JEE Adv. Critical Question Bank - Physics @OnDemandMaterial

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 2mv0 1  qE0  3 15. =2  t2 21. Clearly, PM = 2 cm qB0  m   1 37º > sin–1 n0  a(3 / 2) 4m 2v 0 q 2E0B0 = t2 31 > 5 3a 2m v0 n0  2 x = v0 × t  x = v0 q E0B0 16. Velocity of efflux at section (4) is v = 2gh 9a 3n0 + 2 > 5 Applying Bernoulli’s equation between section (3) and (4) 9a 2 1 v32 1 v42 >1 ; a> 9 2 2 2 P3 + = P4 + 1 3  1  22.  – 1 f  2   15  1 2gh )2 = P0 + 1 2gh )2     P3 + 2 (2 2 ( f = 30 cm.  P3 = P0 – 3gh. 11 1 11 1 1 v  36  30 v  30 – 36  180 17. Since toy is not accelerating so net external force on toy is zero. So (A) v = 180 cm F  (m  M)gsin   Pf Pi 180 cm 18. a =  (m  M)  t = 6 sec t = 0 36 cm 1cm   60 cm 60 cm 5 cm So, f = ma cos   Ans. (B) Ii F  (m  M)gsin   = (m  M)  mcos   19. a =  3  1 ×A= A 23. Let the focal length of each piece be f  2  2  1 11 Then f1  f  f W =  3 / 2  1 A = A 1 11  4 / 3  8 f2  f  f  air 4  f1 = f2 = For the third arrangerment the liquid forms a concave lens which has a diverging effect. So f3 > f1 = f2  water 1  cos glass 3  1 11 24. ' cos air putting the values we have’ = 2  20. 40 –  60 = f ....(1) 11 1 ....(2) 25. Electric field on surface of a uniformly charged sphere is given v –  60 = 2f Q R 1 1 1 1 1 by 40R3  30 + = .    v 60 2  40 60  Electric field at outside point is given by Q R3 E = 40r2  30r2 1 1 1 100  r0 3  v + 60 = 2 . 40  60  2  11 1 | Er | = r0 –  17r0  – v = 48 – 60 30  = 540 left  v = 240 cm 30  3r0 2  2     RESONANCE Page - 48

Click here to join our telegram channel, @OnDemandMaterial PHYSICS  V = –Ex x – Ey y 32. Point of intersection of two curves. y = x2 and y = x + 3 is, 26. E  Exˆi  Eyˆj , for A and B 4 16 – 4 = – Ex (– 2 – 2) – Ey (2 – 2) (–2, 1) Ex = 3 V/m for B and C dy x  dy  = 2 = –1 = i.e. 12 – 16 = –Ex {2– (– 2)} – Ey (4 – 2) dx 2  dx  2 Ey = –4 V/m.  x2   E  (3ˆi – 4ˆj)V / m 1 (stress)2 27. U = 2Y Tension tan = slope of incident ray RP = 1 or  = 45º Stress = area UA TA2 F2  dy  Umiddle = (Tmiddle )2 = (F / 2)2 = 4 and tan =  dx  = 1   = 45º  x 2  1 Q2 1 Q2  From geometry we can show that i = r = 0º i.e., the ray is    incident normally. 28. Ui =  2 80R 2 80R     Hence, the desired unit vector : A = 1 ˆi  ˆj 1 Q2 2 Uf = 2 80R 1 Q2     Heat = 16 0R . 33. sin = 4y 2 , = sin–1  4y02  0   29. Both open. 34. For earth and sun system RA = 1, RB = 3, RC = 6, RD = 1 T02 = 42 R3 ..............(i) Current in upper circuit when S1 is open is 1 = 1A GMs Null point is mid point of B or C when S2 and S1 both open When both closed, current in upper circuit For trinary star system 1 = 2A Current in lower circuit 2 = 1A V = 2 – 2 (1) = 1V null point is mid point of A.  Q 2  30.  2  = 1 Li02   2 2C Q_ Q_ Q_ Q_ Q_ Q_ 2 Gm2 3 = m2 a 3Gm 2 2 –2 2 –2 2 a2 2 3 2 = a3 T2 = 42 42 = a3 .............(ii) 2 3Gm  i L Given, T = 3T0 a = 2R Q 42 i0 = 2 LC 9T02 = 3Gm 8R3 P A A 9 42 R3 8 42 R3 31. 2 + 2A = 90° GMS 3 Gm 90° 90°  = A A 5A 2A  m= 8MS 2 = 90° A /2 27 R 2A Q A = 36°  Angles (36°, 72°, 72°) 8 Total mass = 3m = 9 MS . RESONANCE Page - 49

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 35. Let P = power radiated by the sun, R = radius of planet. Power received by planet = P  R2 . vnet = 7 4d2 3. Power radiated by planet = (4R2) T4. 40. aB = 2 m/s2 (  ) For thermal equilibrium, P  R2  4R2T4. 4d2 1 or T  d–1/2. or T4  d2 1 or T  d1/2  Ans. (C) 36. Wein’s displacement law is, ( – x3) + (x2 – x3) = k1 mT = b & x3 + 3x4 = k2 So, when m becomes half  T doubles ui = (T4 – T04) ui =  T 4   T 4  a2 – 2a3 = 0  a3 = a2 12   2  2   = = 6 m/s2     2   [a3  acceleration of P1 pulley]  T 4   2  uf =  (2T )4     a3 6     a4 = 3 = 3 = 2 m/s2 .    & – a3 + 3a4 = 0  1  1 15 [a4  acceleration of block B] 16 255 ui = = uf 1 41. M is the highest point of ball’s motion. 16  16 Horizontal component of velocity remains the same i.e. u cos . 255 uf = 15 × 30 uf = 510 Jm–2s–1 . 37.  Q = nCPT = 2  5  8.31 5 At point P, let  is the angle formed by Vnet (net velocity) with horizontal 2 v Q = 207.75J  tan = u cos  38. TA = TC where v = vertical velocity component at point P. UBC = – Q Taking vertically downward as positive and motion from M to P UAB = Q v = 0  vertically downward velocity at highest point. 3  n 2 RT = Q Q  (v)2 = (v)2 + 2g (H–h) (3rd equation of motion) WAB = nRT = 2 3  (v)2 = (0)2 + 2g(H–h) 39.  (v) = 2g(H – h) . v 2g(H – h)  tan = ucos  = ucos   2g(H – h)  1 cos 60º = v cos30º – 1 cos60º or  = tan–1  ucos  .   2 v= 3 42. 1st sep. will decrease then sep will increase RESONANCE For slope :- 1st there will be some Vapp, then Vapp will decrease slowly and become zero, after Vsep will increase gradually. Hence slope will decrease till it become zero afterward it will increase. Page - 50

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 43. At t = 3 48. Magnitude velocity of ball A at the end of 4 seconds  vA = uA + a1 (4)  vA = 4a1  vA = 0 + 4a1 d 1  magnitude of velocity of ball B at the end of 3 second  V  Slope =  = –1  vB = uB + a2 (3)  vB = 0 + 3a2 = 3a2 Thereafter, ball B maintains same velocity dt  velocity of B at the end of 4 second 1 dV  vB = 3a2.  – V 2 dt = –1  Velocity of A as seen by B  |vAB| = |–4a1 – 3a2| [Taking positive direction along right] : vAB = vrel = (4a1+ 3a2) m/sec. 2 49. We known that change in velocity = Area under a-t curve.  1  dV dV   dt = 1  a = dt = 3 m/s2  v t2 – v t1 = Area under the graph and t-axis from D to B   3  (which is quarter circle) 44. Initial velocity = 0 + 0 + 2 = 2 m/sec  v t2 – v t1 r 2 22 ( 7 )2 11 4 = = × = dv 4 7 2 for 3  t < 5, a = dt = 2 constant 11 11 17 for 5  t, a = 2t – 6  v t2 – 3 = 2 or v t2 = 2 + 3 = 2 m/sec.  a-t curve is a straight line not passing through origin 50. 45. For t = 2 sec. v = t3 + t2 + 2 x2  dx  dt = (t3 + t2 + 2) or dx  (t3  t2  2) dt 00 t4  t3 2 16 8  4 3 3  x=  4  2t =  4  – [0] v1 v2   x –x  0 =  8 v1 – v1x = xv2 = 8  3  = 32 m . v1   3 x = v1  v2 46. speed of boat in upstream = 15 km/hr 51. mv0R = I + mvR = mR2 v  + mvR  v = v0 Distance = 1.5 km  2  R  1.5  time t1 = 15 = 6 min Ki = 1 mv02 Dist. moved by float in this time 2 vk Kf = 1 mv2 + 1 I2 2 2 boat float = 1  v0 2 1 mR2 v02 = 1 mv 2 2 4R2 4 0 1.5 km m  2   + 2  11 So Wfriction =  1 mv 2 = 5 × 10 = 2 km 4 0 2 km  speed of boat w.r.t. float = 20 km/hr, t2 = 20 = 6 min. 52. WF = KE where WF =  = (F)    2  v0 v0 1  m 2  2 ×1+ 4 2  3  47. Distance = v0 × 1 + × 1 + ............  =   2 (F) 2  = v0 1 1  1  .............. 3F  2 4   = m = 2v0 PF = () () 3F 3F3 . PF = (F) m  PF = m RESONANCE Page - 51

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 53. When the system is about to slide, fricition will be equal to the 57. Situation after long time limiting value.Since acm  0 2t0   m  Mg t0   m  Mg 11 Work done = K = 2 (2m)(v)2 – 2 (2m) (3v)2 = –8mv2 2 58. Extension is 5cm and equilibrium is at the extension of 2 cm so d 2tR 4t Amplitude of SHM is 3cm dt  mR2  mR Also  w  k  10 5 rad / s 2 m  d   4tdt vmax = A = 30 5 ms–1 mR amax = 2A = 15 ms–2 4 . t02 2t02 59. PC – PR = 2BAK S = A sin (kx – t) mR 2 mR   S  S   x  = AK  C KE  1 I2 1 mR2 4t 4 2 22 0   m2R2 Px 1  m  Mg  4 m = t04 = 1  2  C R m    60. x + R = constant dx Rd dt  dt = 0   = d v = 54. dt R v R  N  x 2R sin   =A  2   = d/dt = a/R At = 60º, Na a  N  2 N  = R      A=2    sin  2  a 3 v2   a 2  vP = v and aP =        2 R  2    55. Area of cross-section of the tube is increased 61. Using KVL – I1 + 10 – 10 – 2I1 = 0 56. Heat released by 4.5 kg of water when its temperature falls from 40ºC to 0ºC is, 10V I1 Q1 = ms = (4.5) (1) (40 – 0) = 180 Kcal A4 when 5 kg ice at –40ºC comes to a temperature of 0ºC, it takes A3 I1+ 2 5A an energy 2A I1+ 5 A5 Q2 = ms = (5) (0.5) (40) = 100 Kcal The remaining heat 8V Q = Q1 – Q2 = 80 Kcal will melt mass m of the ice, where 5V 80 A1 m = 80 = 1 kg I1 A2 10V So, the temperature of the mixture will be 0ºC, I1 = 0 mass of water in it is 4.5 + 1 = 5.5 kg and mass of ice is 5 – 1 = 4 kg.  Ans. (B), (C), (D). RESONANCE Page - 52

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 62. If x(1 – ) < R then vA > vB if x is increased then 0 current may decrease, become zero or may increase. 66. B = 2a by the direction of force particle is positively charge. Now 63. (A) – 1 k(x2 – x 2 )  1 mv2 0 0 2 0 2 mg + 2a × q × v = 2mg 0 2mga k 2 x2 2a qv = mg  q = 0V m 0  v= (x – ) k 69. (A) m  P = F.v = kx (x 2 – x2 ) 0 (B) P = k k (x 2 x2 – x4 ) m 0 2 x2 – x4 (C) 0 (C) y= x dy x0 (D) Image is inverted  It should be real dx = 0  x = 2 x0 70. due to a convex mirror of focal length 2.5 cm 2 Pmax is at x = due to a concave mirror having its pole at (2, 0) real virtual pair 64. 71. The above lens mirror combination is an equivalent converging mirror. 72. PV = nRT ...............(1) Mass of piston is 9kg & area of piston is 0.09 m2 9g So P = P0 + 0.09 P = 1.06 × 105 N/m2 As mass of piston is not changed & Piston is frictionless So process is isobaric P = Pi = 1.06 × 105 N/m2  P = Pi = 1.06 × 105 N/m2  dQ = n Cp dT .................(2) or P2V2 – P1V1 = n R dT .................(3) (3)  (2) P(V2 – V1) = 2 dQ 5 65. 2 5 P (V2 – V1 ) = × 2.5 ×104 Q 1 et / RC  Q(t) =2 W = PV = 104 J & by equation (1) 1.06 105  0.0027 n= R  300 1.06  270 n = 2500 = 0.11448 or by dQ = n Cp dT i(t) = Q et / RC (from B to A) 2.5 104 2RC  T2 – T1 = 5R 2 (Q / 2)2 0.11448   heat = 2C . on solving T2 = 10782.18 K RESONANCE Page - 53

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 73. S1 : Isobaric process must pass through origin. Similarly , MA > T1 S2 : Isochoric process must pass through origin. MB T2 S3 : Straight line on PV does not represent isothermal process. KAd d If gas A is O2 and gas B is N2, then MA > MB. 74. H = dr = K (2rL) dr (Vmp)T1 < (Vmp)T2  T1 < T2  T2 > MB MA MB T1 MA r2 dr  = 2 KL(2 – 1) = 80 H r r1 dm 10ms–1 dt  L = 80 A 80. dm 8 80  g  dt = = = Kg/second. L 80  4200 4200 75. f = mg sin 20 and PE sin = mg sin. R 20 q 2R E sin = mg sin R 15  mg  Ans. (A) & (C) 5 E = 2q g Distance = 5 13 × 2 = 10 13 76. 81. v = ucos ˆi + usin ˆj – gt ˆj a = –g ˆj a  v = u cos  g kˆ = constant 4R K  rdr  R 2 (A) v = r 2  (3R)2 = 4 0 0 4r K  rdr  R At highest point 2 (B) v = = 2 0 . a = –g ˆj 0 r v = ucos ˆi 77. In space (gravity not zero) net force on + q is vertically upward.  a.v = 0 So that equilibrium can be achieved in the space. If charge is –q then net force on it is vertically downward so 82. x = 24 = u cos.t equilibrium not achieved. 24 1 78. External electric field and induced charge on outer surface of  t = 24 cos  = cos  the sphere will produce a net zero field inside it. Net field at P must be vertically upward. 79. At constant temperature, decrease in molecular mass causes 1 flattening of the graph. For same molecular mass of gas, y = 14 = u sint – gt2 increase in temperature causes flattening of the graph. (Vmp)T1 < (Vmp)T2 2 T1 T2 T2 MB usin 5  MA < MB  T1 > MA  14 = cos   cos2  MB  14 = u tan – 5 sec2 But, MA can be less than or greater than 1  5 tan2 – 24 tan + 19 = 0  tan = 1, 19/5. Ans RESONANCE Page - 54

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 83. Average velocity of particle from t = 0 to t = 20 sec. is zero. V 2gh cos  Acceleration of particle from t = 10 sec. to t = 15 sec. is =  =  sec  positive. Average acceleration of particle from t = 0 to t = 20 is zero. r 84. 1  m2 2   mg /2 = mg (/2) cos  + 2  3 ...(1)  ....(2) mg cos  – N = m2 (/2) ....(3) m2 ....(4) 2gh  2 3 ....(5)  (mg sin ) /2 = 2gh cos2 =      =    2  h2    mg sin  – fs = m (/2)   fs  N  = 2gh 2  h2 .    85. (A) L = mvr = m 2gh ()  (D)  = Fr  = (mg) (). y (B) & (C) 86. Tc= Ic  tan  =   1 cos  1 = 1(1)2 d2  2  12 dt 2    d = 6 cos d 00 2  2 = 6 sin    sin sec2  d 1  dy  TC = 1 cos  positive so  increases as t increases. So rod dt 2 =  dt     will be parallel to y-axis in finite time . C.O.M. moves on y-axis  d 1 2gh / 2 d  = dt =  sec2  [if necessary use  2.6 ]. 0 sin  2gh 87. x = (SA + AP) – SP = (SA + AB) – SP = 2gh =   h 2  =  2gh = SB – SP  (1 tan2 ) 1     2  h2 = (500x)2  (3x )2  (500x)2  x 2        Alternate Solution : Wrong solution : V 2gh = r =  Correct solution : RESONANCE Page - 55

Click here to join our telegram channel, @OnDemandMaterial PHYSICS  1 9 1 90. S1 : The focal length of a concave mirror depends only on its 5002 radius of curvature.  = 500x   – 1 5002  1 1  R2 S2 : 1 = (nrel – 1)  –  f  R1   = 500x  1  9  1  nrel  2  5002  – 1 2  5002  nrel = nsurrounding   500x  3 .... nsurrounding  nrel  f  = 2  5002 [8] =, – dv 2 2 S3 : Since E = dr : if E = 0 x = 62.5187.5 88. Due to self induced emf current through B2 and B3 remains V = constant not necessarily equal to zero unchanged but current through B1 increased. 91. S1  True Both are falling under gravity. S2  False Function has maxima at x = x1 , not necessary maximum value also. S3  True W T = V to minimize T, V will be maximum. i.e whole effort of swimmer must towards opposite bank. Just Before 92. S1 : When pressure is increase density also increases in same proportion S2 : True v S3 : 1 = v  u  u v 2 =   v    v uv =  1 – 2  v u v     89. Faraday’s law induced e.m.f. = – rate of change of magnetic v2  v2  u2  u2  flux Inside the solenoid, flux = B × Area =  (v  u) v  = (v  u) v >0 where B is the magnetic flux density   d dB = 2 dB 93. At t = 0 All the capacitors are short circuited. So current at t = Induced e.m.f. =  dt (B × Area) = –Area × dt –R dt 0 can be calculated as below. since area of triangle ABC is R2. (b) the important point is that a changing magnetic field induces O a circulating electric field (from which we derive the induced V e.m.f.). in other words, there are no radial components of electric field (radiating outwards or inwards) so that no e.m.f VS is actually induced in the length AC of the circuit. Induced Electric Field R R R R A OC 2 2 6  R R 2  R 3 R 2 B Hence the induced e.m.f is shared out between AB and BC. 2V So for AB, induced e.m.f. =  1 R2 dB i= 2 dt R Now at t = . Current through each capacitor reduce to zero. So at t =  circuit is RESONANCE Page - 56

Click here to join our telegram channel, @OnDemandMaterial PHYSICS VV 20 2 V 0V 2R 2R V R and R = 7 a R V ·V V Put x = 2 a , y = Hmax 2 VV VV 2 2 2R 2R 2 V 94. & 95. Let the initial temperature, pressure and volume of gas in ‘A’ be 7 28 1 = 7 a . T0, P0, V0 and the area of the position A and B be a, 2a. Hmax = 2 a × 3 3 Now gas in chamber ‘A’ undergoes adiabatic compression  2  2 whereas gas in chamber ‘B’ undergoes isothermal   expansion. 98. Now solving for gas in ‘A’ P1A V1A = P2A V2A  P0 V = P2A (0.25V0 )  P2A = 8P0 0 Solving for gas is chambers ‘B’ P1BV1B = P2BV2B  P1BV0 = P2B x 2.5V0  P2B = 0.4 P1B Also (P2A x a) = (P2B x 2a) for gas in ‘B’  8P0a = P2 x 2a  P2 = 4P0  0.4 P1 = 4P0  P1 = 10P0 Now comparing the moles of gas in A and B nA  P0V0 , At t1 energy A, B, C is zero but at t2 it is more for B so energy RT0 flow from A to B & C to B during t1 & t2, so towards right of A. 99. At B amplitude is maximum so energy is maximum. nB  10P0V0 = 10nA 100 & 101. RT0 (a) Since wall is smooth time of flight remains unchanged mB  10 mA = 10 x 100gm = 1kg. T 2uy  2 100  20s g 10 Again for gas is A, Q = U + W The particle reaches the wall  0 = U + W at t  600 s  10 s 60  V0   P0V0  8P0  4  y Q   0 = U +  (  1)  80m/s  30m/s  O x  U = 2P0V0 = 2 x 105 x 102 = 2000J 100 F = 8P0a = 8 x 105 x 102 = 8000 N 600m 60 80 R 96 & 97. P z At t = 10s, the particle is at its maximum height. Hence, t = 10 s and  = 0° (b) Let Q be the point of impact on the wall. So its X - co-ordinate = 80 × 10 = 800 m Y - co-ordinate = uy2 (100)2  500 m 2g  2 10 Z - co-ordinate = 0 y = xtan 1 x  (c) Since only velocity component along line of impact changes  R  so, just after collision with wall.  V Q  80ˆi  e  60kˆ  (80ˆi  30kˆ )m / s Put co–ordinate of A & B And, time to reach the XZ- plane is 10 s from Q, so, for R, 3 X - co-ordinate = 80 × 20 =1600 m to get cos = 31 Y - co-ordinate = 0 Z - co-ordinate = 30 × 10 = 300 m RESONANCE Page - 57

Click here to join our telegram channel, @OnDemandMaterial PHYSICS (f  x) 102. The image distance of A is vA = – xf ( x < f) [(2f  x)  f ] u The image distance of B is vB = (2f  x)  x u1 GM 2R pole A FB C x x v GM 2R  Ans. (B) f 106. 1 mu2 – 0   – 1 P cos   Solving | vA | + | vB | = 4f we get x = 2 2 Q 0 40 r2  xf    required time t = u  2u v= –QPcos  103. At time t = 0, velocity of image of A and B are  uˆi and 20mr 2  uˆi respectively. Therefore magnitude of their relative 107. Tangential acceleration of bead at point ‘A’ is zero velocity is 2u. 104. All area of earth with be covered other than shaded area. 108 & 109. The direction of electric field is in x-y plane as shown in shaded area = 2 2R2 (1 – cos ) figure = 4R2 1– 3 = 2R2 2 – 3  O y    N 2   60°  E x O The magnitude of electric field is 2R E = E2x  E2y = 3  1 = 2V/m. area covered 4R2 – 2 2 – 3   The direction of electric field is given by 2R Ey 1 = 2R2 3  = tan–1 E x = tan–1 3 = 30° 105. 1 mu12 – GMm  0 Hence electric field is normal to square frame LMNO as shown 2 2R in figure. v1  GM  R  electric flux = E . A = E A cos0 = 2 × 1 = 2 V/m GM  u2  GM  C is correct option of Q. 15. 2R R Since flux is maximum at  = 60° , rotation by 30° either way would lead to decrease in flux.  D is correct option of Q.16.  Lines ON and NM are both normal to uniform electric field E . Hence work done by electric field as a point charge 1 C is taken from O to M is zero. RESONANCE Page - 58

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 110 to 112. 113 to 115. Let mass of bead = m Total magnetic flux at any position   Bzr02  L to complete vertical circle VH > 0 Since, R = 0, so   B0 (1  z)r02 – L = constant Apply WE therom From initial condition (z = 0, I = 0), the value of constant is   B0r02 Using the above equation the current in the ring I= 1 B0 r02 z L The lorentz force acting on the ring (which can only be vertical, because of the symmetry of the assembly) can be expressed as 1 Fz  Br (z)2r0  2B20  2r04 z  kz –mg (3) = 2 m (V2 – V2) L V2 = VH2 + 6g Equation of motion of the ring is maz = Fz – mg = –kz – mg V2 > 6g Equilibrium position z0 = –mg/k Vmin. = 60 k 0  m Inside water forces acting are gravity, Buoyant force & Normal reaction. 2 x 2 1.25 Wa + WB + WN = KE = 1.25 , tan = 1 y   1 –mg(1) + 2mg(1) + 0 = 2 m (V2 – V02) 117. (A) 2g = 1.25 V02 = V2 – 2g > 40 2 x 2 1.25 1 = 1.25 , tan = 5  4 V0 min = 40 = 2 10 m/s y   (B) 2g a 1 (C) y = 5 tan = 5 g = 2.5 m, tan = g  2 Inside water at = g cos as it moves up, with time t,  decreases (D)  at increases i.e. speed increases and rate of increase of speed (at) also increases Just outside water V = 60 at = gsin 60 = 5 3 m/s2 Equating the pressure at the junction a = 900  75 > 3g   2  1   1 2  1 2 tan     118. (Moderate) The electric field due to one dipole at centre of other dipole is parallel to that dipole. Hence torque on all given dipoles is zero. In case B and C the electric field at second dipole due to first is along the second dipole. hence electrostatic potential energy of second dipole is negative. In case A and B x-axis is line of zero potential. In case B and C electric field at origin is zero. RESONANCE Page - 59

Click here to join our telegram channel, @OnDemandMaterial PHYSICS X – ++ ++ – ++ ++ 119. (B) Phase difference = tan1  R  = /4 for 1 – ++ – – ++ –   – ++ – – ++ – – ++ – – ++ –  (B) – ++ – – ++ – = – 4 for 2 – ++ – – ++ – – ++ – – ++ – – ++ – – ++ – – – – – – – X (C) Phase difference  = tan1  R  = /4, for 1 final   Initial = /4, for 2 A0 C= d X (D) Phase difference  = tan1  R  = /4 for both   1 1 Uf = 2 c2  2 c2  kc22 120. VP, P  V2ˆi  25ˆj  V1kˆ F13 = 2A0  a  2ˆi  12.5ˆj P, P  c 22 VP,P = Velocity of particle relative to platform  2  25 F13 = 2A0 Time = 12.5 = 4 sec. 1 W by battery is –ve 8  V2 × 4 – 2 × 2 × 42  16 –Q QQ –Q –Q Q Q –Q 6  V2  8 – + + – 16  V1 × 4  24 – ++ – – + + – 4  V1  6 – ++ – – + + – – ++ – – + + – 1 – ++ – – + + – Y = 25 × 4 – 2 × 10 × 42 (C) – ++ – – + + – – ++ – – + + – = 100 – 80 = 20m – ++ – – + + – – ++ – – + + – – ++ – d1 d1 dd 121. (A ) Q2 Q2 Ui  2c  2c – + + – – + + – – + + – Q2 Q2 – + + – – + + –  – + + – – + + – Uf = 2c1 2c1 C1 < C – + + – – + + – – + + – – + + – Q2 – + + – – + + – F13 = 2A0 – + + – – + + – – + + – – + + – – + + – Initial Q2 F13 = 2A0 final A0 1 2c 2 1 c= d Uf = 23  2 2c2 –Q Q –Q/2 Q/2 Q/2 –Q/2 + – + – ++ – – + – ++ – 1 c2 1 c2 4 c2 – + – ++ – Ui = 2  2 = 3 = c2 – + – ++ – + – ++ – – + – ++ – – + – ++ – 2c (D) – + – ++ – 3 – ++ – .2c – – F13 = 2A0 c 2 2 2c 2 2 Q2 Ui = 2c   Q2 Q2 Q2 F13 = 2A0 = 3A0 Uf = 8c  8c  4c Work by battery = 2c .  2 c2 F13 = 0 3 3 Q2 / 4 F13 = 2A0 RESONANCE Page - 60

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 122. At t = 0 126. Since A and B are at same potential we can redraw the circuit as V0 R 12V 9V 1A 10V 86 3 8 2 RR 9V 18V 36V 9 E 6 Pconsume = V02 = 3 12 18 2R / 3 2 P0 12 1A After a long time : V0 Hence current through 8 is 1A and VB – VD = (VB – VE) + (VE – VD) = – 8 + 12 = 4V. x+y = 1+ 0 = 1. RR 127. q1 x – q2 Pconsume = V02 = P0 47 2R 2 Since current in B1 decreases with time so its brightness q1 q2 =0 q1 q2 =0 decreases. x4 4 x7 7 Initially brightness of B2 is less than B1 but later on B2 will q1 x  7 be brighter. q2 = 7 123. WNA + WNG + WG = 1 mB V 2 0 q1 = x  4 2 q2 4 1 x4 x7 WNA + 0 + 0 = 2 (2) (1)2 – 0 = 1 = F 124. at = m sin 47 7x – 28 = 4x + 28 3x = 56 Rd 2(2 ) F 56 q1 56  7 11 dt 2 = m sin x= 3 q2 3 =3  = 7 d 2 F sin ...(1) 12 12 11 ...(ii) | q2 | = + 11 c  q1 = 11 × 3 = 4 c  dt 2 = 2mR F 129. Only effective pairs will be body diagonals for this interaction ac = m cos  d 2 F 4kq2  dt m R (2 ) = cos electrostatic potential energy is = 3 .  d 2  nQ 3Q  Q 3Q  Q  – –   130.  0    dt 2  50 50 50 50    d 2 = 2 tan = 2 Ans. n = 1   dt    125. From diagram 131. 2A 6V 5A q  2(1 cos 37º )  q  2(1 cos 53º )       A 10V 0  4  0  4  5A   q 1  1 1 4   1 1 3  0  2  5  2  5  Current through ammeter   = (2 + 5) = 7A q 1 1 2 7q  0 10  10  10 0  . RESONANCE Page - 61

Click here to join our telegram channel, @OnDemandMaterial ///////////////////////// PHYSICS 132. F = e  B This force produces an acceleration, a, perpendicular to the line from the electron gun to the screen. The force remains in this direction since d is assumed to be small. F d D Thus, a = F  eB and d 1 at2 m m 2 D Where t = is the time it takes for the electrons to reach the  screen. 0 R2 0 1 eB  D 2 eBD2 2 (R2  3R2 )3 / 2 = 16R  Hence, d =  or d = B1 = 2 m  2m B2 = B12 + B12 + 2B1 B1 cos 120 Substituting the numerical values : 0 30 1.6  1019  5  105  (0.2)2  B = B1 = 16R = 48R d = 2  9.1 1031  3  107  d = 5.9 10–3 m i.e., d = 6 mm. 133.  = | (v    | B).  135. P0 + v12 + 0 = PA + gh + v 2 2 2 2 ( V2 = 2g  5 = 10 >> V1) A  = VB (PQ) V1  3m 1m = VB 2 R2   vt 2 = VB 4R2  V 2t2 4m  2     V2  = 4 × 0.25 4  25  16  4  v12  v 2   2 2  – gh = (PA – P0) = 6 volt   (vQ > vP) 2 134. Method-I v 2 (PA – P0) = – gh – 2 2 B 103 103 10 10 7 2 2 2 = – 2 × 104 =–  10  2  r 20 cm d V 77 m, –e = – 20 × 105 N/m2 = – 20 atm r= mV  3.4m 136. The minima will be heard at P when a crest from S1 and a eB trough from S2 reach there at the same time. This will happen if L1 – L2 is /2 or  + (/2) or 2+ (/2) and so on. Hence, the r sin  = 20 cm increase in L1 between consecutive minima is 1 and from the 1 data we see that = 0.40 m. Then from = v/f  f = 340/0.40   17 radian = 850 Hz. r – r cos  = d P02 Power r2 137. 2V = I = 4r 2 2 Put the values to get the answer. d  6mm Method-II 138. (10 × 11 – 10 × 6 ) × 10–4 × 2T = w The maximum magnetic force is perpendicular to the velocity 50 × 10–4 × 2T = 3 × 10–4 and has a magnitude 3  T = 100 = 3 × 10–2 N/m. RESONANCE Page - 62

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 145. at plane surface 139. (n = 2) image of object is = –  xR  4 distance 2v ———(i) 12 f1 = 2L ———(i) at current surface n=6 u= –  xR 4  R 6v  12 f2 = 2L v=  ROC = – R r  i r i R =v –u f2 6v / 2L 1 4 4 f1 = 2V = 3 R =   xR4  R 2L  12 f2 = 3f1 = 3 × 100 = (300 Hz) x=1  140.xCM  dm  x 2L dT dm  3 146. H = – K. 2rl dr Mass of rod ; R M = L  103 2 0x dx 0 Torque about ‘O’ ; L 2L 2R FB 2 cos 37° – Mg 3 cos 37° = TL cos 37° 1 1 104 2 T 103   2 3    × 10 2 –  R2 H dr T2 d T R1 2rl T1 104 =– K T= N 2l k(T1  T2 ) l nR2 6 H= Hi = Hf  Ans. n = 4 141. V0 sin30° = V1 cos30° (i) R1 eV0 cos30° = V1 sin30° (ii)  Dividing (i) & (ii) 147. QAB = nCpT =  1 nRT =  1 [3P0V0 – P0V0] 1 11  e tan 30° = tan 30 e = 3 = 2PV0 ×  1 142. 1 .q  1 q   QAC = U + w 40 R 40 R nR 1 = T + × 3V0 [P0 + 4P0] 2 q  1 20R   = [16P0V0  P0V0 ] + 15P0V0  Ans. K = 1  1 2 144. XBF = 1 + 2  56 = 2P0V0 ×  1 d d 2 dxBF = dt + dt   1  dt 360 = 15P0V0  2(  1)    4 360 15 ( 1) 3VBF = (– 8 + 2) + (4 – 2) 56 = 4  4 12 = 7 + 7 3= × (–6) + 2 = – 8 + 2 = –6 ms–1 72  = 5 = 1 + f  f = 5 RESONANCE Page - 63

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 148. 152. dNA = –1NA, dNB = 21NA – 2NB dt dt NB=maximum  dNB =0 dt N 21NA = 2 Bmax N =Bmax 21 NA 2 N4 N =Bmax 21 N0e–1t = 2. 5 = 6 mg 2  7.5 mg 153. 8 = 6 tan 1 6   R  39  R = 7.5 mg 5 = 4.5 mg = 2 mg 8 6 = 8 tan 1 R    AP  149. 8 6  R  6  16 R  6 6=8   = R8    R  8   9 1g  16R – 16 × 8 = 9R – 9 × 6  7R = 16 × 8 – 9 × 6 = 128 – 54 = 74 N cos 37° = 1 g N sin 37° = 1a ... (1) 37  2 ... (2) R= 7 m a g = tan 37° 154. From figure (S3 – S2) – (S2 – S1) = ad.d 3 S1 + S3 – 2S2 = ad2 a = 10 × 4 = 7.5 m/s2 P = 2a P = 15 N v a= S3  S1  2 S1S3 150. 10 3 .cos30º = 2 d2 v = 30. 2 g  S3  S1 151. Acceleration of wedge equal to g tan30º = 3 towards left. = mg mg sin 30º + 3 cos 30º = marel d2 N  Ans. n = 1 m mg 155. |S1| + |S2| = H 3 1 gt2 + ut – 1 gt2 =H mg arel 2 2 w.r. to wedge ut = H . ....(I) gg u2 ....(II) are; = 2  2 = g H = 2g gg u 3 3 = m (arel cos 30º – 3 ) t = 2g  m = 6 Kg. S2 = ut – 1 gt2 = 4 u  1 g u2  3u2 = 3H 2 2g 2 4g2 8g 4 RESONANCE Page - 64

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 156. a || v 2mg  mg    23 3  2 ˆi  3 ˆj  xkˆ  3 ˆi  4.5ˆj  6kˆ x = 4. 157. 3L 3 15 159. gsin   10  sin 30 = 3s 160. The plate is free to rotate about vertical axis yy’. Let v, vocfmmaasnsd  be the velocity of particle, velocity of centre of plate and angular velocity of plate just after collision. p   From conservation of angular momentum about vertical Ft axis passing though O is  =m (10 – 0) ....(i) a a ma2 ....(ii) mu 2 = mv 2 + 3    R 2 mR2 ....(1) L Ft 2 5 ....(iii) centre  = ( – 0) ....(iv)  d sUnz  R 2 mR2 since the collision is elastic, the equation of coefficient of L Ft 2 5 restitution is = ( – 0) e = vcm  v = 1 ....(2) u Equation (ii) / equation (i) 2 R2 R a ....(3) 5 2 But vcm = 2 = 10 solving equation (1), (2) and (3) we get  12 u Vcm = 10 & R = 12.5 kinetic friction acts  = 7 a = 5 rad/s 161. Apply Snell's law on various surfaces one by one : 1 sin 90° = 1 sin r1  sin r1 = 1  r1 = 45° 2 mg 1 acm = m = g 1 cos r1 = 2 sin r2 sin r2 = 2 2 2 cos r2 = 3 sin r3  sin r3 = 2 1 sin2 r2 Tc = – kmgR = 5 mR2  3 Let after time t, V’cm = ’R 2  1 3 cos r3 = 1 = 2 10 + kgt = 12.5  2.5 kg  R 3 R   R  sin2 r3 + cos2 r3 = 1   10 + kgt = 12.5 – 2.5kg 2  1 + 1 =1  22 = 3  3.5 kgt = 2.5 3 2  3 11  2 = 3 t= 7  t =7 158. Tthe FBD of any one rod is ...(i) 1  3  1 1 1 ...(ii) 162. fL =  2     T  N   20 20  mg = N  FL = 20 cm Taking torque about point 'P' 11 2 1 T f =  10 – 20 = – 5 N 1 1 1 111 v +  30 =  5  v = 30 – 5 mg 370 NP 11 1 v –  30 = 20 mg L cos 370 for lens 2 = TL sin 37º mg 4 2mg T= 2 3  3 RESONANCE Page - 65

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 11 1 25 169. From lenz's law the current i in the loop is clockwise For equilibrium 163. f = 25  25 = v +  25  V = 2 BIa V 1 Mg = 2 m= u = 2 PQ 164. T  length free length becomes  So time period 4 T AB becomes T1 = 2 3 a/4 Half oscillation is with free length  and half with free length  R 4 2mg 2´ 2´ 10 T T1 3T I = Ba = 10´ 1 = 4 Total time = + = v  22 4 170. || B 165. f = 2N m/s2 So, F = 0 a1kg = 2 1 171. If the block has moved by distance x. S= 2 ×2×4=4m Wfriction = – 2 × 4 = –8J Ma = 2µMg – µMg  µMg  µMg ( –   3 3 3 x) 166. Energy entering in the windmill = 1 mv2 a = 2µg –  µg  2µg  3µg ( – x)  2  3 3 3    PIn  dE   1 v2  dm  da1 3 µg dt  2  dx 3   dt  = [for 2nd and 3rd cells similarly]  PIn   1 v2  AV   1 AV 3 tan 1 : tan 2 : tan 3. = 3 : 2 : 1  2  2   172. For first lens Electrical power output Pout  1  1 AV3  11 1 3  2  v –u = f   Pout  1 AV 3  1  1.2  10   20 3 11 1 1 6 6  v = 20 – 30  60  y Pout = 16 kW. 167. acart = 50  k (20) 10 mm x (–30, 0) (0, 0) (90, 0) 53 10 mm abrick = k10 90cm Optical axis 150cm 20mm 50  20k arel = 8 – k(10)  v = 60 cm For second lens 1  50  100k  u = 30 cm, f = +20 cm  0.4 = 2  8  (0.8)2   1 11  1 = 5 – 10 k f = v –u k = 0.4 E 2 11 1 R  v = 20 – 30 168. P =   Rx     1 3–2 1  v = 60  60 dp E2 dR  dx = R2 dx  v2 = 60 cm  m=–2 E2 0x So coordinates are (150 cm, – 30 mm) = 02L4 LA 4A 2L2 RESONANCE Page - 66

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 173. hv = 13.6(3)2 1 1 = 2.75 eV 179. Let at any time t charge flown through the plate B to plate A is  42  52  q and instantaneous current is I. for n = 4 to n = 3 –Q0+q – q –Q0+q – q hv = (13.6) × 2 1 1 AB AB  32  42  I I (3) = 5.95 eV R R for shorter wavelength 3.95 = 5.95 –    = 2eV From loop theorem for longer wavelength  2q  Q0   IR    0 eVs = 2.75 – 2 = 0.75 eV  2C    (1 r)  cos2  174. Pressure = C 3  R dq   2q  2C  Q0 Where,  is the angle with normal = 8C . dt 2C 175. PQ  152  202  25m dq dt  2C  Q0  2q  2RC 25  VP2 sin 2(45)  VP  5 10m / s Now for charge on plate A to be zero q = Q0. g  VA2 sin2   2 10 12.5  (5 5)2  VA sin  5 15 m / s Q0 dq t dt VA cos  VP cos 45  5 5 m / s and   60 2C  Q0  2q  0 2RC  AB  VA2 sin 2  25 3  Integrating g 0 = t  RC ln 2C  Q0   2C  Q0   176. Consider equation of torque about C Putting the value of C, Q0,  and R We get t = 2 seconds. mg 4R  = mR2  180. No current passes through capacitors in  3  2 steady state. Assume potential at point '4' to be zero.  8g 8g 8g  = 3R = 32R = 30R 3v 2V Then points '1' and '2' are at same potential 3 . 177. f = 4L  0.6r  Hence C1 and C2 can be taken in parallel. df 3v [–0.6 dr ] dt dt V  = The potential at point 3 is 3 . 4 L 0.6r 2    Equivalent circuit of all three capacitors is shown  Hence potential difference across capacitor C3 is dr 1 = m/s dt 72  F  0.02 178.    A   10  103kg / ms  102poise   2  dv   dx  1   2C  2V V  2V =     = 2C C  3 3  9  RESONANCE Page - 67

Click here to join our telegram channel, @OnDemandMaterial PHYSICS 181. Applying work energy theorem we can write 185. Since the rod is in translation so,  = 0  about O,   R  V2  0 R g 2R  R  2  2  2   4  Na    O gR  8  10  F V2 =   = 900 2    182. mg F  sin   N  cos   0 2 2  F sin   Ncos   tan   N ....... (1) F ....... (2) Also, N = Mg So,  tan   Mg  40  2 F 60 3 Wb = U + H    tan 1 2   3  4  4 × 3 = 0 + H X=3 16  H = 3 J 16 x 16  12 3 = 2 x= 3 = 64 (r 2 ) 2gy  x 2   dy  183.   dt    (r 2 ) 2gy  x 2   yx4 Hence vr%, n = 4. 184. AB : Ft + Nx t = mv1 ....(1) Ft(/2) – Nx t (/2) ....(2) ....(3) = (m2/12)1 BC : (Nxt)t = (m2/3)2 2  = (/2) 1 – v1 ....(4) Solving we get 2 = 6(Ft)/7ml = 6 rad/s. RESONANCE Page - 68

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Station, Nerul (W), Navi Mumbai-6 INDORE Tel.: 011-31922222, 60606602 Tel.: 022-31922227, 60606605 1st Floor, Gravity Tower, Opp. Agrawal Stores, VISAKHAPATNAM (ICCP) e-mail: [email protected] e-mail: [email protected] Indore - 452003 (MP) Vikas Vidyaniketan, Door No.: #1-1-1/1, Sheela Tel: 0731-3192222, 4274200 Nagar, NH-5, Visakhapatnam, (AP) - 530012 North West Campus Panvel Campus e-mail: [email protected] Tel: 0891-2757575 10, Ground Floor, Central Market, Punjabi Bagh Deep Dhara, Plot No.08, Road No.18, Sector-1, e-mail: [email protected] (West), New Delhi - 110026 Opp Geetanjali Building, New Panvel - 410 206 SIKAR Tel.: 011-31922224, 60606601 Tel.: 022-31922232, 60606606 Vishwakarma Apartment, Piprali Road, Sikar-332001 BHAVNAGAR (ICCP) e-mail: [email protected] e-mail: [email protected] Tel: 01572-319222, 606066 & 258222 Sardar Patel Educational Institute, e-mail: [email protected] Kalvibid, Bhavnagar 364002 (Gujarat) East Campus Thane Campus Mob.: 8000816767 / 8000716868 14, Pratap Nagar, Mayur Vihar Phase-1, Paranjpe Udyog Bhavan, 2nd floor, Above Khandelwal AGRA New Delhi - 110096 Sweets, Opp. station Thane (W) - 400607 1, Kailash Vihar, Behind St. Conrad's college, DEHRADUN (ICCP) Tel.: 011-31922223 / 5, 60606603 Tel.: 022-31922228, 32191498, 6060 6607 Khandari, By-Pass Road, Agra-282005 SelaQui International School, Chakrata Road, e-mail: [email protected] e-mail: [email protected] Tel: 0562-3192222 Dehradun - UK, INDIA e-mail: [email protected] Tel: 91-135-3051000 | Mob.: 9873743433 LUCKNOW Kharghar Campus Hazaratganj Campus 5/6/7/8, Swami Samarth Bldg., Sector-12, NANDED 13/1, Moti Chamber, Rana Pratap Marg, YMCA Opp. Kendriya Vihar Colony, Behind Syndicate Bank, Beside Akola Urban Cooperative Bank, V.I.P Road, Complex, Ground Floor, In Front of Gauri Apartment, Behind Mercantile Bank, Kharghar Paras Square, Nanded.- 431602 R.F. Bahaduri Marg, Hazratganj, Lucknow (U.P.) - 1 Tel.: 022-32191499, 60606608 Tel: 02462-250220, 606066 Tel.: 0522-3192222, 3205854, 6060660/1 e-mail: [email protected] e-mail: [email protected] Mob.: 9369225676 e-mail: [email protected] CORPORATE OFFICE: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Rajasthan) - 324005 Reg. Office: J-2, Jawahar Nagar Main Road, Kota (Raj.) - 324005 | Tel. No.: 0744-3192222, 3012222, 3022222 | Fax : 022-39167222, 0744-2427144 To know more: sms RESO at 56677 | Toll Free : 1800 200 2244 | Website: www.resonance.ac.in | e-mail: [email protected]