then u2 sin2 q Maximum height h = 2g or 2gh = u sin q (i) The time of flight of the stone is 2u sin q 2h (using (i)) T= = gh = g g g 42 (d) : Here, RC = 500 W, ICRC = 0.5 V, VCC = V, a = 0.96 As ICRC = 0.5 V V V \ IC = = = × 10−3 A = mA RC 500 W The current gains a and b are related as a 0.96 b= = = 24 − a − 0.96 The base current is mA I IB = C = = mA b 24 24 43 (a) : Here, Distance between the slits, d = 0.15 mm = 0.15 × 10–3 m Distance of screen from the slits, D = 50 cm = 50 × 10–2 m Since there are three and a half fringes contained within mm (i.e from first maxima to fifth minima), \ b = mm (where b is the fringe width) 14 mm or b = = mm lD As b = d −3 −3 bd (2 × 10 m)(0.15 × 10 m) \ l= = D (50 × 10−2 m) = × 10–7 m = 600 × 10–9 m = 600 nm ( nm = 10−9 m) 44 (a) 45 (a) : Here, m = kg, r = m, v = m s–1, T = N Let the string make an angle q with the vertical Then mv r (1 kg )(4 m s −1 )2 mv mg cos q = T − =6N− r 1m = –10 N T − mg cos q = or or cos q = −10 N mg = −10 N (1 kg )(10 m s −2 ) = −1 or q = 180° It means the stone is at the top of the vertical circle nn 125 V = 10 W 12.5 A V 125 X L = wL = puL = = = 12.5 I 10 12.5 12.5 or pL = = = 0.25 u 50 \ X′L = 2pL × u′ = 0.25 × 40 = 10 W Impedance of the circuit (a) : R = Solution Set-30 (b) : As batteries are connected in series, so e +e I= r1 + r2 + R When resistor R is connected to battery of emf e1, e I1 = r1 + R When resistor R is connected to battery of emf e2, E I2 = r2 + R For required condition to be fulfilled I < I1 and I < I2 e1 + e2 e \ < r1 + r2 + R r1 + R e1 + e2 e and < r1 + r2 + R r2 + R r r +R e2 e ⇒ < and > r1 e1 r1 + R e1 Thus option (b) is correct mv0 v x (c): r = = , \ = = sin q B0q B0a r ⇒ q = 60° p T 2p = t0 A = = × 6 B0a 3B0a y Z = R2 + X L′2 = 10 W 100 = 10 / A \ Current = 10 (c): Taking torque about C, F × OCcos q = N × OCsin q or F = ma = mgtan q ( N = mg) or a = gtan q C = centre of mass O X I z dz x x Therefore, x-coordinate of particle at any time p t> will be 3B0a p  v0  cos 60 x= + v0  t − B0a  3B0a  = F ( ) Hence, V = k n − Q × r = (n − 1) r E r kQ The magnetic field at point P is  A O v0 v0  p  + t − B0a  3B0a  mg (b) : At the centre, the intensity is effectively due to one charge and the potential is due to (n – 1) charges k (n − 1) Q kQ \ E= and V = r r2 v0  C N  Z Y B y z  P dz Z  I Y Physics for you | February ‘16 73 B= µ0 2p I \ y +z The magnetic flux through the shaded strip in figure is µ I sin q dφ = (W dz ) 2p y + z z where sinq = y + z2 \ Total magnetic flux through rectangular loop is µ0 I0 sin wtWzdz 2p y2 + z2  y + L2  µ = W ln   I0 sin wt 4p  y2  \ Induced emf in the loop is e=  L2 + y  dφ µ0 I0W w cos wt ln  =  dt p  y2  The net torque acting on the sphere is   m I y  B x     τ = µ × B = NIAj × ( Bi) = − NIABk   ( A = pR2) or τ = −N pR2 IBk The sphere is free to rotate, it must rotate about the centroidal axis   τ − N pR2 IB   2 \ a= = k   Ic = mR  Ic mR2  5N pIB  a= k 2m At t = 0, the stone was going up with a velocity of 5.0 m s–1 After that it moved as a freely falling particle with downward acceleration g Take vertically upward as the positive X-axis If it reaches the ground at time t, x = – 50 m, u = m s–1, a = – 10 m s–2 As x = ut + at 2 ( 74 ) Physics for you | February ‘16 × (–10)t2 ± 41 s or, t = –2.7 s or 3.7 s Negative t has no significance in this problem The stone reaches the ground at t = 3.7 s During this time, the balloon has moved uniformly up The distance covered by it is m s–1 × 3.7 s = 18.5 m Hence, the height of the balloon when the stone reaches the ground is 50 m + 18.5 m = 68.5 m or, L φ=∫ – 50 = 5t + t= At poles, the apparent weight is same as the true weight Thus, 98 N = mg = m(9.8 m s–2) or m = 10 kg At the equator, the apparent weight is mg′ = mg – mw2R The radius of the earth is 6400 km and the angular speed is p rad w= = 7.27 × 10−5 rad s −1 24 × 60 × 60 s Thus, mg′ = 98 N – (10 kg) (7.27 × 10–5 rad s–1)2 (6400 × 103 m) = 97.66 N 10 Suppose the deceleration of the block is a The linear deceleration of the rim of the wheel is also a The angular deceleration of the wheel is a = a/r If the tension in the string is T, the equations of motion are as follows: Mg sinq – T = Ma and Tr = Ia = Ia/r Eliminating T from these equations, Mg r sinq a Mg sinq – I = Ma or a = I + Mr r The initial velocity of the block up the incline is v = wr Thus, the distance moved by the block before stopping is x= v w2r ( I + Mr ) ( I + Mr ) w2 = = 2a Mg sin q Mgr sin q Solution Sender of Physics Musing Set-30 Harsh Sharma (uP) Samrat Gupta (Wb) Pradnesh amod Patil (Maharashtra) Set-29 Subrata Dutta (Wb) anoop Verma (Delhi) nn chapterwise McQs for practice Useful for All National and State Level Medical/Engg Entrance Exams ElEctric chargEs and fiElds Which of the following figures cannot possibly represent electrostatic field lines? + + (a) + (b) + + (c) (d) + A metallic spherical shell has an inner radius R1 and outer radius R2 A charge is placed at the centre of the spherical cavity The surface charge density on the inner surface is (a) (c) q π R12 q2 (b) (d) −q π R12 q π R22  An electric dipole of moment p is placed normal  to the lines of force of electric intensity E , then the work done in deflecting it through an angle of 180° is (a) pE (b) +2pE (c) –pE (d) zero 4π R22 A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of × 104 N C–1 Then, the magnitude of charge is (a) × 104 C (b) × 10–10 C (c) × 10–6 C (d) × 10–9 C Two equal and opposite charges × 10–8 C are placed × 10–2 cm away, form a dipole If this dipole is placed in an external electric field of × 108 N C–1, the value of maximum torque and the work done in rotating it through 180° will be (a) 64 × 10–4 N m and 64 × 10–4 J (b) 32 × 10–4 N m and 32 × 10–4 J (c) 64 × 10–4 N m and 32 × 10–4 J (d) 32 × 10–4 N m and 64 × 10–4 J A charged ball B hangs from a silk thread S, which makes an P angle q with a large charged conducting sheet P as shown in figure The surface charge density of the sheet is proportional to (a) cos q (b) cot q (c) sin q (d) tan q  S B A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it The net  electric field E at the centre O is j (a) q 2π2 e (c) − r2 j q π2 e O r2 (b) j i q π2 e (d) − 0r j q 2π2 e 0r j Two positive ions, each carrying a charge q are separated by a distance d If F is the force of repulsion between the ions, the number of electrons Physics for you | February ‘16 75 missing from each ion will be (e being the charge of an electron) (a) (c) πe0 Fd e2 (b) πe0 Fd e2 (d) πe0 Fd q2 πe0 Fd q2 A point charge +10 µC is at a distance cm directly above the centre of a square of side 10 cm as shown in figure What is the magnitude of the electric flux through the square? 13 Two spherical conductors A and B of radii mm and mm are separated by a distance of cm and are uniformly charged If the spheres are connected by a conducting wire then, in equilibrium position, the ratio of the magnitude of electric fields at the surface of the spheres A and B is (a) : (b) : (c) : (d) : 14 The tracks of three charged particles in a uniform electrostatic field is shown in the figure Which particle has the highest charge to mass ratio? cm +q 10 10 cm (a) (b) (c) (d) cm 1.80 × 105 N m2 C–1 × 105 N m2 C–1 1.92 × 105 N m2 C–1 1.88 × 105 N m2 C–1 10 In a hydrogen atom, the distance between the electron and proton is 2.5 × 10–11 m The electrical force of attraction between them will be (a) 2.8 × 10–7 N (b) 6.2 × 10–7 N –7 (c) 3.7 × 10 N (d) 9.1 × 10–7 N 11 Two small charged spheres A and B have charges 10 µC and 40 µC respectively, and are held at a separation of 90 cm from each other At what distance from A, electric intensity would be zero? (a) 22.5 cm (b) 18 cm (c) 36 cm (d) 30 cm 12 Four charges are arranged at the corners of a square as shown in the figure The direction of electric field at the centre of the square is along +3q D +4q A +2q (a) DC (c) AB 76 B C +q (b) BC (d) AD Physics for you | February ‘16 (a) (c) A C (b) B (d) A and B 15 If there were only one type of charge in the universe, then   E (a)  ∫ S ⋅ dS ≠ on any surface   (b)  E ∫ S ⋅ dS = if the charge is outside the surface   q (c)  E ∫ S ⋅ dS ≠ e0 if charge of magnitude q were inside the surface (d) both (b) and (c) are correct ElEctrostatic PotEntial and caPacitancE 16 The equivalent capacitance for the network shown in the figure is 1200 (a) pF 1000 pF (b) 1800 pF (c) 1300 (d) pF 17 A cube of side x has a charge q at each of its vertices The potential due to this charge array at the centre of the cube is 4q 4q (a) (b) 3πeo x 3πeo x (c) 3q πeo x (d) 2q 3πeo x 18 A charge +q is placed at the origin O of x-y axes as shown in the figure The work done in taking a charge Q from A to B along the straight line AB is qQ  a − b  qQ (a) (b)   πe0  ab  πe (c) qQ  b  − πe0  a2 b  (d) y B(0, b) O A(a, 0) x b −a    ab  qQ  a  − πe0  b2 b  19 Two charges q1 and q2 are placed 30 cm apart, as shown in the figure A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D 22 A charged oil drop of mass 2.5 × 10–7 kg is in space between the two plates, each of area × 10–2 m2 of a parallel plate capacitor When the upper plate has a charge of × 10–7 C and the lower plate has an equal negative charge, the oil remains stationary The charge of the oil drop is (Take g = 10 m s–2) (a) (c) × 10–1 C 8.85 × 10–13 C (b) × 10–6 C (d) 1.8 × 10–14 C 23 Three charges, each + q, are placed at the corners of an isosceles triangle ABC of sides BC and AC = 2a D and E are the mid points of BC and CA, as shown in figure The work done in taking a charge Q from D to E is A E B (a) q The change in the potential energy is k , where 4πe0 k is (a) 8q1 (b) 6q1 (c) 8q2 (d) 6q2 20 Two materials of dielectric constants k1 and k2 are filled between two parallel plates of a capacitor as shown in figure Area = A/2 k1 k2 d The capacitance of the capacitor is (a) Ae0 (k1 + k2 ) 2d (c) Ae0 d  k1k2   k +k   2 (b) Ae0  k1 + k2    2d  k1k2  (d) Ae0 d  k1k2   k +k   2 21 Work done in placing a charge of × 10–18C on a condenser of capacity 100 àF is (a) 16 ì 1032 J (b) 31 × 10–26 J –10 (c) × 10 J (d) 32 × 10–32 J qQ 8πe0 a (c) zero C (b) qQ πe0 a (d) 3qQ πe0a 24 Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities + s, – s, + s respectively If VA, VB and VC denote the potentials of the three shells, then for c = a + b, we have (a) (c) Area = A/2 D VC = VB = VA VC = VB ≠ VA (b) VC = VA ≠ VB (d) VC ≠ VB ≠ VA 25 If a charged spherical conductor of radius 10 cm has potential V at a point distant cm from its centre, then the potential at a point distant 15 cm from the centre will be (a) V (b) V 2 (c) V (d) V 3 26 A parallel plate air capacitor is charged to a potential difference of V After disconnecting the battery, distance between the plates of the capacitor is increased using an insulating handle As a result the potential difference between the plates (a) decreases (b) increases (c) becomes zero (d) does not change Physics for you | February ‘16 77 27 The velocity v acquired by an electron starting from rest and moving through potential difference V is shown by which of the following graphs ? v v (a) (b) V V v v (c) (d) V V 28 In the circuit shown in figure, C = µF The charge stored in the capacitor of capacity C is C (a) (c) zero 40 µC 2C 10 V (b) 90 µC (d) 60 µC 29 A ball of mass g and charge 10–8 C moves from a point A, where potential is 600 V to the point B where potential is zero Velocity of the ball at the point B is 20 cm s–1 The velocity of the ball at the point A will be (a) 22.8 cm s–1 (b) 228 cm s–1 –1 (c) 16.8 m s (d) 168 m s–1 30 The potential at a point distant x (measured in µm) due to some charges situated on the x-axis is given 20 V by V (x ) = x −4 The electric field at x = µm is given by (a) V µm–1 and in positive x direction 10 (b) V µm–1 and in negative x direction 10 (c) V µm–1 and in positive x direction (d) V µm–1 and in negative x direction solutions (b) : Positive charge tends to move normally away from the conductor i.e starts or ends only at 90° to the surface of the conductor Thus, the lines of force are parallel and normal to the surface of conductor So curve (b) cannot represent electrostatic field lines 78 (b) : When a charge +q is placed at the centre of spherical cavity as shown in figure, Physics for you | February ‘16 Charge induced on the inner surface of shell =–q Charge induced on the outer surface of shell =+q \ Surface charge density on the inner surface −q = πR12 (d) : Here, q1 = 90° q2 = 90° + 180° = 270° \ Work done = q2 =270° ∫ pE sin q dq q1 =90° 270° = [− pE cos q]90 ° =0 (b) : We know that, qE = mg mg \ q= E × 10−6 × 10 30 × 10−10 = = × 104 = × 10–10C (d) : Maximum torque is given by τ max = pE ( sin q = 1) ( p = q × 2a) = (q2a) E = (4 × 10–8 × × 10–4) × × 108 = 32 × 10–4 N m If q = 180°, then W = pE(1 – cos 180°) = pE[1 –(–1)] = 2pE = × 32 × 10–4 = 64 × 10–4 J (d) :  T T cos  S  P T sin  q B F = qE mg According to diagram, T sin q = qE = q(s/e0) = qs/e0 and T cos q = mg qs Hence, tan q = e0mg or Symmetry of six faces of a cube about its centre ensures that the flux fS through each square face is same when the charge q is placed at the centre \ Total flux, q f E = × fS = e0 q = × 10 × 10−6 × π × × 109 or fS = 6e = 1.88 × 105 N m2 C–1 10 (c) : F = = × 109 ×  e mg  s=  tan q or s ∝ tan q  q  11 (d) : or or π qdq   k  qdq    sin q  as dq = π  ∫ r 0 π  kq q sin q dq = 2 π e0 r πr ∫0  E=− q 2π2 e r2  π   as ∫ sin q d q =       j (as E is directed along –Y-axis) (c) : According to Coulomb’s law, qq F= 22 4πe0 d As, q1 = q2 = ne (ne)(ne) \ F= πe0 d πe0 Fd e2 (d) : We can imagine the square as face of a cube with edge 10 cm with the charge of +10 µC placed at its centre, as shown in figure (ii) n= 40 C P B 0.90 – x or or 10 × 10−6 40 × 10−6 ⋅ = ⋅ πe0 πe0 (0.90 − x )2 x = x (0.90 − x )2 0.90 – x = 2x or x = 0.30 m = 30 cm 12 (a) : The resultant fields due to the diagonally opposite charges will act as shown in the figure Here, the resultant electric field at the centre of the square is along DC +3q (ii) D A +4q B +2q C +q 13 (d) : When joined by a wire, the two spheres attain common potential V qA V = \ Electric field, E A = πe0 RA2 RA Similarly, EB = (i) EA At point P, EA = EB π = EB 10 C x dq ⇒ E = ∫ k sin q r = 1.6 × 10 −19 × 1.6 × 10 −19 = 3.7 × 10 −7 N (2.5 × 10 −11 )2 A π (d) : E = dE sin q ∫ π q1q2 ⋅ πe0 r V RB \ E A RB = = E B RA 14 (c) : Particles A and B have negative charges because they are being deflected towards the positive plate of the electrostatic field Particle C has positive charge because it is being deflected towards the negative plate Physics for you | February ‘16 79 \ Deflection of charged particle in time t in y-direction 1 qE ( ma = qE ) h = × t + at = t 2 m i.e h ∝ q/m As the particle C suffers maximum deflection in y-direction, so it has highest charge to mass q/m ratio 15 (d) : According to Gauss’s theorem in electrostatics   q ∫ E ⋅ dS = e0 Here q is charge enclosed by the surface If the charge is outside the surface, then qinside =   Also,  ∫ E ⋅ dS = So, both (b) and (c) are correct 16 (a) : 4q 8q = πeo x 3πeo x 18 (a) : Potential at point A is q VA = πe0 a Potential at point B is q VB = πe0 b Work done in taking a charge Q from A to B is \ V= W = Q(VB − VA ) = 19 (c) : q3 C q1 A C1 = C4 = 100 pF, C2 = C3 = 400 pF Supply voltage, V = 400 V Capacitors C2 and C3 are connected in series, Equivalent capacitance 1 = + = or C ′ = 200 pF C ′ 400 400 400 Capacitors C1 and C′ are in parallel Their equivalent capacitance C ′′ = C ′ + C1 = 200 + 100 = 300 pF Capacitors C ′′ and C4 are connected in series 1 1 Equivalent capacitance, = + = + Ceq C ′′ C4 300 400 = 1200 pF Ceq 1200 \ Ceq = 17 (b) : The length of diagonal of the cube of each side x is 3x = x \ Distance between centre of cube and each vertex, x r= q Now, potential, V = 4πeo r Since cube has vertices and charges each of value q 80 Physics for you | February ‘16 Qq  1  Qq  a − b  − =  πe0  b a  πe0  ab  q2 B D The potential energy when q3 is at point C  q2 q3  q1q3 + U1 =   πe0  0.40 (0.40)2 + (0.30)2  The potential energy when q3 is at point D  q1q3 q2 q3  + U2 = πe0  0.40 0.10  Thus change in potential energy is DU = U2 – U1 q3  q1q3 q2 q3 q1q3 q2 q3  k = ⇒ + − − 4πe0 πe0  0.40 0.10 0.40 0.50  5q − q 4q ⇒ k = 2 = = 8q2 0.50 0.50 20 (a) : The arrangement is equivalent to a parallel combination of two capacitors, each with plate area A/2 and separation d Total capacitance is e0 A k1 e0 A k2 2 C = C1 + C2 = + d d Ae (k + k ) = 2d 21 (d) : Here, q = × 10–18C, C = 100 µF = 10–4 F ( ) ( ) q × 10 −18 = = × 10 −14 V 10 −4 C Work done = qV = × × 10−18 × × 10−14 = 32 × 10–32 J V= 22 (c) : We know that qE = mg e0 Amg qQ = mg , q = Q e0 A − 12 − 8.85 × 10 × × 10 × 2.5 × 10−7 × 10 = × 10−7 = 8.85 × 10–13 C 23 (c) : According to figure, AC = BC C q E D A q q B Potential at D = potential at E q q i.e VD = VE = × + πe0 a πe0 DB 2q q = + × πe0 a πe0 2a sin 60° q   q 2 +  2+ =   πe0a  × /  πe0 a   \ Work done in taking charge Q from D to E = QVE – QVD = Q (VE – VD) = Q × = 24 (b) : It is clear from figure that = VC VB VA a b c sa sb sc s − + = (a − b + c) e0 e0 e0 e0 As c = a + b s \ VA = (2a) e0 s  s  VB =  a − b + c  =  a − b + a + b  e0  b e b    VA = VB = s  a2  sa (a + b) s0ac = + a = e0  b e 0b  e0 b VC = s  a b2  sa  a2 − b2 + c  c − +  = e   e0  c c c = s  (a + b)(a − b) + (a + b)2   e0  c s  (a + b)  s2a (a − b + a + b) = e0  c  e0 We found that, VA = VC ≠ VB Choice (b) is correct 25 (c) : The potential at any point inside a charged conductor is same as on its surface, q q V= = ( r = 10 cm) πeo r πe0 × 10 q ⇒ … (i) = 10V πe0 At r = 15 cm 10 V q V′ = \ V′ = = V (using (i)) πeo × 15 15 VC = 26 (b) : Potential on parallel plate capacitor Q V= C Also, capacity of parallel plate capacitor is given by e A Qd C= \ V= d e0 A ⇒ V∝d So, on increasing the distance between plates of capacitor, the potential difference between plates also increases 27 (b) : K.E gained by the electron, mv = eV \ v2 ∝ V Thus, the graph between v and V must be a parabola Only option (b) is correct 28 (c) : Two capacitors as shown in figure are in series Therefore, charge on each capacitor is same 1 1 2C = + = + = , Cs = Cs C1 C2 C 2C 2C 2C 2×6 q = Cs ì V = V= ì 10 = 40àC 3 29 (a) : By using, m(v12 − v22 ) = QV × 10−3[v12 − (0.2)2 ] = 10−8 (600 − 0) v1 = 22.8 cm s–1 20 = 20(x − 4)−1 x2 − 40 x dV = −20(−1)(x − 4)−2 (2 x ) = E=− (x − 4)2 dx At x = àm, E = 40 ì = 160 = 10 Vµm −1 (42 − 4)2 144  E is along the positive x-direction 30 (c) : Here, V (x ) = nn Physics for you | February ‘16 81 Y U ASK WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough The best questions and their solutions will be printed in this column each month Q1 (a) Why does every celestial object rotate on its axis? (b) Why almost every object in this space has a spherical shape? –Ruqaiyah Rahman, Madhya Pradesh Ans (a) In general, almost every celestial object has a spin motion around its axis The major reason behind this, is conservation of their angular momentum As gravity is the central force in the universe because it is only one which has a significant pull over large distances When things collapse under their own gravity in space i.e., clouds of gas and dust, any small amount of asymmetry in the collapse will be enough to start it spinning Even if it spins by a tiny amount, as it collapses, angular momentum conservation will make it spin more and more quickly-just like an spinning ice-skater pulling his arms close to his body and hence spinning more quickly This means that all coherent masses are spinning, e.g asteroids, neutron stars, galaxies, quasars (b) As we know that stars, planets, moon and other celestial bodies are made up of gases, rocks, ice and water We assume a planet like earth having larger part covered by water The water molecules on the north pole are pulling towards south pole The ones on the left are pulling towards right Gravity and surface tension pull it in and molecular forces are pushing it outward Due to inward pulling towards centre of mass, it gets roughly a spherical shape Q2 What is coherent source? –Archisman Das, West Bengal Ans Two sources are said to be coherent if they produce waves of same frequency with a constant phase difference Unlike sound waves, two independent sources of light cannot be coherent Since sound is a bulk property of matter, therefore two independent 82 Physics for you | february ‘16 sources of sound can be identical in all respects and can produce coherent waves On the contrary, light is not a bulk property of matter, it is a property of each individual atom As the individual atoms emit light randomly and independently, therefore two independent sources of light cannot be coherent Coherent sources can be obtained by splitting a light beam from source into two This can be done in two ways : (a) Division of wavefront, (b) Division of amplitude Q3 What is the reason behind Earth’s magnetism? –Bidhan Banerjee, West Bengal Ans The actual process by which the magnetic field is produced in this environment is extremely complex However, for magnetic field generation to occur several conditions must be met - there must be a conducting fluid - there must be enough energy to cause the fluid to move with sufficient speed and with appropriate flow pattern - there must be a seed magnetic field All these conditions meet in the outer core Molten iron is a good conductor There is sufficient energy to drive convection and the convection motion, coupled with the Earth’s rotation, produce the appropriate flow pattern Even before the Earth’s magnetic field was first formed, magnetic fields were present in the form of the Sun’s magnetic field Once the process is going, the existing field acts as the seed field As a stream of molten iron passes through the existing magnetic field, an electric current is generated through a process called magnetic induction The newly created electric field will in turn create a magnetic field The generated magnetic field can reinforce the initial magnetic field As long as there is sufficient fluid motion in the outer core the process will continue Q4 Why rain drops or water droplets are spherical in shape? –Nayan Sengupta, West Bengal Ans The reason why a water droplet takes spherical shape is the surface tension of the water that tends to minimize the surface area of the drop as this minimizes the potential energy For larger amount of water surface tension is however, too weak to overcome the force of gravity, that tends to distribute water as flat as possible on the surface as this minimizes the potential energy Same thing happens in case of rain drops also Although, as the raindrops fall down, they start losing the rounded shape due to their speed nn Courtesy - The Times of India Physics for you | FEBRUARY ‘16 83 Readers can send their responses at editor@mtg.in or post us with complete address by 25th of every month to win exciting prizes Winners' name with their valuable feedback will be published in next issue across A shallow crater with a complex, scalloped edge (6) Droplets formed by condensation of water vapour on surfaces (3) A prefix denoting 10–15 (5) A small, wide-field telescope attached to a larger telescope (6) A term used to describe a large, circular plain (4) 10 To increase the abundance of a particular isotope in a mixture of the isotopes of an element (6) 12 A measure of radiation received by a material (3) 13 A radar like technique employing pulsed or continuous wave laser beams for remote sensing (5) 17 19 20 22 24 Process of forming ions from molecules (10) The shortened version of linear accelerator (5) A unit of heat and internal energy (7) Subatomic particles composed of three quarks (6) The idea of creating a place or object that is free from the force of gravity (4, 7) 25 A gauge boson with no electric charge (1, 5) 26 A hypothetical scalar field that could drive the period of inflation that took place in the early universe (8) 27 A metric system unit of volume, usually used for liquids (5) down The shortest path between two points in curved space (8) Icy region of planet, specifically the north and south poles (5, 3) An extraneous low pitched droning noise heard in sound reproduction systems (3) Long zirconium alloy tubes containing fissionable material used in a nuclear reactor (4, 3) Cut Here 11 10 12 13 14 15 16 17 18 19 20 21 22 23 25 24 A 26 27 The outer edge or border of a planet or other celestial body (4) A giant star whose surface temperature is relatively low, so that it glows with a red colour (3, 5) An electric line that conveys electric power from a generating station to a point of a distributing network (6) 11 Term refers to any one of four mesons with nonzero strangeness (4) 14 Two or more sounds that when together sound unpleasant (10) 15 The superpartner of graviton (9) 16 Area of the sun’s surface that are cooler than surrounding areas (7) 18 A film that transmits only polarized light (8) 21 An active galaxy with very active and highly variable radio, electromagnetic and optical emissions (6) 23 Label that distinguish one type of measurable quantity from other types (4)  Physics for you | February ‘16 85 86 Physics for you | February ‘16 ... 20 16 crossword 1 v  Mv = M   + Mv ? ?2 2 ? ?2? ?? v2 − v2 = v ? ?2 ⇒ v ′ = v 48 (d) : mv = mv′ cos q + mv′cos q v v v′ = ⇒ v′ > cos q 49 (b) : m1v – m2v = (m1 + m2)v /2 2m1 – 2m2 = m1 + m2 m1 = 3m2... are connected with an 18 Physics For you | FEBRUARY ‘16 \ (c) : vT /4 = v 2xT /4 + v 2yT /4 ≈ 145 m s −1 2u2 sin q cos q u2 sin2 q =2? ? g 2g tan q = ⇒ sin q = / \H= 2u2 5g (a) : 980 = 9.8t ⇒ t... (20 ) (20 ) 20 F (x) 20 V (0) (0) ∑ qx = −50 mC (initial charge on isolated part marked) ⇒ 5(x – 20 ) + 20 (x – 0) = –50 ⇒ x=2V \ Charge on mF, q5 = 5 (20 – 2) = 90 mC Charge on 20 mF, q20 = 20 (2