Title 18. Current Electcity Eas 1 Ans.pdf ✅

1. (a) Number of electrons flowing per second e i t n = 19 19 = 4.8 / 1.6 10 = 310 − 2. (c) ne J vd =  v J d  (current density) A i J1 = and 2 1 2 2 J A i A i J = = = ;  v v v ( d )1 = ( d )2 = 3. (b) Order of drift velocity 10 m /sec 10 cm /sec −4 −2 = = 4. (b) Density of 3 3 Cu = 9  10 kg / m (mass of 1 m3 of Cu)  6.0  1023 atoms has a mass = 63  10–3 kg  Number of electrons per m3 are 3 28 3 23 9 10 8.5 10 63 10 6.0 10   =    = − Now drift velocity neA i v = d = 28 19 3 2 8.5 10 1.6 10 (0.5 10 ) 1.1 − −       =  0.1 10 m /sec −3 =  5. (c) Because 1 H.P. = 746 J/s = 746 watt 6. (a) 2 R  l  l l R R  =  2  % 2 0.1 0.2% 0 =  =  R R 7. (a) =    = =   − − − − 6 2 2 2 8 10 (50 10 ) 50 10 50 10 A l R  8. (a) Resistivity of some material is its intrinsic property and is constant at particular temperature. Resistivity does not depend upon shape. 9. (d) (1 0.00125 ) (1 0.00125 27) 2 1 (1 ) (1 ) 2 1 2 1 t t t +  +   = + + =      t = 854 C  T = 1127 K 10. (c) A l i e R A l R A l R1   2  . . 2  2 2  R1 = R2 11. (d) In case of stretching of wire 2 R  l  If length becomes 3 times so Resistance becomes 9 times i.e. R' = 920 =180 12. (d) Resistivity is the property of the material. It does not depend upon size and shape. 13. (a) Because with rise in temperature resistance of conductor increase, so graph between V and i becomes non linear. 14. (c) Because V-i graph of diode is non-linear. 15. (a)  l V m e vd =  or  l El m e vd = . (Since V = El) vd  E 16. (a) Resistance of conductor depends upon relation as  1 R  . With rise in temperature rms speed of free electron inside the conductor increase, so relaxation time decrease and hence resistance increases 17. (c) ampere t q i 2 2 4 = = = 18. (b) l Al A 3 Volume = = 3  = Now 3 / 3 3 2 l l l A l R    =  =  =   2 9 3  l = = 19. (c) ampere t ne i 10 1 62.5 10 1.6 10 18 19 =    = = − 20. (b) In twisted wire, two halves each of resistance 2 are in parallel, so equivalent resistance will be = 1 2 2 . 21. (d) In stretching of wire 4 1 r R  22. (b) 3 2 (1 10 ) 7 22 1 0.7 −   =  =   A L R 6 2.2 10 −  =  ohm-m. 23. (b) 2 2 1 1 1 r d R A R     [d = diameter of wire] 24. (b) i = q 19 15 = 1.6 10  6.6  10 − 10.56 10 A 1mA 4 =  = − 25. (d) 2 r l R   l m r l r l l R R 20 1 5 2 1 1 2 2 2 2 1 2 2 2 1 2 1   =      =   =  26. (c) 27. (b) In semiconductors charge carries are free electrons and holes
28. (b) Net current = (+) + (−) i i i nett n q t n(+)q(+) (−) (−) = + e t n e t n =  +  (+) (−) 2 = 3.2  1018  2  1.6  10–19 + 3.6  1018  1.6  10–19 = 1.6 A (towards right) 29. (b) In the absence of external electric field mean velocity of free electron (Vrms) is given by V T m KT Vrms =  rms  3 30. (a) With rise in temperature specific resistance increases 31. (c) For metallic conductors, temperature co-efficient of resistance is positive. 32. (d) 33. (b) Length l = 1 cm m 2 10 − = Area of cross-section A = 1 cm  100 cm = 100 cm2 = 10–2 m2 Resistance R = 3  10–7 2 2 10 10 − −  = 3  10–7  34. (c) 29 6 19 10 10 1.6 10 20 − −    = = nAe i vd 1.25 10 m / s −3 =  35. (b) Specific resistance j E k = 36. (b) 2 d l A l R    1 2 4 2 2 1 2 2 1 2 1  =      =         =  d d L L d d l l R R  R2 = R1 = R. 37. (c) 6 19 22 10 1.6 10 8.4 10 1.344     = = − − nAe i vd 0.01cm / s 0.1mm / s 10 1.6 8.4 1.344 = =   = 38. (a) Internal resistance Temperatur e 1  39. (d) Charge = Current × Time =5 × 60 = 300 C 40. (b) By R = l / A 41. (b) 42. (a) 43. (b) a l R  = for first wire and R’= 4 4 R a l =  for second wire. 44. (c) For semiconductors, resistance decreases on increasing the temperature. 45. (a) A l ne n A l R . 2  =  = 46. (b) Because as temperature increases, the resistivity increases and hence the relaxation time decreases for conductors . 1            47. (b) In VI graph, we will not get a straight line in case of liquids. 48. (c) A l R =  49. (a) Since 2 R  l  If length is increased by 10%, resistance is increases by almost 20% Hence new resistance R' = 10 + 20% of 10 10 12 . 100 20 = 10 +  =  50. (c) [1 (500 )] [1 (150 )] 500 150   + + = R R . Putting R150 = 133  and  = 0.0045 / C, we get R500 = 258  51. (c) 2 6 7 22 64 10 198 7 r A l R    =  = −   r = 0.024 cm 52. (b) Current density 2 r i A i J  = =  2 1 2 2 2 1 2 1 r r i i J J =  But the wires are in series, so they have the same current, hence 1 2 i = i . So 9 : 1 2 1 2 2 2 1 = = r r J J 53. (b) As R i V = and R  temperature inet +2e + – – e 100 cm 1 c m 1 cm
54. (b) 2 R  l  If l doubled then R becomes 4 times. 55. (a) Temperature coefficient of a semiconductor is negative. 56. (a) The reciprocal of resistance is called conductance 57. (a) Current Potential difference Resistance = 58. (c) Ohm’s Law is not obeyed by semiconductors. 59. (c) Drift velocity d d v l n e V v ;  = does not depend upon diameter. 60. (a) Using [1 ( )] RT2 = RT1 + T2 − T1  [1 (100 50)] R100 = R50 + −  7 = 5[1 + (  50)]  C o 0.008 / 250 (7 5) = −  = 61. (b) This is because of secondary ionisation which is possible in the gas filled in it. 62. (b) 63. (c) (1 3.92 10 ) (1 3.92 10 20) 76.8 50 (1 ) (1 ) 3 3 2 1 2 1 t t t R R − − +  +    = + + =    t = 167 C 64. (c) From 2 i v A i v r neA i vd =   d   d 65. (d) Resistivity depends only on the material of the conductor. 66. (a) A particular temperature, the resistance of a superconductor is zero  = = =  0 1 1 R G 67. (b) Net current t n q t n q i i i ( )( ) ( )( ) + + − − = + + − = +  e t n e t n i =  +  + − ( ) ( ) 18 19 18 19 2.9 10 1.6 10 1.2 10 1.6 10 − − =    +     i = 0.66 A 68. (d) If E be electric field, then current density j = E Also we know that current density A i j = Hence j is different for different area of cross-sections. When j is different, then E is also different. Thus E is not constant. The drift velocity d v is given by ne j vd = = different for different j values. Hence only current i will be constant. 69. (d) 70. (a) A l R =  and mass m = volume (V)  density (d) = (A l) d Since wires have same material so  and d is same for both. Also they have same mass  Al = constant  A l 1   4 1 2 2 1 2 1 2 2 1 2 1         =         =  = r r A A A A l l R R  4 2 2 34       = r r R  R2 = 544  71. (a) L A A R R A l R ( , 1 2 2 1 =   =  constant) 2 1 2 2 1  = = R R A A Now, when a body dipped in water, loss of weight = V L g = AL L g So, 2; (Loss of wight) (Loss of weight) 2 1 2 1 = = A A so A has more loss of weight. 72. (c) Q = it = 20 × 10–6 × 30 = 6× 10–4 C 73. (b) Ge is semiconductor and Na is a metal. The conductivity of semiconductor increases and that of the metals decreases with the rise in temperature. 74. (b) t ne i =  16 19 3 10 1.6 10 1.6 10 1 =    = = − − e it n . 75. (c) Drift velocity 2 1 or 1 d v A v neA i vd =  d  d  P Q P Q Q P v v d d d d v v 4 1 4 / 2 1 2 2  =  =      =          = . 76. (c) Human body, though has a large resistance of the order, of K (say 10k ), is very sensitive to minute currents even as low as a few mA. Electrons, excites and disorders the nervous i Ne+ – + e –
system of the body and hence one fails to control the activity of the body. 77. (c) (1 ) 0 R R t t = +  4.2 = 0 4 0 R (1 + 0.004 100 ) =1. R  R0 = 3 . 78. (d) m l R 2   2 3 3 2 2 2 2 1 1 1 2 3 : : : :                         = m l m l m l R R R 5 1 : 3 9 : 1 25 = 125 : 15 : 1 5 1 = 25 : 3 :  . 79. (b) 80. (a) 4 2 4 2 4 1 2 2 1 n R R r nr R R r r R R   =       =         = . 81. (d) (1 100 ) (1 50) 6 5 (1 ) (1 ) 2 1 2 1 +  +   = + + =     t t R R C o per 200 1  = Again by (1 ) 0 R R t t = + 50 4 . 200 1 5 0 1   0 =        = R +  R 82. (b) Q mA T Q i 1.6 10 5 10 0.8 1 9 1 5 = = =    = −  . 83. (b) 2.4 1.7 10 1 10 8 7 copper iron Copper iron    = = − −   r r . 84. (c) 1.6 10 6.8 10 1.1 10 . 19 15 3 i e amp − − =  =    =  85. (b) Resistivity of the material of the rod 1 3 10 (0.3 10 ) −3 −2 2   = =   l RA =    m −  9 27 10 Resistance of disc (Area of cross section) (Thickness) R = = 2 2 3 9 (1 10 ) (10 ) 27 10 − − −       2.7 10 . 7 =   − 86. (c) By using (1 ) 0 R R t t = + 3 (1 4 10 ) 3 0 0 R R t −  = +  t C o  = 500 . 87. (a) 6 10 1.6 10 0.96 . 15 19 i =    = mA − 88. (a) 1 1.6 10 16 10 19 3 − −   =   = n t ne i 17  n = 10 89. (d) =     = = 10 0.25 10 0.2 100 0.5 i V R . 90. (a) A l i V R = =   3 2 2 (1 10 ) 50 10 4 2 − −   =   =  m −6  1 10 . 91. (c) 92. (b) C R Vt Q t Q R V i 240 10 20 2 60 =   = =  = = . 93. (a) 2 2 1 2 2 1           = l l R R R l 4 4 /2 2 2 2 R R l l R R =  =          = . 94. (b) 95. (b) 2 9 6 1 9 10 10 1.6 10 40 − −    = = neA i Vd = 2.5 10 m/sec −3  . 96. (c) 2 8 6 1 9 8.4 10 10 1.6 10 5.4 − −     = = nAe i Vd = 0.4 10 m/sec 0.4mm/sec 3  = − . 97. (a) 2 2 1 2 1         = l l R R  = = =        = 160 16 1 20 10 5 2 2 2 R R . 98. (c) ; 1  R  where  = Relaxation time. When lamp is switched on, temperature of filament increase, hence  decrease so R increases 99. (d) R = 91  10  9.1k 2 . 100. (c) 101. (a) 102. (d) 3 2 3 2 2 2 1 2 1 1 2 3 2 : : : : m l m l m l R R R m l R   = 27 : 6 :1 3 1 : 2 4 : 1 9 : :  R1 R2 R3 = = . 103. (b) 15 19 3 6.25 10 1.6 10 1 10 =    = − − n . 104.(b) 2 1 2 2 1 2 2 '         =    =  r r i i v v r i v ne r i vd d  2 ' v  v = .