CUET Physics

174 PHYSICS 102. In parallel grouping of cells, the sizes of the 122. Copper has low electrical resistivity, hence electrodes will increase without affecting the high electrical conductivity. emf of the cell. 123. For balanced Wheatstone bridge, we have 104. 1 kWh = 103 Wh P 3=R 6 1J × 3600 s Q = 2 S′ = S′ 1s = 103 × = 3.6 × 106 J ⇒ S′ = 4 W 106. If each resistance is r, If a resistance r is used for shunting S to obtain S′ (= 4 W), we have RP = r and Rs = nr 4 = 4r n ⇒ 4+r ⇒ RP = r × 1 = 1 r = 8 W Rs n nr n2 1 βt2 1 γt3  110. P = V2/R 124. q = αt − 2 + 6 ... (i) ... (ii) When R′ = R/m I = ddqt = α − βt + γt2  2 V2 V2 mV 2 P′ = R′ = R/m = R Initial current, t = 0 gives, I = a ⇒ P′ = mP 125. R = ρl = r (Given) 111. P = VI = 1.5 × 0.3 W = 0.45 W A 112. Heat energy produced = VI t Volume of the wire (V) = A × l = 6 × 0.5 × 1J = 3J = A × 2l 2 113. Because H ∝ I2, hence for heating effect both a.c. and d.c. can be used. \\ R′ = ρ 2l = 4r A/2 114. In parallel, power of n bulbs = n × 60 W So, increase in resistance = 4r – r n × 60 Current, 9 = 220 = 3r or n = 220 × 9 = 33 126. Resistivity, r = RA 60 l 115. Current capacity = current × time = V A I l 116. Heat produced by d.c. is the same as by r.m.s = ML2 T − 2 × L recorded by the ammeter in a.c. A2T 117. Brightness of a bulb in a house fitting is directly proportional to the power of the bulb. = [ML3 T– 3 A– 2] Also, resistance ∝ 1 127. Conductance = 1 Power Resistance 120. Internal resistance, = I = A2T V ML2T − 2 V ML2T − 2 r = I = (AT) (A) = [M– 1 L– 2 T3 A2] = [ML2 T– 3 A– 2] 128. No. of free electrons in a conductor = 1029 m– 3 121. A potentiometer is an ideal voltmeter.

ANSWERS   175 129. Conductance = 1 Pot. drop across the resistance Resistance = Reading of the voltmeter 1 1000 Ω− 1 = I × (2.5 × 1000) 2m Ω 2 = = = 6 × 2500 = 5V 3000 = 500 S 136. 1 = 1+1+ 1 130. Current through arm ABC = R 4 6 12 Current through arm ADC = 1 A = 3 +2+ 1 12 VA – VB = 1 × 2 = 2V VA – VD = 1 × 3 = 3V = 1 VB – VD = (VA – VD) – (VA – VB) 2 = 3V – 2V or R = 2 W Also, r = 2 W = 1V \\ I = 4A (2 + 2) 131. The resistivity of metal increases when it is converted into an alloy. = 1 A 2W 3V 1W 137. 1 = 1+ 1 CD rp 4 12 133. A B 2A Pot. diff. between A and C, ...(i) = 3+1 = 1 VA – VC = 2 × 2 = 4V ...(ii) 12 3 ...(iii) Pot. diff. between C and D, Rp = 3 W Also, VC – VD = 3V I = 3V Pot. diff. between D and B, 3Ω VD – VB = 2 × 1 = 2V = 1 A We have, 138. Resistance = 22 × 10° = 22 W VA – VB = (VA – VC) + (VC – VD) + (VD – VB) = 4V + 3V + 2V 139. In the given network, a balanced Wheatstone Bridge circuit is connected across r. = + 9V The resistance of the Wheatstone Bridge 1 1+ 1 circuit is r. rp 8 16 134. = 1 1 1 2 rp r r r 2+1 \\ = + = 16 = 3 ⇒ rp = r 16 2 = ⇒ rp = 136 Ω Here, rp is resistance between P and Q. 1.5 A 135. Total resistance = (2.5 + 0.5) kW 140. I1 1A I¢ 1.3 A = 3000 W 2A 0.5 A I Current, I = 6 A 3000 I1 = (1.5 + 2 + 0.5) A = 4.0 A

176 PHYSICS Also, I′ = (4.0 – 1.0) A = 3 A 145. Equivalent circuit is given by, I = (3 – 1.3) A 10 W 10 W 10 W A = 1.7 A B 141. The resistivity of a conductor is independent 10 W 10 W of the dimensions of the conductor. 142. R = 100 W 10 W = r l RAB = 10 Ω + 30 Ω + 10 Ω A 4 R′ = ρ 2l = 4×ρ l = 10 W + 7.5 W + 10 W A/2 A = 4R  1 = 1 + 1 = 4  rp 30 10  = 4 × 100 W 30  = 400 W  rp = 30 = 7.5 Ω  ⇒ 4    143. R = ρ l = 27.5 W A 146. In the given circuit, 6 W is parallel to 6 W and Volume = A × L gives equivalent resistance of 3 W. = 2A × 3L 3 W and 6 W in series gives 9 W resistance. 3 2 \\ RAB = 6 W + 3.6 W = 9.6 W R′ = ρ l × 9  1 = 1 + 1 = 6+4 = 10  A 4  rp 6 9 36  36  % Increase in resistance = R′ − R × 100  ⇒ rp = 3.6 Ω  R    9 1 147. 1 = 1+1 4 rp 48 = − × 100 2 + 1 3 = 8 = 8 = 125% 144. 4.5 A = I1 ⇒ rp = 8 Ω = 2.67 W 3 Also, I2 = (4.5 – 0.5) A = 4.0 A 148. Equivalent resistance between A and B can written as, Again, I3 = (4 – 2.5) A = 1.5 A 2W \\ I = (1.5 – 1.2) A 2W 2W 3W A B = 0.3 A 2W 3A 0.5 A 1 1 1 1 1.5 A rp 2 2 2 = + + I1 3 I2 2 2.5 A = 1.2 A I3 I 2 ⇒ rp = 3 Ω

ANSWERS   177 \\ RAB = 2 Ω + 2 Ω + 3 Ω So, 2 = 3 3 X 6 = 17 Ω ⇒ X = 4 W 3 155. Since p.d. between B and D is zero, it is a case 149. 1 = 1 +1 of balanced Wheatstone Bridge circuit. rp 18 12 12 X+ 6 \\ 1/ 2 = 1 2+3 5 = 36 = 36 36 ⇒ 24 × 1 = X + 6 5 ⇒ rp = = 7.2 W ⇒ X = (24 – 6) W = 18 W 150. Potential gradient = V 156. Let the equivalent resistance between 100 W l and X be r. ML2 T − 2 In case of a balanced Wheatstone Bridge = AT × L circuit, = [ML T– 3 A– 1] P = R Q S 151. V = IR 100 200 r 40 = Iρ l ⇒ = =5 A Since I is constant and the wire has a uniform ⇒ r = 20 W area of cross-section, therefore, V ∝ l or V/l = const. Also, 1 + 1 = 100 + X 100 X 100 X ⇒ y ∝ x, it is a straight line passing through the origin. ⇒ 100 X = 20 100 + X 152. Potential gradient is a scalar physical quantity ⇒ 5 X = 100 + X while others are vector physical quantities. ⇒ X = 25 W Potential gradient = dV 157. P= V2 dx R or 153. When X is shunted, let the equivalent or R = V2 = 100 × 100 resistance be R. P 500 We have, R = 0.625 × ρ = 625 R = 20 W 2 0.375 × ρ 375 V12 (r is resistance per unit length of the wire) Also, P′ = 20 = 150 × 150 20 ⇒ R = 625 × 2 = 10 Ω or P′ = 1125 W Also, 375 3 158. H = V2/R, where H is heat produced in R per R = 5× r = 10 , second. When wire is cut into 3 parts, each 5× r 3 part has resistance of R/3. where r is connected (in parallel) across 5 W. Effective resistance, ⇒ r = 10 W 154. Since no current flows through CO, it becomes 1 = 3 + 3 + 3 a case of balanced Wheatstone Bridge circuit. r R R R ⇒ VC = VO i.e. VC – VO = 0 = 9 R

178 PHYSICS ⇒ r = R Resistance, R = ρl = ρ l2d 9 A M So, heat H′ produced per second is given by, Also, heat produced (H) ∝ R ∝ l2 M V2 9V 2 H′ = r = R = 9H (5)2 (3)2 (1)2 ⇒ 1 3 5 H′ = 9H \\ H1 : H2 : H3 = : : 159. Equivalent resistance of the circuit = 25 : 3 : 1 5  3 × 6 1 =  6 + 3 +  Ω = 125 : 15 : 1  162. Fuse wire has high resistance and low melting = (2 + 1) W = 3 W point. Current I in the circuit = 12 =4A 163. R = 10 W 3 V = 22000 Volt Potential drop across 1 W resistor (R3) = 4 × 1V = 4V P = 2200 W Remaining potential, 12V – 4V = 8V exists \\ I = P = 2200 V 22000 across the parallel combination of R1 and R2. 1 I2 = 8V = 8A = 10 A 3Ω 3 Power dissipated = I2R \\ Heat produced per minute in R2 = (I22 × R2 × 60)J = 1 × 10 100 2 =  8  × 3 × 60 J = 0.1 W  3 164. = 64 ×3× 60 J 10 A 9 3A = 1280 J I1 = 3 A 160. Resistance of each light bulb 7 A I1 + 3 A = 6 A = V2 = (120)2 4 A I2 I 60 60 = 240 W 240 Algebraic sum of currents at a junction (point) 2 is zero i.e. sum of currents entering a junction Effective resistance of circuit = 240 + is equal to the sum of the currents leaving the junction. = (240 + 120) W 10 A = I1 + 7 A = 360 W ⇒ I1 = 3 A Also, 7 A + 4 A = I2 ∴ Power dissipated = (120)2 = 40 W ⇒ I2 = 11 A 360 ∴ I = I1 + I2 = 6 A + 11 A = 17 A 161. Mass = volume × Density 165. V = IR ⇒ V ∝ I i.e. M = Al d, It is a straight line passing through the origin. or A = M , ld where d is density of the material of the wire.

ANSWERS   179  1  173. Rate of flow of charge is current.  2  166. Given,     R1 = Ω 174. Current travels at the speed of light i.e. 3 × 108 ms– 1.  1  R2 = 4 Ω 175. It is a case of balanced Wheatstone Bridge  1  circuit, therefore, P = R .  6  Q S R3 = Ω ⇒ 12 = X+ 6 1 1 In parallel, 1 = 1 + 1 + 1 rp R1 R2 R3 22 = 2 + 4 + 6 = 12 ⇒ X = 6 Ω \\ Equivalent conductance 176. I = Q = ne t t = Gp = 1 = 12 S rp n × 1.6 × 10− 19 3 mA = 60 167. Potential difference across 300 W = (60 – 30) V 3 × 10− 3 × 60 = 30 V ⇒ n = 1.6 × 10− 19 Since 60 V is equally divided between = 18 × 1018   1018 two parts, therefore, effective resistance of 16 voltmeter resistance R and 400 W in parallel will be equal to 300 W. eE t m \\       300 = 400 R 177. Drift velocity, vd = 400 + R or  300 R + 120000 = 400 R =  e t  V  m l or          R = 1200 W 168. Internal resistance of the battery is zero. ⇒ vd ∝ V 178. Resistance of a semiconductor (Germanium) \\ Current through 5 W resistance = 10 V =2A 5Ω decreases with increase in temperature. 169. Since galvanometer (G) shows zero deflection, 179. Thermistors are usually prepared from the no current flows through it. So, it is a case of oxides of metals. balanced Wheatstone Bridge circuit. 10 A 2A So, 10 Ω = 40 Ω 180.      I1 I3 7Ω X 6A ⇒ X = 40 × 7 = 28 W I 10 8 A I2 170. At balanced point, no current flows through 10 A = I1 + 6 A the galvanometer. ⇒ I1 = 4 A 171. R′ = ρ  2l  = 4  ρl  = 4 × 10 = 40 W Also, I2 = 6 A + 8 A = 14 A  A/2   A    I1 = 4 A = 2 A + I3 172. R = ρ l , hence resistance will be least for ⇒ I3 = 2 A A \\ I = I2 + I3 = 14 A + 2 A minimum l and maximum A. = 16 A

180 PHYSICS 181. When in series, equivalent resistance = 6 Ω 186. V = Resistance (R) = slope of the graph. In When in parallel, equivalent resistance I = 2Ω = 0.67 Ω case of a conductor, resistance (hence slope of 3 V–I graph) increases with increase of When two in parallel are in series with one temperature, hence T1 > T2 > T3. resistance, we have equivalent resistance Work done Charge 2×2 187. Potential difference = 2×2 = + 2 = 3 Ω So, (b) cannot be possible. = [ML2 T − 2 ] [ AT ] 182. Here each of 1 Ω resistance will be in parallel ⇒ [Pot. Diff.] = [ML2 T– 3 A– 1] combination between points P and Q. 188. Kilowatt hour (kWh) is associated with energy. 1Ω \\ Effective resistance = 3 189. The equivalent circuit is shown as follows: 183. Effective resistance across P and Q is given r by, r r 1 1 1 6r R 2r 2r = + r 4r r 1 = r ⇒ R = r E, 2 W r ⇒ P = R Q Q S rr r 2r 1 2r 4r 2 i.e. = = Pr T It is a case of a balanced Wheatstone Bridge circuit Current does not pass through 6 Ω 184. Effective resistance across P and T is given by, resistance. 1 = 1 + 1 1 = 1 + 1 = 2+1 = 1 R r 3r rp 3r 6r 6r 2r = 4 ⇒ rp = 2r 3r For maximum power to the network, rp should be equal to the internal resistance of ⇒ R = 3r the battery. 4 So, 2r = 2 Ω(Given) 185. Given, I = 50 = 4.5 ⇒ r = 1 Ω 10 + r 190. Equivalent circuit can be written as, ⇒ 10 + r = 2 × 50 9 rr 100 10 rrr 9 9 PQ ⇒ r = − 10 = ⇒ r = 1.1 Ω 1 = 1 + 1 = 3 rp r 2r 2r

ANSWERS   181 ⇒ rp = 2r So, P − P′ × 100 = 8 × 100 3 P 100 \\ RPQ = r + 2r + r = 8% 3 198. P = V2/R = 8r When V′ = V −5 V = 19 V 3 100 20 191. e.m.f. is energy per unit charge.  19 I 2 192. Imax = e.m.f. P′ = (V ′)2 =  20  Internal resistance R R = 1.5 V % decrease in power =  P − P′  × 100 0.05 Ω  P  = 30 A   19 2  1  20   DV =  −  × 100 DI 193. Resistance, R = Corresponding to portion CD of the curve, =  −  1 − 1 2  × 100 DV is positive but DI is negative, hence R is 1 20   negative for the portion CD of curve.   194. The upper segment of the circuit is a = 2 × 100 = 10% balanced Wheatstone Bridge circuit, hence its 20 equivalent resistance is r. [Using Binomial Theorem] 1 1+1 2 199. H = mL = V 2t rPQ rr r RJ \\ = = (210)2 × 1 rPQ = r = 20 × 4.2 2 210 × 210 195. Both the bulbs 1 and 5 are in series in the ⇒ m = 20 × 4.2 × 80 = 6.56 g/s circuit, hence max. current flows through them. So, melting of ice per second = 6.56. 196. Let R be the resistance of each resistor. 200. Energy produced = V I t V2 V2 = 6 × 0.5 × 1 = 3.0 J 2r r So, P =  ⇒ = 2P 201. When a current flows through a conductor, heating of the conductor takes place which V2  V 2  raises the temp. of the conductor. r/2   Also, P′ = = 2 r 202. H=VIt = 2 × 2P = 4P = V ×V ×t = V2 t r r 197. P = I2R ⇒ H ∝ 1 2 r  4I  P′ =  I − 100 R 203. Appliances work on both a.c. and d.c. = I2R  1 − 4×2  206. R = r l 100  A [Using Binomial Theorem, (1 + x)n  1 + nx] Volume of the wire = l × A = I 2 R  1 − 8  = nl × A  100  n

182 PHYSICS nl n2  rl  = n2R \\ Resistance between any two corners A/n  A  ∴ R′ = r × = 1 1 1 rp 2 2 = + 207. Resistivity of a conductor increases when ⇒ rp = 1 W some impurity is added to it. 216. Given circuit can be written as, 208. Volume of wire = l × A r A 2r PQ = l × nA n So, R′ = r l′ = r × l × nA r 2r A′ n 1  l  B n2  A  = r Points B and Q are common. = R So, 2r in arm AB and 2r in AQ are in parallel n2 between A and Q. 209. A conductor will behave as a superconductor Effective resistance = 2r × 2r =r at very low temperature. 2r + 2r 210. If we have n different conductors, then the The given circuit reduces to, number of possible combinations are 2n. r PA Note: In case of n identical conductors, each of rr equal resistance then the number of combinations we can have using all at a time is 2n– 1. 211. The number of closed circuits in Wheatstone Bridge is 7. Q 212. Potentiometer is an ideal voltmeter. So, effective resistance between P and Q is given by, 213. The circuit has a balanced Wheatstone Bridge circuit, and a resistance of 2 W. Two resistors, R′ = r × (r + r) each 2 W, are connected in series. r + (r + r) rs = 2 W + 2 W + 2 W = 6 W 2r2 2r 3r 3 Also, 1 = 1 + 1 = 1+ 3 = 4 = = Rp 6 2 6 6 ⇒ Rp = 3 Ω 217. The effective circuit can be written as follows. 2 r 214. All the n resistors are connected in parallel. P Q r 1 = 1+ 1+ 1 + ... n times Rp R R R rr = n R ⇒ Rp = R So, effective resistance between P and Q is n given by, 215. Across all the sides of the triangular network,  r  × (2r) r2 2r we have a balanced Wheatstone Bridge  2  2r 5r 5 circuit, which gives a resistance of 2 W. Reff = = = r+ 22

ANSWERS   183 218. In case of a balanced Wheatstone bridge, V = 10−3 I 1000 P R, 5. R= Ω = 10–6 W Q S = in which P, Q, R and S may not have the same value. Resistance/metre = 10−6 × 100 Ω m −1 10 So, if Q and S are interchanged, we have = 10 mW m–1 P≠R i.e. the network is not balanced. Case Study–2 SQ 3. I = q = ne 219. I= 6 = 6A tt 6+4+1 11 or  I = n × 1.6 × 10−19 t Also, voltmeter reading = Pot. Diff. across (6 + 4) W or   n = 6.25 × 1018 s−1 t = I × 10 V = 6 × 10 = 60 V 4. Current, I = 0.50 A ; Time, t = 10 s. 11 11 Each Cu++ ion carries charge of two protons. If n be the number of copper ions, then charge, 220. Cu, Ag and Au are good conductors of electricity. q = n × 2e 221. If r is the resistance of each wire, total Now, I = q t resistance in series = 2r, and in parallel = r . 2 or q = It So, heat produced per second  V2  will be or n × 2e = It = R    It 2e four times in parallel than in series. or n = 222. Resistance of electric bulb = V2 = 2 0×.510.6A× 1×01−019sC P = (100)2 = 50 W = 1.5625 × 1019 200 Let r be the resistance used in series of circuit 5. Current, when a voltage of 200 V is applied. I = 60 W = 0.5 A (or C s−1 ) 120 V So, new power is given by, 200 = (200)2 Now, 0.5 = ne [ t = 1 second] r + 50 1 ⇒ r = 150 W or n = 0.5 × 10−19 Case Study–1 1.6 4. I= (1.0 × 1018 + 2.7 × 1018 ) × 1.6 × 10−19 A = 3.125 × 1018 1 Case Study–3 = 0.592 A l × 4 (2l) × 4 Resistance, R = V = 230 Ω 3. R1 = r πd2 , R2 = r π(2d)2 I 0.592 R1 = 388.5 W Clearly, R2 = 2  or R1 = 2R2

184 PHYSICS Q1 = V2  and Q2 = V2 And t′ ∝ 3l ; t′ = 3 R1 R2 4 t4 Now, Q1 = R2 = 1 or t′ = 3 t = 3 × 10 min Q2 R1 2 4 4 \\ Q2 = 2Q1 = 7.5 min. So, the longer coil will produce double the heat produced by the shorter coil. 5. R= V2 = 220 × 220 Ω = 484 Ω P 100 4. Q= V2 t = V 2a t ; t ∝ l R ρl UNIT 3:  MAGNETIC EFFECTS OF CURRENT AND MAGNETISM 1. (a) 2. (c) 3. (d) 4. (b) 5. (c) 6. (d) 7. (a) 8. (c) 9. (a) 10. (a) 11. (a) 12. (c) 13. (c) 14. (d) 15. (a) 16. (c) 17. (b) 18. (d) 19. (a) 20. (b) 21. (d) 22. (c) 23. (a) 24. (a) 25. (d) 26. (c) 27. (d) 28. (b) 29. (d) 30. (d) 31. (c) 32. (b) 33. (b) 34. (d) 35. (c) 36. (d) 37. (a) 38. (a) 39. (a) 40. (b) 41. (c) 42. (b) 43. (c) 44. (b) 45. (a) 46. (b) 47. (a) 48. (a) 49. (d) 50. (d) 51. (b) 52. (b) 53. (b) 54. (d) 55. (b) 56. (c) 57. (a) 58. (b) 59. (b) 60. (c) 61. (d) 62. (a) 63. (c) 64. (a) 65. (a) 66. (d) 67. (d) 68. (b) 69. (d) 70. (b) 71. (c) 72. (a) 73. (b) 74. (c) 75. (a) 76. (b) 77. (b) 78. (b) 79. (c) 80. (b) 81. (d) 82. (b) 83. (b) 84. (d) 85. (c) 86. (b) 87. (d) 88. (b) 89. (a) 90. (c) 91. (d) 92. (b) 93. (b) 94. (a) 95. (d) 96. (d) 97. (a) 98. (a) 99. (d) 100. (a) 101. (d) 102. (d) 103. (d) 104. (a) 105. (d) 106. (d) 107. (d) 108. (a) 109. (d) 110. (a) 111. (d) 112. (b) 113. (b) 114. (c) 115. (a) 116. (b) 117. (b) 118. (a) 119. (d) 120. (c) 121. (b) 122. (a) 123. (a) 124. (d) 125. (c) 126. (a) 127. (a) 128. (b) 129. (b) 130. (a) 131. (d) 132. (c) 133. (a) 134. (b) 135. (a) 136. (a) 137. (b) 138. (d) 139. (d) 140. (a) 141. (a) 142. (a) 143. (b) 144. (b) 145. (d) 146. (d) 147. (d) 148. (c) 149. (d) 150. (a) 151. (d) 152. (d) 153. (a) 154. (c) 155. (a) 156. (a) 157. (d) 158. (b) 159 (b) Case Study–1 4. (b) 5. (b) 1. (a) 2. (a) 3. (a) 4. (c) 5. (a) Case Study–2 4. (c) 5. (d) 1. (a) 2. (d) 3. (c) Case Study–3 1. (c) 2. (b) 3. (a)

ANSWERS   185 HINTS/SOLUTIONS → B2 = µ0 × 2pI 4p r 1. Current element of length d l carrying current I is given by, ∴ B0 = µ0 × 2I (1 + p) = µ0I (1 + p) 4p r 2pr → I d l , and is expressed in Am. 2. Magnetic field at a point on the axis of a 7. For a point on a linear conductor carrying circular coil carrying current is, current we have, magnetic field induction B = µ4p0 × 2pn Ia2 2 given by, (a2 + x2 )3/ →→ d l × r = 0 where a is radius of the coil, x is distance of → the point on the axis from the centre of the coil. \\ B = 0 8. B= µ0I  or B ∝ 1 4pr r Here, n = 1 1 x >> a It is of the form, y ∝ x and represents a \\ B =  µ0  2M , rectangular hyperbola shown in (c).  4p  x3 9. A toroid can be considered as a closed where M = IA, is called soleniod of a small radius. Magnetic field induction at point well inside the solenoid magnetic dipole moment of the loop. carrying current I is given by, So, B ∝ x–3. 3. From the above expression, B = µ0 NI 2pr  µ0  2M B =  4p  × x3 So, the magnetic field inside the toroid varies ⇒ B ∝ 1 as 1 and hence is uniform. When area of x3 r So, B does not depend upon the radius of the cross-section is very small as compared to r, coil at far off points along its axis. variation in r can be neglected. 4. Magnetic field (B) of an infinitely long So, (N/2pr) will be constant and equals the conductor is given by, number of turns per unit length. B =  µ0  × 2I We have, B = m0 nI,  4p  a which is same as for a long solenoid. So, (a) is correct. ⇒ B ∝ 1 10. Since the total current linking the loop L” is a zero, we have 5. When direct current is passed through a L¢ metallic pipe, the magnetic field exists only outside the pipe. However, magnetic fields Open space are cancelled out inside the pipe. L¢¢ 6. Magnetic field at the centre of the coil, B B0 = B1 + B2 B1 = µ0 2I 4pr

186 PHYSICS ∫ → ⋅ d → = 0 18. Neutrons are chargeless i.e. neutral. So, q = 0 B l ⇒ B = 0 ⇒ F = qv B sin q 11. For any point outside the toroid, the net = 0 current threading the loop L” through that point is zero. This is because each turn of 19. Currents in the same direction, force is the winding passes twice through the area attractive. bounded by the path, carrying equal currents in opposite directions. Currents in the opposite directions, force is repulsive ∫ → → 20. Fm = qv B sin 90° So, B ⋅ d l = 0 = qv B ⇒ B = 0 21. Magnetic dipole moment of a current loop does not depend on the magnetic field in L →→ → I which it is kept. Pm = I A , where A is area r vector. 12. I I S TO P Q 22. Force on the proton moving along the direction of magnetic field, Point O lies on the axis of the linear current carrying conductor ST and PQ, hence the F = qv B sin 0° = 0 current through ST and PQ will not contribute to the magnetic field at O. Hence, path as well as velocity remain unchanged. However, the current through the semicircular →→ loop TLP only contributes to the magnetic 23. When v makes an angle q with B , the field  = µ0I  charged particle follows a helical path whose  4r  . → So, the correct choice is (c). axis is parallel to B . → →→ The speed of the particle does not change. 13. F = ( v × B) When the charged particle is moving → →→ perpendicular to the magnetic field, the Lorentz force, F=m q( v × B) changes only magnetic force (known as Lorentz magnetic force) will be maximum. →→ → v not v i.e. it changes the direction of v and not its magnitude. Fm = qv B sin 90° So, (a) is correct. = qv B 24. Fm = Bqv ⇒ 14. When the charged particle is moving parallel B = Fm or antiparallel to the magnetic field, i.e. q = 0° qv or 180°, magnetic force will be zero. [MLT − 2 → →→ ⇒ [B] = [AT]× [LT ] 1] 15. F = ( v × B) = qv B sin q, where q is angle − →→ = [MT– 2 A– 1] between v and B . → → →→ → 16. 1 T = 104G 25. As F m = q( v × B), F is perpendicular to v i.e. or 1 tesla = 104 gauss direction of motion of the charged particle. So, it can’t do work on the particle. Hence, no 17. SI unit of magnetic dipole moment (m): work is done on a charged particle moving in Am2 (ampere metre2) a uniform magnetic field. or J/T (joule/tesla)

ANSWERS   187 26. B acts along the Z-axis. → →→ The particle will move in a helix instead of a 31. F = q(F × B) = qv B sin q circle. Magnitudes of charge of electron and proton The particle will have a speed v cos q, along are equal, hence magnitudes of forces will be equal. → 32. The horizontal component of the earth’s the direction of B and v sin q, along the magnetic field will deflect it towards east. direction which is perpendicular to B (hence 33. A cyclotron cannot accelerate uncharged describes uniform circular motion in the particles like x-rays and g-rays. plane). So, no change of momentum along +Z axis i.e. 34. F= µ0I1I2 l 2pd → along the direction of B . 27. When an electric charge is moving in free ⇒ F ∝ I1I2 ∝ I2 space, both electric and magnetic fields are d r produced, whereas a static charge produces only electric field. A stationary charge cannot Also, F′ ∝  I 2 × 1 produce a magnetic field.  2  2r 28. Force on a conductor of length l, carrying ∝ 8I2r = F8 current I2 and placed at a distance b parallel to another infinitely long conductor carrying 35. We have, B= µ0aI current I1 is given by, 4pr F = µ0 × 2I1I2l = µ0I ( a = 2p, n = 1) 4p b 2r Here, I1 = I2 = I Also, B′ = µ0 In 2× r ∴ Force per unit length (= F/l) µ0 I 2 2 2pb = = µ0I × 2 ( n = 2) r →→ ⇒ B′ = 4 B 29. When v makes an angle q with B , the charged particle follows a helical path whose ⇒ B′ = 4B → 36. At the neutral point, H = 0 axis is parallel to B . \\ v = 1 = 1 MH =0 T 2p I 30. The electric force experienced by a charge q in 38. A cyclotron can’t be used to accelerate → neutrons, which have no charge. an electric field E is given by, → →→ →→ 39. F = q( v × B) F e = q E → →→ → 40. F = q( v × B) F e is independent of whether the charge q Using the definition of cross product of two itself is moving or not. → →→ →→ vectors, F is perpendicular to v and F is So, when v is parallel to E , the charged → → also perpendicular to B . particle experiences a force, q E in the → direction of v . Hence, it follows a straight line path.

188 PHYSICS 41. When a moving charged particle is subjected QP to a perpendicular uniform magnetic field then: I I I1 (i) there is no change in the magnitude RS of its momentum but the direction of momentum continuously changes 48. We have, (ii) However, K.E. of the particle remains F = µ0I1I2l constant 2pd (iii) Also, work done on the charged particle F ∝ I1 I2, other parameters being constant. is zero. F′ ∝ 2I2 × 2I1 42. When a magnet is supended freely under the ⇒ F′ = 4F combined action of two uniform magnetic fields of intensities B1 and B2 acting at 90° to 49. F= µ0I1I2 each other, the magnet comes to rest making 2pd an angle q with the direction of B2, such that B1 = B2 tan q. ∴ F ∝ 1 d B2 It is of the form, y ∝ 1 , and the correct N x q variation is shown by (d). B1 50. Current sensitivity of the galvanometer, S IS = θ = nBA I k This is tangent law. A tangent galvonometer is based on the tangent law. So, IS ∝ n The tangent law applies when there are two ∝ B uniform magnetic fields acting perpendicular to each other. ∝ A → →→ Hence, (d) is correct choice. 43. F = q( v × B) 51. We have, t = nIBA sin a = MB sin a, where M = nIA. Here a is the angle which a normal ∴ Fm = qv B sin 90° drawn on the plane of the coil makes with the = qv B( sin 90° = 1) direction of magnetic field. 44. Apply Fleming’s left hand rule. Given, a = 90° – q → \\ t = MB sin (90° – q) = MB cos t 45. F = B I l sin q So, t ∝ cos q, and the correct choice is (b). → 52. S= IgG For q = 0°, F = BIl sin 0° = 0 I − Ig →→ → Putting the given values we have, 46. We have, t = Pm × B S = Ir I = r nI − (n − 1) → →→ Torque t is perpendicular Pm as well as B . 47. Force of attraction ion the arm SP of loop due to conductor will be stronger than the force of repulsion on arm QR of the loop. Hence the loop will get attracted towards the conductor.

ANSWERS   189 53. We know, R = V −G 57. Magnetic dipole momentum is a vector Ig quantity directed from south to north. Putting the given values we have, 58. The permeability (m) of diamagnetic substances is less than one and their R = nV −G  Ig = V  susceptibility is negative, which does not V G  change with temperature. G 59. The permeability (µ) of paramagnetic substances is greater than one and their = nG – G ( G = r) susceptibility is positive. or R = (n – 1)r 60. At a neutral point (also called null point), 54. In order to increase the range of voltmeter n field due to the earth gets balanced by the times, the value of resistance to be connected magnetic field due to magnetic (s). in series with the galvanometer is, 61. At magnetic poles, the angle of dip (d) is 90°. R = (n – 1) G 62. Initial magnetic moment of each magnet = ml So, (d) is the correct choice. From figure it is clear that N1 and S2 neutralize each other. 55. Given, Effective distance between S1 and N2 = l2 + l2 Ig = I 2 = 2 l S = 20 Ω ∴ Magnetic moment of the combination G = ? We have, = 2 ml G = (I – Ig) × S = 2 p Ig S1 =  I − I  × 20  2  I l Ö2l 2 = 20 Ω 56. Force per unit length on Q due to P, N1 N2 S2 l F1 = µ0 × 10 × 20 towards left 2pr 63. When the magnet is cut into two parts there Force per unit length on Q due to R, will be two magnets, each with its own north and a south pole. The magnetic moment F2 = µ0 × 20 × 30 towards left of each part will be half that of the original 2pr magnet. N et magnetic force on Q due to P and R, 64. When the magnet is cut into two parts, there will be two magnets, each with its own north Fnet = F1 + F2 and a south pole. The pole strength of each part will remain the same as that of the = 2µp0r × (800) towards left in the original magnet. plane of paper 65. The reduction factor (K) is given by, Though we have assumed that the distance between the current-carrying conductors I = K tan θ is same (= r), the result will still hold if the distance is not same. ⇒ I = K, if θ = 45°

190 PHYSICS The reduction factor of a tangent galvanometer 76. Due to large permeability of soft iron, magnetic is equal to the amount of current required to lines of force have a greater tendency to pass produce a deflection of 45° in it. through it. The concentration of magnetic lines of force in soft iron bar increases as 66. Iron, cobalt and steel are ferromagnetic but shown in (b). copper is diamagnetic substance, hence is repelled by the external magnetic field. 77. Steel is preferred for making permanent magnets because of large coercivity. 67. (a), (b) and (c) represent SI units of magnetic 78. Magnetic dipole moment, p = I A = A m2 constant, Km = µ0 i.e. NA–2 or Wb A–1 m–1. 4p 79. Iron is ferromagnetic, hence µr is largest and is of the order of 104. We can use, F = qvB and φm = BA 80. Diamagnetic substances are repelled by the i.e. N = AmT and Wb = Tm2 external magnetic field. The bar aligns itself perpendicular to the field lines. (c) Wb2 J–1m–1 = Wb2 81. Ferromagnetic materials show all the Nm2 properties of paramagnetic substances to a much greater degree.At a certain temperature, = Wb × Wb = Wb ×  T  called curie temperature, ferromagnetics pass Nm2 N over to paramagnetics. = Wb A–1 m–1 → So, (d) is different from others. 82. When a bar magnet of magnetic moment M 68. Magnetic moment = pole strength × magnetic is held at ∠ θ with the direction of a uniform length magnetic field B, a torque (t = MB sin q) acts on the magnet. It tries to align in the direction Since pole strength is halved, magnetic of the magnetic field. So, (b) is correct. moment (p) is also halved. However, in a non-uniform magnetic field, the magnetic needle experiences both, a force 69. F =  µ0  m1m2 and a torque. 4p r2 F′ =  µ0  m1 / 2 × m2 / 2 →→ → 4p (2r)2 83. (a) p = I A , where A is area vector =  µ0  m1m2 × 1 = F/16 = Am2 4p r2 16 →→ ( c) Work done (W) = – p ⋅ B = – pB cos θ 70. Cabin should be made of iron as iron has → large permeability. p = J T– 1 71. Work done gets stored as potential energy. → W = U = −p ⋅ B]θθ12== 180° (d) p = J T–1 0° = N mT–1 = –pB (cos 180° – cos 0°) → = –pB (–1 –1) = –pB(–2) = 2 pB So, (b) is not the unit of p . 72. The strength of the magnet considerably 84. Magnetic field due to a magnetic dipole, decreases on heating. B ∝ 1 ∴ d3 73. Liquids and gases cannot be ferromagnetic. n = – 3 1 74. c = 85. We have, tan d = V =1 H µ0ε0 So, µ0ε0 has the dimensions of 1 . ⇒ tan d = tan 45° velocity ⇒ d = 45°

ANSWERS   191 86. We have, Fm = Bqv sin θ, 92. Time of oscillation of a magnet in the vibration When v = 0 magnetometer does not depend on the length and breadth of the magnet. Fm = 0 So, (b) is correct. 93. At equator, R = H As H = R cos d, cos d = 1 ⇒  d = 0° 87. Fm = Bqv sin θ Also, V = R sin d = R sin 0° = 0 \\ B = Fm θ H qv sin d [B] = [ [MLT − 2 ] ] VR AT][LT − 1 = [MT– 2 A– 1] 88. 1 = ML2 So, the vertical component of the earth’s MB AL2 × (ML− 2 A− 1) magnetic field is zero at the magnetic equator. = T 2 94. Domain theory is the most essential characteristic of ferromagnetism. = T 95. When a magnet is placed with its N-pole 89. We have, T= I towards geographic north, the neutral points MH lie on equatorial line of the magnet. Freq. of oscillation, When a magnet is placed with its N-pole towards geographic south, the neutral points v = 1 lie on axial line of the magnet. ⇒ T µ0 v ∝ 1 96. We have, 4p = 10– 7 (in SI units) I When I becomes 4 times, v will become half. \\ m0 = 4p × 10– 7 SI units 90. Magnetic field intensity (B1) due to a bar 97. Magnetisation (M) of a material is defined magnet at any point on axial line of the magnet is given by, as magnetic moment per unit volume of that material. SI unit of M is Am– 1. B1 =  µ0  2Mr M = p/volume  4p  2 − l2 (r )2 = q × length = q Area × length A For a short magnet, r2 >> l2 ∴ B1 =  µ0  2m = B (given) (i) 98. A static charge produces only electric field.  4p  r3 99. As per Curie’s law, cm ∝ 1 . Also, B2 along the equatorial line at a distance T r is given by,  µ0  M χm χm  4p  r3 B2 = (ii) From (i) and (ii), B2 = B 2 O 1/T    O 91. Neutral points are the points where net field         (a)          (b) T intensity due to the field of the bar magnet and field of the earth is zero.

192 PHYSICS 100. Inside the rod, magnetic field (B) ∝ r. Outside 108. For motion of a charged particle in a magnetic the rod, magnetic field (B) ∝ 1 . field, Bqv = mv2 r r 101. Cu, Au and Hg are diamagnetic substances. Co is a ferromagnetic substance. ⇒ Bq = mv r 102. Pt, Al and Mn are paramagnetic substances. Sb is a diamagnetic substance. ⇒ v ∝ r →→ So, with increase in velocity, radius will increase. 103. F e = q E → →→ 109. Given, Ig = 10% I F m = q( v × B) = I 10 → →→ We know, S = IgG = I × 99 \\ F = F e + F m I − Ig 10 → →→ = q E + q( v × B) → →→ I  1 − 1 10  = q(E + v × B) 104. For isotropic materials, M ∝ H = I × 99 × 10 10 9I or M = cmH ⇒ S = 11 Ω where cm is a dimensionless quantity and is 110. When an electric current is passed through a called magnetic susceptibility. conductor, only magnetic field is associated with the conductor but there is no electric For diamagnetic substances, susceptibility is field, because the conductor is electrically negative. neutral. So, graph between I and M is correctly shown 111. Magnetic effect of current was given by by (a). Oersted. 105. In case of ferromagnetic, cm is positive. 112. Magnetic flux density (B) has Tesla (T) as the Ferromagnetic materials show all properties SI unit. of paramagnetic materials to a much greater degree. Initially, I increases with increase in 113. Magnetic field induction (B) at a point due M and becomes constant after saturation is to long current carrying wire is related with reached. The correct variation is shown by (d). distance r by the relation, 106. In case of paramagnetic materials, the graph B∝ 1 between M and H or I and H is a straight line r passing through the origin. 1 107. S= IgG It is of the form y ∝ x . I − Ig So, the graph is rectangular hyperbola which Ig = 2% of I is correctly shown by (b). = 2I = I 114. B1 = r2 100 50 B2 r1 IG G ⇒ 0.6T = 2r 50 50 B2 r ∴ S = I− I =  1  50  1 − 50  ⇒ B2 = 0.3 T 115. A straight wire carrying current produces or S = G 49 circular magnetic field or magnetic lines of force.

ANSWERS   193 116. SI unit of pole strength is Am, hence ⇒ 2 × 10–7 = 10–7 × 2 I1 I2 dimensions are [M0LT 0 A]. ⇒ I1 I2 = 1 117. Magnetic dipole moment, ⇒ I1 = I2 = 1 A [p] = Am2 124. The resistance of an ideal voltmeter is infinity. = [M0L2 T0A] The voltmeter is always connected across the resistor. 118. Magnetic field due to a long straight current carrying conductor at a distance r from it is 125. If K is figure of merit of the galvanometer, given by, B = µ0I we have, I = 50 K 2pr Ig = 10 K B ∝ I S = 12 W 119. If r is radius of the circle, then Also, L = 2pr S = Ig G/(I – Ig) ⇒ 12 = 10 K × G/(50 K – 10 K) or r = L 2p Area of the circle = pr2 ⇒ G = 12 × (50 K − 10 K) 10 K L2 = p × 4p2 = 12 × 4 W ⇒ A = L2 = 48 W 4p 126. The resistance of an ideal ammeter is zero. So, magnetic moment, p = IA The ammeter is always connected in series in the circuit. = I× L2 4p 127. A current loop in a magnetic field behaves as magnetic dipole, whose magnetic dipole 120. When a charged particle moves in a moment acts along the axis of current loop. magnetic field, it experiences a force which is perpendicular to velocity, due to which 128. The magnetic effect of current was discovered the magnitude of velocity does not change. by Oersted. Hence there will be no change in its kinetic energy. 129. Area of circular orbit of electron, A = pr2 121. The charged particle describes a helical path Current due to motion of electron is given by, of radius = mv sin θ . It moves along the i == et 2=pe r ev Bq 2pr direction of magnetic field with speed v cos q. v 122. In a moving coil galvanometer, the radial ∴ Magnetic moment = iA magnetic field is used to make the scale of the galvanometer linear i.e. I ∝ q. = ev × pr2 = evr 2pr 2 123. Force between two parallel current carrying 130. f = F/l = µ0 × 2IAIB conductors is given by, 4p r F = µ0 × 2I1I2  = µ0IAIB 4p d 2pr F/l =  µ0  × 2 × I1I2 131. For a point inside the pipe, we consider a  4p  closed path for ampere circuital law. The current enclosed is zero, hence B = 0

194 PHYSICS 132. Applying Fleming’s left hand rule we find 140. Energy gained = qV that the force on +Q will be towards west. = eV 133. fB = BA = e × 1V Weber = tesla (metre)2 = 1eV So, weber per metre2 is equal to tesla. Here e is magnitude of charge of proton. 134. In the adjoining loops of spring, the current |e| = 1.6 × 10–19 C being in the same direction, there will be attraction, hence the spring gets compressed. 141. Magnetic moment = nIA 135. Given, Ig = 2 = 1A 142. In case of a short dipole (magnetic moment m), 2000 1000 Baxial =  µ0  2m , also called end on position New range = 10 V  4p  r3 So, R = V − G Bequatorial =  µ0  m , also called broad side on Ig  4p  r3 10 position. / 1000) = (1 − 2000 ∴ Baxial = 2Bequatorial 143. Platinum is a paramagnetic material, as it = (10,000 – 2,000)Ω = 8000 Ω is weakly magnetised in the direction of 136. We know, magnetic field in which it is placed. mv2 144. Neutral point r Bqv = 145. Magnetic permeability, ⇒ Bqr = mv m = B H  m  ⇒ r =  Bq  v SI unit is Hm–1.   So, increase in velocity increases the radius of 146. A voltmeter is a galvanometer of very high the circle traced by the particle. resistance. It is always connected in parallel so that it draws minimum current from the 137. We have, main circuit and the potential difference to be measured is not affected materially. IS = IG = 1× 8 G+S 8+2 147. Weber = tesla × m2 = 0.8 A ∴ Webemr ×etarme pere 138. Magnetic moment is given by, p = IL2 = (tesla × m2 ) × A 4p m 1× (2)2 = = T m A = T  m  (As)  s  4p =  1  Am2 = T (ms–1) C  p  = newton 139. When an electric current is passed through a Note: F = Bvq conductor, only magnetic field gets associated with the conductor but there is no electric 148. Force on a moving charge while moving in a field because the conductor is electrically → → neutral. So, the net charge on the conductor is zero. magnetic field is given by, F = q( v × B ), → where F is perpendicular to v.

ANSWERS   195 →→ As mass of electron (out of the given particles) is least, hence K.E. of electron is maximum. ∴ Work done per second = F ⋅ v 156. Ammeter gets damaged due to excess of = Fv cos 90° current. = 0 157. Force acting per unit length on P due to 149. When the charged particle is moving along currents in Q and R is given by, the axis of the solenoid, angle θ between V and B is 0° or 180°, so sin θ = 0. F = µ0  2 × 30 × 10 − 2 × 20 × 10  4p  3 × 10−2 2 × 10−2    Fm = QVB sin θ = 0 = µ (2 × 104 – 2 × 104) 4p 150. We have, Bqv = mv2 = 0 ⇒ r ⇒ mv2 Bqr = mv 158. We have, r = Bqv r ∝ m , if v and B are fixed. ⇒ r = mv  or r ∝ v q Bq B 151. Due to high resistance of voltmeter connected \\ r1 = 2v / B / 2 = 4 in series, the effective resistance of the circuit r v/B increases, hence the current in the circuit decreases. So, r1 = 4r Hence, both the ammeter and voltmeter will 159. We have, Ig = 30 mA = 3 × 10–2 A not be damaged. G = 20 Ω 152. In a moving coil galvanometer, the turning V = 3 volt effect of the coil (θ) ∝ I. V 153. The cyclotron can be used to accelerate the So, R = Ig −G heavy charged particles like α-particles. 154. Two parallel beams of positions moving in =  3 − 20  Ω the same direction attract each other.  3 × 10−2    155. K.E (K) = p2 i.e. K ∝ 1 = 80 Ω 2m m UNIT 4:  ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS 1. (c) 2. (c) 3. (b) 4. (c) 5. (a) 6. (d) 7. (b) 8. (b) 9. (a) 10. (b) 11. (b) 12. (d) 13. (d) 14. (d) 15. (a) 16. (c) 17. (a) 18. (d) 19. (a) 20. (c) 21. (a) 22. (d) 23. (a) 24. (b) 25. (d) 26. (b) 27. (a) 28. (d) 29. (b) 30. (a) 31. (d) 32. (a) 33. (b) 34. (c) 35. (c) 36. (a) 37. (a) 38. (b) 39. (a) 40. (b) 41. (b) 42. (b) 43. (a) 44. (c) 45. (a) 46. (a) 47. (c) 48. (b) 49. (d) 50. (d) 51. (c) 52. (a) 53. (c) 54. (c) 55. (b) 56. (c) 57. (d) 58. (a) 59. (b) 60. (c) 61. (c) 62. (c) 63. (c) 64. (c) 65. (a) 66. (c) 67. (a) 68. (a) 69. (c) 70. (a) 71. (a) 72. (c)

196 PHYSICS 73. (a) 74. (b) 75. (c) 76. (c) 77. (c) 78. (a) 79. (d) 80. (d) 81. (a) 82. (a) 83. (b) 84. (b) 85. (d) 86. (c) 87. (c) 88. (b) 89. (c) 90. (d) 91. (a) 92. (c) 93. (a) 94. (a) 95. (c) 96. (a) 97. (b) 98. (d) 99. (b) 100. (a) 101. (a) 102. (b) 103. (b) 104. (c) 105. (b) 106. (c) 107. (d) 108. (a) 109. (b) 110. (c) 111. (d) 112. (c) 113. (b) 114. (c) 115. (c) 116. (b) 117. (d) 118. (b) 119. (a) 120. (d) 121. (b) 122. (a) 123. (a) 124. (a) 125. (c) 126. (a) 127. (c) 128. (c) 129. (a) 130. (d) 131. (b) 132. (c) 133. (c) 134. (d) 135. (d) 136. (d) 137. (a) 138. (b) 139. (b) 140. (a) 141. (a) 142. (c) 143. (a) 144. (c) 145. (a) 146. (a) 147. (a) 148. (b) 149. (d) 150. (b) 151. (a) Case Study–1 3. (c) 4. (c) 5. (c) 1. (c) 2. (a) 3. (c) 4. (c) 5. (d) Case Study–2 1. (c) 2. (c) HINTS/SOLUTIONS 1.        f = LI 10. f = LI ⇒ weber = henry × ampere or L = f I ⇒ henry = weber ampere 11. Joule’s heating is independent of the direction of current in the circuit. 2.        eS = −M dI p H = I2Rt dt 12. Since the primary circuit has 2V battery, there 3. Charge developed is proportional to the rate is a constant current flowing through it. of change of magnetic flux. Therefore, the primary circuit has no change in magnetic flux linked with it. Accordingly, 4. Split ring commutator is used in a d.c. no induced e.m.f. is set up in the primary, generator/dynamo. hence no induced e.m.f. in the secondary circuit. The current (Is) in the secondary is 5. Lenz’s law is in accordance with the principle zero. of conservation of energy. Work done in moving the magnet w.r.t. the coil changes into 13. A transformer cannot work on d.c. electrical energy producing induced current. 6. Fm = Bqv So, (d) is correct Since magnitude of charge of electron and 14. Let I = I0 sin wt, where w = ....... proton is same, magnetic force (Fm) will be same. \\ Average value over a complete cycle 7. Fleming’s right-hand rule ∫ ∫ = 0TTIddtt = 0T I0 sTin wt dt ∫0 8. A cell is a source of constant d.c. = IT0  − cows wt T0 9. Greater the number of turns of the coil greater the magnetic flux linked with the coil So, f ∝ N

ANSWERS   197 = −wIT0 [cos wT − cos 0] = 2TI0 − cows wt T/2 = −wIT0 (1 − 1) =0 = 2wIt0 − cos w2T + 1 15. Let I = I0 sin wt or Iavg = 2I0 × 2 wt Ieff or Irms = I0 = 2pI0   w = 2p  2 T  21. Eddy currents do not cause sparking. Iavg = 2pI0 22. The value of L depends on the geometry of the coil and is given by,  ∫=0T /2 I0Ts/2indtwt dt  = µ0 N 2 A = Iavg ∫0 l 2I0  L  p  where l is length of the coil (solenoid), N is  total number of turns of solenoid and A is area of cross-section of the solenoid. \\ Form factor = Irms Iavg So, L ∝ N2 i.e., L becomes four times. I0 23. Using Fleming’s right hand rule, the direction = 2 = =p 3=.14 1.1 → 2I0 22 2.28 of magnetic induction B in the region P is downward into the plane of the paper. p 24. In order to oppose the motion of the rectangular coil ABCD, the direction of 16. In a purely inductive circuit, the current lags the induced current in the coil must be anticlockwise. behind the voltage by one-fourth of a cycle or p/2. 25. E.m.f. would last in the circuit as long as magnetic flux linked with the circuit changes. 17. Power factor, cos f = R Z 26. Restoring force is provided by the inductor as Here, f is angle of lead or lag, R is resistance it acts as a store of energy. and Z is impedance of the AC circuit. 27. Ohm is not connected with the theory of 18. Ieff or Irms = 10 A electromagnetic induction. I = I0 sin wt, where I0 is peak value of current. 28. Magnetic flux, linked with the coil is directly proportional to all the given factor viz. I0 We know, Irms = 2 (a) magnetic field (b) area of cross section (c) number of turns. So, I0 = 2 Irms 29. The negative sign in E = – df shows that the = 1.414 × 10 A dt = 14.14 A induced e.m.f. opposes the changes in 19. In case of a resistor, the current and applied magnetic flux. voltage are in phase with each other 30. We have, E = E0 sin wt = E0 sin q, where q is 20. I = I0 sin wt the angle between the normal to the coil and the magnetic field. T/2 I dt T/2 When q = q°, the normal to the coil makes an ∫ ∫ Iavg = 0 I0 sin wt dt angle of 90° with the magnetic field lines. 0 = T/2 ∫0T/2 I dt So, Emax = E0 sin 90° = E0

198 PHYSICS 31. Here, (a), (b) and (c) are vector quantities. So, p.d across L varies inversely with time Magnetic flux (fB) is a scalar quantity. which is correctly shown in (a). fB = BA 32. Induced current opposes the motion and not 40. emf. = work done charge the induced e.m.f. When a magnet is falling freely along the axis of a complete ring, its MLT −2 × L2 = ML2T −3 A−1  acceleration is less than the acceleration due = to gravity i.e., a < g. However, if the ring is AT × LT −1 incomplete, induced current is zero. 41. The induced emf does not depend upon the So, a = g. composition of the wire. 33. The end P of the solenoid behaves like a north 42. A choke coil is a pure inductor used for pole to oppose the change in magnetic flux. controlling current in an A.C. circuit. 34. Transformers do not make use of eddy I0 = E0 = E0 currents, whereas (a), (b) and (d) are based on XL wL the application of eddy currents. Therefore, for reducing low frequency 35. The induction furnace makes use of eddy alternating current, choke coils with currents. laminated soft iron cores are used, so that L becomes large. 36. The transmission of A.C. over long distances, at extremely high voltages, reduces the energy Hence, self induction helps in the operation losses and also the cost of transmission. of a choke coil. 37. It is the effective e.m.f. which sends a current 43.        I = 2 sin (100 pt + p/4) (I) through the armature coil of resistance R. Comparing it with, I = V − ∈ , where ∈ is back emf. I = I0 sin wt R I0 = 2 38. If I is the current flowing in the circuit at any I0 2 =1A time (t), self induced e.m.f. in the coil = − L dI \\ Irms = 2 = 2 dt This induced emf opposes E (applied) and the 44. Absence of moving parts in a transformer enhances its efficiency. dI net emf (i.e., E – L dt ) in the circuit and sends 45.    f = (2t2 + t + 5) Wb a current I through R. e = df = d (2t2 + t + 5) = 4t + 1 dt dt So, E – L ddIt = RI For  t = 1s, When the current attains it max. value (I0), its e = 4 × 1 + 1 = 5 Volt dI growth stops i.e., when I = I0, dt = 0. The 46. A choke coil behaves like a pure L. So, phase difference between E and I is 90° growth of current is exponential. Average power consumed/cycle I = I0 (1 – e–t/t), = Erms × Irms × 90° where t = L = 0 (Q cos 90° = 0) R Theoretically, the current in the circuit will 47. Q = 1L RC attain its max. value only after infinite time. So, (b) is correct graph. In case of LCR circuit at resonance, 39. If I is current flowing in the circuit at any time 1 wrC (t), self-induced emf in the coil = − L dI WrL = dt

ANSWERS   199 or C = 1 of the loop directed inwards. As per Lenz’s wr2 L law, induced current will be anticlockwise. \\ Q = 1 1 53. When the contact is suddenly broken, self R 1 induced current will flow along the direction of main current. Hence, the bulb will become wr2 L suddenly bright. So, (c) is correct because the intensity of the bulb becomes very high = 1 wr2 L2 = wr L instantly before becoming zero. R R 54. There will not be any appreciable change 48. Impedance of the AC circuit having L and R in the magnetic flux linked with the circuit, in series is given by, hence no induced current, and intensity of the bulb remains the same. Z = R2 + XL2 = R2 + L2w2 55. L = [ML2 T–2 A–2] So, admittance of the circuit = 1 Z R = wchoarkrgdeo×nIe = MAL22TT −2 = 1 R2 + L2w2 or R = [ML2 T–3 A–2] 49. In case of series resonance circuit, the circuit \\       L = [ML2 T −2 A−2 ] = [T ] R [ML2 T −3 A−2 ] admits maximum current for which XL = XC, 1 which gives f0 = 2p LC i.e., frequency of A.C. T hus, L has dimensions of time. R fed to the circuit becomes equal to the natural frequency of energy oscillations in the circuit. 56. There will be no induced e.m.f. set up in the square loop PQRS, because the metallic square Impedance Z is minimum for, Z = R is neither leaving nor entering the magnetic field. In such a case, the flux linked with it is So, I0 = ER0 , will be maximum not changing. So, the induced current in the square loop is zero. 57. The dimensions of R/L, 1 and 1 have RC LC the dimensions of frequency. 50. For two coupled coils, So, R does not have the dimensions of LC M = K L1 L2 frequency. where K is coefficient of coupling between two coils. When the coils are tightly coupled, 58. The iron core decreases the loss of energy due such that there is no leakage of flux, K = 1 and to flux leakage. M is maximum. 59.  Power factor = True power So, M = L1 L2 Apparent power 51. Apply Fleming’s right hand rule. = cos q = R 52. The conventional current will be along Q to Z P. This will set up magnetic field through the It is a unitless and dimensionless quantity loop which will be perpendicular to the plane whose value lies between 0 and 1.

200 PHYSICS 60. In a pure inductor, f = 90° 71. Resistance is independent of frequency of A.C. Average power (P) dissipated in the A.C. However, XL = Lw = 2pfL circuit, XC = 1= 1 Cw 2pfC P = Eeff Ieff cos f = Eeff Ieff cos 90° Z = R2 + (XL − XC )2 = 0 (Q cos 90° = 0)  1 2  Cw  61. In case of series resonant LCR circuit, XL = XC = R2 + Lw − Z = R2 + (XL − XC )2 = R2 +  2pfL − 1 2 Z will be minimum if XL = XC, such that Z = R.  2pfC    E0 E0 \\ Imax = Zmin = R 72. In a D.C. line, the inductive reactance of the choke i.e., XL = 0. When an iron core is 62. To measure A.C. we use hot wire meters which inserted in the choke, L changes but XL does are based on the heating effect of current, not change. Thus, the lamp shines brightly which does not depend on the direction of due to large current through it depending on current. E and R. 63. Hot wire instruments measure r.m.s. voltage. 64.       Vrms = 200 V 73.       XL = Lw = L(2pf) or XL ∝ f \\ V0 = 2 × 200 V It is of the form, y = mx, and represents a I0 2 straight line passing through the origin. 65.        Irms = 74.        XC = 1 wC ⇒ 5 = I0 2 = 1 2pfC ⇒ I0 = 5 2 A 1 f 66. 50 Hz means 50 complete cycles in one second. \\ XC ∝ In one cycle, voltage becomes zero twice. It is of the form, y ∝ 1 , and represents 67. A rectifier converts AC to DC. x 68. Z = R2 + (XL − XC )2 rectangular hyperbola shown in (b). For XL = XC, Z = R i.e., the circuit becomes 75. We have, I = I0 (1 – e–t/t), where t = L . resistive. R The voltage and current for a resistive circuit L/R is called inductive time constant of the LR are in phase with each other. circuit. So, phase difference is 0°. The growth of current is exponential. For 69. If the core is made of the laminated sheets decay of current, I = I0e–t/t and the current will of soft iron insulated from each other, it will become zero after infinite time. greatly reduce loss of energy due to eddy currents. 76. Mechanical motion does not cause any loss of energy in the transformer. 70. The transformer varies the form of electrical energy e.g., for converting low voltage A.C. 77. When charging of the capacitor builds into high voltage A.C. or vice versa. up, the current in the RC circuit decreases exponentially.

ANSWERS   201 78. Charging of the capacitor in the RC circuits is 87. Impedance, Z = R2 +  Lw − 1 2 when given by,  Cw  q = q0(1 – e–t/t) frequency of A.C. is doubled,  Lw − 1 2 where t = RC,  Cw  called capacitive time constant of the circuit. dq = I0 e–1/t further increases (more than its double value). dt I = 88.         V0 = 400 Volt where I0 = q0 , Irms = Vrms t R which is maximum value of current flowing = 400 = 2 through the circuit. 2 × 0.2 × 1000 In fact, charging and discharging are or Irms = 1.414 A exponential curves. During discharging, the 89. Power factor, current decreases from maximum to zero. cos f = R Z Z = R2 + (XL − XC )2 For a pure resistor, 79.  XC = 1 , which implies XC ∝ 1 . XL = XC 2pfC f ⇒ Z = R 80.         h = Output × 100 \\ cos f = R =1 Input R 90. Average power over a complete cycle is zero. = 100 × 100 I = I0 sin wt 220 × 0.5 T I dt or h = 90% ∫I0 T sin wt=dt 0 81. Slip ring arrangement is used in A.C. ∫Iavg = 0T=dt ∫0 T0 generator. 91. We have,   fr = 1 82. In a D.C. circuit, 2p LC XL = wL = 2pfL = 0 fr’ = 2p 1 \\ Z = R = 10 W L × 4C 83. A motor starter is a variable resistance which = 1 reduces the current as desired. 2 × 2p LC 84. Magnetic flux linked with the coil is NLI i.e., = f f = NLI. 2 85. As resistance of 1 H coil is zero, the entire 92. We know,   f = BA current flows through the coil. Therefore, \\   Weber/m2 = tesla current through 10 W resistance is zero. 86. In an inductance coil, alternating voltage leads 93.       1 T = 104 G the alternating current by 90°. Therefore, (c) is 94. Lenz’s law is consequence of law of correct variation shown. conservation of energy.

202 PHYSICS 95. Pole strength (m) has SI unit as Am. Reqd. 102. Lenz’s law ratio is 102 103. An inductor has a magnetic field around it, in 96. We have,   X= 1 = 1 which energy gets stored. wC 2pfC dI 1 104.    e = − L dt Also, X’ = 2p(2 f) (2C) dI dt |e| = L = 1 × 1 dI 4 2pfC dt Given  = 1 X = 4 \\ |e| = L 97. Given,     L = 2 mH = 2 × 10–3 H 105. We have, L= µ0 N 2 A l dI = 1 mAs–1 = 10–3 As–1 dt Here, N is total number of turns in the solenoid. We have, |e| = L dI dt So, L ∝ N2 = 2 × 10–3 × 10–3 V 106. We have, = 2 µV |e| = dφ 98. Power factor, dt cos f = R = R I = |e| = 1 dφ Z R2 + L2 w2 R R dt 99. The three coils are in parallel. ⇒ I dt = 1 dφ R 1 1 1 1 =3 So, Lp = 2 + 2 + 2 2 ⇒ dq = 1 dφ R ⇒ Lp = 2 H Q dq = 1 φ2 dφ 3 R 0 φ1 ∫ ∫ 100. Induced current opposes the motion and not Q = 1 (φ2 − φ1 ) the induced emf. When a magnet is falling R along the axis of a complete ring, its acc. <g. If the ring is in complete, induced current is So, the total charge induced in a loop depends zero. So, a = g. on resistance and change in magnetic flux 101. When current is increasing, induced current linked with the loop. in the loop is clockwise, so as to oppose the increase in magnetic flux in the loop. How 107. Faraday ever, if current in PQ decreases, induced current in the loop will be anticlockwise so as 108. Given, dI = – 2 As–1 to oppose the decrease in magnetic flux. dt L = 5 H \\ emf developed across the coil = − L dI dt Current increasing Current decreasing = 5 × 2 = 10 V

ANSWERS   203 109. We have, 1 = 1+1 \\ XL = 1 Lω = ω2 LC Lp LL XC / ωC = 2 118. Average power dissipated in LCR circuit, L P = Irms Erms cos f ⇒ Lp = L P ∝ cos f 2 R 110. Energy gets stored in a coil during the growth ∝ Z of current through the coil, which is supplied back during the decaying of current. Hence, ⇒ Power dissipated is directly proportional the self induction can be compared to inertia. to R. V Work 119. XL = XC I charge × Current 111. [R] = = 2pnnL = 1 2πvnC = [ ML2T −2 ] ⇒ Vn = 1 [ A2T ] LC 2π = [ML2 T–3 A–2] 120. Given, V = 4 cos wt = 4 sin (wt + p/2) [L] = [ML2 T–2 A–2] I = 3 sin wt    \\  R  =  ML2T −3 A−2  = [T −1 ] P = Vrms Irms cos f  L   ML2T −2 A−2    4× 3 × cos π 2 22 = [Frequency] = 112. A choke coil limits the current in the circuit = 0 (Q cos p/2 = 0) without much power dissipation. 121. Given, R = 8 W 113. We have, |e| = L dI XL = 6 W dt Here, L is a measure of inertia (i.e., mass) and \\ Impedance = R2 + XL2 I is a measure of drift velocity, hence dI = (8)2 + (6)2 = 64 + 36 = 10 W dt relates to acceleration. 122. AC reverses its direction after every half cycle, hence it can’t be used for carrying out So, |e| corresponds to mass × acceleration i.e., chemical effects. force. 114. Momentum = mv 123. Power factor, cos f = R Z v corresponds to current (I) and m corresponds to inertia i.e., L. If Z = R (i.e., XL = XC), cos f = 1 So, momentum in mechanics corresponds to 124. ∫Iavg = T I0 sin ωt dt =0 IL in electricity. 0 115. Mass is a measure of inertia, which relates to ∫0T dt the coefficient of self inductance (L). 116. The restoring force gets provided by the 125. We have, fr = 2π 1 inductor as it acts as a storehouse of energy. LC 117. XL = Lw So, fr will increase when L decreases. XC = 1 ωC

204 PHYSICS 126. We have, Z = R2 + XL2 132. In a purely inductive AC circuit, current lays behind voltage by 90° (or leads by – 90°). = (4)2 + (3)2 = 16 + 9 = 5 W 133. eS = − M dIP dt 127. Irms = 5A [eS ] work done × time charge × current We know, [M] =  dIP  =  dt  I0 Irms = 2 [ML2T −2 ]× [T ] [ A2T ] = ⇒ I0 = 2 Irms = 2 × 5A = 5 2 A = [ML2T–2 A–2] 128. XC = 1 = 1 134. In a pure capacitive A.C. circuit, current leads Cω 2πfC the voltage by p/2. Given, C = 1 F P = Irms Erms cos p/2 = 0 \\ XC = 1 1 = 10–2 W ∴ cos π = 0  50 2  2π × × π 129. We know, fr = 1 135. Impedance, Z = R2 + (XL − XC )2 2π LC At resonance, XL = XC      f  ′r = 1 =f So, Z = R 2π L (2C) 136. There will be 25 pulses present in the half wave rectifier (i.e., other 25 negative half 2 So, C changes to 2G. cycles get eliminated). 130. Efficiency, 137. Power factor, h = Output cos f = R Input Z ES IS At resonance, Z = R EP I P = , with 30% power loss \\ cos f = 1 \\ IP = ES × 1 138. The component of alternating current which IS EP η does not contribute to any power loss is defined as the wattless current. Such a current = 110 = 5 leads the alternating emf by a phase angle of 220 × (70 / 100) 7 90°. 131. NP = 1 139. In a choke coil XL >> R, hence the A.C. circuit NS 10 becomes almost pure inductive. This ensures negligible power dissipation. We know, 140. Given VL = VC = 100 Volt IP NS V = 200 Volt IS NP = We have, V = VR2 + (VL − VC )2 ⇒ IP = 10 ⇒ 200 = VR2 + (100 − 100)2 = VR IS 1 \\ VR = 200 V

ANSWERS   205 141. XC = 1 = 1 XL = Lw = 2pnL ωC 2πfC = 2p × 100 × 25 × 10−3 = 5W π 1 ⇒ XC = ∝ f We have, It is of the form, y ∝ 1 , which is rectangular tan f = XL = 5Ω = 1 = tan 45° hyperbola. x R 5Ω ⇒ f = 45° 142. Frequency of A.C. remains unchanged in a 151. Max. power gets imparted at resonance, transformer. 1 143. In a step down transformer, voltage decreases Vr = 2π LC and corresponding current increases. 144. A transformer works on the principle of ⇒ C = 1 νr2 mutual induction. 4π2L 145. The best material for the core of a transformer or = 1 × 50 × 10 is soft iron as it involves minimum loss of 4 × (3.14)2 energy. 146. At resonance, C  10–6 F = 1µF XL = XC Case Study–1 The circuit becomes pure resistive A.C. circuit, 5. From Gauss’s theorem, where current and applied voltage are in phase f = q = 2 with each other. So, phase difference is 0°. ε0 8.85 × 10−12 147. A transformer does not change the frequency = 2.26 × 105 Nm2/C leaving the surface of A.C. Case Study–2 148. In case of a step down transformer, voltage decreases hence the corresponding current 1. 1R2 t V increases. R 149. We have, I = |Re| = dφ R = t ;  V = 12 V, dt vm = 2 V = 1.414 × 12 volt dφ = d (4t2 − 4t − 3) = 16.968 volt dt t = 0.5s dt = 8t – 4 = 8 × 1 –4=0 5. Np = 3000, Vp = 2200 V, 2 Vs = 220 V, Ns = ? \\ I = 0 = 0 Ns = Vs  or Ns = Vs Np 20 Np Vp Vp 150. Given, R = 5 W or Ns = 3000 × 220 = 300 2200 L = 0.025 H and v = 100 Hz π

206 PHYSICS UNIT 5:  ELECTROMAGNETIC WAVES 8. (d) 16. (d) 1. (a) 2. (c) 3. (b) 4. (c) 5. (c) 6. (b) 7. (b) 24. (c) 9. (b) 10. (b) 11. (a) 12. (c) 13. (b) 14. (c) 15. (a) 32. (a) 17. (a) 18. (a) 19. (c) 20. (b) 21. (b) 22. (a) 23. (c) 40. (a) 25. (a) 26. (b) 27. (d) 28. (b) 29. (a) 30. (c) 31. (b) 48. (d) 33. (d) 34. (c) 35. (a) 36. (b) 37. (a) 38. (b) 39. (a) 56. (c) 41. (b) 42. (b) 43. (a) 44. (b) 45. (c) 46. (d) 47. (a) 64. (b) 49. (d) 50. (d) 51. (d) 52. (d) 53. (c) 54. (c) 55. (a) 72. (c) 57. (b) 58. (a) 59. (c) 60. (d) 61. (b) 62. (b) 63. (c) 80. (d) 65. (d) 66. (a) 67. (b) 68. (a) 69. (d) 70. (c) 71. (c) 88. (d) 73. (c) 74. (c) 75. (a) 76. (b) 77. (a) 78. (c) 79. (b) 96. (a) 81. (c) 82. (d) 83. (d) 84. (d) 85. (c) 86. (d) 87. (b) 89. (c) 90. (a) 91. (d) 92. (d) 93. (c) 94. (d) 95. (a) 97. (c) 98. (b) 99. (b) 100. (b) 101. (a) 102. (b) Case Study–1 3. (c) 4. (d) 1. (d) 2. (c) 3. (c) 4. (b) Case Study–2 3. (c) 4. (d) 1. (d) 2. (d) 3. (b) 4. (a) Case Study–3 1. (d) 2. (a) Case Study–4 1. (d) 2. (a) HINTS/SOLUTIONS 1. Considering a plane electromagnetic wave 3. F = Bqv propagating along x-axis. the electric and magnetic field in a plane e.m wave can be ⇒ B = F/qv written as, ⇒ [B] = [MLT −2 ] [AT][LT −1 ] E = E0 sin w  t − x    c  ⇒ [B] = [ML0T–2A–1] B = B0 sin w  t − x →→    c  4. B . d l µ0 If C is speed of light, c = E = E0 B B0 We have, B = [MT–2A–1] 2. Maxwell’s modified form of Ampere’s   m0 = [MLT–2 A–2] ∫circuital law is, →→ = m0I + m0e0 dfE →→ = [ML−2 A−1] [L] = [A] dt [MLT −2 A−2 ] B.dl \\  B . d l µ0

ANSWERS   207 →→ The average energy density of magnetic field is, B.dl Thus, µ0 has the dimension of current. uB = 2Bµ20  B0 2 × 1 B02  2  2µ0 4µ0 →→ =   = ...(ii) 7. ∫ B .dS = 0, suggests that the poles of a   From (i) and (ii), uE = uB   =c  magnet can’t be separated. In other words, E=0 1 isolated magnetic poles do not exist. B0 µ0ε0  8. We have, c = 1 19. The accelerated charged particle produces µ0ε0 electromagnetic waves. ⇒ c2 = 1 20. Direction of e.m. wave is perpendicular to the µ0ε0 → ⇒ [m0e0] =  1  = [M0L–2T2]  c2  variation of electric field E as well as to the → variation of magnetic field B . 9. ∫ →→ = 0. 21. The displacement current is set up in a region where the electric field (= p.d/distance with B.ds time) is varying s 22. (a) Ex, By and (b) Ey, Bx As the direction of e.m. waves is given by 10. ∫ →→ = 0, will have to be modified. →→ B.ds E.B. s ∫ 11. →→ q , suggests that electric field lines 23. Height of transmitting antenna (h) is given by the relation, E.ds = ε0 s do not form closed loops. d = 2hR 12. The displacement current is given by, where d is the radius of the circle on the surface of earth within which the transmitted ID = e0 dfE signal from the transmitting antenna can be dt received and R is radius of the earth. where e0 = absolute permittivity of space, Area covered = pd2 ddftE = rate of change of electric flux. = p(2hR) The conduction current and displacement current are equal at an instant i.e. IC = ID. Like 24. Electromagnetic waves are transverse waves conduction current, the displacement current in which there are sinusoidal variation of is also a source of magnetic field. electric and magnetic fields at right angles to each other as well as at right angles to the 14. Microwaves have frequencies greater than direction of wave propagation. those of TV signals. Microwaves are used in telecommunication much better than radio Ey = E0(wt – kx) waves. They do not bend or spread around corners of any obstacle. ⇒ Ex = Ez = 0 25. w = 2p/T = 2pv, where v is frequency of 15. Electromagnetic waves are transverse and travel with the speed of light in free space. oscillation. 16. Electromagnetic waves do not travel with the 26. k = 2p same speed in all media. l 17. The average energy density of electric field is, 27. Bz = B0 sin   t − x  suggests Bx =0 and w  c  By = 0.  uE = 21 ε0E2 = e0  E0 2 = 1 e0 E02...(i)   4  2

208 PHYSICS 28. The the average density of electric field 38. We know, refractive index (m) = c associated with e.m. wave is, v ue = 1 e0E02 = 1× µ∈ µ∈ 4 µ0 ∈0 1= µ0∈0 29. The average density of magnetic field 40. n = 8.2 × 106 Hz associated with e.m. wave is, l = ? um = B02 c = 3 × 108 ms–1 4µ0 We know, c = nl 30. We have, ue = um     or n = c = 3 × 108 = 36.6 m Total average energy density = ue + um ν 8.2 × 106 = 2ue = 2um = B02 41. E0 = cB0 2µ0 42. The oscillating electric and magnetic field 31. c = vl vectors of electromagnetic wave are oriented along the mutually perpendicular directions     or  l = c = 3 × 108 = 1.5 × 10–2 m and are in phase. v 2 × 1010 43. Maximum rate of energy flow (S) is equal 32. Radio waves travelling directly following to the product of amplitudes of electric and the surface of the earth are known as ground magnetic field vectors. waves. These waves have frequency upto 1500 kHz. S = 100 V/m × 0.265 A/m = 26.5 W/m2 34. w = 2p 44. B0 = E0 T c k = 2p 18 Vm−1 l = 3 × 108 ms−1 ⇒ w = l = c = 6 × 10–8 T k T 47. As the capacitor is connected in the circuit, Also, c = 1 initially the rate of flow of charge to the µ0ε0 capacitor is maximum. After charge gets stored to the maximum, it will repel the \\  wk = c = 1 similar charge further. So, the rate of flow of µ0ε0 charge in the circuit will decrease. 48. By definition, 35. Direction of propagation of the wave is given e0 ddt fE has the dimensions of displacement current. →→ by, E × B 36. If monopoles existed, then the equation, 50. Electromagnetic waves have the same speed (= 3 × 108 m/s) in vacuum. ∫ →→ = 0, would require some B.ds modification. 52. Radio waves of constant amplitude can be produced by an oscillator. 37. g-rays, x-rays and ultraviolet light have frequencies in decreasing order. 55. (a) When an electron is placed in the path of e.m. wave, it will experience force due to

ANSWERS   209 electric field vector but not due to magnetic 75. l for infrared varies from 8 × 10–5 cm to field vector. 3 × 10–3 cm. Fe = eE \\ lm ≈ = 10– 4 cm 76. a-rays are not the part of e.m. spectrum. → →→ 79. Ampere circuital law is inconsistent and also |Fm | = |Bm |e| v | = Be v becomes consistent as defined by Maxwell. → 80. Linear momentum (p) carried by the portion ( |v |= 0) of wave having energy U is given by, = 0 56. We have, c = 1 [ML2T −2 ] µ0ε0 [LT −1] ⇒     µ0e0 = 1 = 1 p = U = = [MLT–1] c2 (ms−1 )2 c SI unit of (µ0e0) will be m–2s2. 81. No. of waves = 4mm 57. Electromagnetic waves do not transport any 4000 Å charge. They transport momentum, energy 10−3 m and information from one place to another.         = 4 × 10−7 m = 2500. 58. Radio waves are reflected by the ionosphere. Energy U Velocity c 59. The penetrating power of X-rays depends 82. Momentum of the wave = = . on its energy which further depends on its frequency. 84. When a wave of energy U is totally reflected 60. Amongst the given e.m. waves, radio waves from the surface, the momentum delivered to have the longest wave length. the surface is 2U . c 62. g-rays have the shortest wavelength among the given waves. 87. In case there is no atmosphere, there would be no green house effect i.e., no reflection 66. An accelerated electron would produce e.m. (bouncing back) of infrared radiations from waves. the atmospheric particles that are emitted by the earth. So, the temp. of the earth would 67. When light travels from one medium to decrease. another, frequency remains same. 68. During the day time, the radiations from 91. Mean wavelength of visible light, the sun reach the earth. At night, the earth’s atmosphere prevents the infrared radiations l = 6000 Å of the earth passing through it and thus helps in keeping the earth’s atmosphere warm. This \\ Frequency, phenomenon is called ‘Green House Effect.’ n = c l 69. Velocity of light can be changed by changing 3 × 108 = 5 × 104 ≈ 1015 Hz. the medium only. = 6000 × 10−10 72. k = 2p , w = 2p 95. Hydrogen atom does not emit X-rays because lT it has single electron. \\  k = 2p 2p/T = T =c 97. b-rays are not an electromagnetic wave. w l l 98. According to Maxwell, a changing electric →→ field is a source of magnetic field. 73. E × B 99. Due to the absence of atmosphere, there would have been no green house effect, 74. Heat radiations travel with the speed of light. hence the temperature of earth would have decreased.

210 PHYSICS 100. Ozone layer absorbs ultraviolet radiations. =  6.6340××101.−634××130−×13108   101. g-rays have max. energy out of the given rays. MeV E = hn = hc  ⇒ E ∝ 1 (Q  1 MeV = 1.6 × 10–13 J) l l So, wavelength of g-rays is least. = 3.1 × 10–14 MeV 102. We have, I = dQ Case Study–3 dt 3. l = 19 m, c = 3 × 108 ms–1, c = 3 × 108 ms–1, n = ? = d (CV ) c = nl  dt or n = c = 3 × 108 λ 19 = C dV dt = 15789.5 × 103 Hz = 15.8 MHz I = 2 × 10–12 × 1012 A 4. Microwaves are used in the analysis of very fine details of atomic and molecular structure. ⇒ I = 2 A Case Study–4 Case Study–1 2. The windows of the car allow the sunlight to enter into the car. This light is then partially 3. Here l = 40 m,  c = 3 × 108 m/s,  h = 6.63 × 10–34 converted into heat. However, these same Js,  E = ? windows do not allow the heat inside the car to pass through as easily as light. E = hn = hc =  6.63 × 10−34 × 3 × 108  J λ  40  UNIT 6:  OPTICS 1. (a) 2. (a) 3. (d) 4. (d) 5. (c) 6. (a) 7. (c) 8. (c) 9. (d) 10. (a) 11. (b) 12. (a) 13. (b) 14. (d) 15. (d) 16. (b) 17. (a) 18. (d) 19. (c) 20. (a) 21. (c) 22. (a) 23. (a) 24. (c) 25. (c) 26. (c) 27. (c) 28. (a) 29. (d) 30. (b) 31. (b) 32. (a) 33. (a) 34. (d) 35. (c) 36. (c) 37. (d) 38. (b) 39. (a) 40. (c) 41. (c) 42. (a) 43. (a) 44. (a) 45. (a) 46. (b) 47. (b) 48. (d) 49. (a) 50. (b) 51. (b) 52. (a) 53. (b) 54. (a) 55. (c) 56. (c) 57. (a) 58. (c) 59. (d) 60. (b) 61. (d) 62. (c) 63. (d) 64. (a) 65. (a) 66. (a) 67. (b) 68. (c) 69. (a) 70. (c) 71. (c) 72. (c) 73. (a) 74. (a) 75. (a) 76. (a) 77. (a) 78. (c) 79. (d) 80. (a) 81. (b) 82. (d) 83. (a) 84. (a) 85. (c) 86. (b) 87. (c) 88. (a) 89. (c) 90. (a) 91. (d) 92. (a) 93. (b) 94. (a) 95. (d) 96. (a) 97. (b) 98. (b) 99. (b) 100. (c) 101. (d) 102. (c) 103. (c) 104. (d) 105. (b) 106. (a) 107. (c) 108. (d) 109. (a) 110. (b) 111. (b) 112. (a) 113. (b) 114. (c) 115. (d) 116. (d) 117. (c) 118. (d) 119. (b) 120. (a) 121. (d) 122. (a) 123. (c) 124. (b) 125. (d) 126. (b) 127. (d) 128. (a) 129. (b) 130. (b) 131. (c) 132. (d) 133. (a) 134. (a) 135. (b) 136. (b) 137. (b) 138. (b) 139. (c) 140. (b) 141. (b) 142. (d) 143. (d) 144. (a)

ANSWERS   211 145. (b) 146. (a) 147. (a) 148. (d) 149. (a) 150. (c) 151. (d) 152. (d) 153. (a) 154. (b) 155. (d) 156. (d) 157. (a) 158. (c) 159. (a) 160. (a) 161. (d) 162. (d) 163. (c) 164. (d) 165. (b) 166. (d) 167. (c) 168. (c) 169. (b) 170. (a) 171. (b) 172. (b) 173. (a) 174. (c) 175. (c) 176. (b) 177. (a) 178. (d) 179. (d) 180. (d) 181. (d) 182. (d) 183. (a) 184. (b) 185. (b) 186. (b) 187. (c) 188. (c) 189. (c) 190. (a) 191. (d) 192. (d) 193. (b) 194. (b) 195. (d) 196. (c) 197. (c) 198. (c) 199. (b) 200. (c) 201. (d) 202. (c) 203. (c) 204. (c) 205. (b) 206. (d) 207. (c) 208. (c) 209. (a) 210. (c) 211. (b) 212. (d) 213. (d) 214. (c) 215. (d) 216. (b) 217. (c) 218. (d) 219. (a) 220. (b) 221. (d) 222. (d) 223. (c) 224. (d) 225. (c) 226. (b) 227. (d) 228. (c) 229. (b) 230. (d) 231. (c) 232. (a) 233. (b) 234. (c) 235. (d) 236. (c) 237. (c) 238. (d) 239. (a) 240. (a) 241. (a) 242. (a) 243. (d) 244. (b) 245. (b) 246. (d) 247. (d) 248. (d) 249. (a) 250. (d) 251. (a) 252. (d) 253. (b) 254. (d) 255. (c) 256. (a) 257. (b) 258. (d) 259. (d) 260. (b) 261. (a) 262. (b) 263. (d) 264. (c) 265. (b) 266. (a) 267. (a) 268. (c) 269. (d) 270. (a) 271. (c) 272. (d) 273. (d) 274. (a) 275. (c) 276. (c) 277. (a) 278. (d) 279. (a) 280. (a) 281. (d) 282. (d) 3. (d) 4. (b) Case Study–1 3. (d) 4. (d) 1. (c) 2. (c) 3. (d) 4. (c) Case Study–2 3. (d) 4. (c) 1. (b) 2. (a) 3. (c) 4. (d) Case Study–3 3. (d) 4. (c) 1. (b) 2. (c) 3. (b) 4. (b) Case Study–4 3. (c) 4. (d) 1. (d) 2. (a) 3. (b) 4. (d) Case Study–5 1. (d) 2. (a) Case Study–6 1. (a) 2. (b) Case Study–7 1. (a) 2. (c) Case Study–8 1. (c) 2. (b) Case Study–9 1. (b) 2. (a)

212 PHYSICS HINTS/SOLUTIONS 1. On reflection at a denser medium, there Since l becomes smaller, fringe width (b) will occurs a phase difference of p or 180°. decrease. So, the diffraction fringes become narrower and crowded. 3. Light does not require a material medium for its propagation. 18. White light comprises mainly seven colours. 4. At same distance, light from a source arrives at 20. As per the given condition, the reflected and the same time in all the directions. Therefore, refracted rays are mutually perpendicular to the phase difference is zero. each other. 5. There is no change of phase during refraction. So, i = ip = 60° (Given) However, in case of reflection at a denser medium, there is a phase change of p. So, the But µ = tan ip, as per Brewster’s law phase difference between the reflected and refracted waves is 180°. \\ µ = tan 60° 6. A wave front travels parallel to itself and = 3 perpendicular to the rays. = 1.732 8. µ = tan q, where q is polarizing angle. This 21. Phase difference 2p corresponds to path relation is called Brewster’s law. difference l. 9. The plane in which vibrations of polarized Since on reflection from a denser medium light are confined is called plane of vibration. A plane perpendicular to the plane of phase difference of p is introduced, therefore, vibration is called plane of polarization. the corresponding path difference of l is 10. A parallel beam of light gives rise to a plane 2 wavefront. introduced. 11. According to Brewster’s law, when unpolarised light is incident at polarizing 22. Diffraction is the phenomenon of bending angle on an interface separating a rarer of light around corners of an obstacle or medium from a denser medium, then the light aperture in the path of light. On account of reflected in the rarer medium gets completely this bending, light is able to penetrate into the polarized. geometrical shadow of an obstacle. 12. A point source of light generates a plane 23. The penetration of light into the region of wavefront at a very large distance from the geometrical shadow refers to diffraction. source. 24. aa21 = 3 13. Velocity of light in vacuum is maximum and 5 is an absolute constant given by Imax ∝ (a1 + a2)2 c = 3 × 108 m/s Imin ∝( a1 – a2)2 14. When one of the slits is closed (say 1 covered \\ Imax =  a1 + a2 2 with black opaque paper), no interference Imin  a1 − a2  occurs and we obtain uniform illumination.   15. No constant phase difference between the  a1 + 1   3 + 1 2 lights coming from the two slits.  a2    =  a1   5 16. We know, lb < lr  a2  =  3  Also, the fringe width is proportional to   − 1 5 − 1  wavelength.  or Imax = 16 Imin 1

ANSWERS   213 25. Velocity of all colours of light in vacuum is =  2211 +− 11 2 = 91 same. In any other transparent medium, different colours travel with different velocities. v = c Imin 1 µ Imax 9 \\ = As µv > µy > µg > µr, therefore, red colour travels fastest and violet travels slowest 31. In the interference pattern, the bright fringes through glass. are yellow and the dark fringes are black. 26.    l = 10 m 33. When Young’s double slit experiment is repeated in water instead of air, then l would c = 3 × 108 ms–1 decrease, hence b decreases. In simpler words, fringes would shrink or become narrower. n = c l 34. It must be noted that there is no loss of 3 × 108 = 10 energy due to interference. However, there is a redistribution of light energy = 3 × 107 Hz due to superposition of two waves during 28. For constructive interference, path difference interference. l 2 xd = nl, where n = 0, 1, 2, 3..... for first, 2nd, 35. For minima, path difference = (2n + 1) D So, phase difference of minima = (2n + 1)p 3rd, ..... bright fringes. 37. Velocity of light in vacuum is absolute constant. Here, n = 0 refers to bright centre. For destructive interference, path difference 38. Coherence is a measure of capability of producing interference by waves. Coherent xDd = (2n + 1) l , where 1, 2, 3, ..... for first, 2nd, sources are an essential requirement for 2 interference of light. 3rd, ..... dark fringes. 39. Any transparent material would become invisible when its refractive index is equal 29. Coherent sources are those which emit to the refractive index of the surrounding continuous light waves of same amplitude, medium i.e., vacuum. in case of vacuum,µ = 1. same wavelength/frequency in the same phase or are having a constant phase difference. Two 40. If x is the real path length, the optical path independent sources can never be coherent. length (in a medium of refractive index µ) = µx. 30. Intensity ∝ (amplitude)2 41. t= S c a12 1 \\ a22 = 4 3 × 10−2  = 3 × 108 = 10–10s or a1 = 1 42. Given,   aµw = 1.3 a2 2 aµg = 1.5 Also, Imax =  a1 + a2 2 wµg = aµg = 1.5 Imin  a1 − a2  a µw 1.3    a1 + 1 2 43. The speed of light will decrease in water.  a2   a1  =  a2  cair 3 × 1010 = 2 × 1010 cm s–1.   44. cmed = µ = 1.5  − 1 

214 PHYSICS 46. 1 Å = 10–10 m = 10–13 km For n = 1,2..., the missing wavelengths are 47. If the sodium light in Young’s double slit given by, b2 , b2 ,... experiment is replaced by a red light, l would d 3d increase, hence fringe width would increase. 49. We have, 58. d1 = 7l1 D d Imin = I1 + I2 + 2 I1I2 cos f d2 = 7λ2 D d = I + 4I + 2 I × 4I cos 180° d1 λ1 = 5I – 4I = I (cos 180° = – 1) \\ d2 = λ2 Imax = I1 + I2 + 2 I1 I2 cos f 59. When one of the slits is covered with red and other with green filter, no interference pattern = I + 4I + 2 I × 4I cos 0° is obtained. = 9I (cos 0° = 1) cair νλa cmedium νλg \\ Imin = I = 1 60. m = = Imax 9I 9 50. Order of film thickness ≈ 10–6 m = 1 µm ⇒ lg = λa = 7200 Å = 4800 Å µ 1.5 lD 51. b = d ca cd 61. m= So, b ∝ 1 ca 3 × 1010 d µ 2 ⇒ cd = = 53. Thickness   (t) = ct = 3 × 108 × 10–10 = 1.5 × 1010 cm/s = 3 × 10–2 m 63. According to Brewster’s law, m = tan ip = 3 cm ⇒ ip = tan– 1 (1.54) = 57° (from tables) 55. l corresponds to phase difference of 2p. \\  Phase difference 5p corresponds to path As is clear from the figure, difference of 5l . i = 90° – 30° 2 = 57° = ip 57. Path difference at the point in front of one of So, the reflected light along OB is plane the slits, Dx = D2 + d2 − D ∼ d2 polarized. When the Nicol prism is rotated, 2D intensity gradually reduces to zero and then it increases again. In case of missing wavelengths, λD (2n − 1) l =d2 64. b = d 2 2D Dx = 1 x In this case, It is of the form, y ∝ and represents a D = d and d = b rectangular hyperbola as shown in (a). \\  (2n – 1) l = b2 65. In case of interference, the resultant intensity 2 2 varies from 0 to 4 I0 (i.e. 4 times the intensity due to an individual wave) and the average or l = b2 1) value of intensity (Iavg) is 2 I0. d(2n −

ANSWERS   215 Intensity 71. 4Io Iav = 2Io –f –4p –2p O 2p 4p +f 66. Consider E1 and E2 as electric vectors starting from slits A and B, such that E1 = E0 sin wt Angle subtended at the fish = q + q = 2q Beyond this value of q, light suffers total E2 = E0 sin (wt + f), internal reflection. where f is phase difference between the two 72. We know, R2 = a2 + b2 + 2ab cos f interfering waves. \\ Imax = I1 + I2 + 2 I1 I2 cos 0° So, E = E0 sin wt + E0 sin (wt + f) = 2E0 cos φ sin  ωt + φ  = I1 + I2 + 2 I1 I2  2  2 73. Given, Amplitude of the resultant wave is 2E0 m = 1 = ca =2 cos f/2. Since intensity at a point (say P) is sin C cm proportional to the square of the resultant electric vector (E), ⇒ sin C = 1 = 1 µ 2 I ∝ 4E20 cos2 (f/2) Also, I0 ∝ E20 ⇒ sin C = sin 30° \\ I = 4I0 cos2 (f/2) For maximum, ⇒ C = 30° f = 0, 2p, 4p, ... 74. 1 = sin i = sin r = sin r ( r′ = 90° – r) µ sin r′ sin r′ cos r So, I = 4I0 = maximum \\ 1 = tan r = sin C For minimum, f = 0, 3p, 5p, ... µ So, I = 0 = minimum or C = sin– 1 (tan r) 67. The diffraction pattern due to a single 75. According to Cauchy’s formula, slit consists of central bright band having alternate dark and bright bands of decreasing m = A + B + ... intensity on both sides. λ2 69. m = tan (ip) Therefore, as l decreases, m increases, hence (a) is correct. ⇒ ip = tan– 1 (m) 1−1 1 = tan– 1 (1.33) 76. vµ = f 70. Relative speed of the image 1 1 + 1 v µ f distance or = time w.r.t. the man = It is of the form, y = mx + c, and represents a straight line as shown in (a). = (3 + 3) = 6 ms– 1 1s 77. We have, ca 3 × 108 m = cg = Cg

216 PHYSICS \\ cg = 3 × 108 = 2 × 108 m/s = 1 = 1 1.5 5000 Å 5000 × 10− 10 Required time = 2 = 10– 8 s or ν = 2 × 106 m– 1 2 × 108 84. sin A + δm 78. P = P1 + P2 or 2 m= sin A = + 6D – 4D = + 2D 2 \\ Focal length = 1 = 0.5 m sin A 2 sin A cos A 2 sin A 22 m = = 1 1 sin A 79. m= sin C = sin 60° 22 or m = 1 or m = 2 cos A 3/2 2 \\ m = 2 \\ cos A = 1.5 = 3 cos 40° 3 2 2 4 80. Angle of deviation is minimum (dm) for one or A = 80° angle of incidence (im). Near im, d varies 85. dm = 2i – A slowly with i. = 2 × 60° – A d But A = 60° \\ dm = 120° – 60° = 60° dm 86. In case of normal adjustment, i M = f0 = 20 im fe 81. Refractive index (n) of the material of the prism,  A + δm  ⇒ fe = f0  2  20 sin 82. We have, 60 n =  A  = 20 cm = 3 cm 2 Since sin 87. Given, f = – 50 cm 1 = (µ − 1)  1 − 1  Power (P) = 100 f  R1 R2  f   100 R1 = R2 = − 50 1 = 0 = – 2D f 88. Length of the telescope tube in normal ⇒ f = infinity adjustment (L) = f0 + fe = 100 cm So, a plane glass sheet has zero power or |M| = f0 = 24 infinite focal length. ⇒ fe 83. Wave number (ν) = 1 f0 = 24 fe wavelength

ANSWERS   217 \\ 24fe + fe = 100 cm 93. We know, 1 = (µ − 1)  1 − 1  f  R1 R2  ⇒ fe = 4 cm   Also, f0 = 96 cm 1 (1.5 − 1)  1 − 1  20 ∞ R 89. We have, \\ = m = ca = vλa which gives, R = – 10 cm ⇒ cg vλ g Also, refraction from rarer to denser medium λa = m leads to, λg – µ1 + µ2 = µ2 − µ1 , where u = ∞, v = f u v R 1 100 90. P = f (in m) = 50 = + 2D \\ 0+ 1.5 = 1.5 − 1 = 1 f 10 20 91. The focal length (f ′) of each part of a double convex lens cut by a vertical plane into two or f = 30 cm equal parts is twice the focal length (f) of the given lens. 94. Angular dispersion, dv – dr = (mv – mR)A = (1.54 – 1.52) 10° Y = 0.02 × 10° = 0.2° 95. PC = P1 + P2 = + 6D – 4D = + 2D O Also, focal length of the combination, fC = 1 = 100 = + 50 cm PC 2 Y¢ 96. For a prism in the minimum deviation, angle f ′ = 2f of incidence is equal to angle of emergence, i1 = i2. However, the intensity of light transmitted by each part is the same as that in the case of a You must remember that for prism in given lens. minimum deviation position, i1 = i2; r1 = r2; but i1 ≠ r1. Also, 1 = 1 + 1 = 1 fc 2f 2f f A N2 ⇒ fc = 20 cm N1 d 92. The focal length (f ′) of each part of a double i1 i2 convex lens cut by a horizontal plane into two r2 equal parts is equal to the focal length (f) of r1 the given lens. O BC f ′ = f 97. AB = AC, ∠A = ∠C = 45° At each reflection, ∠i = 45° = C, critical angle However, intensity of each part is reduced to half. \\ mmin = 1 = 1 = 2 sin C sin 45° Focal length of the combination (inverted prisms) is zero. X¢ X 98. Outer space looks black to an astronaut O because there is nothing to scatter light. 99. The final image will be inverted with respect to the object and the final image will lie between C1 and C2.

218 PHYSICS 100. The sun appears elliptical when it sets 112. We know, and rises. This happens on account of the phenomenon of refraction of light. 1 = (µ − 1)  1 − 1  f  R1 R2    101. Refractive index (m) of a medium depends on the nature of medium, temp. (T) of medium  1 1   ∞ R   µ ∝ 1  , density (r) of medium (m ∝ r) and = (1.5 − 1) −  T  100 wavelength (l) of incident light  µ ∝ 1  . = – 0.5 × 50 D  λ2  = – 0.1 D The variation of m with l is given by, 113. Given, u = – 50 cm m v = –30 cm f = ? We have, 1 = 1 − 1 f v u Ol = – 1 + 1 102. According to Cauchy’s formula, 30 50 m = A + B + C , where A, B, C are arbitrary 5+3 1 λ2 λ4 100 75 = – =– constants. 106. Focal length in water (fw) So, P = – 100 = – 1.33 D = 4 × Focal length in air (fa) 75 or fw = 4fa 114. Radio waves will undergo maximum = (4 × 8) cm diffraction because they have the largest wavelength. = 32 cm 115. Actually there is no significant difference 107. Huygen’s wave theory fails to explain between an interference pattern and a polarization of light. diffraction pattern. In case of interference there is superposition of only a few waves, say 108. b = λD two or three waves, but in case of diffraction d very large number of waves superpose. Since b ∝ l, the fringe width is least for blue, 116. The resolving power of eye for a healthy out of the given colours. sin A + Dm A person is 1 minute  = 1  . The pupil of 2 2  60°  111. m = sin A = cot human eye is 2 mm in normal sight. Two 2 illuminated pin holes lying close can appear \\ sin A + Dm = cos A as separate if they subtend an angle (Dq) of 2 2 abut 3 × 10– 4 rad (about one minute of arc). The reciprocal of ‘Dq’ is the resolving power = sin  90° − A of the eye.  2  118. n = 4 × 1014 Hz or A + Dm = 90° − A l = 5000 Å 2 2 cm = nl or Dm = 180° – 2A = 4 × 1014 × 5 × 107

ANSWERS   219 = 2 × 108 ms– 1 1.5 = 1 − 1 (1.5 − 1) 9 µl 5 10 ca = 3 × 108 ms– 1 ⇒ = \\ m = ca = 3 × 108 = 1.5 ⇒ ml = 1.5 × 10 = 5 cm 2 × 108 9 3 119. Red colour has largest wavelength, hence 129. We have, 1 =  n1 − 1  1 − 1  f  n2   R1 R2  minimum scattering. Intensity of scattered    light, I ∝ 1 . When n2 = n1, 1 =0 λ4 f This is called Rayleigh’s law of scattering. ⇒ f = ∞ So, the convex lens does not act as a lens. 120. We have, sin  A + δm  2 m = 130. We have,  1 =  n1 − 1  1 − 1   A  f  n2  R1 R2  2    sin  60° + 30°  = n1 − n2  1 − 1   2  n2  R1 R2  sin   = sin 30° = sin 45° ⇒ f ∝ 1 sin 30° n1 − n2 = 1 ×2 131. Given, P1 = 1 21 f1 ⇒ m = 2 P2 = – 1 f2 So, m = 1.41 or – P2 = 1 121. In case of a convex lens, the minimum distance f2 between the object (O) and real image (I) is four times the focal length of the lens. We have, P1 > P2 or  1 > 1 , for a converging combination. f1 f2 123. We have, 1 = (µ − 1)  1 − 1  1 f  R1 R2  f   132. Given, P1 = = P (say) = (1.5 − 1)  1 − 1  P2 = – 1  ∞ (− R)  f   1 = 0.5 × 1 = 1 or – P2 = 1 =–P ⇒ f R 2R f f = 2R So, power of the combination is P – P = 0, hence focal length is infinity.  µg − 1   124. Given, Pa =  µa = +5 =–5 133. A glass slab can be supposed to be made of Pl − 100 two prisms which are joined in opposition.  µg − 1 100 So, the net dispersion produced is zero i.e.   dispersive power is zero.  µl ⇒ − 5  µg −  =  µg  134. We know, m = 1 , where C is critical angle  µl 1  µa − 1 sin C ⇒ sin C = 1 µ

220 PHYSICS A 144. We have, d = (m – 1)A 45° = ( 2 − 1)30° BC = (1.414 – 1)30°  12° C for blue and green are less than 45° while 145. Magnifying power of a microscope decreases, for red colour, it is more. The red ray gets if length (which equals v0 + ue) increases. refracted through the face AC while blue ray and green rays are totally internally reflected 146. Distance travelled = v × t towards BC. They fall normally on BC and will both emerge. So, only the red colour gets = µc × t separated from blue and green colours. = 3 ×1.1508 × 10–9 135. E = hv = hc = 0.2 m = 20 cm λ 147. The splitting of white light is due to refraction ⇒ E ∝ 1 of light on passing through a prism. λ However, the phenomenon is called It is of the form y ∝ 1 , which is equation of dispersion. x 148. Resolving power (R.P) of a telescope depends rectangular hyperbola, shown in (b) on diameter (D) of the objective lens and wavelength (l) of light used. 136. Frannhoffer spectrum is a line absorption spectrum. 137. The correct deviation and dispersion are R.P. = 1 = D shown in (b) only. dθ 1.22l 138. In case of minimum deviation position, the where dq is angular separation of two stars refracted ray inside the prism is parallel to base of prism, which is correctly shown in (b). which are just resolved in the telescope. 139. The colour of a star is an indication of 149. A ray of light incident normally on a plane temperature of star. mirror, will retrace its path, i.e. angle of reflection will be 0°. 140. Optical fibres are based on the phenomenon ∠i = ∠r = 0° of total internal reflection. 150. In case the ray suffers minimum deviation, it 143. We know, becomes parallel to base of prism A. Since B and C are of identical shape and of the same sin A + dm  material, the ray continues to be parallel to   the base of the prisms B and C. Therefore, the m = sin 2 deviation remains the same as before. A 151. Since the surfaces of all the media are parallel 2 and PQ || ST, we have, sin A + (180° − 2A) m1 = m4 152. Given, i1 = 60° = 2 sin A A = 30° 2  90° − A  d = 30°  2  sin cos A A i2 = ? 2 2 2 We know, i1 + i2 = A + d = sin A = = cot \\ i2 = A + d – i1 sin A 22 = 30° + 30° – 60° = 0°

ANSWERS   221 So, the angle of emergent ray with the second 163. Frequency of light remains unchanged. face of the prism = 90° – 0° = 90°. 165. We have, a sin q = n l = l, for n = 1 153. Angle of refraction at the 2nd face, i2 = 0° 154. We have, r1 + r2 = A ll ⇒ sin q = a = d , as a = d \\ r1 = A – r2 = 30° – 0° when q is small, sin q  q = l d = 30° Because principal maximum is on either side 155. We have, of the centre, therefore, angular width = ±l d sin i1 m = sin r2 169. For path difference l, Imax = 4I = K …(i) = sin 60° lp sin 30° For path difference 4 , phase difference φ = 2 . \\ Resultant intensity (I) is given by, = 3×2 = 3 I = I1 + I2 2 I1I2 cos φ 21 ⇒ m = 1.732 = I + I + 2I cos p 2 156. Sun’s spectrum is continuous with absorption = 2I   cos p = 0  lines.  2  157. A converging lens collects (i.e., converges) = K  (from (i)) rays. 2 158. Given, m = f 170. Energy is conserved during interference. f +u 171. Since water sticks to glass, the shape of water So, – n = f f drop is as shown in (b). +u 173. On heating red glass to high temperature, it ⇒ f = – nf – nu will glow with green light. ⇒ nu = –(n + 1)f 175. Magnifying power of Astronomical telescope, ⇒ – u =  n + 1  f |M| = f0  n  fe 160. We know, b = lD 176. We have, M = m0 × me d vo ×  1 + d  ⇒ b ∝ l = −u0  fe    When l decreases, b decreases. ⇒ 100 = 5 × m So, the interference fringes will becomes narrower. ⇒ m = 20 177. The prism made of rock salt is employed to 161. Path difference, S1P – S2P = 11 l , which is 2 observe infrared spectrum of light. 178. ∠i = ∠0° = ∠r l Here, i and r respectively denote angles of odd integral multiple of 2 . incidence and reflection respectively. 162. We know, c = 1 179. In case of a convex lens, the distance between µ0 ∈0 O and I (real) is 4f.

222 PHYSICS 180. The sky appears black because there is 192. Green and red colours are complimentary nothing to scatter light. colours, so the interference pattern will be invisible. 181. We know, d = (m – 1)A 193. Angular separation q ∝ 1 , where d is Out of the given colour, m is maximum for d blue colour, hence d is maximum. separation between the slits. 184. For minimum deviation, QS is horizontal, i.e. parallel to the base of the prism. So, q decreases as d increases 185. Velocity of light changes because of change in 194. As ll > la wavelength. \\ bl = llD would decrease 186. Given, I1 = I2 = I = a2 (say, amplitude is ‘a’) d Imin = (a – a)2 = 0 Imax = (a + a)2 = 4a2 So, the fringe pattern would shrink. = 4I 195. Colours are seen in soap bubbles because of thin film interference. 187. Path difference (say x) 196. We know, b = lD d = (SS2 + S2O) – (SS1 + S1O) i.e., b ∝ 1 If x = nl, the central fringe at O will be bright. d If x = (2n + 1) l , the central fringe at O will be So, b is too small to be detected. 2 dark. 197. We know, b = lD d 188. E = hv = hc =E∝ 1 When two coherent sources are infinity close l l (i.e. d = 0), b = ∞. It is of the form, y ∝ 1 , which is rectangular In such a case, a single fringe would occupy x the entire screen and the interference bands would not be seen. hyperbola, as shown in (c). 189. We know, m = c or v = c . Out of the given 198. Diffraction of waves is observed only when v µ the size of the obstacle is comparable to the wavelength of the waves. options, m is minimum for vacuum. 190. Fringe with, b’ = b = 0.8 mm 199. The separation of bright bands depends upon µ 4/3 the ratio of wavelength (l) to the width (a) of the slit i.e. l/a. = 0.6 mm lD 191. We know, b’ = l’D’ 202. We have, b = d d’ l2D 3l × 3D b2 l2 = d/3 \\ b1 = 2d = l1 l1D = 27 lD d d ⇒ b2 = l2b1 = 7500 Å × 0.8 mm 9 2l1 2 × 6000 Å Number of fringes = 5 b = 0.5 mm = 9×d 203. We know, m = tan ip, where ip is polarizing 5 × 27lD angle and is also called the Brewster’s angle. For glass ip is about 57.5° and for water it is = d about 37°. 15lD

ANSWERS   223 204. Intensities due to two slits are: lD d (i) proportional to their widths 210. We know, b = (ii) proportional to the squares of the ⇒ b ∝ D amplitudes of the waves that are coming out of these slits It is of the form y ∝ x. So, I1 = a12 = 1 211. b= lD I2 a22 16 d a1 1 lD lD a2 4 2 4d ⇒ = b′ = 2d = Also, Imax (a1 + a2 )2  a1 2 ⇒ b’ = b Imin (a1 − a2 )2  a2 + 1 4 = =    a1 − 12 212. We have, b= lD  a2  d  =  1 + 12 =  5 2 = 25 ⇒ b ∝ 1  4 − 12  4  9 d   − 3 2  1  4  It is of the form of y 1 , which is rectangular 4 x 205. We have, I ∝ a2 hyperbola, as shown by (d). + a2 )2 213. y1 = a1 cos wt − a2 )2 Imax = (a1 = a1 sin (wt + 90°) Imin (a1 y2 = b1 sin wt 2  a1 + 1  \\ Phase difference, φ = 90° ⇒ cos 90° = 0  a2 =   We have, R = a12 + b12 + 2a1b1 cos f  a1  a2 − 12 = a12 + b12 + 2a1b1 × 0   = (10 + 1)2 = a12 + b12 (10 − 1)2 ⇒ Resultant intensity, = 121 ≈ 3 I = Rt = (a12 + b12 ) 81 2 214. Energy associated with each photon 206. Given, l = 600 × 10–9 m = 6 × 10–7 m = hn a = 4 m 6.6 × 10−34 × 3 × 108 5000 × 10−10 D q = 1.22  l  = 1.22 × 6 × 10−7 = hc =  a  4 l = 1.83 × 10–7 red = 3.96 × 10–19 J 207. Shift towards the longer wavelength shows Power = Total energy per second that the star is moving away from the observer such as earth. = 100 W = 100 J/s We have, \\ Number of photons per second VS = cDl = 3 × 108 × 0.03 = 9 × 104 m/s = 100 −19 = 2.5 × 1020 ≈ 1020 l 100 3.96 × 10