Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

Single Correct Answer Type

1. In the electric circuit shown each cell has an emf of 2V and internal resistance of 1Ω. The external
resistance is 2Ω. The value of the current I is(in ampere)
2V ,1

a) 2 b) 1.25 c) 0.4 d) 1.2

2. A, B, C and D are four resistances of 2Ω, 2Ω, 2Ω and 3Ω respectively. They are used to form a Wheatstone
bridge. The resistance D is short circuted with a resistances R in order to get the bridge balanced. The
value of R will be
a) 4 Ω b) 6 Ω c) 8 Ω d) 3 Ω

3. The arrangement as shown in figure is called as

Total P.D.

Variable P.D.

a) Potential divider b) Potential adder c) Potential substracter d) Potential multiplier

4. If the balance point is obtained at the 35th cm in a meter bridge, the resistances in the left and right gaps
are in the ratio of
a) 7 : 13 b) 13 : 7 c) 9 : 11 d) 11 : 9

5. Two electric bulbs rated P1 watt V volts and P2 watt V volts are connected in parallel and V volts are
applied to it. The total power will be
a) P + P watt b) P P watt c) P1P2 watt d) P1+P2 watt
1 2 1 2 P1+P2 P1P2
6. In a meter bridge a 30Ω resistance is connected in the left gap and a pair of resistances P and Q in the right
gap. Measured from the left, the balance point is 37.5 cm, when P and Q are in series and 71.4 cm when
they are parallel. The values of P and Q (in ohm) are
a) 40, 10 b) 35, 15 c) 30, 20 d) 25, 25

7. In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance

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 point is at a length of 2m when the cell is shunted by a 5Ω resistance; and is at a length of 3m when the cell
is shunted by a 10Ω resistance. The internal resistance of the cell is, then
a) 1.5Ω b) 10Ω c) 15Ω d) 1Ω

8. Two electroplating cells, one of silver and another of aluminium are connected in series. The ratio of the
number of silver atoms to that of aluminium atoms deposited during time t will be
a) 1 : 3 b) 3 : 1 c) 1 : 9 d) 9 : 1

9. A coil of wire of resistance 50 Ω is embedded in a block of ice and a potential difference of 210 V is applied
across it. The amount of ice which melts in 1 sec is
a) 0.262 g b) 2.62 g c) 26.2 g d) 0.0262 g

10. The resistance of 1 A ammeter is 0.018Ω. To convert it into 10 A ammeter, the shunt resistance required
will be
a) 0.18 Ω b) 0.0018 Ω c) 0.002 Ω d) 0.12 Ω

11. When current flows through a conductor, then the order of drift velocity of electrons will be

a) 1010m/sec b) 10-2cm/sec c) 104cm/sec d) 10-1cm/sec

12. Which of the following statements is wrong

a) Voltmeter should have high resistance

b) Ammeter should have low resistance

c) Ammeter is placed in parallel across the conductor in a circuit

d) Voltmeter is placed in parallel across the conductor in a circuit

13. A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a
wire made of A. Then for the two wires to have the same resistance, the ratio lB/lA of their respective
lengths must be
a) 1 b) 1/2 c) 1/4 d) 2

14. In the circuit shown below, the power developed in the 6Ω resistor is 6 watt. The power in watts
developed in the 4Ω resistor is

a) 16 b) 9 c) 6 d) 4

15. The value of internal resistance of an ideal cell is

a) Zero b) 0.5 Ω c) 1 Ω d) Infinity

16. If the electronic charge is 1.6×10-19 C, then the number of electrons passing through a section of wire per
second, when the wire carries a current of 2 A is
a) 1.25×1017 b) 1.6×1017 c) 1.25×1019 d) 1.6×1019

17. Two bulbs are working in parallel order. Bulb A is brighter than bulb B. If RA and RB are their resistance
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 respectively then
a) R > R b) R < R c) R = R d) None of these
A B A B A B

18. The amount of chlorine produced per-second through electrolysis in a plate which consumes 100 KW
-3 -1
power at 200 V is (Given, electrochemical equivalent of chlorine = 0.367×10 gC )
a) 18.35 g b) 1.835 g c) 183.5 g d) 0.1835 g

19. Three resistors each of 2 ohm are connected together in a triangular shape. The resistance between any
two vertices will be
a) 4/3ohm b) 3/4 ohm c) 3 ohm d) 6 ohm

20. Two different conductors have same resistance at 0°C. It is found that the resistance of the first conductor
at t1℃ is equal to the resistance of the second conductor at t2℃. The ratio of the temperature coefficients of
α
resistance of the conductors, 1 is
α2
a) t1 b) t2-t1 c) t2-t1 d) t2
t2 t2 t1 t1
21. Which of the following set up can be used to verify the Ohm’s law?

a) b) c) d) V

22. The resistance of a galvanometer is 25 ohm and it requires 50 μA for full deflection. The value of the shunt
resistance required to convert it into an ammeter of 5 amp is
a) 2.5×10-4ohm b) 1.25×10-3ohm c) 0.05 ohm d) 2.5 ohm

23. The resistivity of a potentiometer wire is 40×10-8ohm - m and its area of cross-section is 8×10-6m2. If 0.2
amp current is flowing through the wire, the potential gradient will be
a) 10-2volt/m b) 10-1volt/m c) 3.2×10-2volt/m d) 1 volt/m

24. In the circuit shown here, the readings of the ammeter and voltmeter are

a) 6 A, 60 V

b) 0.6 A, 6 V

c) 6/11 A, 60/11 V

d) 11/6 A, 11/60 V

25. A thermocouple of negligible resistance produces an e.m.f. of 40μV/℃ in the linear range of temperature. A
galvanometer of resistance 10 ohm whose sensitivity is 1μA/div, is employed with the thermocouple. The
smallest value of temperature difference that can be detected by the system will be
a) 0.1℃ b) 0.25℃ c) 0.5℃ d) 1℃
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26. The resistance across R and Q in the figure.
A
r r
rr r

P B r C Q

a) r/3 b) r/2 c) 2r d) 6r

27. When a current Iflows through a wire, the drift velocity of the electrons is v. When current 2I flows
through another wire of the same material having double the length and double the area of cross-section,
the drift velocity of the electrons will be
a) v b) v c) v d) v
8 4 2
28. A wire is broken in four equal parts. A packet is formed by keeping the four wires together. The resistance
of the packet in comparison to the resistance of the wire will be
a) Equal b) One fourth c) One eight d) 1 th
16
29. In an electroplating experiment, m gm of silver is deposited when 4 ampere of current flows for 2 minute.
The amount (in gm) of silver deposited by 6 ampere of current for 40 second will be
a) 4 m b) m/2 c) m/4 d) 2 m

30. Which of the following relation is wrong?

a) 1 ampere ×1 ohm=1 volt b) 1 watt ×1 sec=1 joule

c) 1 newton per coulomb =1 volt per metre d) 1 columb ×1 volt=1 watt

31. To convert a 800 mV range milli voltmeter of resistance 40 Ω into a galvanometer of 100 mA range, the
resistance to be connected as shunt is
a) 10 Ω b) 20 Ω c) 30 Ω d) 40 Ω

32. The effective resistance between points A and B is

10 10
A B
10

10 10

a) 10 Ω b) 20 Ω

c) 40 Ω d) None of the above three values

33. 1 2
If the total emf in a thermocouple is a parabolic function expressed as E = at + bt , which of the
2
following relation does not hold good?
a) Neutral temperature t = - a b) Temperature of inversion, t = - -2a
n
b i
b
c) Thermoelectric power P = a + bt d) t = a
n
b
34. The plot represents the flow of current through a wire at three different times.

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 The ratio of charges flowing through the wire at different times is
a) 2 : 1 : 2 b) 1 : 3 : 3 c) 1 : 1 : 1 d) 2 : 3 : 4

35. When the resistance of 9 Ω is connected at the ends of a battery, its potential difference decreases from 40
volt to 30 volt. The internal resistance of the battery is
a) 6 Ω b) 3 Ω c) 9 Ω d) 15 Ω

36. A cylindrical metal wire of length l and cross sectional area S, has resistance R, conductance G, conductivity
σ and resistivity ρ. Which one of the following expressions for σ is valid
a) GR b) ρR c) GS d) Rl
ρ G l S
37. The heat developed in an electric wire of resistance R by a current I for a time t is
2 2 2
a) I Rt cal b) I t cal c) I R cal d) Rt cal
2
4.2 4.2R 4.2 t 4.2 I
38. In the circuit of adjoining figure the current though 12 Ω resistor will be

a) 1 A b) 1 A c) 2 A d) 0 A
5 5
39. An electric bulb is designed to draw power P0 at voltage V0. If the voltage is V it draws a power P. Then

() () () ()
2 2
a) P = V0 P b) P = V P c) P = V P d) P = V0 P
V 0 V0 0
V 0
0
V 0
40. When two resistances R1 and R2 are connected in series and parallel with 120 V line power consumed will
be 25 W and 100 W respectively. Then the ratio of power consumed by R1 to that consumed by R2 will be
a) 1 :1 b) 1 :2 c) 2 :1 d) 1 :4

41. For which of the following the resistance decreases on increasing the temperature

a) Copper b) Tungsten c) Germanium d) Aluminium

42. The effective resistance between the points A and B in the figure is

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 D
3 3

6
A C

3 3

a) 5Ω b) 2Ω c) 3Ω d) 4Ω

43. How much energy in kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a
month (30 days)
a) 1500 b) 5,000 c) 15 d) 150

44. Express which of the following setups can be used to verify Ohm’s law

a)

b)

c)

d)

45. If in a voltaic cell, 5 g of zinc is consumed, we will get how many ampere hour (given that ECE of zinc is
-7 -1
3.38×10 kgC )
a) 2.05 b) 8.2 c) 4.1 d) 5×3.338 ×10-7

46. The resistance of a conductor is 5 ohm at 50℃ and 6 ohm at 100℃. Its resistance at 0℃ is

a) 1 ohm b) 2 ohm c) 3 ohm d) 4 ohm

47. A metallic wire of resistance 12 Ω is bent to from a square. The resistance between two diagonal points
would be
a) 12 Ω b) 24 Ω c) 6 Ω d) 3 Ω

48. A piece of metal weighing 200 g is to be electroplated with 5% of its weight in gold. How long it would take
to deposits the required amount of gold, if the strength of the available current is 2 A?
-4 -1
(Given, electrochemical equivalent of H = 0.0104×10 gC atomic weight of gold = 197.1, atomic weight
of hydrogen = 1.008)
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 a) 7347.9 s b) 7400.5 s c) 7151.7 s d) 70 s

49. In the circuit shown in figure, the heat produced by the 6 Ω resistance is 60Ω cal s-1. What heat per second
is produced across 3Ω resistance?
2Ω 3Ω

6Ω 4Ω

a) 30 cal b) 60 cal c) 100 cal d) 120 cal

50. Thirteen resistance each of resistance R ohm are connected in the circuit as shown in the figure below. The
effective resistance between A and B is
R R

R R R R

A R B

R
R R R

R R

a) 2R Ω b) 4R Ω c) 2 R Ω d) R Ω
3 3
51. In the shown circuit, what is the potential difference across A and B
20 V

A B

a) 50 V b) 45 V c) 30 V d) 20 V

52. The internal resistance of a cell is the resistance of

a) Electrodes of the cell b) Vessel of the cell

c) Electrolyte used in the cell d) Material used in the cell

53. In potentiometer a balance point is obtained, when

a) The e.m.f. of the battery becomes equal to the e.m.f. of the experimental cell

b) The p.d. of the wire between the +ve end to jockey becomes equal to the e.m.f. of the experimental cell

c) The p.d. of the wire between +ve point and jockey becomes equal to the e.m.f. of the battery

d) The p.d. across the potentiometer wire becomes equal to the e.m.f. of the battery

54. A conductor wire having 1029 free electrons/m3 carries a current of 20A. If the cross-section of the wire is
2
1mm , then the drift velocity of electrons will be
a) 6.25×10-3ms-1 b) 1.25×10-5ms-1 c) 1.25×10-3ms-1 d) 1.25×10-4ms-1

55. Figure shown three similar lamps A,B and C connected across a power supply. If the lamp C fuses, how will
the light emitted by A and B change?

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 a) No change Brilliance of A decreases and that of B increases

b)

c) Brilliance of both A and B increases d) Brilliance of both A and B decreases

56. Bulb B1 (100 W-250 V) and bulb B2 (100 W-200 V) are connected across 250 V. What is potential drop
across B2?

a) 200 V b) 250 V c) 98 V d) 48 V

57. The amount of charge required to liberate 9 gm of aluminium (atomic weight =27 and valency = 3) in the
process of electrolysis is (Faraday’s number =96500 coulombs/gm equivalent)
a) 321660 coulombs b) 69500 coulombs c) 289500 coulomb d) 96500 coulomb

58. In the circuit shown below, the reading of the voltmeter V is

1A 4 A 16

V
2 C

1A 16 B 4

a) 12 V b) 8 V c) 20 V d) 16 V

59. If each resistance in the figure is of 9 Ω then reading of ammeter is

+
9V
–

a) 5 A b) 8 A c) 2 A d) 9 A

60. 160W-60V lamp is connected at 60 V DC supply. The number of electrons passing through the lamp in 1
-19
min is (the charge of electron e = 1.6×10 C)
a) 1019 b) 1021 c) 1.6×1019 d) 1.4×1020

61. The smallest temperature difference that can be measured with a combination of a thermocouple of
thermo e.m.f. 30 μV per degree and a galvanometer of 50 ohm resistance, capable of measuring a minimum
-7
current of 3×10 amp is
a) 0.5 degree b) 1.0 degree c) 1.5 degree d) 2.0 degree

62. The magnitude and direction of the current in the circuit shown will be

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 1 e 2
a b
10V 4V

3
d c

a) 7/3A from a to b through e b) 7/3A from b to a through e

c) 1A from b to a through e d) 1A from a to b through e

63. If the cold junction is held at 0℃, the same thermo-emf V of a thermocouple varies as V = 10×10-6
1 -6 2
t - ×10 t , where t is the temperature of the hot junction in ℃. The neutral temperature and the
40
maximum value of thermo-emf are respectively
a) 200℃;2 mV b) 400℃;2 mV c) 100℃;1 mV d) 200℃;1 mV

64. A voltmeter has a range 0 - V with a series resistance R. With a series resistance 2R, the range is 0 - V'. The
correct relation between Vand V' is
a) V' = 2V b) V' > 2V c) V' ≫ 2V d) V' < 2V

65. A potentiometer wire of length L and resistance 10 Ω is connected in series with a battery of e.m.f. 2.5 V
and a resistance in its primary circuit. The null point corresponding to a cell of e.m.f. 1V is obtained at a
L
distance . If the resistance in the primary circuit is doubled then the position of new null point will be
2
a) 0.4 L b) 0.5 L c) 0.6 L d) 0.8 L

66. The ratio of voltage sensitivity (VS) and current sensitivity (Is) of a moving coil galvanometer is

a) 1 b) 12 c) G d) G2
G G
67. Find the power of the circuit

a) 1.5 W b) 2 W c) 1 W d) None of these

68. Five conductors are meeting at a point x as shown in the figure. What is the value of current in fifth
conductor?

a) 3 A away from x b) 1 A away from x c) 4 A away from x d) 1 A towards x

69. A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the

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 same source. The energy now liberated per second is
a) 200 J b) 400 J c) 25 J d) 50 J

70. For comparing the e.m.f.'s of two cells with a potentiometer, a standard cell is used to develop a potential
gradient along the wires. Which of the following possibilities would make the experiment unsuccessful
a) The e.m.f. of the standard cell is larger than the E e.m.f.'s the two cells

b) The diameter of the wires is the same and uniform throughout

c) The number of wires is ten

d) The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells

71. Two different metals are joined end to end. One end is kept at constant temperature and the other end is
heated to a very high temperature. The high depicting the thermo e.m.f. is
a) E b) E c) E d) E

t t t t

72. In the circuit element given here, if the potential at point B, VB = 0, then the potentials of A and D are given
as
1 amp 1.5 Ω 2.5 Ω 2V

A B C D

a) V = -1.5V, V = +2V b) V = +1.5V, V = +2V

A D A D

c) V = +1.5 V,V = +0.5 V d) V = +1.5V, V = -0.5 V

A D A D

73. If in the circuit shown below, the internal resistance of the battery is 1.5 Ω and VP and VQ are the potentials
at P and Q respectively, what is the potential difference between the points P and Q
20 V 1.5 Ω
+ –

3Ω P 2Ω

2Ω Q 3Ω

a) Zero b) 4 volts (V > V ) c) 4 volts (V > V ) d) 2.5 volts (V > V )

P Q Q P Q P

74. Two resistance of 10 Ω and 20 Ω and an inductor of inductance 5 H are connected to a battery of 2 V
through a key k as shown in the figure. At time t = 0, when the key k is closed the initial current through
the battery is

a) 0.2 A b) 2 A c) 1 A d) 0
15 15
75. Find the equivalent resistance across the terminals of source of e.m.f. 24 V for the circuit shown in figure

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 a) 15 Ω b) 10 Ω c) 5 Ω d) 4 Ω

76. Twelve cells, each having emf E volts are connected in series and kept in a closed box. Some of these cells
are wrongly connected with positive and negative terminals reversed. This 12-cell battery is connected
with an ammeter, an external resistance R ohm and a two-cell battery (two cells of the same type used
earlier, connected perfectly in series). The current in the circuit when the 12-cell battery and 2-cell battery
aid each other is 3A and 2A when they oppose each other. Then, the number of cell in 12-cell battery that
are connected wrongly is
a) 4 b) 3 c) 2 d) 1

77. In hydrogen atom, the electron makes 6.6×1015 revolutions per second around the nucleus in an orbit of
-10
radius 0.5×10 m. It is equivalent to a current nearly
a) 1 A b) 1 mA c) 1 μA d) 1.6×10-19A

78. Two conductors made of the same material are connected across a common potential difference.
Conductor A has twice the diameter and twice the length of conductor B. The power delivered to the two
conductors PA and PB respectively is such that PA/PB equals to
a) 0.5 b) 1.0 c) 1.5 d) 2.0

79. Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat
produced in the two cases is
a) 1 : 4 b) 4 : 1 c) 1 : 2 d) 2 : 1

80. Consider the following statements regarding the network shown in the figure.
1. The equivalent resistance of the network between point A and B is independent of value of G.
2. The equivalent resistance of the network between points A and B is
4
R
3
3. The current through G is zero.
Which of the above statements is/zero true?
R R

A B
G

2R 2R

a) 1,2 and 3 b) 2 and 3 c) 2 alone d) 1 alone

81. In a copper voltmeter experiment, current is decreased to one-fourth of the initial value but is passed for
four times the earlier duration. Amount of copper deposited will be

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 a) Same b) One-fourth the previous value

c) Four times the previous value d) 1 th the previous value

16
82. A strip of copper and another of germanium are cooled from room temperature to 80 K. The resistance of

a) Each of these increases b) Each of these decreases

c) Copper strip increases and that of germanium d) Copper strip decreases and that of germanium
decreases increases
83. The resistance of the following circuit figure between A and B is
E
2Ω 2Ω
C D
2Ω

2Ω 2Ω
2 2
F

A B
2Ω

a) (3/2) Ω b) 2 Ω c) 4 Ω d) 8 Ω

84. The amount of charge Q passed in time t through a cross-section of a wire is Q = 5t2 + 3t + 1.The value
of current at time t = 5 s is
a) 9A b) 49A c) 53A d) None of these

85. Silver and copper voltameter are connected in parallel with a battery of e.m.f. 12 V. In 30 minutes, 1g of
silver and 1.8g of copper are liberated. The power supplied by the battery is
-4 -4
(ZCu = 6.6×10 g/C and ZAg = 11.2×10 g/C)
a) 24.13 J/sec b) 2.413 J/sec c) 0.2413 J/sec d) 2413 J/sec

86. In a copper voltmeter, the mass deposited in 30 s is m gram. If the current-time graph is as shown in
-1
figure, the electrochemical equivalent of copper, in gC is

a) 0.1 m b) 0.6 m c) m d) m
2
87. The electron in a hydrogen atom circles around the proton in 1.5941×10-18s. The equivalent current due to
motion of the electrons is
a) 127.37 mA b) 122.49 mA c) 100.37 mA d) 94.037 mA

88. The effective resistance between points P and Q of the electrical circuit shown in the figure.

a) 2Rr b) 8R(R+r) c) 2R + 4r d) 5R + 2R
R+r (3R+r) 2
89. In a thermo-couple, one junction which is at 0℃ and the othe at t℃ the emf is given by E = at2 - bt2. The

P a g e | 12
 neutral temperature is given by
a) a/b b) 2 a/3b c) 3a/2b d) b/2a

90. In the arrangement shown in figure, the current through 5Ω resistor is

2 2
12V 5Ω 12V

a) 2A b) Zero c) 12 A d) 1A
7
91. A straight conductor of uniform cross-section carries a current i, If s is the specific charge of an electron,
the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is
a) is b) i/s c) i/s d) (i s)2

92. When a copper voltmeter is connected with a battery of emf 12V, 2 g of copper is deposited in 30 min. If
the same voltmeter is connected across 6 V battery, the mass of copper deposited in 45 min would be
a) 1 g b) 1.5 g c) 2 g d) 2.5 g

93. A resistor R and 2μF capacitor in series is connected through a switch to 200 V direct supplies. Across the
capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulk light up 5 s alter
the switch has been closed (log10 2.5=0.4)
a) 1.7×105 Ω b) 2.7×106 Ω c) 3.3×107 Ω d) 1.3×104 Ω

94. In above question, if length is doubled, the drift velocity

a) Is doubled b) Is halved c) Remains same d) Becomes zero

95. Out of five resistances of resistance R Ω each 3 are connected in parallel and are joined to the rest 2 in
series. Find the resultant resistance

()
a) 3 R Ω
7 ()
b) 7 R Ω
3 ()
c) 7 R Ω
8 ()
d) 8 R Ω
7
96. If the resistivity of an alloy is ρ' and that of constituent metal is ρ then

a) ρ’> ρ b) ρ’< ρ

c) ρ' = ρ d) There is no simple relation between ρ and ρ'

97. The mass of a substance liberated when a charge 'q' flows through an electrolyte is proportional to

a) q b) 1/q c) q2 d) 1/q2

98. The resistance of a discharge tube is

a) Ohmic b) Non-ohmic c) Both (a) and (b) d) Zero

99. If the resistance of a conductor is 5Ω at 50℃ and 7Ω at 100℃ then the mean temperature coefficient of
resistance of the material is
a) 0.008/℃ b) 0..006/℃ c) 0.004/℃ d) 0.001/℃

100. The resistance of a galvanometer coil is R, then the shunt resistance required to convert it into a ammeter
of range 4times, will be
a) 4R b) R/3 c) R/4 d) R/5

101. All bulbs in figure, are identical. Which bulb lights brightly?

P a g e | 13
 a) 1 b) 2 c) 3 d) 4

102. An ammeter gives full scale deflection when current 1.0 A is passed in it. To convert it into 10 A range
ammeter, the ratio of its resistance and the shunt resistance will be
a) 1 : 9 b) 1 : 10 c) 1 : 11 d) 9 : 1

103. Same current is being passed through a copper voltmeter and a silver voltmeter. The rate of increase in
weights of the cathode of the two voltmeters will be proportional to
a) Atomic masses b) Atomic number c) Relative densities d) None of the above

104. For measurement of potential difference, potentiometer is preferred in comparison to voltmeter because

a) Potentiometer is more sensitive than voltmeter

b) The resistance of potentiometer is less than voltmeter

c) Potentiometer is cheaper than voltmeter

d) Potentiometer does not take current from the circuit

105. The resistance of an ideal ammeter is

a) Infinite b) Very high c) Small d) Zero

106. In the given circuit the current I1 is

30 

I1
40 

I3
I2 40V
40 

80V

a) 0.4 A b) -0.4 A c) 0.8 A d) -0.8 A

107. The chemical equivalent of copper and zinc are 32 and 108 respectively. When copper and silver
voltmeters are connected in series and electric current is passed through for sometime, 1.6 g of copper is
deposited. Then, the mass of silver deposited will be
a) 3.5 g b) 2.8 g c) 5.4 g d) None of these

108. When current is passed in antimony-bismuth couple, then

a) The junction becomes hot when the current is from bismuth to antimony

b) The junction becomes hot when current flows from antimony to bismuth

c) Both junctions becomes hot

d) Both junctions becomes cold

109. The current inside a copper voltameter

a) Is half the outside value

P a g e | 14
 b) Is the same as the outside value

c) Is twice the outside value

d) Depends on the concentration of CuSO

4

110. I - V characteristic of a copper wire of length L and area of cross-section A is shown in figure. The slope of
the curve becomes

a) More if the experiment is performed at higher b) More if a wire of steel of same dimension is used
temperature
c) More if the length of the wire increased d) Less if the length of the wire increased

111. A heater of 220 V heats a volume of water in 5 min time. A heater of 110 V heats the same volume of water
is
a) 5 min b) 8 min c) 4×104 min d) 20 min

112. Two wires having resistance of 2Ω and 4Ω are connected to same voltage. Ratio of heat dissipated at
resistance is
a) 1 :2 b) 4 :3 c) 2 :1 d) 5 :2

113. A group of N cells whose emf varies directly with the internal resistance as per the equation EN = 1.5rN are
connected as shown in the figure. The current I in the circuit is

a) 5.1A b) 0.51A c) 1.5A d) 0.15A

114. For the network shown in the figure the value of the current i is
2

4 4
3

6

a) 9V b) 5V c) 5V d) 18V
35 18 9 5
115. Which of the following is not a correct statement

a) Resistivity of electrolytes decreases on increasing temperature

b) Resistance of mercury falls on decreasing its temperature

P a g e | 15
 c) When joined in series a 40 W bulb glows more than a 60 W bulb

d) Resistance of 40 W bulb is less than the resistance of 60 W bulb

116. For a certain thermocouple the emf is E = aT + bT2, where t (in℃) is the temperature of hot junction, the
-6 -6
cold junction is at 0℃. The value of contants a and b are 10×10 and 0.02×10 respectively, then the
temperature of inversion (in℃) will be
a) 150 b) 250 c) 500 d) 750

117. In the given figure, potential difference between A and B is

10K D
A
30 V
10K 10K

a) 0 b) 5 volt c) 10 volt d) 15 volt

118. A cell of emf E is connected across a resistance R. the potential difference between the terminals of the cell
is found to be V volt. Then the internal resistance of the cell must be
a) (E-V) b) (E-V) R c) 2(E-V)R d) 2(E-V)V
V E R
119. Electric field (E) and current density (J) have relation

a) E ∝ J-1 b) E ∝ J c) E ∝ 1 d) E2 ∝ 1
2
J J
120. In a network as shown in the figure, the potential difference across the resistance 2R is (the cell has an emf
of E volt and has no ingternal resistance)
4R

2R

a) 2E b) 4E c) E d) E
7 7
121. Two identical conductors maintained at same temperatures are given potential differences in the ratio 1 :
2. Then the ratio of their drift velocities is
a) 1 : 2 b) 3 : 2 c) 1 : 1 d) 1/2
1 :2
122. A 100 W bulb produces an electric field of 2.9 V/m at a point 3 m away. If the bulb is replaced by 400 W
bulb without disturbing other conditions, then the electric field produced at the same point is
a) 2.9 V/m b) 3.5 V/m c) 5 V/m d) 5.8 V/m

123. The neutral temperature tn = 285°C is constant for a Cu-Fe thermocouple. When the cold junction is at
0°C, the value of inversion temperature is ti = 570°C but if the cold junction is at 10°C, the inversion
temperature (ti) will be
a) 550°C b) 560°C c) 570°C d) 580°C

124. When a battery connected across a resistor of 16 Ω, the voltage across the resistor is 12 V. When the same
battery is connected across a resistor of 10 Ω, voltage across it is 11V. The internal resistance of the
battery (in ohm) is

P a g e | 16
 a) 10 b) 20 c) 25 d) 30
7 7 7 7
125. For obtaining chlorine by electrolysis a current of 100 kW and 125 V is used. (Electro chemical equivalent
-1
of chlorine is 0.367 ×kgC ). The amount of chlorine obtained in one minute will be
a) 1.7616 g b) 17.616 g c) 0.17161 g d) 1.7616 kg

126. V - i graphs for parallel and series combination of two identical resistors are as shown in figure. Which
graph represents parallel combination

a) A

b) B

c) A and B both

d) Neither A nor B

127. As the temperature rises the resistance offered by metal

a) Increase b) Decrease c) Remains same d) None of these

128. A wire 100cm long and 2.0 mm diameter has a resistance of 0.7 ohm, the electrical resistivity of the
material is
a) 4.4×10-6 ohm×m b) 2.2×10-6ohm×m

c) 1.1×10-6 ohm×m d) 0.22×10-6 ohm×m

129. In the Wheatstone’s bridge (shown in figure) X = Y and A > B. The direction of the current between ab
will be
a
A B
c d

X Y
b

a) From a to b b) From b to a

c) From b to a through c d) From a to b through c

130. The chemical equivalent of silver is 108. If the current in a silver voltmeter is 2 amp, the time required to
deposit 27 grams of silver will be
a) 8.57 hrs b) 6.70 hrs c) 3.35 hrs d) 12.50 hrs

131. By increasing the temperature, the specific resistance of a conductor and a semiconductor

a) Increases for both b) Decreases for both c) Increases, decreases d) Decreases, increases

132. Two electric bulbs whose resistances are in the ratio of 1 :2 are connected in parallel to a constant voltage
P a g e | 17
 source. The powers dissipated in them have the ratio
a) 1 :2 b) 1 :1 c) 2 :1 d) 1 :4

133. Two cells of same emf E but of different internal resistances r1and r2are connected in series with an
external resistance R. The potential drop across the first cell is found to be zero. The external resistance R
is
a) r + r b) r - r c) r - r d) r r
1 2 1 2 2 1 1 2

134. A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot
and the other is kept cold, then, an electric current will
a) Flow from Antimony to Bismuth at the hot junction

b) Flow from Bismuth to Antimony at the cold junction

c) Not flow through the thermocouple

d) Flow from Antimony to Bismuth at the cold junction

135. The resistivity of a wire

a) Increase with the length of the wire

b) Decreases with the area of cross-section

c) Decreases with the length and increases with the cross-section of wire

d) None of the above statement is correct

136. In the circuit shown figure potential difference between X and Y will be
40 X Y

120 V 20

a) Zero b) 20 V c) 60 V d) 120 V

137. Heat produced (cals) in a resistance R when a current I amperes flows through it for t seconds is given by
the expression
2 2 2
a) I Rt b) IR t c) 4.2IR
2
d) IRt
4.2 4.2 t 4.2
138. Kirchoff’s second law for the analysis of circuit is based on

a) Conversion of charge b) Conversion of energy

c) Conversion of both charge and energy d) Conversion of momentum of electron

139. A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in
it. The heat developed is doubled if
a) Both the length and radius of wire are halved b) Both the length and radius of wire are doubled

c) The radius of wire is doubled d) The length of wire is doubled

140. If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance

P a g e | 18
 will be
a) 0.1% b) 0.2% c) 1% d) 2%

141. The figure here shows a portion of a circuit. What are the magnitude and direction of the current i in the
lower right-hand wire

a) 7 A b) 8 A c) 6 A d) 2 A

142. The current flowing through a wire depends on time as I = 3t2 + 2t + 5. The charge flowing through the
cross-section of the wire in time from t = 0 to t = 2 sec. is
a) 22 C b) 20 C c) 18 C d) 5 C

143. We have a galvanometer of resistance 25Ω. It is shunted by a 2.5 Ω wire. The part of total current that
flows through the galvanometer is given as
a) I = 1 b) I = 1 c) I = 3 d) I = 4
I0 11 I0 10 I0 11 I0 11
144. A current of 3 amp. flows through the 2Ω resistor shown in the circuit. The power dissipated in the 5Ω
resistor is

a) 1 watt b) 5 watt c) 4 watt d) 2 watt

145. One junction of a certain thermoelectric couple is at a fixed temperature Tr and the other junction is at

[ 1
temperature T. The thermo-electromotive force for this is expressed by E = k(T-Tr) T0- (T+Tr) . At
2 ]
1
temperature T = T0, the thermoelectric power is
2
a) 1 kT b) kT c) 1 kT2 d) 1 k (T -T )2
r
2 0 0
2 0 2 0

146. In a given network, each resistance has value of 6Ω. The point X is connected to point A by a copper wire of
negligible resistance and point Y is connected to point B by the same wire. The effective resistance
between X and Y will be

6 6 A 6
X Y
B

a) 18Ω b) 6 Ω c) 3 Ω d) 2 Ω

P a g e | 19
147. Faraday’s laws of electrolysis are related to

a) The atomic number of positive ion b) The equivalent weight of electrolyte

c) The atomic number of negative ion d) The velocity of positive ion

148. A cell having emf of 1.5V, when connected across a resistance of 14 Ω, produces a voltage of only 1.4V
across this resistance. The internal resistance of the cell must be
a) 1 Ω b) 14 Ω c) 15 Ω d) 21 Ω

149. Two similar accumulators each of emf E and internal resistance r are connected as shown in the following
figure. Then, the potential difference between x and y is

X y
| |

a) 2E b) E c) Zero d) None of these

150. In a meter bridge experiment, the ratio of the left gap resistance to right gap resistance is 2:3, the balance
point from left is
a) 60 cm b) 50 cm c) 40 cm d) 20 cm

151. A conductor wire having 1029 free electrons/m3 carries a current of 20A. If the cross-section of the wire is
2
1mm , then the drift velocity of electrons will be
a) 6.25×10-3ms-1 b) 1.25×10-5ms-1 c) 1.25×10-3ms-1 d) 1.25×10-4ms-1

152. A potentiometer wire of length 10 m and resistance 20 Ω is connected is series with a 15V battery and an
external resistance 40 Ω. A secondary cell of emf E in the secondary circuit is balanced by 240 cm long the
potentiometer wire. The emf E of the cell is
a) 2.4V b) 1.2V c) 2.0V d) 3V

153. In circuit shown below, the resistances are given in ohm and the battery is assumed ideal with emf equal
to 3V. The voltage across the resistance R4 is

R1 R3 R4
R2
3V
R5

a) 0.4V b) 0.6V c) 1.2V d) 1.5V

154. Constant current is flowing through a linear conductor of non-uniform area of cross-section. The charge
flowing per second through the area of conductor at any cross-section is
a) Proportional to the area of cross- section b) Inversely proportional to the area of cross-section

c) Independent of the area of cross-section d) Dependent on the length of conductor

155. Total surface area of a cathode is 0.05m2 and 1 A current passes through it for 1 hour. Thickness of nickle
-4
deposited on the cathode is (Given that density of nickle = 9g/cc and it’s E.C.E. = 3.04×10 g/C)
a) 2.4 m b) 0.24 μm c) 2.4 μm d) None of these
P a g e | 20
156. An AC generator of 220 V have internal resistance r = 10 Ω and external resistance R = 100 Ω. What is
the power developed in the external circuit?
a) 227 W b) 325 W c) 400 W d) 500 W

157. In the circuit shown here, what is the value of the unknown resistor R so that the total resistance of the
circuit between points P and Q is also equal to R
10

3
P Q
3 R

a) 3 ohm b) 39 ohm c) 69 ohm d) 10 ohm

158. The resistance of a wire is R. If the length of the wire is doubled by stretching, then the new resistance will
be
a) 2R b) 4R c) R d) R
4
159. By ammeter, which of the following can be measured

a) Electric potential b) Potential difference c) Current d) Resistance

160. The maximum power drawn out of the cell from a source is given by (where r is internal resistance)

a) E2/2r b) E2/4r c) E2/r d) E2/3r

161. The emf is thermocouple changes sign at 600 K. If the neutral temperature is 210℃, the temperature of
cold junction is
a) 180 K b) 117 K c) 93℃ d) 90℃

162. How many minimum number of 2 Ω resistance can be connected to have an effective resistance of 1.5 Ω?

a) 3 b) 2 c) 4 d) 6

163. Equal potentials are applied on an iron and copper wire of same length. In order to have the same current
flow in the two wires, the ration r (iron)/r (copper) of their radii must be (Given that specific resistance of
-7 -8
iron = 1.0×10 ohm - m and specific resistance of copper = 1.7×10 ohm - m)
a) About 1.2 b) About 2.4 c) About 3.6 d) About 4.8

164. A fuse wire of circuit cross-section and having diameter of 0.4 mm, allows 3 A of current to pass through it.
But if another fuse wire of same material and circular cross-section and having diameter of 0.6 mm is
taken, then the amount of current passed through the fuse is

() ()
3/2
a) 3 A b) 3× 3 A c) 3× 3 d) 3 × 3 A
A
2 2 2
165. Two identical cells weather connected in parallel or in series gives the same current when connected to an
external resistance 1.5 Ω. Find the value of internal resistance of each cell.
a) 1 Ω b) 0.5 Ω c) Zero d) 1.5 Ω

166. In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of
galvanometer is x, what would be its value if the radius of the wire AB is doubled?

P a g e | 21
 | |

R1 R2

A
B
X C

a) x b) x/4 c) 4x d) 2x

167. A current of two ampere is flowing through a cell of e.m.f. 5 volt and internal resistance 0.5 ohm from
negative to positive electrode. If the potential of negative electrode is 10V, the potential of positive
electrode will be
a) 5 V b) 14 V c) 15 V d) 16 V

168. Two bulbs 25 W, 220 V and 100 W, 220 V are given. Which has higher resistance?

a) 25 W bulb b) 100 W bulb

c) Both bulbs will have equal resistance d) Resistance of bulbs cannot be compared

169. The temperature of cold, hot junction of a thermocouple is 0℃ and T℃ respectively. The thermo-emf
1 2
produced is E = AT - BT . If A = 16, B = 0.080, the temperature of inversion will be
2
a) 100℃ b) 300℃ c) 400℃ d) 500℃

170. The equivalent resistance across A and B is

4Ω 4Ω

A 10 B
Ω

4Ω 4Ω

a) 2Ω b) 3Ω c) 4Ω d) 5Ω

171. The length of a potentiometer wire is 5m. An electron in this wire experiences a force of 4.8×10-19N, emf of
the main cell used in potentiometer is
a) 3 V b) 15 V c) 1.5 V d) 5 V

172. When a piece of aluminium wire of finite length is drawn through a series of dies to reduce its diameter to
half its original value, its resistance will become
a) Two times b) Four times c) Eight times d) Sixteen times

173. A voltmeter of resistance 1000 Ω is connected across a resistance of 500 Ω in the given circuit. What will
be the reading of voltmeter

P a g e | 22
 a) 1 V b) 2 V c) 6 V d) 4V

174. A resistance of 4Ω and a wire of length 5 metres and resistance 5Ω are joined in series and connected to a
cell of e.m.f. 10 V and internal resistance 1Ω. A parallel combination of two identical cells is balanced
across 300 cm of the wire. The e.m.f. E of each cell is
4 10V
1

3m
5, 5m
E
G
E

a) 1.5 V b) 3.0 V c) 0.67 V d) 1.33 V

175. Current flows through a metabolic conductor whose area of cross-section increases in the direction of the
current. If we move in this direction,
a) The carrier density will change b) The current will change

c) The drift velocity will decrease d) The drift velocity will increase

176. The resistance will be least in a wire with dimension

a) L/2,2A b) 2L, A c) L, A d) None of these

177. At room temperature, copper has free electron density of 8.4×1028 per m3. The copper conductor has a
-6 2
cross-section of 10 m and carries a current of 5.4 A. The electron drift velocity in copper is
a) 400 m/s b) 0.4 m/s c) 0.4 mm/s d) 72 m/s

178. A battery is charged at a potential of 15 V in 8 hours when the current flowing is 10 A. The battery on
discharge supplies a current of 5 A for 15 hours. The mean terminal voltage during discharge is 14 V. The
"Watt - hour" efficiency of battery is
a) 80% b) 90% c) 87.5% d) 82.5%

179. A combination of two resistance of 2 W and 2/3 W connected in parallel is joined across a battery of emf of
3 V and of negligible internal resistance. The energy given out per sec will be
a) 1 ×3×3 J b) 1 × 1 ×3×3 J c) 2×3 J d) 3×3×2 J
2 2 3
180. The length of the wire is doubled. Its conductance will be

a) Unchanged b) Halved

c) Quadrupled d) 1/4 of the original value

181. In the circuit as shown in the figure, the heat produced by 6 ohm resistance due to current flowing in it is
P a g e | 23
 60 calorie per second. The heat generated across 3 ohm resistance per second will be
2 3

6 4

a) 30 calorie b) 60 calorie c) 100 calorie d) 120 calorie

182. Two identical incandescent light bulbs are connected as shown in the figure. When the circuit is an AC
voltage source of frequency f, which of the following observations will be correct?

a) Both bulbs will glow alternatively

b) Both bulbs will glow with same brightness provided frequency f = 1 (1/LC)
2π
c) Bulb b will light up initially and goes off, bulb b will be ON constantly
1 2

d) Bulb b will blink and bulb b will be ON constantly

1 2

183. A wire of length 5m and radius 1 mm has a resistance of 1 ohm. What length of the wire of the same
material at the same temperature and of radius 2 mm will also have a resistance of 1 ohm
a) 1.25 m b) 2.5 m c) 10 m d) 20 m

184. It is possible that any some constant value of emf, but the potential difference between the plates is zero?

a) Not, possible

b) Yes, if another identical battery is joined in series

c) Yes, if another identical battery is joined in opposition

d) Yes, possible, if another similar battery is joined in parallel

185. Six equal resistances are connected between point s P,Q and R as shown in the figure. Then the net
resistance will be maximum between

a) P and Q b) Q and R c) P and R d) Any two points

186. When a 12Ω resistor is connected with a moving coil galvanometer then its deflection reduces from 50
divisions to 10 divisions. The resistance of the galvanometer is
a) 24 Ω b) 36 Ω c) 48 Ω d) 60 Ω

187. A galvanometer having a resistance of 8 ohm is shunted by a wire of resistance 2 ohm. If the total current
is 1 amp, the part of it passing through the shunt will be
P a g e | 24
 a) 0.25 amp b) 0.8 amp c) 0.2 amp d) 0.5 amp

188. Resistors of 1, 2, 3 ohm are connected in the form of a triangle. If a 1.5 volt cell of negligible internal
resistance is connected across 3 ohm resistor, the current flowing through this resistance will be
a) 0.25 amp b) 0.5 amp c) 1.0 amp d) 1.5 amp

189. A galvanometer of resistance 50 Ω is connected to a battery of 3V along with a resistance of 2950 Ω in

series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this
deflection to 20 divisions, the resistance in series should be
a) 5050 Ω b) 5550 Ω c) 6050 Ω d) 4450 Ω

190. If a 30 V, 90 W bulb is to be worked on a 120 V line, a resistance of how many ohms should be connected in
series with the bulb
a) 10 ohm b) 20 ohm c) 30 ohm d) 40 ohm

191. n equal resistors are first connected in series and then connected in parallel. What is the ratio of the
maximum to the minimum resistance
a) n b) 12 c) n2 d) 1
n n
192. Figure shows a simple potentiometer circuit for measuring a small e.m.f. produced by a thermocouple. The
meter wire PQ has a resistance 5 Ω and the driver cell has an e.m.f. of 2 V. If a balance point is obtained
0.600m along PQ when measuring an e.m.f. of 6.00 mV, what is the value of resistance R

a) 995 Ω b) 1995 Ω c) 2995 Ω d) None of these

193. The ammeter has range 1 ampere without shunt. The range can be varied by using different shunt
resistances. The graph between shunt resistance and range will have the nature

a) P b) Q c) R d) S

194. In a Wheatstone’s network P = 2Ω, Q = 2Ω, R = 2Ω and S = 3Ω. The resistance with which S is to be
shunted in order that the bridge may be balanced is
a) 1 Ω b) 2 Ω c) 4 Ω d) 6 Ω

P a g e | 25
195. In the Wheatstone’s bridge shown, P = 2Ω, Q = 3Ω, R = 6Ω and S = 8 Ω. In order to obtain balance,
shunt resistance across 'S' must be
P Q

S R

a) 2 Ω b) 3 Ω c) 6 Ω d) 8 Ω

196. If an observer is moving with respect to a stationary electron, then he observes

a) Only magnetic field b) Only electric field c) Both (a) and (b) d) None of the above

197. If 2.2kW power is transmitted through a 100Ω line at 22,000V, the power loss in the form of heat will be

a) 0.1 W b) 1 W c) 10 W d) 100 W

198. The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficient of
resistance be 0.005 per°C, its resistance will become 200 Ω at a temperature of
a) 300°C b) 400°C c) 500°C d) 200°C

199. Find out the value of current through 2Ω resistance for the given circuit

5 10 20V
10V

2

a) 5 A b) 2 A c) Zero d) 4 A

200. Which of the plots shown in figure may represent the thermal energy produced in a resistor in a given time
as a function of the electric current?

a) a b) b c) c d) d

201. Two resistors of resistances 200 kΩ and 1MΩ respectively form a potential divider with outer junctions
maintained at potentials of +3V and -15V. Then, the potential at the junction between the resistors is
a) +1 V b) -0.6 V c) 0 V d) -12 V

202. In a Wheatstone bridge, P = 90Ω,Q = 110Ω, R = 40Ω and S = 60Ω and a cell of 4 V emf. Then the
potential difference between the diagonal along which a galvanometer is connected is
a) -0.2 V b) +0.2 V c) -1 V d) +1 V

203. A thermo couple develops 200μ V between 0℃ and 100℃. If it develops 64 μ V and 76 μ V respectively
between (0℃ - 32℃) and (32℃ - 70℃) then what will be the thermo emf it develops between 70℃ and
100℃
a) 65 μ V b) 60 μ V c) 55 μ V d) 50 μ V

204. In a potentiometer arrangement, a cell of emf 1.5V gives a balance point at 27cm length of wire. If the cell

P a g e | 26
 is replaced by another cell and balance point shifts to 54cm, the emf of the second cell is
a) 3V b) 1.5V c) 0.75V d) 2.25V

205. Which of the following is not reversible

a) Joule effect b) Peltier effect c) Seebeck effect d) Thomson effect

206. The equivalent resistance of the circuit shown in the figure is

2
2 2

2

a) 8 Ω b) 6 Ω c) 5 Ω d) 4 Ω

207. A battery having e.m.f. 5V and internal resistance 0.5 Ω is connected with a resistance of 4.5 Ω then the
voltage at the terminals of battery is
a) 4.5 V b) 4 V c) 0 V d) 2 V

208. If the electric current through an electric bulb is 3.2 A, the number of electrons flow through it in one
second is
a) 2×109 b) 2×1019 c) 3.2×1019 d) 1.6×1018

209. Two tangent galvanometer A and B are identical except in their number of turns. They are connected in
series. On passing a current through them, deflections of 60° and 30° are produced. The ratio of the
number of units A and B is
a) 1:3 b) 3:1 c) 1:2 d) 2:1

210. A solenoid is at potential difference of 60 V and current flowing through it is 15 ampere, then the
resistance of coil will be
a) 4Ω b) 8Ω c) 0.25Ω d) 2Ω

211. A 50 ohm galvanometer gets full scale deflection when a current of 0.01 A passes through the coil. When it
is converted to a 10 A ammeter, the shunt resistance is
a) 0.01 Ω b) 0.05 Ω c) 2000 Ω d) 5000 Ω

212. Sensitivity of potentiometer can be increased by

a) Increasing the e.m.f. of the cell

b) Increasing the length of the potentiometer wire

c) Decreasing the length of the potentiometer wire

d) None of the above

213. In the circuit shown the value of I in ampere is

4Ω

P a g e | 27
 I
4Ω 4Ω

4
V

a) 1 b) 060 c) 0.4 d) 1.5

214. A moving coil galvanometer of resistance 100Ωis used as an ammeter using a resistance 0.1Ω. The
maximum deflection current in the galvanometer is 100μA. Find the minimum current in the circuit so that
the ammeter shows maximum deflection
a) 100.1mA b) 1000.1mA c) 10.01mA d) 1.01mA

215. When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the
rate of heat produced in R is J1. When the same batteries are connected I parallel across R, the rate is J2. If
J1 = 2.25 J2 then the value of R in Ω is
a) 4 b) 6 c) 4.8 d) 5.16

216. A metal wire is subjected to a constant potential difference. When the temperature of the metal wire
increases, the drift velocity of the electron in it
a) increases, thermal velocity of the electron decreases

b) Decreases, thermal velocity of the electron decreases

c) increases, thermal velocity of the electron increases

d) Decreases, thermal velocity of the electron increases

217. A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2Ω each
and leaves by the corner R. Then the current I1 and I2 are

a) 2A,4A b) 4A,2A c) 1A,2A d) 2A,3A

218. There is a current of 0.21 A in a copper wire whose area of cross-section is 10-6m2. If the number of free
3 28
electrons per m is 8.4×10 , then find the drift velocity,
-19
(e = 1.6×10 C)
a) 2×10-5 ms-1 b) 1.56×10-5 ms-1 c) 1×10-5 ms-1 d) 0.64×10-5 ms-1

219. If 2.2 kilowatt power is transmitted through a 10 ohm line at 22000 volt, the power loss in the form of heat
will be
a) 0.1 watt b) 1 watt c) 10 watt d) 100 watt

220. In the circuit shown in figure the heat produced in the 5Ω resistor due to the current flowing through it is
-1
100Js .The heat generated in the 4Ω resistor is

P a g e | 28
 4 6
i i
5

a) 10 Js-1 b) 20 Js-1 c) 30 Js-1 d) 40 Js-1

221. Two copper wires have their masses in the ratio 2 : 3 and the lengths in the ratio 3 : 4 the ratio of their
resistance is
a) 4 : 9 b) 27 : 32 c) 16 : 9 d) 27 :128

222. A battery of emf E and internal resistance r is connected to an external resistance R the condition for
maximum power transfer is
a) r<R b) r>R c) r=1/R d) R=R

223. The cold junction of a thermocouple is maintained at 10℃. No thermo e.m.f. is developed when the hot
junction is maintained at 530℃. The neutral temperature is
a) 260℃ b) 270℃ c) 265℃ d) 520℃

224. An electric heater boils 1 kg of water in a time t1. Another heater boils the same amount of water in a time
t2. When the two heaters are connected in parallel, the time required by them together to boil the same
amount of water is
a) t + t b) t t c) t1+t2 d) t1t2
1 2 1 2
2 t1+t2
225. A voltmeter having a resistance of 998 ohm is connected to a cell of e.m.f. 2 volt and internal resistance 2
ohm. The error in the measurement of e.m.f. will be
a) 4×10-1volt b) 2×10-3volt c) 4×10-3 volt d) 2×10-1 volt

226. In the circuit shown, if the 10 Ω resistance is replaced by 20 Ω then what is the amount of current drawn
from the battery?
3Ω 3Ω

10 Ω

4A
4Ω 4Ω

a) 2.5A b) 3A c) 3.5A d) 4A

227. The figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 2.5 V
cell. The balance point of the cell in the open circuit is 75cm. When a resistor of 10 Ωis used in the external
circuit of the cell, the balance point shifts to 65 cm length of potentiometer wire. Then the internal
resistance of the cell is

P a g e | 29
 2.0 V

j' j
--
--
-
G- - -
-- G
--
--
--
1.5 V --
-

10 Ω

a) 2.5 Ω b) 2.0 Ω c) 1.54 Ω d) 1.0 Ω

228. A battery of 24 cells, each of emf 1.5 V and internal resistance 2Ω is to be connected in order to send the
maximum current through a 12Ω resistor. The correct arrangement of cells will be
a) 2 rows of 12 cells connected in parallel b) 3 rows of 8 cells connected in parallel

c) 4 rows of 6 cells connected in parallel d) All of these

229. A cell of emf E and internal resistance r supplies currents for the same time t through external resistance
R1 =100 Ω and R2 = 40 Ω separately. If the heat developed in both the cases in the same, then the internal
resistance of the cell is given by
a) 28.6 Ω b) 70 Ω c) 63.3 Ω d) 140 Ω

230. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the
increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances
R100, R60 and R40, respectively, the relation between these resistances is
a) 1 = 1 + 1 b) R = R + R c) R > R > R d) 1 > 1 > 1
R100 R40 R60 100 40 60 100 60 40 R100 R60 R40
231. In a potentiometer circuit there is a cell of e.m.f. 2 volt, a resistance of 5 ohm and a wire of uniform
thickness of length 1000 cm and resistance 15 ohm. The potential gradient in the wire is
a) 1 V/cm b) 3 V/cm c) 3 V/cm d) 1 V/cm
500 2000 5000 1000
232. Equal amounts of a metal are converted into cylindrical wire of different lengths L and cross-sectional area
A.The wire with the maximum resistance is the one, which has
a) Length=L and area = A

b) lengths = L and area = 2A

2
c) lengths = 2L and area = A
2
d) All have the same resistance, as the amount of the metal is the same

233. Amount of electricity required to pass through the H2O voltmeter so as to liberate 11.2 litre of hydrogen
will be
a) 1 faraday b) 1 faraday c) 2 faraday d) 3 faraday
2
234. Five cells each of internal resistances 0.2Ω and emf 2 V are connected in series with a resistance of 4Ω. The
current through the external resistance is
a) 4 A b) 2 A c) 1 A d) 0.5 A

235. 3 identical bulbs are connected in series and these together dissipate a power P. If now the bulbs are

P a g e | 30
 connected in parallel, then the power dissipated will be
a) P b) 3P c) 9P d) P
3 9
236. How many coulombs of electric charge must pass through acidulated water in order to release 22.4 L Of
hydrogen at NTP?
a) 96500 Faraday b) 193000 coulomb c) 196500 Faraday d) 96500 coulomb

237. Two identical cells connected in series send 1.0 A current through a 5Ω resistor. When they are connected
in parallel, they send 0.8 A current through the same resistor. What is the internal resistance of the cell?
a) 0.5 Ω b) 1.0 Ω c) 1.5 Ω d) 2.5 Ω

238. Which of the following is not equal to watt

a) (Amp)2×ohm b) Amp/Volt c) Amp×Volt d) Joule/sec

239. The current in the given circuit is

RA = 3 RB = 6
4.8V

RC = 6

a) 8.31 A b) 6.82 A c) 4.92 A d) 2 A

240. A lamp having tungsten filament consumes 50 W. Assume the temperature coefficient of resistance for
-3 -1
tungsten is 4.5×10 ℃ and temperature of the surrounding is 20℃. When the lamp burns, the
temperature of its filament becomes 2500℃ , then the power consumed at the moment switch is on, is
a) 608 W b) 710 W c) 215 W d) 580 W

241. A wire of resistance R is divided in 10 equal parts. These parts are connected in parallel, the equivalent
resistance of such connection will be
a) 0.01 R b) 0.1 R c) 10 R d) 100 R

242. In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance 0.1
ohm/cm. The value of unknown resistance X and the current drawn from the battery of negligible
resistance is
X 6

G
40 cm 60 cm
A B

5V

a) 6 Ω, 5 amp b) 10 Ω, 0.1 amp c) 4 Ω, 1.0 amp d) 12 Ω, 0.5 amp

243. Emf is most closely related to

a) Mechanical force b) Potential difference c) Electric field d) Magnetic field

244. An electron (charge = 1.6×10-19coulomb) is moving in a circle of radius 5.1×10-11m at a frequency of

15
6.8×10 revolutions/sec. The equivalent current is approximately
a) 5.1×10-3amp b) 6.8×10-3amp c) 1.1×10-3amp d) 2.2×10-3amp

P a g e | 31
245. The drift velocity of free electrons in a conductor is 'v' when a current 'i' is flowing in it. If both the radius
and current are doubled, then drift velocity will be
a) v b) v c) v d) v
2 4 8
246. If a steady current of 100 A is passed then how much time is taken to deposit 0.254 kg of copper on the
cathode of copper voltmeter. Use the known value of Faraday constant and relative atomic mass of copper
is 63.5.
a) 15440 s b) 7720 s c) 3760 s d) 5480 s

247. Two similar cells, whether joined in series or in parallel, have the same current through an external
resistance of 2 Ω. The internal resistance of each cell is
a) 1 Ω b) 2 Ω c) 0.5 Ω d) 1.5 Ω

248. The current in the 1Ω resistor shown in the circuit is

4
ΩI

1Ω 6V

4Ω

a) 2 A b) 3A c) 6A d) 2A
3
249. An electric heater of 1.08 Kw is immersed in water. After the water has reached a temperature of 100℃,
how much time will be required to produce 100 g of steam?
a) 420 s b) 210 s c) 105 s d) 50 s

250. How many calories of heat will be produced approximately in a 210 W electric bulb in 5 min?

a) 80000 cal b) 63000 cal c) 1050 cal d) 15000 cal

251. Four resistances are connected in a circuit in the given figure. The electric current flowing through 4 ohm
and 6 ohm resistance is respectively
4 6

4 6

20V

a) 2 amp and 4 amp b) 1 amp and 2 amp c) 1 amp and 1 amp d) 2 amp and 2 amp

252. The current passing through the ideal ammeter in the circuit given below is
2Ω
1Ω , 4V
2Ω
+ - 2Ω

4Ω

P a g e | 32
 a) 1.25A b) 1A c) 0.75A d) 0.5A

253. The potential difference between point A & B is

8 B 6

4 3

10 V

a) 20 V b) 40 V c) 10 V d) 0
7 7 7
254. According to Joule’s law, if the potential difference across a conductor having a material of specific
resistance remains constant, then the heat produced in the conductor is directly proportional to
a) ρ b) ρ2 c) 1 d) 1
ρ ρ
255. If power dissipated in the 9 Ω resistor in the circuit shown is 36 Watt, the potential difference across the 2
Ω resistor is

a) 2 volt b) 4 volt c) 8 volt d) 10 volt

256. The resistor of resistance R is connected to 25 V supply and heat produced in it is 25 Js-1. The value of R is

a) 225 Ω b) 1 Ω c) 25 Ω d) 50 Ω

257. The resistance between the points A and C in the figure below is
RΩ
A B
RΩ
RΩ
RΩ E RΩ

R RΩ

D C
RΩ

a) R Ω b) 4 Ω c) 2 RΩ d) 8R
3 3 3
258. The deflection in a moving coil galvanometer is reduced to half when it is shunted with a 40 Ω coil. The
resistance of the galvanometer is
a) 15 Ω b) 20 Ω c) 40 Ω d) 80 Ω

259. The resistance of a cell does not depend on

a) Current drawn from the cell b) Temperature of electrolyte

P a g e | 33
 c) Concentration of electrolyte d) The e.m.f. of the cell

260. In the given figure the steady state current in the circuit is

2Ω

3Ω

C = 0.2 μF 4Ω

6V 2.8Ω

a) Zero b) 0.6A c) 0.9A d) 1.5A

261. Twelve wires of equal length and same cross-section are connected in the form of a cube. If the resistance
of each of the wires is R, then the effective resistance between the two diagonal ends would be

a) 2 R b) 12 R c) 5 R d) 8 R
6
262. The length of a conductor is doubled and its radius is halved, its specific resistance is

a) Unchanged b) Halved c) Doubled d) Quadrupled

263. Four identical resistors of 4 Ω each are joined in circuit as shown in figure. The cell B has emf 2V and its
internal resistance is negligible. The ammeter reading is
4

B
+ -
A
4
4

a) 3 A b) 2A c) 1 A d) 1 A
8 2 8
264. Variation of current and voltage in a conductor has been shown in the diagram below. The resistance of
the conductor is

P a g e | 34
 V 6
5
4
3
2
1
1 2 3 4 5 6
i

a) 4 ohm b) 2 ohm c) 3 ohm d) 1 ohm

265. Resistance of rod is 1 Ω.It is bent in form of square. What is resistance across adjoint corners?

a) 1 Ω b) 3 Ω c) 3 Ω d) 3 Ω
16 4
23 -19
266. The Avogadro’s number is 6×10 per gm mole and electronic charge is 1.6×10 C. The Faraday’s number
is
23
a) 6×1023×1.6×10-19 b) 6×10
-19
1.6×10
-19
c) 2 d) 1.6×10
23 -19 23
6×10 ×1.6×10 6×10
267. In the network shown in the figure, each of the resistance is equal to 2 Ω. The resistance between the
points A and B is

A
B

a) 1 Ω b) 4 Ω c) 3 Ω d) 2 Ω

268. The temperature of cold junction of thermo-couple is 0℃. If the neutral temperature is 270℃, then the
inversion temperature is
a) 540℃ b) 520℃ c) 640℃ d) 580℃

269. The mobility of free electrons (charge e, mass m and relaxation time τ) in a metal is proportional to

a) e τ b) m τ c) e d) m
m e mτ eτ
270. In the figure shown, the total resistance between A and B is
2 C 1 1 1 1 1
A
8 8 4

B 2 D 1 1 1 1 1

a) 12 Ω b) 4 Ω c) 6 Ω d) 8 Ω

271. In the electrical network shown in the figure, the potential difference across 3Ω resistance will be

P a g e | 35
 3Ω

6Ω
18A

a) 12V b) 2.4V c) 24 V d) 36 V

272. If n,e,τ and m respectively represent the density, charge relaxation time and mass of the electron, then the
resistance of a wire of length l and area of cross-section A will be
2 2 2
a) ml 2
b) mτ A c) ne τA d) ne A
2
ne τA ne l 2ml 2mτl
273. A ring is made of a wire having a resistance R0 = 12Ω. Find the points A and B as shown in the figure, at
which a current carrying conductor should be connected so that the resistance R of the sub circuit between
8
these points is equal to Ω
3

a) l1 = 5 b) l1 = 1 c) l1 = 3 d) l1 = 1
l2 8 l2 3 l2 8 l2 2
274. When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found
to be 540 cm. If the balancing length becomes 500 cm when the cell is short circuited with 1 Ω, the
internal of the cell is
a) 0.08 Ω b) 0.04 Ω c) 1.0 Ω d) 1.08 Ω

275. What is the equivalent resistance between the points A and B of the network
2Ω 3Ω 2Ω
A
2Ω
4Ω 1Ω
10Ω 1Ω

1.8Ω 5Ω
2.2Ω
B

a) 57 Ω b) 8 Ω c) 6 Ω d) 57 Ω
7 5
276. An electric water kettle rated 2.1 kW is filled with 1.5 kg of water at 20°C. How many seconds does it take
to reach the boiling point of water? Assume that there are no heat losses from the kettle. Specific heat
-1 -1
capacity of water is 4200 Jkg K
a) 60 b) 120 c) 240 d) 480

277. A cell can be balanced against 110cm and 100cm of potentiometer wire, respectively with and without
being short circuited through a resistance of 10 Ω. Its internal resistance is
a) 1.0 Ω b) 0.5 Ω c) 2.0 Ω d) Zero

P a g e | 36
278. Kirchhoff’s I law and II law of current, prove the

a) Conservation of charge and energy b) Conservation of current and energy

c) Conservation of mass and charge d) None of these

279. The resistivity of iron is 1×10-7ohm - m. The resistance of a iron wire of particular length and thickness is 1
ohm. If the length and the diameter of wire both are doubled, then the resistivity in ohm - m will be
a) 1×10-7 b) 2×10-7 c) 4×10-7 d) 8×10-7

280. In an experiment, a graph was plotted of the potential difference V between the terminals of a cell against
the circuit current i by varying load rheostat. Internal conductance of the cell is given by

a) xy b) y c) x d) (x - y)
x y
281. For a thermocouple the neutral temperature is 270℃ when its cold junction is at 20℃. What will be the
neutral temperature and the temperature of inversion when the temperature of cold junction is increased
to 40℃
a) 290℃, 580℃ b) 270℃, 580℃ c) 270℃, 500℃ d) 290℃, 540℃

282. In the following circuit, 5Ω resistor develops 45 J/s due to current flowing through it. The power
developed per second across 12 Ω resistor is
i2 9 6

12 

i1 5

a) 16 W b) 192 W c) 36 W d) 64 W

283. 2
A current of
3 ()
A produces a deflection of 60° in a tangent galvanometer. The reduction factor is

a) 2 A
3 () b) 2A c) 2 A
3 () d) 2 A
3 ( )
284. Two sources of equal emf are connected to an external resistance R. The internal resistances of the two
sources are R1 and R2(R2>R1). If the potential difference across the source having internal resistance R2 is
zero, then
a) R = R R /(R + R ) b) R = R R /(R - R )
1 2 1 2 1 2 2 1

c) R = R ×(R + R )/(R - R ) d) R = R - R
2 1 2 2 1 2 1

285. Which of the following has a negative temperature coefficient

a) C b) Fe c) Mn d) Ag

286. A cell supplies a current i1through a resistance R1 and a current I2 through a resistance R2. The internal
resistance of a cell is

P a g e | 37
 a) R - R b) (i1+i2) R R c) i1R2-i2R1 d) i2R2-i1R1
2 1 i1-i2 1 2 i1-i2 i1-i2
287. Two ends of a conductor are at different temperatures the electromotive force generated between two
ends is
a) Seebeck electro motive force (e.m.f.) b) Peltier electro motive force (e.m.f.)

c) Thomson electro motive force (e.m.f.) d) None of these

288. What is the equivalent resistance between A and B

C
A 2R 2R D R B

a) 2 R b) 3 R c) R d) 2R
3 2 2
289. The effective resistance between points A and B in figure

4
6
12
3 5
A 24 B

a) 10 Ω b) 12 Ω c) 9.85 Ω d) 10.85 Ω

290. The current density (number of free electrons per m3) in metallic conductor is of the order of

a) 1022 b) 1024 c) 1026 d) 1028

291. The potential difference between points A and B of adjoining figure is

5 5
A B

2V
5 5

5 5
D C

a) 2 V b) 8 V c) 4 V d) 2V
3 9 3
292. The potential gradient along the length of a uniform wire is 10Vm-1.The length of the potentiometer wire is
4 m. What is the potential difference across two points on the wire separated by 50cm?
a) 2.5 V b) 5.0 V c) 1.25 V d) 4.0 V

293. In the circuit shown, the value of each resistance is r, then equivalent resistance of circuit between points
A and B will be
r

r r r
r

r r
A B
C

P a g e | 38
 a) (4/3)r b) 3r/2 c) r/3 d) 8r/7

294. For the circuit shown in the figure

2k R1
I 24 V

24 V 6k R2 RL 1.5k

a) The current I through the battery is 7.5mA b) The potential difference across R is 18 V
L

If R1 and R2 are interchanged, magnitude of the

c) Ratio of powers dissipated in R and R is 3 d) power dissipated in R will decrease by a factor of
1 2 L
9
295. A brass rectangular plate 12cm×3cm is to be electroplated with copper. If we wish to coat it with a layer of
0.02 mm thick both sides, how much time will it take with a constant current of 5A? Given ECE of copper is
-5 -1 -3
33×10 g C and density of copper is 8.9 g cm .
a) 388 s b) 776 s c) 400 s d) 800 s

296. You are given several identical resistances each of value R = 10Ω and each capable of carrying maximum
current of 1 ampere. It is required to make a suitable combination of these resistances to produce a
resistance of 5Ω which can carry a current of 4 ampere. The minimum number of resistances of the type R
that will be required for this job
a) 4 b) 10 c) 8 d) 20

297. A 12 V lead accumulator is being charged using 24 V supply with an external resistance 2Ω. The internal
resistance of the accumulator is 1Ω. Find the time in which it will store 360 W-hour energy
a) 1 hr b) 7.5 hr c) 10 hr d) None of these

298. As the switch S is closed in the circuit shown in figure, current passed through it is

10 V 4 2 5V

a) Zero b) 1 A c) 2 A d) 1.6 A

299. If nearly 105C liberate 1 g equivalent of aluminium, then the amount of aluminium (equivalent weight 9)
deposited through electrolysis in 20 min by a current of 50 A will be
a) 0.09 g b) 0.6 g c) 5.4 g d) 10.8 g

300. Potential gradient is defined as

a) Fall of potential per unit length of the wire

b) Fall of potential per unit area of the wire

P a g e | 39
 c) Fall of potential between two ends of the wire

d) Potential at any one end of the wire

301. The current flowing in a copper voltmeter is 3.2 A. The number of copper ions (Cu2+) deposited at the
cathode per minute is
a) 0.5×1020 b) 1.5×1020 c) 3×1020 d) 6×1020

302. The specific resistance of manganin is 50×10-8 ohm×m. The resistance of a cube of length 50cm will be

a) 10-6 ohm b) 2.5×10-5 ohm c) 10-8 ohm d) 5×10-4 ohm

303. The total current supplied to the given circuit by the battery is

a) 9 A b) 6 A c) 2 A d) 4 A

304. Combination of two identical capacitors, a resistor R and a DC voltage source of voltage 6 V is used in an
experiment onC - R circuit. It is found that for a parallel combination of the capacitor the time in which the
voltage of the fully charged combination reduces to half its original voltage is 10s. For series combination
the time needed for reducing the voltage of the fully charged series combination by half is
a) 200s b) 10s c) 5s d) 2.5s

305. If 400Ω of resistance is made by adding four 100Ω resistance of tolerance 5%, then the tolerance of the
combination is
a) 20 % b) 5 % c) 10 % d) 15 %

306. A 5℃ rise in temperature is observed in a conductor by passing a current. When the current is doubled the
rise in temperature will be approximately
a) 16℃ b) 10℃ c) 20℃ d) 12℃

307. In the following circuit, 18Ω resistor develops 2J/sec due to current flowing through it. The power
developed across 10Ω resistance is
12 9
9

10
12 9
18

a) 125 W b) 10 W c) 4 W d) 25 W
5
308. 1 2
e = α t - β t , if temperature of cold junction is 0℃ then temperature of inversion is
2
(if α = 500.0μV/℃,β = 5.0μV/square℃)
a) 100 b) 200 c) 300 d) 400

309. In Cu - Fe couple, the flow of current at the temperature of inversion is

P a g e | 40
 a) From Fe to Cu through the hot junction

b) From Cu to Fe through the hot junction

c) Maximum

d) None of the above

310. Two wires A and B of same material and same mass have radii 2r and r respectively. If resistance of wire A
is 34Ω, then resistance of B will be
a) 544Ω b) 272Ω c) 68Ω d) 17Ω

311. Three electric bulbs with same voltage ratings of 110 volts but wattage ratings of 40, 60 and 100 watts
respectively are connected in series across a 220 volt supply line. If their brightness are B1,B2,B3
respectively, then
a) B > B > B
1 2 3

b) B > B < B
1 2 3

c) B = B = B
1 2 3

d) Bulbs will burn out due to the high voltage supply

312. Consider a rectangular slab of length L and area of cross section A. A current I is passed through it. If the
length is doubled, the potential drop across the end faces
a) Becomes half of the initial value b) Becomes one-fourth of the initial value

c) Becomes double the initial value d) Remains same

313. Two cells having emf 4V, 2V and internal resistances 1 Ω, 1 Ω are connected as shown in figure below.
Current through 6 Ω resistance is
4 V 1Ω 2 V1Ω

3Ω
2Ω
6Ω

a) 1 A b) 2 A c) 1A d) 2 A
3 3 9
314. Kirchhoff’s first law i.e. ∑i=0 at a junction is based on the law of conservation of

a) Charge b) Energy c) Momentum d) Angular momentum

315. The equivalent resistance of the following infinite network of resistance is

2 2 2
A

2 2 2

B
2 2 2

a) Less than 4Ω b) 4Ω

c) More than 4Ω but less than 12Ω d) 12Ω

P a g e | 41
316. The resistance in which the maximum heat is produced is given by
4

12
2V

a) 2Ω b) 6Ω c) 4Ω d) 12Ω

317. We are able to obtain fairly large currents in a conductor because

a) The electron drift speed is usually very large

b) The number density of free electrons is very high and this can compensate for the low values of the
electron drift speed and the very small magnitude of the electron charge
c) The number density of free electrons as well as the electron drift speeds are very large and these
compensate for the very small magnitude of the electron charge
d) The very small magnitude of the electron charge has to be divided by the still smaller product of the
number density and drift speed to get the electric current
318. A battery of emf E has an internal resistance r. A variable resistance R is connected to the terminals of the
battery. A current i is drawn from the battery. V is the terminal potential difference. If R alone is gradually
reduced to zero, which of the following best describes i and V?
a) i approaches zero,V approaches E b) i approaches E , V approaches zero
r
E
c) i approaches ,V approaches E d) i approaches infinity,V approaches E
r
319. Kirchhoff’s second law is based on the law of conservation of

a) Charge b) Energy

c) Momentum d) Sum of mass and energy

320. The temperature at which thermal electric power of a thermo couple becomes zero is called

a) Inversion temperature b) Neutral temperature

c) Junction temperature d) Null temperature

321. Resistances R1 and R2 are joined in parallel and a current is passed so that the amount of heat liberated is
H
H1 and H2 respectively. The ratio 1 has the value
H2
2 2
a) R2 b) R1 c) R12 d) R22
R1 R2 R2 R1
322. Two wires A and B of same material and mass have their lengths in the ratio 1:2. On connecting them to
the same source, the rate of heat dissipation in B is found to be 5W. The rate of heat dissipation in A is
a) 10W b) 5W c) 20W d) None of these

323. Two electrolytic cells containing CuSO4 and AgNO3 respectively are connected in series and a current is
passed through them until 2 mg of copper is deposited in the first cell. The amount of silver deposited in
the second cell during this time in approximately (atomic weight of copper and silver are 63.6 and 108.0 )
P a g e | 42
 a) 1.7 mg b) 3.4 mg c) 5.1 mg d) 6.8 mg

324.
At neutral temperature, the thermoelectric power ( )
dE
dT
has the value

a) Zero b) Maximum but negative

c) Maximum but positive d) Minimum but positive

325. At steady state, energy stored in capacitor is

2 F
A B

2V

a) 4×10-6 J b) 2 J c) 4 J d) Zero

326. A fuse wire with a radius of 1 mm blows at 1.5 A. If the fuse wire of the same material should blow at 3.0 A,
the radius of the fuse wire must be
a) 1/3 b) 2 mm c) 0.5 mm d) 8.0 mm
4 mm
327. 5 ampere of current is passed through a metallic conductor. The charge flowing in one minute in coulomb
will be
a) 5 b) 12 c) 1/12 d) 300

328. The V - i graph for a good conductor makes angle 40° with V- axis. Here V denotes voltage and i denotes
current. The resistance of the conductor will be
a) sin 40° b) cos 40° c) tan 40° d) cot 40°

329. A galvanometer of resistance, G, is shunted by a resistance S ohm. To keep the main current in the circuit
unchanged, the resistance to be put in series with the galvanometer is
2 2
a) G b) G c) S d) SG
(S+G) (S+G) (S+G) (S+G)
330. A 50V battery is connected across a 10 Ω resistor and a current of 4.5 A flows. The internal resistance of
the battery is
a) 10 Ω b) 0.5 Ω c) 1.1 Ω d) 5 Ω

331. There are two electric bulbs of 40 W and 100 W. Which one will be brighter when first connected in series
and then in parallel
a) 40 W in series and 100 W in parallel

b) 100 W in series and 40 W in parallel

c) 40 W both in series and parallel will be uniform

d) 100 W both in series and parallel will be uniform

332. The conductivity of a superconductor is

a) Infinite b) Very large c) Very small d) Zero

333. Two identical cell send the same current in 2Ω resistance, whether connected in series or in parallel. The
internal resistance of the cell should be
P a g e | 43
 a) 1 Ω b) 2 Ω c) 1 Ω d) 2.5 Ω
2
334. According to Faraday’s law of electrolysis, the amount of decomposition is proportional to

a) 1/time for which current passes b) Electrochemical equivalent of the substance

c) 1/current d) 1/electrochemical equivalent

335. Three similar cells, each of emf 2V and internal resistance r send the same current through an external
resistance of 2Ω,when connected in series or in parallel. The strength of current flowing through the
external resistance is
a) 0.75 A b) 1 A c) 1.5 A d) zero

336. For what value of R the net resistance of the circuit will be 18 ohms
R

10 10

10

10 10 10

A B

a) 8 Ω b) 10 Ω c) 16 Ω d) 24 Ω

337. The resistance of a wire is R Ω.The wire is stretched to double its length keeping volume constant. Now the
resistance of the wire will become
a) 4 R Ω b) 2 R Ω c) R/2 Ω d) R/4 Ω

338. A uniform copper wire of length 1 m and cross-section area 5×10-7m2 carries a current of 1 A. Assuming
28 -3
that there are 8×10 free electron m in copper, how long will an electron take to drift from one end of the
wire to the other?
a) 0.8×103s b) 1.6×103s c) 3.2×103s d) 6.4×103s

339. A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it
into a voltmeter of range nV volt is
a) nG b) G c) (n - 1)G d) G
n n-1
340. To verify Ohm’s law, a student is provided with a test resistor RT, a high resistance R1, a small resistance R2,
two identical galvanometers G1and G2and a variable voltage source V. the correct circuit to carry out the
experiment is
a) G1

R2
G2
RT RT

P a g e | 44
 b) G1

R1
G2
R T R2

c) G1

G2
RT
R2

d) G1
R 2

G2
RT

R1

341. To decrease the range of an ammeter, its resistance need to be increased. An ammeter has resistance
R0and range I. Which of the following resistance can be connected in series with it to decreases its range to
I/n?
a) R0 b) R0 c) R0 d) None of these
n (n-1) (n+1)
342. If a current is allowed to pass through a circuit consisting of two dissimilar metals, there is either evolution
or absorption of heat at the junction, depending upon the direction of the current. The effect is known as
a) Seebeck effect b) Joule effect c) Peltier effect d) Thomson effect

343. Two bulbs when connected in parallel to a source take 60 W each. The total power consumed when they
are connected in series with the same source is
a) 15 W b) 30 W c) 60 W d) 120 W

344. In the circuit shown in figure, find the current through the branch BD

P a g e | 45
 a) 5 A b) 0 A c) 3 A d) 4 A

345. If E = at + bt2, what is the neutral temperature

a) - a b) + a c) - a d) + a
2b 2b b b
346. The resistance of an incandescent lamp is

a) Greater when switched off b) Smaller when switched on

c) Greater when switched on d) The same whether it is switched off or switched

on
347. The voltage V and current I graph for a conductor at two different temperatures T1 and T2 are shown in the
figure. The relation between T1 and T2 is
T1
V
T2

a) T > T b) T ≈ T c) T = T d) T < T
1 2 1 2 1 2 1 2

348. The current in a conductor varies with time t as I = 2t + 3t2 where I is in ampere and t in seconds.
Electric charge flowing through a section of the conductor during t = 2 sec to t = 3 sec is
a) 10 C b) 24 C c) 33 C d) 44 C

349. The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal
resistance r, will be
2 2 2 2
a) E b) E c) E d) E
r 2r 3r 4r
350. When a resistance of 100 Ω is connected in series with a gal vinometer of resistance R, its range is V. to
double its range, a resistance of 1000 Ω is connected in series. Find R
a) 700 Ω b) 800 Ω c) 900 Ω d) 100 Ω

351. A 10 Ω electric heater operates on a 110V line. The rate at which heat is developed in watts is

a) 1310 W b) 670 W c) 810 W d) 1210 W

352. In a galvanometer 5% of the total current in the circuit passes through it. If the resistance of the
galvanometer is G, the shunt resistance S connected to the galvanometer is
a) 19G b) G/19 c) 20G d) G/20

353. A resistance of 2Ω is to be made from a copper wire (specific resistance=1.7×10-8Ω m) using a wire of
length 50cm. The radius of the wire is
a) 0.0116 mm b) 0.367 mm c) 0.116 mm d) 0.267 mm

P a g e | 46
354. When a charged particle of charge e revolves in circular orbit of radius r with frequency n, then orbital
current will be
a) ev2 b) ev c) ev d) ev 2
πr 4πr 2πr 4πr
355. A battery of e.m.f. 10 V and internal resistance 0.5 ohm is connected across a variable resistance R. The
value of R for which the power delivered in it is maximum is given by
a) 2.0 ohm b) 0.25 ohm c) 1.0 ohm d) 0.5 ohm

356. In an electrical cable there is a single wire of radius 9 mm of copper. Its resistance is 5Ω. The cable is
replaced by 6 different insulated copper wires, the radius of each wire is 3mm. Now the total resistance of
the cable will be
a) 7.5 Ω b) 45 Ω c) 90 Ω d) 270 Ω

357. Two identical cells send the same current in 3 Ω resistance, whether connected in series or in parallel. The
internal resistance on the cell should be
a) 1 Ω b) 3 Ω c) 1 Ω d) 3.5 Ω
2
358. In the adjoining figure the equivalent resistance between A and B is

1 2 4
A

6 2

B
1 2Ω 4

a) 5 Ω b) 8 Ω c) 2.5 Ω d) 6.8 Ω

359. 7
If the ratio of the concentration of electron to that of holes in a semiconductor is and the ratio of current
5
7
is , then what is the ratio of their drift velocities
4
a) 4 b) 5 c) 4 d) 5
5 4 7 8
360. Electroplating does not help in

a) Fine finish to the surface b) Shining appearance

c) Metals to become hard d) Protecting metal against conosion

361. Corresponding to the resistance 4.7×106Ω ± 5%, which is order of colour coding on carbon resistors?

a) Yellow, violet, blue, gold b) Yellow, violet, green, gold

c) Orange, blue, green, gold d) Orange, blue, violet, gold

362. For a thermocouple, the neutral temperature is 270℃ and the temperature of its cold junction is 20℃. If
there is no deflection in the galvanometer, the temperature of the hot junction should be
a) 210℃ b) 540℃ c) 520℃ d) 209℃

363. Two cells A and B are connected in the secondary circuit of a potentiometer one at a time and the
balancing length are respectively 400 cm and 440 cm. The emf of the cell A is 1.08V. The emf of the second
cell B is volt is
a) 1.08 b) 1.188 c) 11.88 d) 12.8

P a g e | 47
364. What determines the emf between the two metals placed in an electrolyte?

a) Relative position of metals in the electro chemical b) Distance between them

series
c) Strength of electrolyte d) Nature of electrolyte

365. AB is a potentiometer wire of length 100 cm and its resistance is 10 ohm. It is connected in series with a
resistance R = 40 ohm and a battery of e.m.f. 2 V and negligible internal resistance. If a source of unknown
e.m.f. E is balanced by 40 cm length of the potentiometer wire, the value of E is
R 2V

40 cm
A B

a) 0.8 V b) 1.6 V c) 0.08 V d) 0.16 V

366. In a closed circuit, the current I(in ampere) at an instant of time t(in second) is given by I = 4 - 0.08t. The
number of electrons flowing in 50s through the cross-section of the conductor is
a) 1.25×1019 b) 6.25×1020 c) 5.25×1019 d) 2.55×1020

367. The internal resistance of a cell of emf 2 V is 0.1Ω. It is connected to a resistance of 3.9Ω . The potential
difference across is
a) 0.5 V b) 1.9 V c) 1.95 V d) 2 V

368. The accurate measurement of emf can be obtained using

a) Multimeter b) Voltmeter c) Voltameter d) Potentiometer

369. A potential difference of V is applied at the ends of a copper wire of length l and diameter d. On doubling
only d, the drift velocity,
a) Becomes two times b) Becomes half c) Does not change d) Becomes one-fourth

370. Find equivalent resistance between A and B

a) R b) 3R c) R d) 2R
4 2
371. AB is a wire of uniform resistance. The galvanometer G shows no current when the length AC = 20cm and
CB = 80cm. The resistance R is equal to

P a g e | 48
 R 80 

G
A B
C

a) 2 Ω b) 8 Ω c) 20 Ω d) 40 Ω

372. A heater coil is cut into two parts of equal length and one of them is used in the heater. The ratio of the
heat produced by this half coil to that by the original coil is
a) 2 :1 b) 1 :2 c) 1 :4 d) 4 :1

373. If the cold junction of a thermocouple is kept at 0℃ and the hot junction is kept at T℃, then the relation
between neutral temperature (Tn) and temperature of inversion (Ti) is
a) T = Ti b) T = 2T c) T = T - T d) T = T + T
n n i n i n i
2
374. The reading of the ammeter as per figure shown is
2

2 2V
A
2

2

a) 1 A b) 3 A c) 1 A d) 2 A
8 4 2
375. In the circuit shown, the currents i1 and i2are

i 1 12
2Ω
4Ω
i2

12V,1Ω
a) i = 3A,i = 1A b) i = 1A,i = 3A c) i = 0.5A, i = 1.5 A d) i = 1.5 A, i = 0.5 A
1 2 1 2 1 2 1 2

376. When a potential difference is applied across the ends of a linear metallic conductor

a) The free electrons are accelerated continuously from the lower potential end to the higher potential end
of the conductor
b) The free electrons are accelerated continuously from the higher potential end to the lower potential end
of the conductor
c) The free electrons acquire a constant drift velocity from the lower potential end to the higher potential
end of the conductor
d) The free electrons are set in motion from their position of rest

377. A thermocouple of resistance 1.6Ω is connected in series with a galvanometer of 8Ω resistance. The
thermocouple develops an e.m.f. of 10μV per degree temperature difference between two junctions. When
one junction is kept at 0℃ and the other in a molten metal, the galvanometer reads 8 millivolt. The
temperature of molten metal, when e.m.f. varies linearly with temperature difference, will be
a) 960℃ b) 1050℃ c) 1275℃ d) 1545℃

P a g e | 49
378. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance
of 100 W and 200 lamps, when not in use?
a) 40 Ω b) 20 Ω c) 400 Ω d) 200 Ω

379. The potential difference between the terminals of a cell in open circuit is 2.2V with resistance of 5 Ω across
the terminals of a cell, the terminal potential difference is 1.8V. the resistance of the cell is
a) 9 Ω b) 10 Ω c) 7 Ω d) 12 Ω
10 9 12 7
380. The circuit shown here is used to compare the emf of two cells E1 andE2(E1 > E2).
The null point is at C when the galvanometer is connected to E2. when the galvanometer is connected to E2,
the null point will be

a) To the left of C b) To the right of C c) At C itself d) None where on AB

381. A cell of constant emf first connected to a resistance R1and then connected to a resistanceR2.

a) R R b) R1 c) R1-R2 d) R1+R2
1 2 R2 2 2
382. Find the equivalent resistance between the points a and b
4

2 10 8
a b

4

a) 2 Ω b) 4 Ω c) 8 Ω d) 16 Ω

383. An ammeter with internal resistance 90Ω reads 1.85 A when connected in a circuit containing a battery
and two resistors 700Ω and 410Ω in series. Actual current will be
a) 1.85 A b) Greater than 1.85 A c) Less than 1.85 A d) None of these

384. The current through the circuit shown in figure 1A. If each of 4Ω the resistors is replaced by 2Ω resistor,
the current in circuit will become nearly
15
4Ω
2Ω 15
4Ω
15
10 V

a) 1.11 A b) 1.25 A c) 1.34 A d) 1.67 A

385. A 36Ω galvanometer is shunted by resistance of 4Ω. The percentage of the total current, which passes
P a g e | 50
 through the galvanometer is
a) 8% b) 9% c) 10% d) 91%

386. In the figure a carbon resistor has bands of different colours on its body as mentioned in the figure. The
value of the resistance is

a) 2.2 k Ω b) 3.3 k Ω c) 5.6 k Ω d) 9.1 k Ω

387. In a potentiometer experiment, the galvanometer shows no deflection when a cell is connected across 60
cm of the potentiometer wire. If the cell is shunted by a resistance of 6Ω, the balance is obtained across 50
cm of the wire. The internal resistance of the cell is
a) 0.5 Ω b) 0.6 Ω c) 1.2 Ω d) 1.5 Ω

388. Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance
between two opposite faces, shown by the shaded areas in the figure is

a) Directly proportional to L b) Directly proportional to t

c) Independent of L d) Independent of t

389. When 1 kg of hydrogen forms water, 34×106 cal of heat is liberated. If ECE of hydrogen is
-1
(1/96500,000)kg C , then the minimum voltage requird for decomposition of water is
a) 0.75 V b) 3.0 V c) 1.5 V d) 6.0 V

390. The reciprocal of resistance is

a) Conductance b) Resistivity c) Voltage d) None of the above

391. The resistance of a wire is 20 ohm. It is so stretched that the length becomes three times, then the new
resistance of the wire will be
a) 6.67 ohm b) 60.0 ohm c) 120 ohm d) 180.0 ohm

392. In the circuit shown the cells A and B have negligible resistance. For VA = 12V, R1 = 500Ω and R = 100Ω
the galvanometer (G) shows no deflection. The value of VB is

a) 4V b) 2V c) 12V d) 6V

P a g e | 51
393. Under what condition will the strength of current in a wire of resistance R be the same for connection is n
series or in parallel of n identical cells each of the internal resistance r, when
a) R = nr b) R = r/n c) R = r d) R→∞,r→0

394. Two bulbs 40 W and 60 W and rated voltage 240 V are connected in series across a potential difference of
420 V. Which bulb will work at above its rated voltages?
a) 40 W bulb b) 60 W bulb c) Both will work d) None of these

395. A thermocouple uses Bismuth and Tellurium as the dissimilar metals. The sensitivity of bismuth is -72μ
V/℃ and that of the tellurium is 500μ V/℃. If the difference between hot and cold junction is 100℃, then
the maximum output will be
a) 50 mV b) 7.2 mV c) 42.8 mV d) 57.2 mV

396. The resistance of an ideal voltmeter is

a) Zero b) Very low c) Very large d) Infinite

397. A battery of e.m.f. E and internal resistance r is connected to a variable resistor R as shown here. Which
one of the following is true
E r

a) Potential difference across the terminals of the battery is maximum when R = r

b) Power delivered to the resistor is maximum when R = r

c) Current in the circuit is maximum when R = r

d) Current in the circuit is maximum when R ≫ r

398. A potential difference V is applied to a copper wire of length l and thickness d. If V is doubled, the drift
velocity
a) Is doubled b) Is halved c) Remains same d) Becomes zero

399. Two copper wires of lengths l and 2l have radii r and 2r respectively. What is ratio of their specific
resistances?
a) 1 : 2 b) 2 : 1 c) 1 : 1 d) 1 : 3

400. What is the equivalent resistance between A and B in the given circuit?

2Ω
C D
2Ω
E

2Ω 4Ω
2Ω

4Ω
B

8Ω
A

P a g e | 52
 a) 4Ω b) 2Ω c) 8 Ω d) 3 Ω
3 8
-6
401. A metal wire of specific resistance 64×10 ohm - cm and length 198 cm has a resistance of 7 ohm, the
radius of the wire will be
a) 2.4 cm b) 0.24 cm c) 0.024 cm d) 24 cm

402. A potential difference is applied across the ends of a metallic wire. If the potential difference is doubled,
the drift velocity will
a) Be doubled b) Be halved c) Be quadrupled d) Remain unchanged

403. In the figure shown below, the terminal voltage across E2 is

8 V1Ω 12 V2Ω

E1 E 21

9Ω

a) 12 V b) 12.66 V c) 11.34 V d) 11.66 V

404. A silver voltameter of resistance 2 ohm and a 3 ohm resistor are connected in series across a cell. If a
resistance of 2 ohm is connected in parallel with the voltameter, then the rate of deposition of silver
a) Decreases by 25% b) Increases by 25% c) Increases by 37.5% d) Decreases by 37.5%

405. A tap supplies water at 22℃. A man takes 1 L of water per min at 37℃ from the geyser. The power of the
geyser is
a) 525 W b) 1050 W c) 1575 W d) 2100 W

406. When a current flows through a conductor its temperature

a) May increase or decrease b) Remains same

c) Decrease d) Increase

407. An electric bulb is marked 100 W, 230 V. If the supply voltage drops to 115 V, what is the total energy
produced by the bulb in 10 min?
a) 30 kJ b) 20 kJ c) 15 kJ d) 10 kJ

408. The potentiometer is superior to a voltmeter for measuring a potential difference because

a) The resistance of the voltmeter

b) The potentiometer does not draw any current from the source of the potential

c) The sensitivity of potentiometer is better than that of the voltmeter

d) The voltmeter has a dial and of small size

409. Constantan wire is used in making standard resistances because its

a) Specific resistance is low

b) Density is high

c) Temperature coefficient of resistance is negligible

P a g e | 53
 d) Melting point is high

410. For a given thermocouple neutral temperature

a) Is a constant b) Depends on cold junction temperature

c) Depends on inversion temperature d) Double that of cold junction temperature

411. An electric cable of copper has just one wire of radius 9 mm. Its resistance is 5Ω. This single copper wire of
cable is replaced by 6 different well insulated copper wires each of radius 3 mm. The total resistance of the
cable will now be equal to
a) 7.5 Ω b) 45 Ω c) 90 Ω d) 270 Ω

412. Two resistance R1 and R2 are joined as shown in the figure to two batteries of e.m.f. E1 and E2. If E2 is short-
circuited, the current through R1 is
R1

R2 E2
E1

a) E /R b) E /R c) E /R d) E /(R + R )
1 1 2 1 2 2 1 2 1

413. When two resistances R1 and R2 are connected in series, they consume 12 W powers. When they are
connected in parallel, they consume 50 W powers. What the ratio of the powers of R1 and R2?
a) 1/4 b) 4 c) 3/2 d) 3

414. The lowest resistance which can be obtained by connecting 10 resistors each of 1/10 ohm is

a) 1/250 Ω b) 1/200 Ω c) 1/100 Ω d) 1/10 Ω

415. The electron of hydrogen atom is considered to be revolving round in circular orbit of radius h2/me2with
2
velocity e h, where h = h/2π. The current i is
2 5 2 2 2 2 5 2 5
a) 4π me b) 4π me c) 4π m e d) 4π me
2 3 3 3
h h h h
416. Two wires of same metal have the same length but their cross sections are in the retio 3:1. They are joined
in series. The resistance of the thicker wire is 10Ω. The total resistance of the combination is
a) 5/2 Ω b) 40/3 Ω c) 40 Ω d) 100 Ω

417. A copper wire of length L and radius r is nickel plated till its final radius become R but length remains L. If
the resistivity of nickel and copper be ρn and ρc respectively, the conductance of the nickelled wire is

[ ]
2 2 2 2 2 2
a) πr b) π(R -r ) c) π r + (R -r ) d) Lρc + L.ρn
L.ρc L.ρn L ρc ρn πr
2 2 2
π(R -r )
418. Following figure shows cross-sections through three long conductors of the same length and material, with
square cross-section of edge lengths as shown. Conductor B will fit snugly within conductor A, and
conductor C will fit snugly within conductor B. Relationship between their end to end resistance is
3 a
2 a

A B C

P a g e | 54
 a) R = R = R b) R > R > R
A B C A B C

c) R < R < R d) Information is not sufficient

A B C

419. A 100 V voltmeter of internal resistance 20 kΩ in series with a high resistance R is connected to a 110 V
line. The voltmeter reads 5 V, the value of R is
a) 210 kΩ b) 315 kΩ c) 420 kΩ d) 440 kΩ

420. Three voltmeters A, B and C having resistances R, 1.5R and 3R respectively are used in a circuit as shown.
When a potential difference is applied between X and Y, the readings of the voltmeters are V1, V2and V3
respectively. Then
B

X A Y

a) V = V = V b) V < V = V c) V > V > V d) V > V > V

1 2 3 1 2 2 1 2 3 1 2 3

421. The potential drop across the 3Ω resistor is

3

4

6

3V

a) 1 V b) 1.5 V c) 2 V d) 3 V

422. A conductor wire having 1029 free electrons/m3 carries a current of 20A. If the cross-section of the wire is
2
1mm , then the drift velocity of electrons will be
a) 6.25×10-3ms-1 b) 1.25×10-5ms-1 c) 1.25×10-3ms-1 d) 1.25×10-4ms-1

423. Two wires have resistances R and 2R. When both are joining in series and in parallel, then ratio of heats
generated in these situations on applying the same voltage, is
a) 2 : 1 b) 1 : 2 c) 2 : 9 d) 9 : 2

424. The potential difference in open circuit for a cell is 2.2 volt. When a 4 ohm resistor is connected between
its two electrodes the potential difference becomes 2 volt. The internal resistance of the cell will be
a) 1 ohm b) 0.2 ohm c) 2.5 ohm d) 0.4 ohm

425. The temperature of hot junction of a thermocouple changes from 80℃ to 100℃, the percentage change in
thermo electric power is
a) 25% b) 20% c) 10% d) 8%

426. Two bulbs, one of 50 watt and another of 25 watt are connected in series to the mains. The ratio of the
currents through them is
a) 2 :1 b) 1 :2

c) 1 :1 d) Without voltage, cannot be calculated

427. When a resistor of 11 Ω is connected in series with an electric cell, the current flowing in it is 0.5 A.
Instead, when a resistor of 5 Ω is connected to the same electric cell in series, the current increases by 0.4
A. The internal resistance of the cell is
a) 1.5 Ω b) 2 Ω c) 2.5 Ω d) 3.5 Ω
P a g e | 55
428. A battery of emf 10V and internal resistance 3 Ωis connected to an external resistor. The current in the
circuit is 0.5A. the terminal voltage of the battery when the circuit is close is
a) 10V b) Zero c) 1.5 V d) 8.5 V

429. The relation between voltage sensitivity (σV) and current sensitivity (σi) of a moving coil galvanometer is
(resistance of galvanometer is G).
a) σi = σ b) σv = σ c) G = σ d) G = σ
G v
G i σ v
i
σi V

430. An electric heater of resistances 6 Ω is run for 10 min on 120 V line. The energy librated in this period of
time is?
a) 7.2×105 J b) 14.4 ×105 J c) 43.2 ×105 J d) 28.8 ×105 J

431. A certain piece of silver of given mass is to be made like a wire. Which of the following combinations of
length (L) and the area of cross-section (A) will lead to the smallest resistance
a) L and A b) 2L and A/2

c) L/2 and 2 A d) Any of the above, because volume of silver

remains same
432. The variation between V - i has been shown by V - i graph for heater filament.

a) b) c) d)

433. Thomson coefficient of a conductor is 10μV/K. The two ends of it are kept at 50℃ and 60℃ respectively.
Amount of heat absorbed by the conductor when a charge of 10C flows through it is
a) 1000 J b) 100 J c) 100 mJ d) 1 mJ

434. The junction of Ni-Cu thermo couple are maintained at 0℃ and 100℃. The seeback emf developed in the
temperature is
-6 -1
aNi = 16.3×10 V℃
-Cu
-6 -1
bNi = -0.021×10 V℃
-Cu
a) 2.73×103V b) 1.42×10-3V c) 3.68×10-3V d) 2.23×103V

435. A source of a primary cell is 2V. what is the short circuited it provides 4A current, then the internal
resistance of cell will be
a) 8 Ω b) 2.0 Ω c) 4 Ω d) 0.5 Ω

436. For what value of unknown resistance X, the potential difference between B and D will be zero in the
circuit shown in the figure
B
1
12
1
A C
1
X

6 1
D

a) 4 Ω b) 6 Ω c) 2 Ω d) 5 Ω

P a g e | 56
437. The V - I graph for a wire of copper of length L and cross-section ares A is shown in adjoining figure. The
slope of the graph will be

a) Less if the experiment is repeated at a higher b) More if a wire of silver having the same dimension
temperature is used
c) Doubled if the length of the wire is doubled d) Halved if length of the wire is halved

438. The heat generated through 2 ohm and 8 ohm resistances separately, when a condenser of 200 μF capacity
charged to 200 V is discharged one by one, will be
a) 4 J and 16 J respectively b) 16 J and 4 J respectively

c) 4 J and 8 J respectively d) 4 J and 4 J respectively

439. The circuit shown here is used to compare the e.m.f.'s of two cells E1 and E2(E1>E2). The null point is at C
when the galvanometer is connected to E1. When the galvanometer is connected to E2, the null point will be
B

C
A B
E1

E2 G

a) To the left of C b) To the right of C c) At C itself d) No where on AB

440. The colour code for a resistor of resistance 3.5kΩ with 5% tolerance is

a) Orange, green, red and gold b) Red, yellow, black and gold

c) Orange, green, orange and silver d) Orange, green, red and silver

441. The tangent galvanometer, when connected in series with a standard resistance can be used as

a) An ammeter b) A voltmeter

c) A wattmeter d) Both ammeter and voltmeter

442. Which of the following statement is correct

a) Both Peltier and Joule effects are reversible

b) Both Peltier and Joule effects are irreversible

c) Joule effect is reversible, whereas Peltier effect is irreversible

d) Joule effect is reversible, whereas Peltier effect is reversible

443. Four wires AB, BC, CD, DA of resistance 4 ohm each and a fifth wire BD of resistance 8 ohm are joined to
form a rectangle ABCD of which BD is a diagonal. The effective resistance between the points A and B is
a) 24 ohm b) 16 ohm c) 4 ohm d) 8 ohm
3 3
444. Two resistors are connected (a) in series (b) in parallel. The equivalent resistance in the two cases are 9
P a g e | 57
 ohm and 2 ohm respectively. Then the resistance of the component resistors are
a) 2 ohm and 7 ohm b) 3 ohm and 6 ohm c) 3 ohm and 9 ohm d) 5 ohm and 4 ohm

445. Two wires of the same material and equal length are joined in parallel combination. If one of them has half
the thickness of the other and the thinner wire has a resistance of 8 ohms, the resistance of the
combination is equal to
a) 5 ohm b) 8 ohm c) 3 ohm d) 8 ohm
8 5 8 3
446. In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2 Ω
resistance is

2Ω

3Ω

C = 0.7 μF 4Ω

E = 6V 2.8 Ω

a) 0.6A b) 1.2A c) 0.9A d) 1.5A

447. The equivalent resistance and potential difference between A and B for the circuit is respectively

a) 4 Ω, 8 V b) 8 Ω, 4 V c) 2 Ω, 2 V d) 16 Ω, 8 V

448. A wire of resistance 18Ω is divided into three equal parts. These parts are connected in side of triangle, the
equivalent resistance of any two corners of triangle will be
a) 18Ω b) 9Ω c) 6Ω d) 4Ω

449. The power of heater is 500 W at 800℃. What will be its power at 200℃ ? (Given : temperature coefficient
-4 -1
of resistance, α = 4×10 ℃ )
a) 484 W b) 672 W c) 526 W d) 620 W

450. The equivalent resistance of the arrangement of resistances shown in adjoining figure between the points
A and B is
8
16 20

16
A B
9
6
18

a) 6 ohm b) 8 ohm c) 16 ohm d) 24 ohm

451. Three identical resistances A, B and C are connected as shown in figure.

P a g e | 58
 B

R
A

R C

The heat produced will be maximum

a) In B b) In B and C c) In A d) Same for A, B and C

452. In a conductor 4 coulomb of charge flows for 2 seconds. The value of electric current will be

a) 4 volt b) 4 ampere c) 2 ampere d) 2 volt

453. The main supply voltage to a room is 120 V. The resistance of the lead wires is 6Ω. A 60 W bulb is already
giving light. What is the decrease in voltage across the bulb when a 240 W heater is switched on?
a) No change b) 10 V c) 20 V d) More than 10 V

454. Two bulbs consume same power when operated at 200 V and 300 V respectively. When these bulbs are
connected in series across a DC source of 400 V, then the ratio of power consumed across them is
a) 2/3 b) 3/2 c) 4/9 d) 9/4

455. A wire is stretched so as to change its diameter by 0.25%. The percentage change in resistance is

a) 4.0% b) 2.0% c) 1.0% d) 0.5%

456. The resistor in which maximum heat will be produced is

2Ω 4Ω

3Ω
5Ω
6Ω

a) 2 Ω b) 3 Ω c) 4 Ω d) 6 Ω

457. If an ammeter is connected in parallel to a circuit, it is likely to be damaged due to excess

a) Current b) Voltage c) Resistance d) All of these

458. See the electrical circuit shown in this figure. Which of the following equations is a correct equation for it

P a g e | 59
 a) ε - (i +i )R - i r = 0 b) ε - i r - ε - i r = 0
1 1 2 1 1 2 2 2 1 1 1

c) -ε - (i +i )R + i r = 0 d) ε - (i +i )R + i r = 0
2 1 2 2 2 1 1 2 1 1

459. In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance
of the voltmeter is 80 ohm. The reading of the voltmeter will be
2V
+ –

80 
V

20  80 

a) 0.80 volt b) 1.60 volt c) 1.33 volt d) 2.00 volt

460. 5 cells, each of emf 0.2V and internal resistance 1Ω are connected to an external circuit of resistance of
10Ω. Find the current through external circuit
a) 1 A b) 1 A c) 1 A d) 1 A
2.5 10 15 2
461. A 2V battery, a 990 Ω resistor and a potentiometer of 2m length, all are connected in series of the
residence of potentiometer wire is 10 Ω, then the potential gradient of the potentiometer wire is
a) 0.05Vm-1 b) 0.5Vm-1 c) 0.01Vm-1 d) 0.1 Vm-1

462. A galvanometer of resistance 36 Ω is changed into an ammeter by using a shunt of 4 Ω. The fraction f0 of
total current passing through the galvanometer is
a) 1 b) 1 c) 1 d) 1
40 4 140 10
463. Two cells of equal e.m.f. and of internal resistance r1 and r2(r1>r2) are connected in series. On connecting
this combination to an external resistance R, it is observed that the potential difference across the first cell
becomes zero. The value of R will be
a) r + r b) r - r c) r1+r2 d) r1-r2
1 2 1 2
2 2
464. The current I drawn from the 5 V source will be

10Ω

5Ω 20Ω

10Ω-
I

+ -
5V

a) 0.33A b) 0.5A c) 0.67A d) 0.17A

465. If 2 A of current is passed through CuSO4 solution for 32 s, then the number of copper ions deposited at the
cathode will be
a) 4×1020 b) 2×1020 c) 4×1019 d) 2×1019

466. A steady current is set up in a metallic wire of non-uniform cross-section. How is the rate of flow of
P a g e | 60
 electrons (R) related to the area of cross-section (A)?
a) R ∝ A-1 b) R ∝ A c) R ∝ A2 d) R is independent of A

467. If 96500 coulombs of electricity liberates one gram equivalent of any substance, the time taken for a
current of 0.15 amperes to deposit 20mg of copper from a solution of copper sulphate is (Chemical
equivalent of copper = 32)
a) 5min 20 sec b) 6min 42 sec c) 4min 40 sec d) 5min 50 sec

468. The electron drift speed is small and the charge of the electron is also small but still, we obtain large
current in a conductor. This is due to
a) The conducting property of the conductor

b) The resistance of the conductor is small

c) The electron number density of the conductor is small

d) The electron number density of the conductor is enormous

469. What is the equivalent resistance across the points A and B in the circuit given below?
C

10
B

10 16

12

10 2.5 A

D E

a) 8 Ω b) 12 Ω c) 16 Ω d) 32 Ω

470. The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220
volt and 100 watt is connected to (220×.8) volt sources, then the actual power would be
a) 100×0.8 watt b) 100×(0.8)2 watt

c) Between 100×0.8 watt and 100 watt d) Between 100×(0.8)2watt and 100×0.8 watt

471. Every atom makes one free electron in copper. If 1.1 A Current is flowing in the wire of copper having 1
3 -3
mm diameter, then the drift velocity(approx.) will be (density of copper=9×10 kg m and atomic weight
of copper=63)
a) 0.1 mms-1 b) 0.2 mms-1 c) 0.3 mms-1 d) 0.2 mms-1

472. In a copper voltmeter , if the current (I) and time (t) variations of the type as shown in figure, the mass
deposited in 30 min is [Atomic weight of copper is 63.5 and Faraday constant is 96500 C per g equivalent]
a) 0.078 g b) 0.054 g c) 0.039 g d) 0.0195 g

473. A wire when connected to 220V mains supply has power dissipation P1. Now the wire is cut into two equal
pieces which are connected in parallel to the same supply. Power dissipation in this case is P2. Then P2 :P1
is
a) 1 b) 4 c) 2 d) 3

474. A source of e.m.f. E = 15 V and having negligible internal resistance is connected to a variable resistance
P a g e | 61
 so that the current in the circuit increases with time as i = 1.2 t + 3. Then, the total charge that will flow
in first five seconds will be
a) 10 C b) 20 C c) 30 C d) 40 C

475. Electric bulb 50 W-100 V glowing at full power are to be used in parallel with battery 120 V, 10 Ω.
Maximum number of bulbs that can be connected so that they glow in full power is
a) 2 b) 8 c) 4 d) 6

476. 10 wires (same length, same area, same material) are connected in parallel and each has 1Ω resistance,
then the equivalent resistance will be
a) 10 Ω b) 1 Ω c) 0.1 Ω d) 0.001 Ω

477. In order to pass 10% of main current through a moving coil galvanometer of 99 ohm, the resistance of the
required shunt is
a) 9.9 Ω b) 10 Ω c) 11 Ω d) 9 Ω

478. If the resistivity of a potentiometer wire be ρ and area of cross-section be A, then what will be potential
gradient along the wire
a) Iρ b) I c) IA d) IAρ
A Aρ ρ
479. The current from the battery in circuit diagram shown is
2 A 7

15V

6 1
0.5

8 B 10

a) 1 A b) 2 A c) 1.5 A d) 3 A

480. Two bulbs when connected in parallel to a source take 100 W each. The total power consumed when they
are connected in series with the same source is
a) 25 W b) 50 W c) 100 W d) 200 W

481. A uniform resistance R and length L is cut into for equal parts, each of length L/4, which are then
connected in parallel combination. The effective resistance of the combination will be
a) R b) 4R c) R d) R
4 16
482. In a circuit 5 percent of total current passes through a galvanometer. If resistance of the galvanometer is G
then value of the shunt is
a) 19 G b) 20 G c) G d) G
20 19
483. An electric lamp is marked 60 W, 230 V. The cost of a 1 kWh of energy is Rs. 1.25. The cost of using this
lamp 8 hrs a day for 30 days is
a) Rs. 10 b) Rs. 16 c) Rs. 18 d) Rs. 20

484. Two wires 'A' and 'B' of the same material have their lengths in the ratio 1 :2 and radii in the ratio 2 :1. The
two wires are connected in parallel across a battery. The ratio of the heat produced in 'A' to the heat
produced in 'B' for the same time is
a) 1 :2 b) 2 :1 c) 1 :8 d) 8 :1

485. (1)The product of a volt and a coulomb is a joule

(2)The product of a volt and an ampere is a joule/second
P a g e | 62
 (3)The product of volt and watt is horse power
(4)Watt-hour can be measured in terms of electron volt
State if
a) All four are correct b) (1), (2) and (4) are correct

c) (1) and (3) are correct d) (3) and (4) are correct

486. A copper and a chromium voltmeter are connected in series with a battery. It found that in half an hour
0.475 g of copper and 0.130 g of chromium have been deposited. The ECE ratio of copper and chromium is
a) 0.274 b) 0.523 c) 3.65 d) 1.85

487. Each resistance shown in figure is 2 Ω. The equivalent resistance between A and B is

2Ω 2

A B
V 2Ω

2Ω 2

a) 2 Ω b) 4 Ω c) 8 Ω d) 1 Ω

488. If σ1,σ2 and σ3 are the conductances of three conductors, then their equivalent conductance, when they are
joined in series, will be
a) σ + σ + σ b) 1 + 1 + 1 c) σ1σ2σ3 d) None of these
1 2 3 σ1 σ2 σ3 σ1+σ2+σ3
489. The resistance of a metal increases with increasing temperature because

a) The collisions of the conducting electrons with the electrons increase

b) The collisions of the conducting electrons with the lattice consisting of the ions of the metal increases

c) The number of conduction electrons decrease

d) The number of conduction electrons increase

490. Two electric bulbs A and B are rated as 60 W and 100 W. They are connected in parallel to the same
source. Then
a) B draws more current than A

b) Currents drawn are in the ratio of their resistances

c) Both draw the same current

d) A draws more current than B

491. If an electric current is passed through a nerve of a man, then man

a) Begins to laugh b) Begins to weep

c) Is excited d) Becomes insensitive to pain

492. An electric bulb is rated 220 V-100 W. The power consumed by it when operated on 110 V will be

a) 75 W b) 40 W c) 25 W d) 50 W

493. In the adjacent shown circuit, a voltmeter of internal resistance R, when connected across B and C reads

P a g e | 63
 100
V. Neglecting the internal resistance of the battery, the value of R is
3

a) 100 kΩ b) 75 kΩ c) 50 kΩ d) 25 kΩ

494. A resistor R1 dissipates power P when connected to a certain generator. If the resistor R2 is put in series
with R1, the power dissipated by R1
a) Decreases

b) Increases

c) Remains the same

d) Any of the above depending upon the relative values of R and R

1 2

495. If two wires having resistances R and 2R both are joined in series and in parallel, then ratio of heat
generated in this situation, applying the same voltage is
a) 2 :1 b) 1 :2 c) 2 :9 d) 9 :2

496. For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source
in the circuit is
a) 1.38 V b) 1.68 V c) 2.04 V d) 3.10 V

497. A heater of 220 V heats a volume of water in 5 min. The same heater when connected to 110 V heats the
same volume of water in (minute)
a) 5 b) 20 c) 10 d) 2.5

498. 50 Ω and 100 Ω resistors are connected in series. This connection is connected with a battery of 2.4 volt.
When a voltmeter of 100 Ω resistance is connected across 100 Ω resistor, then the reading of the voltmeter
will be
a) 1.6 V b) 1.0 V c) 1.2 V d) 2.0 V

499. A series combination of two resistors 1 Ω each is connected to a 12 V battery of internal resistance 0.4 Ω.
The current flowing through it will be
a) 3.5 A b) 5 A c) 6 A d) 10 A

500. In the circuit shown in figure, power developed across 1Ω , 2Ω , 3 Ω resistance are in ratio of
1
i 3
2

a) 1 : 2 : 3 b) 4 : 2 : 27 c) 6 : 4 : 9 d) 2 : 1 : 27

501. Four resistances 10 Ω, 5 Ω, 7 Ω and 3 Ω are connected so that they form the sides of a rectangle AB, BC, CD
and DA respectively. Another resistance of 10 Ω is connected across the diagonal AC. The equivalent
resistance between A and B is
a) 2 Ω b) 5 Ω c) 7 Ω d) 10 Ω

P a g e | 64
502. A copper voltmeter and a silver voltmeter are connected in series in a circuit. The rate of the increase in
the weight of the cathode in the two voltmeters will be in the ratio of
a) Atomic weights of Cu and Ag b) Densities of Cu and Ag

c) Half of the atomic weight of Cu to the atomic d) Half of the atomic weight of Ag to half the atomic
weight of Ag weight of Cu
503. In the circuit shown, the cell is ideal, with emf=10V. Each resistance is of 2Ω. The potential difference
across the capacitor is

a) 12 V b) 10 V c) 8 V d) zero

504. A current of 16 ampere flows through molten NaCl for 10 minute. The amount of metallic sodium that
appears at the negative electrode would be
a) 0.23 gm b) 1.15 gm c) 2.3 gm d) 11.5 gm

505. A storage battery has e.m.f. 15 volt and internal resistance 0.05 ohm. Its terminal voltage when it is
delivering 10 ampere is
a) 30 volt b) 1.00 volt c) 14.5 volt d) 15.5 volt

506. There are three voltmeters of the same range but of resistances 10000Ω, 8000Ω and 4000Ω respectively.
The best voltmeter among these is the one whose resistance is
a) 10000 Ω b) 8000 Ω c) 4000 Ω d) All are equally good

507. In the circuit shown in the adjoining figure, the current between B and D is zero, the unknown resistance is
of
B

4 X

12
A C
1
1

3 1
D

a) 4 Ω b) 2 Ω

c) 3 Ω d) e.m.f. of a cell is required to find the value of X

508. In meter bridge or wheatstone bridge for measurement of resistance, the known and the unknown
resistance are interchanged. The error so removed is
a) End correction b) Index error

c) Due to temperature effect d) Random error

509. The potential difference between A and B in the following figure is

6Ω

P a g e | 65
 4V 5Ω
A B
2A 12 V9Ω

a) 32 V b) 48 V c) 24 V d) 14 V

510. Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are α1and
α2. The respective temperature coefficients of their series and parallel combinations are nearly
a) α1+α2 ,α + α b) α α , α1+α2 c) α + α , α1α2 d) α1+α2 , α1+α2
2 1 2 1+ 2
2 1 2
α1+α2 2 2
511. Figure shows a circuit with known resistances R1 .Neglect the internal resistance of the sources of current
and resistance of the connecting wire. The magnitude of electromotive force E1 such that the resistances R
is zero will be

a) ER /R b) ER /R c) E(R +R )/R d) ER /(R + R )

1 2 2 1 1 2 2 1 1 2

512. A current I is passed for a time t through a number of voltmeters. If m is the mass of a substance deposited
on an electrode and z is its electrochemical equivalent, then
a) zIt = constant b) z = constant c) I = constant d) It = constant
m mIt zmt zm
513. The drift velocity of the electrons in a copper wire of length 2 m under the application of a potential
-1 2 -1 -1
difference of 220 V is 0.5ms . Their mobility (in m V s )
a) 2.5×10-3 b) 2.5×10-2 c) 5×102 d) 5×10-3

514. The inversion temperature of a copper-iron thermocouple is 540℃ when the cold junction temperature is
0℃. If the cold junction temperature is increased by 10℃, then the inversion temperature and neutral
temperature of the thermocouple respectively are
a) 270℃ and 530℃ b) 270℃ and 550℃ c) 280℃ and 530℃ d) 280℃ and 550℃

515. The heat produced by a 100 W heater in 2 min will be equal to

a) 12×103 J b) 10×103 J c) 6×103 J d) 3×103 J

516. There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the
2 3 22
wire is 1 mm . If the number of free electrons per cm is 8.4×10 , then the drift velocity would be
a) 1.0 mm/sec b) 1.0 m/sec c) 0.1 mm/sec d) 0.01 mm/sec

517. The neutral temperature of a thermocouple is 350℃ when the cold junction is at 0℃. When the cold
junction is immersed in a bath of 30℃, the inversion temperature is
a) 700℃ b) 600℃ c) 350℃ d) 670℃

518. Ohm's law is true

a) For metallic conductors at low temperature

b) For metallic conductors at high temperature

c) For electrolytes when current passes through them

P a g e | 66
 d) For diode when current flows

519. Which of the following statements is correct

a) Liquids obey fully the ohm's law

b) Liquids obey partially the ohm's law

c) There is no relation between current and p.d. for liquids

d) None of the above

520. The range of a voltmeter of resistance 500 Ω is 10V. the resistance to be connected to convert it into an
ammeter of range 10A is
a) 1 Ω in parallel b) 1 Ω in series c) 0.1 Ω in parallel d) 0.1 Ω in series

521. A uniform wire of resistance R is uniformly compressed along its length, until its radius becomes n times
the original radius. Now resistance of the wire becomes
a) R4 b) R2 c) R d) nR
n n n
522. The rate of increase of thermo e.m.f. with temperature at the neutral temperature of a thermocouple

a) Is negative

b) Is positive

c) Is zero

d) Depends upon the choice of the two materials of the thermocouple

523. What length of the wire of specific resistance 48×10-8Ω m is needed to make a resistance of 4.2 Ω
(diameter of wire = 0.4 mm)
a) 4.1 m b) 3.1 m c) 2.1 m d) 1.1 m

524. In the given current distribution, what is the value of I?

a) 3A b) 8A c) 2A d) 5A

525. A battery is connected to a uniform resistance wire AB and B is earthed. Which one of the graphs below
shows how the current density J varies along AB

– +

A B

a) J

Zero at all
0 points
A B

P a g e | 67
 b) J

0
A B

c) J

0
A B

d) J

0
A B

526. An ammeter and a voltmeter of resistance R are connected in series to an electric cell of negligible internal
resistance. Their readings are A and V respectively. If another resistance R is connected in parallel with the
voltmeter
a) Both A and V will increase b) Both A and V will decrease

c) A will decrease and V will increase d) A will increase and V will decrease

527. When the current i is flowing a conductor, the drift velocity is v. If 2i current is flowed through the same
metal but having double the area of cross-section, then the drift velocity will be
a) v/4 b) v/2 c) v d) 4v

528. In a typical Wheatstone network, the resistances in cycle order are A = 10 Ω, B = 5 Ω, C = 4 Ω and D = 4
Ω. For the bridge to be balanced
A = 10  B=5

D=4 C=4

a) 10 Ω should be connected in parallel with A

b) 10 Ω should be connected in series with A

c) 5 Ω should be connected in series with B

d) 5 Ω should be connected in parallel with B

529. When a resistance of 2ohm is connected across the terminals of a cell, the current is 0.5 ampere. When the
resistance is increased to 5 ohm, the current is 0.25 ampere. The internal resistance of the cell is
a) 0.5 ohm b) 1.0 ohm c) 1.5 ohm d) 2.0 ohm

530. A tap supplies water at 22℃, a man takes of 1 L of water per min at 37℃ from the geyser. The power of
geyser is
a) 525 W b) 1050 W c) 1775 W d) 2100 W

531. For a cell, the graph between the potential difference (V) across the terminals of the cell and the current
(I) drawn from the cell is shown in the figure. The e.m.f. and the internal resistance of the cell are

P a g e | 68
 a) 2V, 0.5Ω b) 2V,0.4Ω c) > 2V,05Ω d) > 2V, 0.4Ω

532. The thermistors are usually made of

a) Metals with low temperature coefficient of resistivity

b) Metals with high temperature coefficient of resistivity

c) Metal oxides with high temperature coefficient of resistivity

d) Semiconducting materials having low temperature coefficient of resistivity

533. A storage cell is charged by 5 amp D.C. for 18 hours. Its strength after charging will be

a) 18 AH b) 5 AH c) 90 AH d) 15 AH

534. The internal resistance of a cell of e.m.f. 12V is 5×10-2Ω. It is connected across an unknown resistance.
Voltage across the cell, when a current of 60 A is drawn from it, is
a) 15 V b) 12 V c) 9 V d) 6 V

535. In the circuit shown in figure, the points F is grounded. Which of the following is wrong statement?

5
B C

2 3

A D
10 V 4 3V
F E

a) D is at 5V b) E is at zero potential

c) The current in the circuit will be 0.5 A d) The potential at E is same whether or not F is
rounded
536. Potentiometer wire of length 1m is connected in series with 490Ωresistance and 2V battery. If 0.2m
-1
Vcm is the potential gradient, then resistance of the potentiometer wire is
a) 4.9Ω b) 7.9Ω c) 5.9Ω d) 6.9Ω

537. The potential difference across the terminals of a battery is 50V when 11A current is drawn and 60V when
1A current is drawn. The e.m.f. and the internal resistance of the battery are
a) 62V, 2Ω b) 63V, 1Ω c) 61V, 1Ω d) 64V, 2Ω

538. Two resistance R1 and R2 are made of different materials. The temperature coefficient of the material of R1
is α and of the material of R2 is -β. The resistance of the series combination of R1 and R2 will not change
with temperature, if R1/R2 equals

P a g e | 69
 2 2
a) α b) α+β c) α +β d) β
β α-β αβ α
539. To deposit one litre of hydrogen at 22.4 atmosphere from acidulated water, the quantity of electricity that
must pass through is
a) 1 coulomb b) 22.4 coulomb c) 96500 coulomb d) 193000 coulomb

540. In ballistic galvanometer, the frame in which the coil is wound is non-metallic to

a) Avoid the production of induced emf b) Avoid the production of eddy currents

c) Increase the production of eddy currents d) Increase the production of induced emf

541. In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal
resistance, the value of the resistor R will be
500
G

A 2V
12 V B R

a) 200Ω b) 100 Ω c) 500 Ω d) 1000 Ω

542. The resistivity of alloys = Ralloy; the resistivity of constituent metals Rmetal. Then, usually

a) R = Rmetal b) R < Rmetal

alloy alloy

c) There is no simple relation between R and R d) R > Rmetal

alloy metal alloy

543. The production of e.m.f. by maintaining a difference of temperature between the two junctions of two
different metals is known as
a) Joule effect b) Seebeck effect c) Peltier effect d) Thomson effect

544. An electric heater is heated respectively by d.c. and a.c. Applied voltage for both the currents is equal. The
heat produced per second will be
a) More on heating by a.c. source b) More on heating by d.c. source

c) Same for both d) None of the above

545. A coil takes 15 min to boil a certain amount of water; another coil takes 20 min for the same process. Time
taken to boil the same amount of water when both coils are connected in series
a) 5 min b) 8.6 min c) 35 min d) 30 min

546. The electron dirft speed is small and the charge of the electron is also small but still, we obtain large
current in a conductor. This is due to
a) The conducting property of the conductor

b) The resistance of the conductor is small

c) The electron number density of the conductor is small

d) The electron number density of the conductor is enormous

547. A small power station supplies electricity to 5000 lamps connected in parallel. Each lamp has a resistance

P a g e | 70
 of 220 and is operated at 220 V. The total current supplied by the station is
a) 2500 A b) 3500 A c) 5000 A d) 10000 A

548. In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will

a) Be zero b) Flow from B to A

c) Flow from A to B d) Flow in the direction which will be decided by the

value of V
549. The effective resistance between points A and B is

a) R b) R c) 2R d) 3R
3 3 5
550. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere
and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1V, the resistance in
Ohm’s needed to be connected in series with the coil will be
a) 103 b) 105 c) 99995 d) 9995

551. A capacitor of capacitance 3μF is first charged by connecting across 10 V battery, then it is allowed to get
discharged through 2 Ω and 4Ω resistor by closing the key Kas shown in figure. The total energy dissipated
in 2Ω resistor is equal to
C= 3 F

2Ω K

4Ω

a) 0.15 m J b) 0.5 m J c) 0.05 m J d) 1.0 m J

552. If each of the resistances in the network in figure. R, the equivalent resistance between terminals A and B
is

a) 5 R b) 2 R c) 4 R d) R
P a g e | 71
553. The tolerance level of a resistor with the colour code red, blue, orange, gold is

a) ±5% b) ±10% c) ±20% d) ±40%

554. A given piece of wire of length l and radius r is having a resistance R. This wire is stretched uniformly to a
r
wire of radius . What is the new resistance?
2
a) 3R b) 8R c) 16R d) 2R
2
555. θ
The expression for thermo e.m.f. in a thermocouple is given by the relation E = 40θ - , where θ is the
20
temperature difference of two junctions. For this, the neutral temperature will be
a) 100℃ b) 200℃ c) 300℃ d) 400℃

556. The steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities
constant along the length of the conductor is/are
a) Current, electric field and drift velocity b) Drift speed only

c) Current and drift speed d) Current only

557. Potential difference between the points P and Q in the electric circuit shown is
P i = 1.5 A
RA = 2 RB = 4

3

RD = 6 RC = 12
Q

a) 4.5 V b) 1.2 V c) 2.4 V d) 2.88 V

558. Who among the following scientists made the statement –“Chemical change can produce electricity”

a) Galvani b) Faraday c) Coulomb d) Thomson

559. A thin wire of resistance 4 Ω is bent to form a circle. The resistance across any diameter is

a) 4 Ω b) 2 Ω c) 1 Ω d) 8 Ω

560. A wire P has a resistance of 20Ω. Another wire Q of same material but length twice that of P has resistance
of 8Ω. If r is the radius of cross-section of P, the radius of cross-section of Qis
a) r b) r c) d) 2r
5r
2
561. A 100 W bulb B1 and two 60 W bulb B2 and B3 are connected to a 250 V source as shown in the figure. Now

W1,W2 and W3 are the out-put powers of the bulbs B1, B2 and B3 respectively. Then
a) W > W = W b) W > W > W c) W < W = W d) W < W < W
1 2 3 1 2 3 1 2 3 1 2 3

562. The electric intensity E, current density j and specific resistance k are related to each other by the relation

a) E = j/k b) E = jk c) E = k/j d) k = jE

563. The kirchoff;s forst law (∑i = 0)and second law (∑iR=∑E)where the symbols have their usual meanings,
are respectively based on
P a g e | 72
 a) Conservation of charge, conversion of momentum

b) Conservation of energy, conservation of charge

c) Conservation of momentum, conservation of charge

d) Conservation of charge, conservation of energy

564. In the circuit shown in figure the potential difference between the points A and B will be

5 5
B

5 2V 5

A
5 5

a) 2 V b) 8 V c) 4 V d) 2V
3 9 3
565. For goldplating on a copper chain, the substance required in the form of solution is

a) Copper sulphate b) Copper chloride

c) Potassium cyanide d) Potassium aurocyanide

566. A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as voltmeter of range 10 volt, the
resistance that must be connected in series with it, will be
a) 999 Ω b) 99 Ω c) 1000 Ω d) None of these

567. Two wires of resistance R1 and R2 have temperature coefficient of resistances α1 and α1 respectively. These
are joined in series. The effective temperature coefficient of resistance is
R1R2α1α2
a) α1+α2 b) α α c) α1R1+α2R2 d)
2 1 2 R1+R2 2
R1+R2
2

568. Conductivity increases in the order of

a) Al, Ag, Cu b) Al, Cu, Ag c) Cu, Al, Ag d) Ag, Cu, Al

569. A current of 1.5 A flows through a copper voltmeter. The thickness of copper deposited on the electrode
2 -3 -7 -1
surface of area 50 cm in 20 min is (density of Cu=9000 kgm ;ECE of Cu = 3.3×10 kgC )
a) 1.3×10-4 m b) 1.3×10-5 m c) 2.6 ×10-4 m d) 2.6 ×10-5 m

570. If resistance of the filament increases with temperature, what will be power dissipated in a 220 V- 100 W
lamp when connected to 110 V power supply
a) 25 W b) < 25 W c) > 25 W d) None of these

571. Resistances of 6 ohm each are connected in the manner shown in adjoining figure. With the current 0.5
ampere as shown in figure, the potential difference VP - VQ is
6 6 6

P 6 Q
0.5 A
6 6

P a g e | 73
 a) 3.6 V b) 6.0 V c) 3.0 V d) 7.2 V

572. What is the volume of hydrogen liberated at NTP by the amount of charge which liberates 0.3175 g of
copper?
a) 224 cc b) 112 cc c) 56 cc d) 1120 cc

573. A coil develops heat of 800 cal/sec. When 20 volts is applied across its ends. The resistance of the coil is (1
cal = 4.2 joule)
a) 1.2 Ω b) 1.4 Ω c) 0.12 Ω d) 0.14 Ω

574. In the given figure, when key K is opened, the reading of the ammeter A will be
10V
+ –

5
E A D

4 A
B C
K

a) 50 A b) 2 A c) 0.5 A d) 10 A
9
575. An electric wire of length 'L' and area of cross-section a has resistance R ohm. Another wire of the same
material having same length and area of cross-section 4a has a resistance of
a) 4R b) R/4 c) R/16 d) 16R

576. The figure shows a network of currents. The magnitude of current is shown here. The current I will be

1A

10 A I

6A

2A

a) 3A b) 9A c) 13A d) 19A

577. An ammeter of 5 ohm resistance can read 5 mA. If it is to be used to read 100 volt, how much resistance is
to be connected in series
a) 19.9995 Ω b) 199.995 Ω c) 1999.95 Ω d) 19995 Ω

578. The thermo-emf of a thermocouple is 25μ V/℃ at room temperature. A galvanometer of 40 Ω resistance,
-5
capable of detecting current as low as 10 A, is connected with the thermocouple. The smallest
temperature difference that can be detected by this system is
a) 16℃ b) 12℃ c) 8℃ d) 20℃

579. An ionization chamber with parallel conducting plates as anode and cathode has 5×107 electrons and the
3
same number of singly-charged positive ions per cm . The electrons are moving at 0.4 m/s. The current
2
density from anode to cathode is 4μA/m . The velocity of positive ions moving towards cathode is
a) 0.4 m/s b) 16 m/s c) Zero d) 0.1 m/s

580. In a potentiometer experiment, when three cells A, B and C care connected in series the balancing length is
found to be 740 cm. if A and B are connected in series balancing length is 540 cm. then the emf of EA, EBand
EC are respectively (in volts)

P a g e | 74
 a) 1,1.2 and 1.5 b) 1,2 and 3 c) 1.5,2 and 3 d) 1.5, 2.5 and 3.5

581. Figure below shows a thick copper rod X and a thin copper wire Y joined in series. They carry a current
which is sufficient to make Y much hotter than X

Which one of the following is correct?

Number density of Mean time between
Conduction collisions of the
electrons electrons
a) Same in X and Y less in X than in Y
b) Same in X and Y same in X and Y
c) Same in X and Y more in X than in Y
d) more in X and Y less in X than in Y
582. A steady current i is flowing through a conductor of uniform cross-section. Any segment of the conductor
has
a) Zero charge b) Only positive charge

c) Only negative charge d) Charge proportional to current i

583. Nine resistors each of 1 kΩ are conneted to a battery of 6 V as shown in the circuit given below. What is the
total current flowing in the circuit

a) 3mA b) 2 mA c) 3 mA d) 2mA
3 2
584. In order to increase the sensitivity of galvanometer

a) The suspension wire should be made stiff

b) Area of the coil should be reduced

c) The magnetic field should be increased

d) The number of turns in the coil should be reduced

585. When the number of turns of the coil is doubled, the current sensitivity of a moving coil galvanometer is
doubled whereas the voltage sensitivity of the galvanometer
a) Remains the same b) Is halved c) Is doubled d) Is quadrupled

586. The reading of a high resistance voltmeter when a cell is connected across it is 2.2 V. When the terminals of
the cell are also connected to a resistance of 5 Ω the voltmeter reading drops to 1.8 V. Find the internal
resistance of the cell
a) 1.2 Ω b) 1.3 Ω c) 1.1 Ω d) 1.4 Ω

587. In copper voltameter, mass deposited in 30 s is m gram. If the time current is as shown in figure, ECE of

P a g e | 75
 copper is

a) m b) m/2 c) 0.6 m d) 0.1 m

588. Two cells with the same emf E and different internal resistances r1and r2 are connected in series to an
external resistance R. the value of R so that the potential difference across the first cell be zero is
a) r r b) r + r c) r - r d) r1+r2
1 2 1 2 1 2
2
589. The resistivity of a wire depends on its

a) Length b) Area of cross-section c) Shape d) Material

590. In the circuit shown, the current through the 5 Ω resistor is

a) 8 A b) 9 A c) 4 A d) 1 A
3 13 13 3
591. A 5.0 A current is setup in an external circuit by a 6.0 storage battery for 6.0 min. The chemical energy of
the battery is reduced by
a) 1.08×104 J b) 1.08×103 J c) 1.8×104 J d) 1.8×103 J

592. A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated
will now be
a) Doubled b) Four times c) One-fourth d) Halved

593. The resistance of a galvanometer is 90 ohm. If only 10 percent of the main current may flow through the
galvanometer, in which way and of what value, a resistor is to be used
a) 10 ohm in series b) 10 ohm in parallel c) 810 ohm in series d) 810 ohm in parallel

594. One junction of a certain thermo-couple is at a fixed temperature Tr and the other junction is at
temperature T. The thermo electric force for this is expressed by
1 2 2
E = K(T-Tr)[ T0 + (T +Tr ) ].
2
At temperature T = T0/2 the thermoelectric power is
a) 1 K T b) 3 K T c) 1 K T2 d) 1 K (T -T )2
r
2 0
2 0
2 0
2 0

595. An electric lamp is marked 60 W, 230 V. The cost of kilowatt hour of power is Rs 1.25. The cost of using
this lamp 8 h a day for 30 days is
a) Rs 10 b) Rs 16 c) Rs 18 d) Rs 20

596. A 5V battery with internal resistance 2 Ω and a 2V battery with internal resistance 1 Ω are connected to a
10 Ω resistor as shown in the figure
P

P a g e | 76
 2

5V 10 2V

2Ω 1Ω

P1

The current in the 10 Ω resistor is

a) 0.27A, P to P b) 0.03A, P to P c) 0.03A, P tp P d) 0.27A, P to P
2 1 1 2 2 1 1 2

597. Calculate the value E, for given circuit, when value of 2A current is either flowing in clockwise or
anticlockwise direction
6Ω

+
E 20 V
-

a) 32 V, 8V b) 38V, 2V c) 32V, 2V d) 30V, 8V

598. The current in the arm CD of the circuit will be

B

i2
i1
O A
i3

C
D

a) i + i b) i + i c) i + i d) i - i + i
1 2 2 3 1 3 1 2 3

599. A battery consists of a variable number 'n' of identical cells having internal resistances connected in series.
The terminals of battery are short circuited and the current i is measured. Which of the graph below slows
the relationship between i and n
a) i b) i c) i d) i

O n O n O n O n

600. Assume that each atom of copper contributes one electron. If the current flowing through a copper wire of
-3
1 mm diameter is 1.1 A, the drift velocity of electrons will be (density of Cu=9 g cm , atomic wt. of Cu=63)
a) 0.3 mm s-1 b) 0.5 mm s-1 c) 0.1 mm s-1 d) 0.2 mm s-1

601. The specific resistance of a wire is ρ, its volume is 3 m3 and its resistance is 3Ω, then its length will be

a) 1 ρ b) 3/ ρ c) 3 /ρ d) ρ/ 3

602. Two similar thermocouples, made of dissimilar metals A and B are connected as shown in figure through a
key K and a sensitive galvanometer G. One of the thermocouples is dipped in a hot bath maintained at
temperature t2 and the other in a cold bath at temperature t1. When the key is pressed, a deflection is seen
in the galvanometer because

P a g e | 77
 a) An emf of the order of a few microvolt is b) An emf is generated the value of which will
generated which is proportional to (t2 - t1) depend upon the temperature of the hot bath only
c) An emf of about one volt is generated which will d) An emf of a few microvolt is generated which will
be proportional to (t2-t1) be proportional to t2 only.
603. 100 cells each of e.m.f. 5 V and internal resistance 1 ohm are to be arranged so as to produce maximum
current in a 25 ohm resistance. Each row is to contain equal number of cells. The number of rows should
be
a) 2 b) 4 c) 5 d) 10

604. A steady current of 1.5 A flows through a copper voltameter for 10 min. If the electrochemical equivalent
-5 -1
of copper is 30×10 g C , the mass of copper deposited on the electrode will be
a) 0.40 g b) 0.50 g c) 0.67 g d) 0.27 g

605. 1kg piece of copper is drawn into a wire 1 mm thick, and another piece into a wire 2 mm thick. Compare
the resistance of these wires
a) 2:1 b) 4:1 c) 8:1 d) 16:1

606. A 2 volt battery, a 15 Ω resistor and a potentiometer of 100 cm length, all are connected in series. If the
resistance of potentiometer wire is 5Ω, then the potential gradient of the potentiometer wire is
a) 0.005 V/cm b) 0.05 V/cm c) 0.02 V/cm d) 0.2 V/cm

607. In charging a battery of motor-car, the following effect of electric current is used

a) Magnetic b) Heating c) Chemical d) Induction

608. In given figure, the potentiometer wire AB has a resistance of 5 Ω and length 10 m. The balancing length
AM for the emf of 0.4 V is
R=45

5V

A M B
0.4V

a) 0.4 m b) 4 m c) 0.8 m d) 8 m

609. When 1 A current flows for 1 min through a silver voltmeter, it deposits 0.067 g of silver on the cathode,
then how much charge will flow to deposit 108 g of silver?
a) 10.6×104Cg-1 b) 9.67× 104Cg-1 c) 8.7 ×104Cg-1 d) 4.3 ×104Cg-1 `
eq eq eq eq
6
610. The voltage of clouds is 4×10 V with respect to ground. In a light ning strike lasting 100 ms, a charge of 4
C is delivered to the ground. The power of lightning strike is
a) 160 MW b) 80 MW c) 20 MW d) 500 Kw

611. A 6V cell with 0.5 Ω internal resistance, a10V cell with 1 Ω internal resistance and a 12 Ω external
resistance are connected in parallel. The current (in ampere) through the 10V cell is
a) 0.60 b) 2.27 c) 2.87 d) 5.14

612. Figure shows three resistor configurations R1, R2 and R3 connected to 3 V batteries. If the power dissipated
P a g e | 78
 by the configuration R1, R2 and R3 IS P1, P2 and P3, respectively, then

1Ω 1Ω
1Ω
1Ω
3 3V 3V
1Ω 1Ω 1Ω

1Ω
1Ω
1Ω

R R R 3

a) P > P > P b) P > P > P c) P > P > P d) P > P > P

1 2 3 1 3 2 2 1 3 3 2 1

613. A cell of internal resistance r is connected to a load of resistance R. Energy is dissipated in the load, but
some thermal energy is also wasted in the cell. The efficiency of such an arrangement is found from the
expression

energy dissipated in the load

.
energy dissipatd in the compete circuit
Which of the following gives the efficiency in this case?
a) r b) R c) r d) R
R r R+r R+r
614. The resistance of a wire is 5Ω at 50℃ and 6Ω at 100℃.The resistance of the wire at 0℃ will be

a) 2Ω b) 1Ω c) 4Ω d) 3Ω

615. To convert a moving a coil galvanometer (MCG) into a voltmeter

a) A high resistance R is connected in parallel with b) A low resistance r is connected in parallel with
MCG MCG
c) A low resistance r is connected in series with MCG d) A high resistance R is connected in series with
MCG
616. A dry cell of emf 1.5 V and internal resistance 0.10Ω is connected across a resistor in series with a very low
resistance ammeter. When the circuit is switched on, the ammeter reading settles to a steady rate of 2A.
Find (i) chemical energy consumption of the cell (ii) energy dissipation inside the cell (iii) energy
dissipation inside the resistor (iv) power output of source is
a) (i) 3 W (ii) 0.4 W (iii) 2.6 W (iv) 2.6 W b) (i) 0.4 W (ii) 3 W (iii) 2.6 W (iv) 2.6 W

c) (i) 2.6 W (ii) 0.4 W (iii) 9 W (iv) 1 W d) None of the above

617. A wire has a resistance of 12 ohm. It is bent in the form of equilateral triangle. The effective resistance
between any two corners of the triangle is
a) 9 ohm b) 12 ohm c) 6 ohm d) 8/3 ohm

618. One end each of a resistance r capacitor C and resistance 2r are connected together. The other ends are
respectively connected to the positive terminals of batteries, P, Q, R having respectively emf’s E,E and 2E.
the negative terminals of the batteries are then connected together. In this circuit, with steady current the

P a g e | 79
 potential drop across the capacitor is
a) E b) E c) 2E d) E
3 2 3
619. Two cells having the internal resistance 0.2 Ω and 0.4 Ω are connected in parallel. The voltage across the
battery terminal is 1.5V. the emf of first cell is 1.2V. the emf of second cell is
a) 2.7 V b) 2.1 V c) 3 V d) 4.2V

620. The equivalent resistance of n resistors each of same resistance when connected in series is R. If the same
resistances are connected in parallel, the equivalent resistance will be
a) R/n2 b) R/n c) n2R d) nR

621. An infinite sequence of resistances is shown in the figure. The resultant resistance between A and B will
be, when R1 = 1 ohm and R2 = 2 ohm
R1 R1 R1 R1 R1
A

R2 R2 R2 R2 R2

a) Infinity b) 1 Ω c) 2 Ω d) 1.5 Ω

622. A 200 W and a 100 W bulb, both meant for operation at 220 V are connected in series. When connected to
a 220 V supply the power consumed by the combination is
a) 33.3 W b) 66.7 W c) 300 W d) 100 W

623. Five resistors of given values are connected together as shown in the figure. The current in the arm BD will
be
B
R R

C
4R
A

R R

a) Half the current in the arm ABC b) Zero

c) Twice the current in the arm ABC d) Four times the current in the arm ABC

624. What is the current (i) in the circuit as shown in figure

i R2 = 2
R3 = 2

3V R1 = 2

R4 = 2

a) 2 A b) 1.2 A c) 1 A d) 0.5 A

625. A galvanometer of resistance 100Ω is converted to a voltmeter of range 10 V by connecting a resistance of

10kΩ. The resistance required to convert the same galvanometer to an ammeter of range 1 A is
a) 0.4Ω b) 0.3Ω c) 1.2Ω d) 0.1Ω

626. The resistance of a 10m long wire is 10Ω. Its length is increased by 25% by stretching the wire uniformly.

P a g e | 80
 Then the resistance of the wire will be
a) 12.5 Ω b) 14.5 Ω c) 15.6 Ω d) 16.6 Ω

627. Resistance of 100 cm long potentiometer wire is 10Ω, it is connected to a battery (2 volt) and a resistance
R in series. A source of 10 mV gives null point at 40 cm length, then external resistance R is
a) 490 Ω b) 790 Ω c) 590 Ω d) 990 Ω

628. In order to quadruple the resistance of a uniform wire, a part of its length was uniformly stretched till the
final length of the entire wire was 1.5 times the original length, the part of the wire was fraction equal to
l

0.5l

a) 1/8 b) 1/6 c) 1/10 d) 1/4

629. A galvanometer has a resistance 50Ω. A resistance of 5Ω is connected parallel to it. Fraction of the total
current flowing through galvanometer is
a) 1 b) 1 c) 1 d) 2
10 11 50 15
630. The thermo emf of copper-constantan couple is 40μV per degree. The smallest temperature difference that
can be detected with this couple and a galvanometer of 100Ω resistance capable of measuring the
maximum current of 1μA is
a) 10℃ b) 7.5℃ c) 5.0℃ d) 2.5℃

631. A parallel combination of two resistors, of 1Ω each, is connected in series with a 1.5Ω resistor. The total
combination is connected across a 10V battery. The current flowing in the circuit is
a) 5A b) 20A c) 0.2A d) 0.4A

632. Dimensions of a block are 1 cm×1 cm×100cm. If specific resistance of its material is 3×10-7ohm - m, then
the resistance between the opposite rectangular faces is
a) 3×10-9ohm b) 3×10-7ohm c) 3×10-5ohm d) 3×10-3ohm

633. Find the equivalent resistance across AB

A
2
2
2
2
2
B

a) 1 Ω b) 2 Ω c) 3 Ω d) 4 Ω

634. In the circuit given E=0.6V, R1=100Ω, R2 = R3 = 50Ω, R4 = 75Ω. The equivalent resistance of the circuit,
in ohm is
R1

i
R 4

R2 R3
E

a) 11.875 b) 26.31 c) 118.75 d) None of these

635. Three resistances 4 Ω each are connected in the form of an equilateral triangle. The effective resistance
P a g e | 81
 between two corners is
a) 8 Ω b) 12 Ω c) 3 Ω d) 8 Ω
8 3
636. The material of wire of potentiometer is

a) Copper b) Steel c) Manganin d) Aluminium

637. A Copper wire of length 1 m and radius 1 mm is joined in series with an iron wire of length 2 m and radius
3 mm and a current is passed through the wires. The ratio of the current density in the wires. The ratio of
the current density in the copper and iron wires is
a) 2:3 b) 6:1 c) 9:1 d) 18:1

638. 1 2
If the emf of a thermocouple, one junction of which is kept 0℃ is given by e = at + bt , then the neutral
2
temperature will be
a) a b) - a c) a d) - 1
b b 2b ab
639. A given carbon resistor has the following colour code of the various strips: orange, red, yellow and gold.
The value of resistance in ohm is
a) 32×104 ± 5% b) 32×104 ± 10% c) 23×104 ± 5% d) 23×104 ± 10%

640. The same mass of copper is drawn into two wires 1 mm and 2 mm thick. Two wires are connected in
series and current is passed through them. Heat produced in the wire is in the ratio
a) 2 :1 b) 1 :16 c) 4 :1 d) 16 :1

641. The resistances of a wire at temperatures t℃ and 0℃ are related by

a) R = R (1 + αt) b) R = R (1 - αt) c) R = R2(1 + αt) d) R = R2(1 - αt)

t 0 t 0 t 0 t 0

642. B1, B2 and B3 are the three identical bulbs connected to a battery of steady emf with key K closed. What
happens to the brightness of the bulbs, B1 and B2 when the key is opened?
B1

K B2

B3

a) Brightness of the bulb B1 increases and that of B2 b) Brightness of the bulbs B and B increases
1 2
decreases
c) Brightness of the bulb B1 decreases and B2 d) Brightness of the bulbs B and B decreases
1 2
increases
643. A new flashlight cell of e.m.f. 1.5 volt gives a current of 15 amp, when connected directly to an ammeter of
resistance 0.04 Ω. The internal resistance of cell is
a) 0.04 Ω b) 0.06 Ω c) 0.10 Ω d) 10 Ω

644. In the circuit shown in figure, the current drawn from the battery is 4A. If 10 Ω resistor is replaced by 20 Ω
resistor, then current drawn from the circuit will be

P a g e | 82
 1 3

10
10

4A 7 21

+ –

a) 1 A b) 2 A c) 3 A d) 0 A

645. Four resistances 40Ω, 60Ω, 90Ω and 110Ω make the arms of a quadrilateral ABCD. Across AC is the battery
circuit, the emf of the battery being 4V and internal resistance negligible. The potential difference across
BD is
B
40 60
A C

90 110
D

4V

a) 1V b) -1V c) -0.2V d) 0.2V

646. An electric current passes through a circuit containing two wires of the same material connected in
parallel. If the lengths of the wires are in the ratio of 4/3 and radius of the wires are in the ratio of 2/3,
then the ratio of the current passing through the wires will be
a) 3 b) 1/3 c) 8/9 d) None of these

647. In the above question if potential difference is applied, the drift velocity at temperature T is

a) Inversely proportional to T b) Proportional to T

c) Zero d) Finite but independent of T

648. If all the resistors shown have the value 2 ohm each, the equivalent resistance over AB is
A B

a) 2 ohm b) 4 ohm c) 1 2 ohm d) 2 2 ohm

3 3
649. A current through a wire depends on time t as i = 10 + 4t. The charge crossing through the section of the
wire in 10 s is
a) 50 C b) 300 C c) 400 C d) 4C

650. A copper and silver voltmeter are connected in parallel. If 2000 C of charge liberates the same mass of
copper and silver, then charge flowing in copper voltmeter is
[Z(Cu=3.36×10 kg C , Z(Ag)=1.008×10 kgC ]
-7 -1 -6 -1

a) 1250 C b) 1500 C c) 1750 C d) 1000 C

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651. Three electric bulbs of 200 W, 200 W and 400 W are shown in figure. The resultant power of the
combination is
200 W
400 W

200 W

( )

a) 800 W b) 400 W c) 200 W d) 600 W

652. What will be the equivalent resistance between the two points A and D
10 10 10
A B

10 10

C D
10 10 10

a) 10 Ω b) 20 Ω c) 30 Ω d) 40 Ω

653. A potentiometer wire, 10 m long, has a resistance of 40Ω. It is connected in series with a resistance box
and a 2V storage cell. If the potential gradient along the wire is (0.1 mVcm ), the resistance unplugged in
-1

the box is
a) 260 Ω b) 760 Ω c) 960 Ω d) 1060 Ω

654. The electrolyte used in Lechlanche cell is

a) Copper sulphate solution b) Ammonium chloride solution

c) Dilute sulphuric acid d) Zinc sulphate

655. In a thermocouple, which of the following statements is not true

a) Neutral temperature depends upon the nature of materials in the thermocouple

b) Temperature of inversion depends upon the temperature of cold junction

c) When the temperature of the hot junction is equal to the temperature of inversion, the thermo emf
becomes zero
d) When the temperature of cold junction increases, the temperature of inversion also increases

656. If current in an electric bulb changes by 1%, then the power will change by

a) 1% b) 2% c) 4% d) 1 %
2
657. Neutral temperature of a thermocouple is defined as the temperature at which

a) The thermo e.m.f. changes sign b) The thermo e.m.f. is maximum

c) The thermo e.m.f. is minimum d) The thermo e.m.f. is zero

P a g e | 84
658. Electric power is transmitted over long distances through conducting wires at high voltage because

a) High voltage travels faster

b) Power loss is large

c) Power loss is less

d) Generator produce electrical energy at a very high voltage

659. The magnitude of I in ampere is

60
Ω
I

15 Ω 5Ω
1A 1A

a) 0.1 b) 0.3 c) 0.6 d) None of the above

660. Six equal resistances each of 4Ω are connected to form a figure. The resistance between two corners A and
B is

4Ω
A B
4Ω
4Ω
O
4Ω
4Ω
D C
4Ω

a) 4 Ω b) 4/3 Ω c) 12 Ω d) 2 Ω

661. A battery of internal resistance 4 Ω is connected to the network of resistance as shown. In order to given
the maximum power to the network, the value of R (in Ω) should be
R R

R 6R R
4

R 4R

a) 4/9 b) 8/9 c) 2 d) 18

662. An electrical cable having a resistance of 0.2 Ω delivers 10 kW at 200 V DC to a factory. What is the
efficiency of transmission
a) 65% b) 75% c) 85% d) 95%

663. A heating coil can heat the water of a vessel from 20℃ to 60℃ in 30 minutes. Two such heating coils are
put in series and then used to heat the same amount of water through the same temperature range. The
time taken now will be (neglecting thermal capacity of the coils)
a) 60 minutes b) 30 minutes c) 15 minutes d) 7.5 minutes

P a g e | 85
664. What is the resistance of a carbon resistance which has bands of colours brown, black and brown

a) 100 Ω b) 1000 Ω c) 10 Ω d) 1 Ω

665. If the ammeter in the given circuit reads 2 A, the resistance R is

a) 1 ohm b) 2 ohm c) 3 ohm d) 4 ohm

666. Faraday’s 2nd law states that mass deposited on the electrode is directly proportional to

a) Atomic mass b) Atomic mass × Velocity

c) Atomic mass/Valency d) Valency

667. Power dissipated across the 8Ω resistor in the circuit shown here is 2 watt. The power dissipated in watt
units across the 3Ω resistor is

a) 0.5 b) 3.0 c) 2.0 d) 1.0

668. When an electric heater is switched on, the current flowing through it (i) is plotted against time (t). Taking
into account the variation of resistance with temperature, which of the following best represents the
resulting curve
a) I b) I c) I d) I

t t t t

669. Which of the following are true, when the cells are connected in series?

a) Current capacity decreases b) Current capacity increases

c) The emf decreases d) The emf increases

670. Two wires have lengths, diameters and specific resistances all in the ratio of 1 : 2. The resistance of the
first wire is 10Ω. Resistance of the second wire in ohm will be
a) 5 b) 10 c) 20 d) Infinite

671. The resistance is connected as shown in the figure below. Find the equivalent resistance between the
points A and B.
7Ω
D C

3Ω 5Ω

P a g e | 86
 a) 205Ω b) 10 Ω c) 3.5 Ω d) 5 Ω

672. The n rows each containing m cells in series are joined in parallel. Maximum current is taken from this
combination across as external resistance of 3 Ω resistance. If the total number of cells used are 24 and
internal resistance of each cell is 0.5 Ω, then
a) m = 8, n = 3 b) m = 6,n = 4 c) m = 12, n = 2 d) m = 2,n = 12

673. A and B are two square plates of same metal and same thickness but length of B is twice that of A. Ratio of
resistances of A and B is

a) 4 : 1 b) 1 : 4 c) 1 : 1 d) 1 : 2

674. In the network of resistors shown in the adjoining figure, the equivalent resistance between A and B is
3 3 3 3 3 3
A B

3 3 3 3 3 3

a) 54 ohm b) 18 ohm c) 36 ohm d) 9 ohm

675. Two bars of radius r and 2r are kept in contact as shown. An electric current i is passed through the bars.
Which one of the following is correct?

a) Heat produced in bar BC is 4 times the heat b) Electric field in both halves is equal
produced in bar AB
c) Current density across AB is doubled that of d) Potential difference across AB is 4 times that of
across BC across BC
676. The alloys constantan and manganin are used to make standard resistance because they have

a) Low resistivity b) High resistivity

c) Low temperature coefficient of resistance d) Both (b) and (c)

677. The internal resistances of two cells shown are 0.1 Ω and 0.3 Ω. If R = 0.2Ω, the potential difference across
the cell
2V, 0.1 2V, 0.3

A B

0.2

P a g e | 87
 a) B will be zero b) A will be zero

c) A and B will be 2V d) A will be > 2V and B will be < 2V

678. In a Wheatstone’s bridge, three resistances P, Q and R connected in the three arms and the fourth arm is
formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will
be
a) P = 2R b) P = R(S1+S2) c) P = R(S1+S2) d) P = R
Q S1+S2 Q S1S2 Q 2S1S2 Q S1+S2
679. In the circuit shown in figure, the heat produced in 5 ohm resistance is 10 calories per second. The heat
produced in 4 ohm resistance is

a) 1 cal/sec b) 2 cal/sec c) 3 cal/sec d) 4 cal/sec

680. In a potentiometer experiment the balancing with a cell is at length 240 cm. on shunting the cell with a
resistance of 2 Ω, the balancing length becomes 120cm. the internal resistance of cell is
a) 4 Ω b) 2 Ω c) 1 Ω d) 0.5 Ω

681. A current I is passing through a wire having two sections P and Q of uniform diameters d and d/2
respectively. If the mean drift velocity of electrons in sections P and Q is denoted by vP and vQ respectively,
then
a) v = v b) v = 1 v c) v = 1 v d) v = 2v
P Q P
2 Q P
4 Q P Q

682. The relaxation time in conductors

a) Increases with the increase of temperature b) Decreases with the increase of temperature

c) It does not depend on temperature d) All of sudden changes at 400 K

683. Two resistances R and 2R are connected in parallel in an electric circuit. The thermal energy developed in
R and 2R are in the ratio
a) 1 : 2 b) 2 : 1 c) 1 : 4 d) 4 : 1

684. In the given figure. A, B and C are three identical bulbs. When the switch S is closed

a) The brightness of bulb A does not change and that of B decreases

b) The brightness of bulb A increases and that of B decreases

c) The brightness of A increases bulb B does not glow

d) The brightness of both bulbs A not B decrease

685. Three resistances each of 4Ω are connected in the form of an equilateral triangle. The effective resistance
between any two corners is
P a g e | 88
 a) (3/8) Ω b) (8/3) Ω c) 8 Ω d) 12 Ω

686. Three resistance A,B and C have values 3R, 6R and R respectively. When some potential difference is
applied across the network, the thermal powers dissipated by A,B and C are in the ratio
3R
A
R

6R C

a) 2 :3 :4 b) 2 :4 :3 c) 4 :2 :3 d) 3 :2 :4

687. 12 cells each having same emf are connected in series with some cells wrongly connected. The
arrangement is connected in series with an ammeter and two cells which are in series. Current is 3 A when
cells and battery aid each other and is 2 A when cells and battery oppose each other. The number of cells
wrongly connected is
a) 4 b) 1 c) 3 d) 2

688. Three resistance P, Q, R each of 2Ω and an unknown resistance S form the four arms of a wheatstone
bridge circuit. When a resistance of 6Ω is connected in parallel to S the bridge gets balanced. What is the
value of S
a) 2 Ω b) 3 Ω c) 6 Ω d) 1 Ω

689. A galvanometer of 25 Ω resistance can read a maximum current of 6mA. It can be used as a voltmeter to
measure a maximum of 6 V by connecting a resistance to the galvanometer. Identify the correct choice in
the given answers
a) 1025 Ω in series b) 1025 Ω in parallel c) 975 Ω in series d) 975 Ω in parallel

690. 1 2
The dimensions of εoE (εo:permittivity of free space; E: electric field) is
2
a) [MLT] b) [ML2T-2] c) [ML-1T-2] d) [ML2T-1]

691. Following figure shows four situations in which positive and negative charges move horizontally through a
region and gives the rate at which each charge moves. Rank the situations according to the effective
current through the region greatest first

a) i = ii = iii = iv b) i > ii > iii > iv c) i = ii = iii > iv d) i = ii = iii < iv

692. An aluminium (resistivityρ = 2.2×10-8Ω - m) wire of a diameter 1.4 mm is used to make a 4Ω esistor. The
length of the wire is
a) 220 m b) 1000 m c) 280 m d) 1 m

693. Voltmeters V1and V2 are connected in series across a DC line. V1reads 80V and has a resistance of
-1
200ΩV and V2 has a total resistance of 32k Ω. The line voltage is
a) 240 V b) 220 V c) 160 V d) 120 V

694. When a current I is passed through a wire of constant resistance, it produces a potential difference V
across its ends. The graph drawn between log I and log V will be

P a g e | 89
 a) b) c) d)

695. A galvanometer of resistance 20 Ω shows a deflection of 10 divisions when a current of 1 mA is passed

through it. If a shunt of 4 Ωis connected and there are 50 divisions on the scale, the range of the
galvanometer is
a) 1A b) 3A c) 10mA d) 30mA

696. For ensuring dissipation of same energy in all three resistors (R1,R2,R3)connected as shown in figure, their
values be related as

a) R = R = R b) R = R and R = 4 R
1 2 3 2 3 1 2

c) R = R and R =R /4 d) R = R + R
2 3 1 2 1 2 3

697. x g of Ag is deposited by passing 4 A of current of for 1 h. How many gram of Ag will be deposited by
passing 6 A for 40 min?
a) 2x g b) 4x g c) x g d) 5x g

698. Resistance in the two gaps of a meter bridge are 10 ohm and 30 ohm respectively. If the resistances are
interchanged the balance point shifts by
a) 33.3 cm b) 66.67 cm c) 25 cm d) 50 cm

699. In the circuit shown below E1 = 4.0 V, R1 = 2 Ω, E2 = 6.0 V, R2 = 4 Ω and R3 = 2 Ω. The current I1 is
R1 = 2 Ω

E1 = 4 V
I1
R3 = 2 Ω

I2
R2 = 4 Ω

E2 = 6 V

a) 1.6 A b) 1.8 A c) 1.25 A d) 1.0 A

700. The amount of charge Q passed in time t through a cross-section of a wire is Q = 5t2 + 3t + 1. The value
of current at time t = 5s is
a) 9 A b) 49 A c) 53 A d) None of the above

701. In water voltmeter, the electrolysis of …… takes place

a) H O b) H SO c) H O and H SO both d) H and O

2 2 4 2 2 4 2 2

702. Which statement is true?

(i) Kirchoff’s law is equally applicable to both AC and DC.
(ii) Semiconductors have a positive temperature coefficient of resistance.
(iii) Meter bridge is greater sensitive when the resistance of all four arms of the bridge is of the same
order.

P a g e | 90
 (iv) The emf of a cell depends upon the size and area of electrodes.
a) (i) and (iv) b) (ii) and (iv) c) (iii) and (iv) d) None of these

703. In the circuit given, the current relation to a balanced Wheatstone’s bridge is
P R

S Q

a) P = R b) P = S c) P = Q d) P = S
Q S Q R S R R Q
704. A thick wire is stretched, so that its length become two times. Assuming that there is no change in its
density, then what is the ratio of change in resistance of wire to the initial resistance of wire?
a) 2 : 1 b) 4 : 1 c) 3 : 1 d) 1 : 4

705. Two electric bulbs marked 40 W, 220 V and 60 W, 220 V when connected in series, across same voltage
P1
supply of 220 V, the effective power is P1 and when connected in parallel the effective power is P2. Then
P2
is
a) 0.5 b) 0.48 c) 0.24 d) 0.16

706. Figure shows a network of eight resistors, each equal to 2 Ω, connected to a 3V battery of negligible
internal resistance. The current I in the circuit is

3V

A B C D

E F

a) 0.25A b) 0.50A c) 0.75A d) 1.0A

707. If the length of filament of a heater is reduced by 10%, the power of the heater will

a) Increase by about 9% b) Increase by about 11%

c) Increase by about 19% d) Decrease by about 10%

708. In the circuit, the potential difference across PQ will be nearest to

100 

48 V
80 
100  Q
20 
P

a) 9.6 V b) 6.6 V c) 4.8 V d) 3.2 V

709. A current of 5 A is passing through a metallic wire of cross-sectional area 4×10-6m2. If the density of charge
26 -3
carries of the wire is 5×10 m , the drift velocity of the electrons will be
a) 1×102ms-1 b) 1.56×10-2ms-1 c) 1.56×10-3ms-1 d) 1×10-2ms-1
P a g e | 91
710. Figure shows a network of three resistance. When some potential difference is applied across the network,
the thermal powers dissipated by A,B and C in the ratio

a) 2 : 3 : 4 b) 2 : 4 : 3 c) 4 : 2 : 3 d) 3 : 2 : 4

711. Two plates R and S are in the form of a square and have the same thickness. A side of S is twice the side of
R Compare their resistances. The direction of current is shown by an arrow head figure.

a) The resistance of R is twice that of S b) Both have the same resistance

c) The resistance of S is four times that of R d) The resistance of R is half that of S

712. A resistor has a colour code of green, blue, brown and silver. What is its resistance?

a) 5600Ω ± 10% b) 560Ω ± 5% c) 560Ω ± 10% d) 56Ω ± 5%

713. The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then

a) The resistance will be doubled and the specific resistance will be halved

b) The resistance will be halved and the specific resistance will remain unchanged

c) The resistance will be halved and the specific resistance will be doubled

d) The resistance and the specific resistance, will both remain unchanged

714. Equivalent resistance between A and B will be

3 3

3 3
3 3

A 3 3 B

a) 2 ohm b) 18 ohm c) 6 ohm d) 3.6 ohm

715. With the rise of temperature the resistivity of a semiconductor

a) Remains unchanged b) Increases

c) Decreases d) First increases and then decreases

716. Two wires of the same material and having same uniform area of cross-section are connected in series in
an electrical circuit. The masses of the wires are m and 2m. When a current Iflows in the circuit, the heats
produced by them in a given time are in ratio
a) 2 : 1 b) 1 : 2 c) 4 : 1 d) 1 : 4

717. When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance
between any two of diametrically opposite points will be

P a g e | 92
 a) R b) R c) 4R d) R
4 8 2
718. The resistance of a 5 cm long wire is 10 Ω. It is uniformly stretched so that its length becomes 20 cm. The
resistance of the wire is
a) 160 Ω b) 80 Ω c) 40 Ω d) 20 Ω

719. The current i1and i2through the resistor R1(=10Ω)and R2(=30Ω)in the circuit diagram with
E1 = 3V,E2 = 3 and E3 = 2Vare respectively.

E1 i1
R1
E2 E 3

R2

a) 02.A, 0.1A b) 0.4A, 0.2A c) 0.1A, 0.2A d) 0.2A, 0.4A

720. A battery has an emf of 15V and internal resistance of 1Ω. Is the terminal to terminal potential difference
less than, equal to or greater than 15V if the current in the battery is (1) from negative to positive
terminal, (2) from positive to negative terminal (3) zero current?
a) Less, grater, equal b) Less, less, equal

c) Greater, greater, equal d) Greater, less, equal

721. Three bulbs B1 ,B¬2 and B3 are connected to the main as shown in figure. How will the brightness of bulb
B1 be affected if B2 or B3 are disconnected from the circuit?

a) Bulb B become brighter b) Bulb B become dimmer

1 1

c) No change occurs in the brightness Bulb B1 becomes brighter if bulb B2 is

d) disconnected and dimmer if bulb B is
3
disconnected.
722. Two wires of same material have length L and 2L and cross-sectional areas 4A and A respectively. The
ratio of their specific resistances would be
a) 1 : 2 b) 8 : 1 c) 1 : 8 d) 1 : 1

723. A steady current of 5 amps is maintained for 45 mins. During this time it deposits 4.572 gm of zinc at the
cathode of a voltmeter. E.C.E. of zinc is
a) 3.387×10-4gm/C b) 3.387×10-4C/gm c) 3.384×10-3gm/C d) 3.394×10-3C/gm

724. The colour sequence in a carbon resistor is red, brown, orange and silver. The resistance of the resistor is

a) 21×103 ± 10% b) 23×101 ± 10% c) 21×103 ± 5% d) 12×103 ± 5%

725. Consider a cylindrical element as shown in the figure. Current flowing through element is I and resistivity

P a g e | 93
 of material of the cylinder is ρ. Choose the correct option out the following

a) Power loss is second half is four times the power loss in first half

b) Voltage drop in first is twice of voltage drop in second half

c) Current density in both halves are equal

d) Electric field in both halves is equal

726. In Wheatstone’s bridge P = 9 ohm, Q = 11 ohm, R = 4 ohm and S = 6 ohm. How much resistance must
be put in parallel to the resistance S to balance the bridge
a) 24 ohm b) 44 ohm c) 26.4 ohm d) 18.7 ohm
9
727. A galvanometer, having a resistance of 50 Ω, gives a full scale deflection for a current of 0.05A. The length
-2 2
in meter of a resistance wire of area of cross-section 2.97×10 cm that can be used to convert the
galvanometer into a ammeter which can read a maximum of 5A current is
-7
(Specific resistance of the wire = 5×10 Ωm)
a) 9 b) 6 c) 3 d) 1.5

728. You are provided three resistances 2 Ω, 3 Ω and 6 Ω. How will you connect them so as to obtain the
equivalent resistance of 4 Ω
a) 3 6 b) 3 2 c) 3 d) None of these
2
2 6 6

729. A uniform wire has resistance 24 Ω. It is bent in the form of a circle. The effective resistance between the
two end points on any diameter of the circle is
a) 6 Ω b) 12 Ω c) 3 Ω d) 24 Ω

730. Above neutral temperature, thermo e.m.f. in a thermocouple

a) Decreases with rise in temperature b) Increases with rise in temperature

c) Remains constant d) Changes sign

731. A galvanometer coil has a resistance of 15 Ω and gives full scale deflection for a current of 4mA. To convert
it to an ammeter of range 0 to 6A
a) 10 m Ω resistance is to be connected in parallel to b) 10 m Ω resistance is to be connected in series with
the galvanometer the galvanometer
c) 0.1 Ω resistance is to be connected in parallel to d) 0.1 Ω resistance is to be connected in series with
the galvanometer the galvanometer
-8 -1
732. When 1 g hydrogen (ECE=1.044×106 kg C ) forms water, 34 kilo cal heat is liberated. The minimum
voltage required to decompose water is
a) 0.75 V b) 1.5 V c) 3.0 V d) 4.5 V

733. A current 2 A flows through a 2Ω resistor when connected across a battery. The same battery supplies a
current 0.5 A when connected across a 9Ω resistor. The internal resistance of the battery is
a) 1Ω b) 0.5Ω c) 1/3Ω d) 1/4Ω

P a g e | 94
734. A cell in secondary circuit gives null deflection for 2.5m length of potentiometer having 10m length of
wire. If the length of the potentiometer wire is increased by 1 m without changing the cell in the primary,
the position of the null point now is
a) 3.5 m b) 3 m c) 2.75 m d) 2.0 m

735. Certain wire has resistance of 10Ω. If its is stretched by 1/10th of its length, then its resistance is nearly

a) 9 Ω b) 10 Ω c) 11 Ω d) 12 Ω

736. A wire of resistance 10 Ω is bent to form a circle. P and Q are points on the circumference of the circle
dividing it into a quadrant and are connected to a battery of 3 V and internal resistance 1 Ω as shown in the
figure. The currents in the two
parts of the circle are

3V
Q
1Ω

a) 6 A and 18 A b) 5 A and 15 A c) 4 A and 12 A d) 3 A and 9 A

23 23 26 26 25 25 25 25
737. In an experiment of meter bridge, a null point is obtained at the centre of the bridge wire. When a
resistance of 10 ohm is connected in one gap, the value of resistance in other gap is
a) 10Ω b) 5Ω c) 1 Ω d) 500Ω
5
738. A conductor wire having 1029 free electrons/m3 carries a current of 20A. If the cross-section of the wire is
2
1mm , then the drift velocity of electrons will be
a) 6.25×10-3ms-1 b) 1.25×10-5ms-1 c) 1.25×10-3ms-1 d) 1.25×10-4ms-1

739. Two resistances R and 2R are connected in parallel in an electric circuit. The thermal energy developed in
in R and 2R is in the ratio
a) 1 : 2 b) 1 : 4 c) 4 : 1 d) 2 : 1

740. Five equal resistances, each of resistance R,are connected as shown in figure below. A bettery of V volt is
connected between A and B. The current flowing in FC will be
C

R R

F R
R A
B

R
E

a) 3V b) V c) V d) 2V
R R 2R R
741. Three moving coil galvanometers A, B and C are made of coils of three different material having torsional
-8 -8 -8
constant 1.8×10 , 2.8×10 and 3.8×10 respectively. If the three galvanometers are identical in all other
respect, then in which of the above cases sensitivity is maximum?
a) A b) C c) B d) Same in each case

742. In the given circuit, the voltmeter records 5V. The resistance of the voltmeter in ohm is

P a g e | 95
 V

100 Ω 50 Ω

( )
10 V

a) 200 b) 100 c) 10 d) 50

743. The emf of the battery shown in figure, is

2 2 1

E 4 2 1
1A

a) 12 V b) 13 V c) 16 V d) 18 V

744. A charge of 2×10-1C move at 30 revolutions per second in a circle of diameter 80 cm. The current linked
with the circuit is
a) 0.02 A b) 20 A c) 0.60 A d) 60 A

745. A capacitor of 10 μF has a potential difference of 40 V across it. If it is discharged in 0.2 s, the average
current during discharge is
a) 2 mA b) 4 mA c) 1 mA d) 0.5 mA

746. An electric bulb is rated 220 V – 100 W. The power consumed by it when operated on 110 V will be

a) 75 W b) 40 W c) 25 W d) 50 W

747. Two batteries A and B each of e.m.f. 2 V are connected in series to an external resistance R = 1 ohm. If the
internal resistance of battery A is 1.9 ohm and that of B is 0.9 ohm, what is the potential difference
between the terminals of battery A
A B

a) 2 V b) 3.8 V c) Zero d) None of the above

748. The effective resistance across the points A and I is

P a g e | 96
 a) 2 Ω b) 1 Ω c) 0.5 Ω d) 5 Ω

749. When the length and area of cross-section both are doubled, then its resistance

a) Will become half b) Will be doubled

c) Will remain the same d) Will become four times

750. The resistance of a conductor increases with

a) Increase in length b) Increase in temperature

c) Decrease in cross-sectional area d) All of these

751. A copper wire of resistance R is cut into ten parts of equal length. Two pieces each are joined in series and
then five such combinations are joined in parallel. The new combination will have a resistance
a) R b) R c) R d) R
4 5 25
752. Which of the following characteristics of electron determines the current in a conductor?

a) Thermal velocity alone b) Drift velocity alone

c) Both thermal velocity and drift velocity d) None of the above

753. Which of the following circuits is correct for verification of Ohm’s law?

d) None of these
a) b) c)

754. If the free electron density be n and relaxation time be τ, the electrical conductivity of a conductor may be
expressed as
2 2 2
a) neτ b) ne τ c) ne d) mee τ
me me τme n
755. Two identical electric lamps marked 500 W, 220 V are connected in series and then joined to a 110 V line.
The power consumed by each lamp is
a) 125 W b) 25 W c) 225 W d) 125 W
4 4 4
756. The charge supplied by source varies with time t as Q = at - bt2. The total heat produced in resistor 2R is

3 3 3
a) a R b) a R c) a R d) None of these
6b 27b 3b
757. A 4μ F conductor is charged to 400 V and then its plates are joined through a resistance of 1 kΩ. The heat
produced in the resistance is
a) 0.18 J b) 0.21 J c) 0.25 J d) 0.32 J

758. In the figure given the value of X resistance will be, when the p.d. between B and D is zero

P a g e | 97
 B
X
6

8 3
15
A C
4
15 6
4
6 4
D

a) 4 ohm b) 6 ohm c) 8 ohm d) 9 ohm

759. A heater is operated with a power of 1000W in a 100V line. It is connected in combination with a
resistance of 10Ω and a resistance R to a 100V line as shown in figure. What should be the value of R so,
that the heater operates with a power of 62.5W

a) 10Ω b) 62.5Ω c) 1 Ω d) 5Ω
5
760. The masses of three wires of copper are in the ratio 1:3:5 and lengths are in the ratio 5:3:1. Then the ratio
of their electrical resistances are
a) 1:3:5 b) 5:3:1 c) 1:15:25 d) 125:15:1

761. In the figure a part of electric circuit has been shown. The value of current i is

a) 1.7 A b) 3.7 A c) 1.3 A d) 1A

762. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the
current, the temperature of the wire due to the current, the temperature of the wire is raised by ∆T is a
time t. A number N of similar cells is now connected in series with a wire of the same material and cross-
section but of length 2L. The temperature of the wire is raised by the same amount ∆T in the same time t.
The value of N is
a) 4 b) 6 c) 8 d) 9

763. A milli voltmeter of 25 milli volt range is to be converted into an ammeter to 25 ampere range. The value
(in ohm) of necessary shunt will be
a) 0.001 b) 0.01 c) 1 d) 0.05

764. Three bulbs of 40 W,60 W, 100 W are arranged in series with 220 volt supply. Which bulb has minimum
resistance
a) 100 W b) 40 W c) 60 W d) Equal in all bulbs

765. A copper voltmeter is connected in series with a heater coil of resistance 0.1Ω. A steady current flows in
the circuit for twenty minutes and mass of 0.99 g of copper is deposited at the cathode. If electrochemical
equivalent of copper is 0.00033 gm/C, then heat generated in the coil is
a) 750 J b) 650 J c) 350 J d) 250 J
P a g e | 98
766. In the circuit shown the equivalent resistance between A and B is

A B
9

a) 27 Ω b) 18 Ω c) 9 Ω d) 3 Ω

767. The relation between Seeback coefficient (or thermo electric power) S and Peltier coefficient π is given by
2
a) S = πT b) S = π c) S = π d) S = π
2
T T T
768. A thermocouple develops 40 μV/kelvin. If hot and cold junctions are at 40℃ and 20℃ respectively, then
then emf developed by a thermopile using such 150 thermocouples in series shall be
a) 150mV b) 80mV c) 144mV d) 120mV

769. The thermo emf produced in a thermo-couple is 3 microvolt per degree centigrade. If the temperature of
the cold junction is 20℃ and the thermo emf is 0.3 millivolt, the temperature of the hot junction is
a) 80℃ b) 100℃ c) 120℃ d) 140℃

770. Battery shown in figure has e.m.f. E and internal resistance r. Current in the circuit can be varied by sliding
the contact J. If at any instant current flowing through the circuit is I, potential difference between
terminals of the cell is V, thermal power generated in the cell is equal to η fraction of total electrical power
generated in it.; then which of the following graphs is correct
+ – r

a) V b) P

I I

c)  d) Both (a) and (b) are correct

771. In the circuit shown in the figure, the current flowing in 2 Ω resistance

10 2

1.4A
G

25 5

a) 1.4 A b) 1.2 A c) 0.4 A d) 1.0 A

772. The current in the given circuit is

10Ω 5
V

P a g e | 99
 A B

2V 20

a) 0.3A b) 0.4A c) 0.1A d) 0.2A

773. Given R1 = 5.0 ± 0.2Ω, R2 = 10.0 ± 0.1Ω. What is total resistance in parallel with possible percentage
error?
a) 15Ω ± 2% b) 3.3 Ω ± 7% c) 15 Ω ± 7% d) 3.3Ω ± 2%

774. 3
The equivalent resistance between the points P and Q in the network given here is equal to (given r = Ω)
2

r r
r
r r
P Q

r
r r

a) 1 Ω b) 1 Ω c) 3 Ω d) 2 Ω
2 2
775. The resistance of a wire at 300 K is found to be 0.3Ω. If the temperature coefficient of
-3 -1
resistance of wire is 1.5×10 K , the temperature at which the resistance becomes 0.6Ω is
a) 720 K b) 345 K c) 993 K d) 690 K

776. Resistance of a resistor at temperature t° C is Rt = R0(1+αt+βt2).

Here Rois the resistance at 0°C. The temperature coefficient of resistance at temperature t°C is
a) (1+αt+βt ) d) (α+2βt) 2
2
b) (α+2βt) c) α+2βt 2
α+2βt (1+αt+βt ) 2(1+αt+βt )
777. If a rod has resistance 4Ω and if rod is turned as half circle, then the resistance along diameter is

a) 1.56Ω b) 2.44Ω c) 4Ω d) 2Ω

778. In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of
10 Ω is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire
then internal resistance r of the battery is
2V

1m

B
A
E r

a) 1 Ω b) 3 Ω c) 10 Ω d) 5 Ω

779. Two wires of equal diameters, of resistivities ρ1 and ρ2 and lengths l1 and l2, respectively, are joined in
series. The equivalent resistivity of the combination is

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 a) ρ1l1+ρ2l2 b) ρ1l2+ρ2l1 c) ρ1l2+ρ2l1 d) ρ1l1-ρ2l2
l1+l2 l1-l2 l1+l2 l1-l2
780. If six identical cells each having an e.m.f. of 6V are connected in parallel, the e.m.f. of the combination is

a) 1 V b) 36 V c) 1 V d) 6 V
6
781. In the circuit shown, the heat produced in the 5Ω resistor due to current flowing in it is 10 cal - s-1. The heat
generated in Ω resistor is

4 6

a) 1 cal - s-1 b) 2 cal - s-1 c) 3 cal - s-1 d) 4 cal - s-1

782. In the figure shown, the capacity of the condenser C is 2 μF. The current in 2Ω resistor is
2Ω

3Ω
2F
4Ω

+ –
6V 2.8Ω

a) 9 A b) 0.9 A c) 1 A d) 1 A
9 0.9
783. To deposit one gm equivalent of an element at an electrode, the quantity of electricity needed is

a) One ampere b) 96000 amperes c) 96500 farads d) 96500 coulombs

784. In the following Wheatstone bridge P/Q = R/S. If key K is closed, then the galvanometer will show
deflection
P Q

R S

a) In left side b) In right side c) No deflection d) In either side

785. We have two wires A and B of same mass and same material. The diameter of the wire A is half of that B. If
the resistance of wire A is 24 ohm then the resistance of wire B will be
a) 12 Ohm b) 3.0 Ohm c) 1.5 Ohm d) None of the above

786. The Petlier coefficient of a thermo-couple of metls A and B at junction temperature T is given by
2 2
a) T2 dE2 b) T dE c) T3 dE d) T4 d E
2
dT dT dT dT
787. A 30, 90 W lamps are to be operated on a 120 V DC line. For proper glow, a resistor of ……. Ω should be
connected in series with the lamp.
a) 40 b) 10 c) 20 d) 30

P a g e | 101
788. The lead wires should have

a) Larger diameter and low resistance b) Smaller diameter and high resistance

c) Smaller diameter and low resistance d) Larger diameter and high resistance

789. A battery of emf 2V and internal resistance 0.1 Ω is being charged by a current of 5A. the potential
difference between the terminals of the battery is
a) 2.5 Ω b) 1.5 Ω c) 0.5 Ω d) 1 Ω

790. A potentiometer circuit shown in the figure is set up to measure e.m.f. of a cell E. As the point P moves
from X to Y the galvanometer G shows deflection always in one direction, but the deflection decreases
continuously until Y is reached. In order to obtain balance point between X and Y it is necessary to
V R

X P
Y

E
G

a) Decreases the resistance R b) Increase the resistance R

c) Reverse the terminals of battery V d) Reverse the terminals of cell E

791. The reading of the ideal voltmeter in the adjoining diagram will be
A

10V 20Ω
V

10Ω 4V

B N C

a) 4 V b) 8 V c) 12 V d) 14 V

792. A wire 20 cm long and 1 mm2in cross-section carries a current of 4A when connected to a 2V battery. The
resistivity of the wire is
a) 2×10-7Ω m b) 5×10-7Ω m c) 4×10-6Ω m d) 1×10-6Ω m

793. The resistance of a galvanometer is 50 Ω and it shows full scale deflection for a current of 1mA. To convert
it into a voltmeter to measure 1V and as well as 10 V (refer circuit digram) the resistance R1 and R2
respectively are
G
R 1 R 2

1V 10 V

a) 950 Ω and 9150 Ω b) 900 Ω and 9950 Ω c) 900 Ω and 9000Ω d) 950 Ω and 9950 Ω

794. The temperature of cold junction of thermocouple is 0°C. If the neutral temperature is 270°C, then the
inversion temperature is
a) 540°C b) 520°C c) 640°C d) 58°C

795. In the circuit shown, if the resistance 5 Ω develops a heat of 42 J per second, heat developed in 2 Ω must be
about (in Js )
-1

P a g e | 102
 6Ω 9Ω

2Ω

5ΩA

a) 25 b) 20 c) 30 d) 35

796. The following four wires are made of the same material and are at the same temperature. Which one of
them has highest electrical resistance?
a) Length =50 cm, diameter=0.5 mm b) Length =100 cm, diameter=1 mm

c) Length=200 cm , diameter=2 mm d) Length=300cm, diameter=3 mm

797. A meter bridge is set-up as shown in figure, to determine an unknown resistance X using a standard 10 Ω
resistor. The galvanometer shows null point when tapping key is at 52cm mark. The end-corrections are
1cm and 2cm respectively for the ends A and B. the determined value of x is

X 10 Ω

A B

a) 10.2 Ω b) 10.6 Ω c) 10.8 Ω d) 11.1 Ω

798. The resistance between the terminal points A and B of the given infinitely long circuit will be
1 1 1
A

1 1 Upto
infinity

B
1 1 1

a) ( 3 - 1) b) (1 - 3) c) (1 + 3) d) (2 + 3)

799. A 60 watt bulb carries a current of 0.5 amp. The total charge passing through it in 1 hour is

a) 3600 coulomb b) 3000 coulomb c) 2400 coulomb d) 1800 coulomb

800. A bulb of 220 V and 300 W is connected across 110 V circuit, the percentage reduction in power is

a) 100% b) 25% c) 70% d) 75%

801. If the temperature of cold junction of thermocouple is lowered, then the neutral temperature

a) Increases b) Approaches inversion temperature

c) Decreases d) Remains the same

802. The resistance of the filament of a lamp increases with the increase in temperature. A lamp rated 100 W
and 200 V is connected across 220 V power supply. If the voltage drops by 10%, then the power of the
P a g e | 103
 lamp will be
a) 90 W b) 81 W

c) Between 90 and 100 W d) Between 81 and 90 W

803. Which factor is immaterial for the wire used in electric fuse?

a) Length b) Radius c) Material d) Current

804. If 1 A current is passed through CuSO4 solution for 10 s, the number of copper atoms deposited at the
cathode will be
a) 8×1019 b) 3.1×1019 c) 6.2×1019 d) 1.6×1020

805. All of the following statements are true except

a) Conductance is the reciprocal of resistance and is measured in Siemen

b) Ohm's law is not applicable at very low and very high temperatures

c) Ohm's law is applicable to semiconductors

d) Ohm's law is not applicable to electron tubes, discharge tubes and electrolytes

806. A galvanometer of resistance 20 Ω is to be converted into an ammeter of range 1 A. If a current of 1 mA

produces full scale deflection, the shunt required for the purpose is
a) 0.01 Ω b) 0.05 Ω c) 0.02 Ω d) 0.04 Ω

807. The net resistance of a voltmeter should be large to ensure that

a) It does not get overheated

b) It does not draw excessive current

c) It can measure large potential difference

d) It does not appreciably change the potential difference to be measured

808. A galvanometer of 50 ohm resistance has 25 divisions. A current of 4×10-4 ampere gives a deflection of one
division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected
with a resistance of
a) 2500 Ω as a shunt b) 2450 Ω as a shunt c) 2550 Ω in series d) 2450 Ω in series

809. A 25 watt, 220 volt bulb and a 100 watt, 220 volt bulb are connected in series across a 220 volt lines.
Which electric bulb will glow more brightly
a) 25 watt bulb b) 100 watt bulb

c) First 25 watt and then 100 watt d) Both with same brightness

810. Twelve wires of resistance 6 Ω are connected to form a cube as shown in the figure. The current centers at
a corner A and leaves at the diagonally opposite corner G. the joint resistance across the corner A and G
are

P a g e | 104
 E F

G
H

D C

a) 12 Ω b) 6 Ω c) 3 Ω d) 5 Ω

811. The total power dissipated in Watts in the circuit shown here is

a) 16 b) 40 c) 54 d) 4

812. The equivalent resistance of resistor connected in series is always

a) Equal to the mean of component resistors

b) Less than the lowest of component resistors

c) In between the lowest and the highest of component resistors

d) Equal to sum of component resistors

813. In the given circuit shown in figure it is observed that the current i is independent of the value of
resistance R6. Then the resistance values must satisfy

a) R R R = R R R b) 1 + 1 = 1 + 1
1 2 5 3 4 6 R5 R6 R1+R2 R3+R4
c) R R = R R d) R R = R R = R R
1 4 2 3 1 3 2 4 5 6

814. By using only two resistances coils-singly, in series or in parallel one should be able to obtain resistance of
3,4,12 and 16 ohm. The separate resistance of the coil are
a) 3 and 4 b) 4 and 12 c) 12 and 16 d) 16 and 13

815. A battery is charged at a potential of 15 V for 8 hours when the current flowing is 10 A. The battery on
discharge supplies a current of 5 A for 15 hours. The mean terminal voltage during discharge is 14 V. The
"Watt - hour" efficiency of the battery is
a) 82.5% b) 80% c) 90% d) 87.5%
P a g e | 105
816. A battery is charged by a supply of 100 V as shown in figure. The charging current is 1.0A. the value of R is

a) 88 Ω b) 68 Ω c) 44 Ω d) None of these

817. Three resistances of values 2Ω, 3Ω and 6Ω are to be connected to produce an effective resistance of 4Ω.
This can be done by connecting
a) 6Ω resistance in series with the parallel combination of 2Ω and 3Ω

b) 3Ω resistance in series with the parallel combination of 2Ω and 6Ω

c) 2Ω resistance in series with the parallel combination of 3Ω and 6Ω

d) 2Ω resistance in parallel with the parallel combination of 3Ω and 6Ω

818. Current provided by a battery is maximum when

a) Internal resistance equal to external resistance

b) Internal resistance is greater than external resistance

c) Internal resistance is less than external resistance

d) None of these

819. Three equal resistances, each of R ohm, are connected as shown in figure. A battery of 2 V and internal
resistance 0.1 Ω is connected across the circuit. The value of R for which the heat generated in the circuit
will by maximum is

a) 0.3Ω b) 0.01Ω c) 0.1Ω d) 0.03Ω

820. For a certain thermocouple, if the temperature of the cold junction is 0℃, the neutral temperature and
inversion temperature are 285℃ and 570℃ respectively. If the cold junction is brought to 10℃, then the
new neutral and inversion temperatures are respectively
a) 285℃ and 560℃ b) 285℃ and 570℃ c) 295℃ and 560℃ d) 275℃ and 560℃

821. Two resistances when connected in parallel across a cell of negligible internal resistance consume 4 times
the power they would consume when connected in series. If one resistance is 5 Ω , the other is
a) 1 Ω b) 2.5 Ω c) 5 Ω d) 10 Ω

822. The maximum current that flows through a fuse wire before it blows out varies with its radius as

a) 3/2 b) r c) 2/3 d) 1/2

r r r
823. At what temperature will the resistance of a copper wire become three times its value at 0℃ (Temperature
-3
coefficient of resistance for copper = 4×10 per ℃)
a) 400℃ b) 450℃ c) 500℃ d) 550℃

824. If N is the Avogadro’s number and e is the electronic charge then the Faraday’s constant F is equal to
P a g e | 106
 a) Ne b) N2e c) Ne2 d) 1
Ne
825. A current of 1.5 A flows through a copper voltameter. The thickness of copper deposited on the electrode
-3
surface of size 50 cm×10 cm is 20 min will be (density of copper = 9000 kg - m and ECE of
-1
copper = 0.00033gC )
a) 3.3×10-6 m b) 6.6 ×10-6 m c) 1.3 ×10-5 m d) 2.6 ×10-5 m

826. Four resistances carrying a current shown in the circuit diagram re immersed in a box containing ice at
0℃. How much ice must be put in the box every 10 min to keep the average quantity of in the box
constant?

10 Ω 5Ω
10A 10A
Q P
5Ω 10 Ω

a) 5 kg b) 1.19 kg c) 3 kg d) 2.29 kg

827. A battery consists of a variable number (n) of identical cells, each having an internal resistance r
connected in series. The terminal of the battery is short-circuited. A graph of current versus the number of
cells will be as shown in figure

a) b) c) d)

828. In a copper voltameter, mass deposited in 6 minutes is m gram. If the current-time graph for the
voltameter is as shown here, then the E.C.E of the copper is
2.0
I ampere

1.0

0 2 4 6
t (min)

a) m/5 b) m/300 c) 5 m d) m/18000

829. In the circuit shown in the figure the potential difference between X and Y will be
40 Ω X Y

|
120 V 20 Ω

a) Zero b) 20 V c) 60 V d) 120 V

830. In the following circuit, bulb rated as 1.5 V, 0.45 W. If bulbs glows with full intensity then what will be the
equivalent resistance between X and Y

P a g e | 107
 a) 0.45 Ω b) 1 Ω c) 3 Ω d) 5 Ω

831. Two wires of same dimensions but resistivities ρ1 and ρ2are connected in series. The equivalent resistivity
of the combination is
a) ρ ρ b) (ρ + ρ ) c) ρ1+ρ2 d) None of these
1 2 1 2
2
832. In a balanced Wheatstone’s network, the resistance in the arms Q and S are interchanged. As a result of
this
a) Network is not balanced

b) Network is still balanced

c) Galvanometer shows zero deflection

d) Galvanometer and the cell must be interchanged to balance

833. On passing the current in water voltmeter, hydrogen

a) Is liberated at anode b) Is liberated at cathode

c) Is not liberated d) Remains in the solution

834. Two electric bulbs have ratings respectively of 25 W, 220 V and 100 W, 220 V. If the bulbs are connected
in series with a supply of 440, which bulb will fuse?
a) 25 W bulb b) 100 W bulb c) Both of these d) None of these

835. A cell of internal resistance 3 ohm and emf 10 volt is connected to a uniform wire of length 500 cm and
resistance 3 ohm. The potential gradient in the wire is
a) 30 mV/cm b) 10 mV/cm c) 20 mV/m d) 4 mV/cm

836. Consider four circuits shown in figure. In which circuit power dissipated is greatest. (Neglect the internal
resistance of the power supply)

a) b) c) d)

837. Consider the following two statements A and B and identify the correct choice given in the answer
(A) Duddells thermo-galvanometer is suitable to measure direct current only
-3
(B) Thermopile can measure temperature differences of the order of 10 ℃
a) Both A and B are true b) Both A and B are false

c) A is true but B is false d) A is false but B is true

838. Three resistors 1 Ω,2Ω and 3Ω are connected to form a triangle. Across 3Ω resistor a 3V battery is
connected. The current through 3 Ωresistor is
a) 0.75A b) 1A c) 2A d) 1.5A

P a g e | 108
839. A silver and a zinc voltmeter are connected in series and a current I is passed through them for a time t,
liberating w gram of zinc. The weight of silver deposited is nearly
a) 1.7 w g b) 2.4 w g c) 3.5 w g d) 1.2 w g

840. In a meter bridge experiment, null point is obtained at 20cm from one end of the wire when resistance X is
balanced against another resistance Y. If X<Y, then where will be the new position of the null point from
the same end, if one decides to, balance a resistance of 4X against Y?
a) 50 cm b) 80 cm c) 40 cm d) 70 cm

841. A galvanometer of resistance 240Ωallows only 4% of the main current after connecting a shunt resistance.
The value of the shunt resistance is
a) 10 Ω b) 20 Ω c) 8 Ω d) 5 Ω

842. The two bulbs, one of 60W and other 200W are connected in series to a 200 volt line, then

a) The potential drop across two bulbs in the same b) The potential drop across the 60 W bulb is greater
than the potential drop across the 200 W bulb
c) The potential drop across the 200 W bulb is d) The potential drop across both the bulbs is 200
greater than the 60 W bulb volt
843. Electromotive force is the force, which is able to maintain a constant

a) Current b) Resistance c) Power d) Potential difference

844. In the following circuit a 10 m long potentiometer wire with resistance 1.2 ohm/m, a resistance R1 and an
accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mV then the
deflection in galvanometer is zero. The current supplied by the accumulator will be

a) 4×10-4A b) 8×10-4A c) 4×10-3A d) 8×10-3A

845. A house wife uses a 100 W bulb 8 h a day , and an electric heater of 300 W for 4 h a day. The total cost for
the month of June at the rate of 0.05 rupee per unit will be
a) Rs 20 b) Rs 25 c) Rs 30 d) Rs 30 paise 50

846. In a potentiometer, the null points are received at 7th wire. If now we have to change the null points at 9th
wire, what should we do?
a) Attach resistance in series with battery b) Increase resistance in main circuit

c) Decrease resistant in main circuit d) Decrease applied emf

847. 12
The effective resistance of two resistors in parallel is
Ω. If one of the resistors is disconnected the
7
resistance becomes 4 Ω. The resistance of the other resistor is
a) 4 Ω b) 3 Ω c) 12 Ω d) 7 Ω
7 12
848. A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The e.m.f. of the cell needed to make
the bulb glow at full intensity is

P a g e | 109
 4.5 W
1.5 V

1Ω

E(r=2.67Ω)

a) 4.5 V b) 1.5 V c) 2.67 V d) 13.5 V

849. The electric current passing through a metallic wire produces heat because of

a) Collisions of conduction electrons with each other

b) Collisions of the atoms of the metal with each other

c) The energy released in the ionization of the atoms of the metal

d) Collisions of the conduction electrons with the atoms of the metallic wires

850. One junction of a thermo-couple is a particular temperature Tr and another is at T. Its thermo emf is
expressed as

{ 1
E = K(T - Tr) T0- (T+Tr)
2 }
T
At a temperature T = 0 , the value of thermo-electric power will be
2
a) 1 KT b) KT c) 1 KT2 d) 1 K(T -T )2.
r
2 0 0
2 0 2 0

851. Two identical heaters rated 220 volt, 1000 watt are placed in series with each other across 220 volt lines.
If resistance does not change with temperature, then the combined power is
a) 1000 watt b) 2000 watt c) 500 watt d) 4000 watt

852. In the adjoining circuit, the battery E1 has an e.m.f. of 12volt and zero internal resistance while the battery
E has an e.m.f. of 2volt. If the galvanometer G reads zero, then the value of the resistance X in ohm is
500 
A G B

E1 X E

D C

a) 10 b) 100 c) 500 d) 200

853. In a potentiometer of one metre length, an unknown e.m.f. voltage source is balanced at 60 cm length of
potentiometer wire, while a 3 volt battery is balanced at 45 cm length. Then the e.m.f. of the unknown
voltage source is
a) 3V b) 2.25V c) 4V d) 4.5V

854. A potentiometer has uniform potential gradient across it. Two cells connected in series (i) to support each
other and (ii) to oppose each other are balanced over 6m and 2m respectively on the potentiometer wire.
The e.m.f.'s of the cells are in the ratio of
a) 1 : 2 b) 1 : 1 c) 3 : 1 d) 2 : 1

855. In a thermocouple, the temperature that does not depend on the temperature of the cold junction is called

a) Neutral temperature b) Temperature of inversion

P a g e | 110
 c) Both the above d) None of the above

856. A current of A enters one corner one corner P of an equilateral triangle PQRhaving 3 wires of resistance 2
Ωeach and leaves by the corner R. then the current I1and I2are
6A

P
I1
I2
2Ω 2Ω

Q
2Ω R

a) 2A, 4A b) 4A, 2A c) 1A, 2A d) 2A, 3A

857. Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is
connected between A and B. The current flowing in AFCEB will be
C

R R R
F
R A
B
R
D E

a) 3V b) V c) V d) 2V
R R 2R R
858. The resistance of a straight conductor does not depend on its

a) Length b) Temperature

c) Material d) Shape of cross-section

859. Consider the circuit shown in the figure. The current I3 is equal to
28 54

6V
I3

8V 12 V

a) 5 amp b) 3 amp c) -3 amp d) -5/6 amp

860. An immersion heater with electrical resistance 7 Ω is immersed in 0.1 kg of water at 20℃ for 3 min. If the
flow of current is 4 A, what is the final temperature of the water in ideal conditions?
(Specific heat capacity of water=4.2×103 Jkg)-1K-1
a) 28℃ b) 48℃ c) 52℃ d) 68℃

861. A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if

a) R = r b) R < r c) R > r d) R = r/2

862. If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a
P a g e | 111
 a) Low resistance in parallel b) High resistance in parallel

c) High resistance in series d) Low resistance in series

863. A nichrome wire 50 cm long and one square millimetre cross-section carries a current of 4A when
connected to a 2V battery. The resistivity of nichrome wire in ohm metre is
a) 1×10-6 b) 4×10-7 c) 3×10-7 d) 2×10-7

864. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of
100 W and 200 V lamp, when not in use?
a) 40Ω b) 20Ω c) 400Ω d) 20Ω
+
865. In a neon gas discharge tube Ne ions moving through a cross-section of the tube each second to the right is
18 18
2.9×10 , while 1.2×10 electron move towards left in the same time; the electronic charge being
-19
1.6×10 C, the net electric current is
a) 0.27 A to the right b) 0.66 A to the right c) 0.66 A to the left d) Zero

866. A battery of e.m.f. 3 volt and internal resistance 1.0 ohm is connected in series with copper voltmeter. The
current flowing in the circuit is 1.5 amperes. The resistance of voltmeter will be
a) Zero b) 1.0 ohm c) 1.5 ohm d) 2.0 ohm

867. If two identical heaters each rated as (1000 W-220 V) are connected in parallel to 220 V, then the total
power consumed is
a) 200 W b) 2500 W c) 250 W d) 2000 W

868. A voltmeter essentially consists of

a) A high resistance, in series with a galvanometer b) A low resistance, in series with a galvanometer

c) A high resistance in parallel with a galvanometer d) A low resistance in parallel with a galvanometer

869. The equivalent resistance between points A and B with switch S open and closed are respectively

a) 4 Ω, 8 Ω b) 8 Ω, 4 Ω c) 6 Ω, 9 Ω d) 9 Ω, 6 Ω

870. In the Wheatstone bridge shown below, in order to balance the bridge, we must have

a) R = 3 Ω ;R = 3 Ω b) R = 6 Ω ;R = 15 Ω
1 2 1 2

c) R = 1.5 Ω ;R = any finite value d) R = 3Ω ;R = any finite value

1 2 1 2

871. A wire of resistance R is cut into 'n' equal parts. These parts are then connected in parallel. The equivalent

P a g e | 112
 resistance of the combination will be
a) nR b) R c) n d) R2
n R n
872. The power dissipated across resistance R which is connected across a battery of potential V is P. If
resistance is doubled, then the power becomes
a) 1/2 b) 2 c) 1/4 d) 4

873. What will happen when a 40 watt, 220 volt lamp and 100 watt, 220 volt lamp are connected in series
across 40 volt supply
a) 100 watt lamp will fuse b) 40 watt lamp will fuse

c) Both lamps will fuse d) Neither lamp will fuse

874. If nearly 105C liberate 1 g equivalent of aluminium, then the amount of aluminium (equivalent weight g)
deposited through electrolysis in 20 min by a current of 50 A will be
a) 0.09 g b) 0.6 g c) 5.4 g d) 10.8 g

875. Two cells, each of e.m.f. E and internal resistance r are connected in parallel between the resistance R. The
maximum energy given to the resistor will be, only when
a) R = r/2 b) R = r c) R = 2r d) R = 0

876. Eels are able to generate current with biological cells called electroplaques. The electroplaques in an eel
are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000
electroplaques. The arrangement is suggestively shown below. Each electroplaque has an emf of 0.15 V
and internal resistance of 0.25 Ω
0.15 V
0.25 
+ – + – + –

+ – + – + –
5000 electroplaques per
100

+ – + – + 12V

500 

The water surrounding the eel completes a circuit between the head and its tail. If the water surrounding it
has a resistance of 500 Ω, the current an eel can produce in water is about
a) 1.5 A b) 3.0 A c) 15 A d) 30 A

877. Time taken by a 836 W heater to heat one litre of water from 10℃ to 40℃ is

a) 50 s b) 100 s c) 150 s d) 200 s

878. What is the equivalent resistance between points A and B in the circuit if figure, if R = 3 Ω?

a) 8 Ω b) 9 Ω c) 12 Ω d) 15 Ω

879. What must be the efficiency of an electric kettle marked 500 W, 230 V, if it was found to bring 1 kg of water

P a g e | 113
 at 15℃ to boiling point in 15 min? (Given specific heat capacity of water = 420 J/kg℃)
a) 79% b) 81% c) 72% d) 69%

880. A metallic resistor is connected across a battery. If the number of collisions of the free electrons with the
lattice is some how decreased in the resistor (for example by cooling it), the current will
a) Remains constant b) Increase c) Decrease d) Become zero

881. Tap supplies water at 20℃. A man takes 1 L of water per minute at 35℃ from a geyser connected to the
tap. The power of geyser is
a) 1050 W b) 2100 W c) 1500 W d) 3000 W

882. The number of free electrons per 100 mm of ordinary copper wire is 2×1021. Average drift speed of
-1
electrons is 0.25mms . The current flowing is
a) 8 A b) 0.8 A c) 80 A d) 5 A

883. On increasing the temperature of a conductor, its resistance increases because the

a) Relaxation time increases b) Mass of electron increases

c) Electron density decreases d) Relaxation time decreases

884. In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a
60 W bulb for use in India is R,the resistance of a 60 W bulb for use in USA will be
a) R b) 2 R c) R/4 d) R/2

885. In the given circuit, with steady current, the potential drop across the capacitor must be
V R

V C

2V 2R

a) V b) V/2 c) V/3 d) 2V/3

886. Two resistances R1 and another R2 of the same material but twice the length and half the thickness are
connected in series with a standard battery E of internal resistance r. The balancing point is
a) 1 b) 1 c) 8 l d) 16 l
8l 4l
887. When a current passes through the junction of two different metals, evolution or absorption of heat at the
junction is known as
a) Joule effect b) Seebeck effect c) Peltier effect d) Thomson effect

888. Three equal resistors are connected as shown in figure. The maximum power consumed by each resistor is
18 W. Then maximum power consumed by the combination is

a) 18 W b) 27 W c) 36 W d) 54 W

889. The length of the resistance wire is increased by 10%. What is the corresponding change in the resistance
of wire?
a) 10 % b) 25 % c) 21 % d) 9 %

890. The number of dry cells, each of e.m.f. 1.5 volt and internal resistance 0.5 ohm that must be joined in series
P a g e | 114
 with a resistance of 20 ohm so as to send a current of 0.6 ampere through the circuit is
a) 2 b) 8 c) 10 d) 12

891. A thermister is dipped in a bath whose temperature is to be measured. When the temperature increase the
current also increase, because of decreases in
a) Capacitance b) Reactance c) Resistance d) Resistivity

892. In the given circuit diagram the current through the battery and the charge on the capacitor respectively in
steady state are

a) 1A and 3 μC b) 17 A and 0 μC c) 6 A and 12 μC d) 11A and 3μC

7 7
893. To sand 10% of main current through a moving coil galvanometer of resistance 9 Ω shut required

a) 9 Ω b) 11 Ω c) 10 Ω d) 9.9 Ω

894. The resistance of a wire of uniform diameter d and length L is R. The resistance of another wire of the
same material but diameter 2d and length 4 L will be
a) 2R b) R c) R/2 d) R/4

895. The potential difference across 8Ω resistance is 48V as shown in figure. The value of potential difference
across points A and B will be

A 3Ω

20 30 60

24 8Ω 48 V

B 1Ω

a) 62 V b) 80 V c) 128 V d) 160 V

896. For obtaining chlorine by electrolysis a current of 100 KW and 125 V is used. (Electro chemical equivalent
6 -1
of chlorine is 0.367×10 kgC ). The amount of chlorine obtained in one min will be
a) 1.7616 g b) 17.616 g c) 0.17161 kg d) 1.7616 kg

897. In a meter bridge, the balancing length from the left end (standard resistance of one ohm is in the right
gap) is found to be 20 cm. The value of the unknown resistance is
a) 0.8 Ω b) 0.5 Ω c) 0.4 Ω d) 0.25 Ω

898. A 25 W and 100 W bulbs are joined in series and connected to the mains. Which bulb will glow brighter?

a) 25 W bulb b) 100 W bulb

P a g e | 115
 c) Both bulb will glow brighter d) None will glow brighter

899. Temperature of cold junction in a thermocouple is 270℃, then the temperature of inversion is

a) 540℃ b) 530℃ c) 280℃ d) 260℃

900. A galvanometer having a coil resistance of 60 Ω shows full scale deflection when a current of 1.0 amp
passes through it. It can be converted into an ammeter to read currents upto 5.0 amp by
a) Putting in parallel a resistance of 240 Ω b) Putting in series a resistance of 15 Ω

c) Putting in series a resistance of 240 Ω d) Putting in parallel a resistance of 15 Ω

901. A voltmeter having resistance of 50×103 ohm is used to measure the voltage in a circuit. To increase the
range of measurement 3 times the additional series resistance required is
a) 105 ohm b) 150 k.ohm c) 900 k.ohm d) 9×106ohm

902. The wiring of a house has resistance 6 Ω. A 100 W bulb is glowing as shown in figure. If a geyser of 1000 W
is switched on, the change in potential drop across the bulb is nearly

a) Nil b) 12 V c) 24 V d) 32 V

903. What is the reading of voltmeter in the following figure

a) 3 V b) 2 V c) 5 V d) 4 V

904. Consider the circuits shown in the figure. Both the circuits are taking same current from battery but
1
current through R in the second circuit is th of current through R in the first circuit. If R is 11 Ω, the
10
value of R1
i i R1
i/10
E R R2 R

(a) (b)

a) 9.9 Ω b) 11 Ω c) 8.8 Ω d) 7.7 Ω

905. A given resistor has the following colour scheme of the various strips on it, brown, black, green and silver.
Its value in ohm is
a) 1.0×104 ± 10% b) 1.0×105 ± 10% c) 1.0×106 ± 10% d) 1.0×107 ± 10%

906. The equivalent resistance of the figure ie, infinite network of resistors between the terminals A and B is

P a g e | 116
 a) Zero b) Infinite

c) R1+R2+R3 [
d) 1 (R +R )+ (R +R )(R +R +4R ) ]
3 2 1 2 1 2 1 2 3

907. If R1 and R2 are respectively the filament resistances of a 200 watt bulb and 100 watt bulb designed to
operate on the same voltage, then
a) R is two times R b) R is two times R c) R is four times R d) R is four times R
1 2 2 1 2 1 1 2

908. The atomic weight of silver and copper are 108 and 64. A silver voltmeter and a copper voltmeter are
connected in series and when current is passed 10.8 gm of silver is deposited. The mass of copper
deposited will be
a) 6.4 gm b) 12.8 gm c) 3.2 gm d) 10.8 gm

909. An unknown resistance R1 is connected in series with a resistance of 10Ω. This combination is connected
to one gap of meter bridge while a resistance R2is connected in the other gap. The balance point is at 50
cm, Now, when the 10Ω resistance is removed the balance point shifts 40cm. The value of R1(in ohm) is
a) 20 b) 10 c) 60 d) 40

910. A wire has a resistance of 6Ω. It is cut into two parts and both half values are connected in parallel. The
new resistance is
a) 3Ω b) 6Ω c) 12Ω d) 1.5Ω

911. A source of emf E=15V and having negligible internal resistance, is connected to a variable resistance, so
that the current in the circuit increases with time as I=1.2t+3. Then, the total charge that will flow in first
5s will be
a) 10C b) 20C c) 30C d) 40C

912. Two resistors of resistance R1 and R2 having R1 > R2 are connected in parallel. For equivalent resistance R,
the correct statement is
a) R > R + R b) R < R < R c) R < R < (R + R ) d) R < R
1 2 1 2 2 1 2 1

913. Identify the incorrect statement regarding a superconducting wire

a) Transport current flows through its surface

b) Transport current flows through the entire area of cross-section of the wire

c) It exhibits zero electrical resistivity and expels applied magnetic field

d) It is used to produce large magnetic field

914. By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A through 7 ohm resistor. The internal
resistance of the cell is
a) 0.5 Ω b) 1.0 Ω c) 1.2 Ω d) 2.0 Ω

915. The value of current required to deposit 0.972 gm of chromium in 3 hours if the E.C.E. of chromium is
0.00018 gm per coulomb, is
a) 1 amp b) 1.5 amp c) 0.5 amp d) 2 amp

P a g e | 117
916. If for a thermocouple Tn is the neutral temperature, Tc is the temperature of the cold junction and Ti is the
temperature of inversion, then
a) T = 2T - T b) T = T - 2T c) T = T - T d) None of these
i n c n i c i n c

917. The temperature of the cold junction of a thermocouple is 0℃ and the temperature of the hot junction is
2
T℃. The emf is E = 16T - 0.04T μV. The inversion temperature Ti is
a) 200℃ b) 400℃ c) 100℃ d) 300℃

918. From the graph between current I and voltage V shown below, identify the portion corresponding to
negative resistance

I E
C
B
D
V
A

a) AB b) BC c) CD d) DE

919. A thermocouple is formed by two metals X and Y, metal X comes earlier to Y in Seebeck series. If
temperature of hot junction increases beyond the temperature of inversion, then direction of current in
thermocouple will be from
a) X to Y through cold junction b) X to Y through hot junction

c) Y to X through cold junction d) Both (b) and (c)

920. A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k
volt/cm and the ammeter, present in the cicuit, reads 1.0 A when two way key is switched off. The balance
points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths
l1 cm and l2cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively,
to

a) kl and kl b) k(l -l ) and kl c) kl and k(l - l ) d) k(l -l ) and kl

1 2 2 1 2 1 2 1 2 1 1

921. The potential difference across the 100Ω resistance in the following circuit is measured by a voltmeter of
900 Ω resistance. The percentage error made in reading the potential difference is
900 
V
10 

100 

a) 10 b) 0.1 c) 1.0 d) 10.0

9
922. A heater coil cut into two equal parts and one part is connected with heater. Now heat generated in heater
will be

P a g e | 118
 a) Twice b) Half c) One-fourth d) Four times

923. Some electric bulbs are connected in series across a 220V supply in a room. If one bulb is fused, then
remaining bulbs are connected again in series across the same supply. The illumination in the room will
be
a) Increase b) Decrease c) Remain the same d) Not continuous

924. A moving coil galvanometer has a resistance of 50Ω and gives full scale deflection for 10 mA. How could it
be converted into an ammeter with a full scale deflection for 1A
a) 50/99 Ω in series b) 50/99 Ω in parallel c) 0.01 Ω in series d) 0.01 Ω in parallel

925. An ammeter reads upto 1A. Its internal resistance is 0.81Ω. To increase the range to 10A the value of the
required shunt is
a) 0.03Ω b) 0.3Ω c) 0.9Ω d) 0.09Ω

926. There are two concentric spheres of radius a and b respectively. If the space between them is filled with
medium of resistivity ρ, then the resistance of the inter gap between the two spheres will be
a) ρ
4π(b+a)
b) ρ 1 - 1
4π b a ( ) ( )
c) ρ 12 - 12
4π a b ( )
d) ρ 1 - 1
4π a b
927. The V-i graphs A and B are drawn for two voltameters. Identify each graph
I I

1.7 V V V
(A) (B)

a) A for water voltameter and B for Cu voltameter

b) A for Cu voltmeter and B for water voltameter

c) Both A and B represents Cu voltameter

d) None of these

928. A certain electrical conductor has a square cross-section, 2.0 mm on side, and is 12 m long. The resistance
between its ends is 0.072Ω. The resistivity of its material is equal to
a) 2.4×10-6Ωm b) 1.2×10-6Ωm c) 1.2×10-8Ωm d) 2.4×10-8Ωm

929. A galvanometer has a resistance of 3663Ω. A shunt S is connected across it such that (1/34) of the total
current passes through the galvanometer. Then the value of shunt is
a) 3663Ω b) 111Ω c) 107.7Ω d) 3555.3Ω

930. In the given figure, the equivalent resistance between the points A and B is
R2 = 4 

R1 = 2 R4 = 2 
A B
R3 = 4 

a) 8 Ω b) 6 Ω c) 4 Ω d) 2 Ω

931. Three resistances of 4Ω, 6Ω and 12Ω are connected in parallel and the combination is connected in series
with 4 V battery with internal resistance of 2 Ω. The battery current is

P a g e | 119
 a) 1 A b) 10 A c) 2 A d) 0.5 A

932. The equivalent resistance between points A and B of an infinite network of resistances each of 1Ω
connected as shown in figure, is
1 1 1

1 1 1

a) Infinite b) Zero c) 2Ω d) (1+ 5 )/2Ω

933. An electric iron draws 5 amp, a TV set draws 3 amp and refrigerator graws 2 amp from a 220 volt main
line. The three appliances are connected in parallel. If all the three are operating at the same time, the fuse
used may be of
a) 20 amp b) 5 amp c) 15 amp d) 10 amp

934. The equivalent resistance between the points A and B in the following circuit is

a) 3.12 Ω b) 1.56 Ω c) 6.24 Ω d) 12.48 Ω

935. A 3℃ rise in temperature is observed in a conductor by passing certain current. When the current is
doubled, the rise in temperature will be
a) 15℃ b) 12℃ c) 9℃ d) 3℃

936. In the given figure, equivalent resistance between A and B will be

3 4

A 7 B

6 8

a) 14 Ω b) 3 Ω c) 9 Ω d) 14 Ω
3 14 14 9
937. If θi is the inversion temperature, θn is the natural temperature, θc is the temperature of the cold junction
then
a) θ + θ = θ b) θ - θ = 2θ c) θi+θc = θ d) θ - θ = 2θ
i c n i c n n c i n
2
938. Potential difference across the terminals of the battery shown in figure is (r = internal resistance of
battery)

10 V r =1Ω

4Ω

a) 8 V b) 10 V c) 6 V d) Zero

P a g e | 120
939. Potentiometer measures the potential difference more accurately than a voltmeter because

a) It has a wire of high resistance b) It has a wire of low resistance

c) It does not draw current from external circuit d) It draws a heavy current from external circuit

940. To get maximum current through a resistance of 2.5 Ω, one can use m rows of cells, each row having n
cells. The internal resistance of each cell is 0.5 Ω. What are the values of n and m, if the total number of cell
is 45?
a) m = 3, n = 15 b) m = 5,n = 9 c) m = 9,n = 5 d) m = 15, n = 3

941. A cold-water pipe and a hot-water pipe are both made of copper and are initially electrically isolated. In
which one of the following arrangements will the galvanometer indicate a thermo-electric current?

a) b)

c) d)

942. 1 2
The emf of a thermocouple, cold junction of which is kept at -300℃ is given by E = 40t + t . The
10
temperature of inversion of thermocouple will be
a) 200℃ b) 400℃ c) -200℃ d) -100℃

943. How much current should be passed through acidified water for 100s to liberate 0.224 L of hydrogen?

a) 22.4 A b) 19.3 A c) 9.65 A d) 1 A

944. The speed at which the current travels, in a conductor, is nearly

a) 3×10-4 ms-1 b) 3×10-5 ms-1 c) 4×106 ms-1 d) 3×108 ms-1

945. A uniform wire of resistance 9 Ω is cut into 3 equal parts. They are connected in the form of equilateral
triangle ABC. A cell of e.m.f. 2 V and negligible internal resistance is connected across B and C. Potential
difference across AB is
a) 1 V b) 2 V c) 3 V d) 0.5 V

946. A bulb rated at (100W - 200V) is used on a 100V line. The current in the bulb is

a) 1 amp b) 4 amp c) 1 amp d) 2 amp

4 2
947. In the circuit figure, the voltmeter reads 30 V. what is the resistance of the voltmeter?
30 V
V

300 Ω 400 Ω

60 V

P a g e | 121
 a) 1200 Ω b) 700 Ω c) 400 Ω d) 300 Ω

948. Forty electric bulbs are connected in series across a 220 V supply. After one bulb is fused, the remaining 39
are connected again in series across the same supply. The illumination will be
a) More with 40 bulbs than with 39 b) More with 39 bulbs than with 40

c) Equal in both the cases d) In the ratio of 492 :392

949. A wire of resistor R is bent into a circular ring of radius r. Equivalent resistance between two points X and
Y on its circumference, when angle XOY is α, can be given by
X

W  O Z

a) Rα (2π - α) b) R (2π - α) c) R(2π - α) d) 4π (2π - α)

2
4π 2π Rα
950. Ampere hour is the unit of

a) Quantity of charges b) Potential c) Energy d) Current

951. In the circuit shown as P ≠ R and the reading of the galvanometer G is same with switch open or closed.
Then

a) I = R b) I = I c) I = I d) I = I
R G P G Q G Q R

952. A bulb of 220 V and 300 W is connected across 110 V circuit. The percentage reduction in power is

a) 100% b) 25% c) 70% d) 75%

953. Peltier coefficient for the junction of a pair of metals is proportional to

a) Absolute temperature of junction T b) Square of absolute temperature of junction

c) 1/T d) 1/T2

954. The following four wires are made of the same material and are at the same temperature. Which one of
them has the highest electrical resistance?
a) Length=50 cm, diameter=0.5 mm b) Length=100 cm, diameter=1 mm

c) Length=200 cm, diameter=2 mm d) Length=300 cm, diameter=3 mm

955. In the circuit shown below the resistance of the galvanometer is 20 Ω. In which of the following alternative
are the currents arranged strictly in the decreasing order
i1
10 ig 100

i2 G

2 20
i

2V 0
P a g e | 122
 a) i,i ,i ,i b) i,i ,i ,i c) i,i , i ,i d) i,i ,i ,i
1 2 g 2 1 g 2 g 1 1 g 2

956. A potentiometer having the potential gradient of 2 mV/cm is used to measure the difference of potential
across a resistance of 10 ohm. If a length of 50 cm of the potentiometer wire is required to get the null
point, the current passing through the 10 ohm resistor is (in mA)
a) 1 b) 2 c) 5 d) 10

957. Two bulbs marked 200 V -100 W and 200 V-200 W are joined in series and connected to a power supply of

( )

200 V
200 V. The total power consumed by the two will be near to
a) 35 W b) 66 W c) 100 W d) 300 W

958. A colour coded carbon resistor has the colours orange, blue, green and silver. Its resistance value and
tolerance percentage respectively are
a) 36×105Ω and 10% b) 36×104Ω and 5% c) 63×105Ω and 10% d) 35×106Ω and 5%

959. A capacitor of capacitance 2μF is connected as shown in figure. The internal resistance of the cell is 0.5Ω.
The amount of charge on the capacitor plates is
2

2 F 10

+ -
2.5 V

a) Zero b) 2μ C c) 4μC d) 6μC

960. A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance
become respectively
a) Both remain the same b) 1.1 times, 1.1 times c) 1.2 times , 1.1 times d) 1.21 times, same

961. Two conductors are made of the same material and have the same length. Conductor A is a solid wire of
diameter 1.0 mm. Conductor B is a hollow tube of outside diameter 2.0 mm and inside diameter 1.0 mm.
The resistance ratio RA/RB will be
a) 1 b) 2 c) 3 d) 4

962. A certain current passing through a galvanometer produces a deflection of 100 divisions. When a shunt of
one ohm is connected, the deflection reduces to 1 division. The galvanometer resistance is
a) 100 Ω b) 99 Ω c) 10 Ω d) 9.9 Ω

963. An expression for rate of heat generated, if a current of I ampere flows through a resistance of R Ω, is

a) I2Rt b) I2R c) V2R d) I R

964. If a 2 kW boiler is used everyday for 1 hour, then electrical energy consumed by boiler in thirty days is

P a g e | 123
 a) 15 unit b) 60 unit c) 120 unit d) 240 unit

965. For the post office box arrangement to determine the value of unknown resistance, the unknown
resistance should be connected between

a) B and C b) C and D c) A and D d) B and C

1 1

966. Six equal resistances are connected between points P, Q and R as shown in the figure. Then the net
resistance will be maximum between
P

Q R

a) P and Q b) Qand R c) P and R d) Only two points

967. Which of the adjoining graphs represents ohmic resistance

a) b) c) d)
V V V V

I I I I

968. A lead acid accumulatory (storage battery) is connected to a battery charge for over night charging. Which
of the following observations will indicate that the battery was partly charged during the next morning
a) The density of acid has decreased b) The density of acid has increased

c) The acid has changed colour d) The acid level has dropped

969. How much current should be passed through a silver voltmeter to deposit 200 gm of silver per hour on the
cathode? (Faraday constant =96500 C/mol and relative atomic mass of silver is 108)
a) 50 mA b) 50 A c) 15 mA d) 15 A

970. All the edges of a block with parallel faces are unequal. Its tangent edge is twice its shortest edge. The ratio
of the maximum to minimum resistance between parallel faces is
a) 8 b) 4 c) 2 d) None of these

971. A battery has e.m.f. 4 V and internal resistance r. When this battery is connected to an external resistance
of 2 ohm, a current of 1 amp. flows in the circuit. How much current will flow if the terminals of the battery

P a g e | 124
 are connected directly
a) 1 amp b) 2 amp c) 4 amp d) Infinite

972. A cell of emf 6 V and resistance 0.5 ohm is short circuited. The current in the cell is

a) 3 amp b) 12 amp c) 24 amp d) 6 amp

973. It is easier to start a car engine on a hot day than on a cold day. This is because the internal resistance of
the car battery
a) Decreases with rise in temperature b) Increases with rise in temperature

c) Decreases with a fall in temperature d) Does not change with a change in temperature

974. The resistance across A and Bin the figure below will be

A B
R
R R

a) 3R b) R c) R d) None of these
3
975. With a potentiometer null point were obtained at 140 cm and 180 cm with cells of emf 1.1 V and one
unknown X volt. Unknown emf is
a) 1.1 V b) 1.8 V c) 2.4 V d) 1.41 V

976. In the given circuit, the potential of the point E is

+ – E 1
A D
8V

C
B
5

a) Zero b) -8 V c) -4/3 V d) 4/3 V

977. Two resistors 400 Ω and 800 Ω are connected in series with 6 V battery. The potential difference measured
by voltmeter of 10k Ω across 400 Ω resistor is
a) 2 V b) 1.95 V c) 3.8 V d) 4 V

978. What is the potential drop between points A and C in the following circuit? Resistances 1 Ω and 2 Ω
represent the internal resistance of the respective cells

- - - -- - - - - - - - - - - - - - - -- - - - - - - - - - - -
- - -
C 1Ω 2V
-
-
-
-
-
- A 4 V2 Ω
-
-
-
-
-
-
-
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - -- - - - - - - - - - - - - -------------- -

5Ω

P a g e | 125
 a) 1.75V b) 2.25V c) 5 V d) 4 V
4 5
979. A long straight wire of a circular cross section (radius a) carries a steady current I and the current I is
uniformly distributed across this cross-section. Which of the following plots represents the variation of
magnitude of magnetic field B with distance r from the centre of the wire

a) b) c) d)

980. A cylindrical conductor has uniform cross-section. Resistivity of its material increases linearly from left
end to right end. If a constant current is flowing through it and at a section distance x from left end,
magnitude of electric field intensity is E, which of the following graphs is correct
a) E b) E c) E d) E

O x O x O x O x

981. The emf of a battery is 2 V and its internal resistance is 0.5 Ω. The maximum power which it can deliver to
any external circuit will be
a) 8 Watt b) 4 Watt c) 2 Watt d) None of the above

982. An electric bulb of 100 watt is connected to a supply of electricity of 220 V. Resistance of the filament is

a) 484 Ω b) 100 Ω c) 22000 Ω d) 242 Ω

983. n identical bulbs, each designed to draw a power p from a certain voltage supply, are joined in series
across that supply. The total power which they will draw is
a) p/n2 b) p/n c) p d) np

984. In the given figure when galvanometer shows no deflection current flowing through 5Ω resistance will be
B
8 2

2.1 A
G
A C
20 5
D

a) 0.5 A b) 0.6 A c) 1.5 A d) 2.0 A

985. If voltage across a bulb rated 220 Volt-100 Watt drops by 2.5% of its rated value, the percentage of the
rated value by which the power would decrease is
a) 20% b) 2.5% c) 5% d) 10%

986. Equivalent resistance between the points A and B is (in Ω)

P a g e | 126
 A 1 1 1 1 1 B

a) 1 b) 1 1 c) 2 1 d) 3 1
5 4 3 2
987. A fuse wire with radius 1 mm blows at 1.5 A. The radius of the fuse wire of the same material to blow at 3
A will be
a) 1/4 b) 1/3 c) 1/2 d) 1/3
3 mm 4 mm 3 mm 2 mm
988. A galvanometer of resistance G can measure 1 A current. If a shunt S is used to convert it into an ammeter
G
to measure 10A current. The ratio of is
S
a) 1 b) 9 c) 10 d) 1
9 1 10
989. Calculate the amount of charge flowing in 2 minutes in a wire of resistance 10Ω when a potential
difference of 20 V is applied between its ends
a) 120 C b) 240 C c) 20 C d) 4 C

990. An electric wire is connected across a cell of e.m.f. E. The current I is measured by an ammeter of
resistance R. According to ohm's law
a) E = I2R b) E = IR c) E = R/I d) E = I/R

991. Drift velocity vd varies with the intensity of electric field as per the relation

a) v ∝ E b) v ∝ 1 c) v = constant d) v ∝ E2
d d
E d d

992. An immersion heater is rated 418 W. It should heat a litre of water from 10℃ to 30℃ in nearly

a) 44 s b) 100 s c) 200 s d) 400 s

993. Current through wire XY of circuit shown is

1Ω X 2Ω

1 2

3Ω Y 4Ω

50V

a) 1 A b) 4 A c) 2 A d) 3 A

994. The figure shows a network of currents. The magnitude of currents is shown here. The current i will be
15A
3A

8A

i
5A

a) 3 A b) 13 A c) 23 A d) -3 A

995. In the circuit shown in the figure, switch S1 is initially closed and S2 is open. Find Va - Vb

P a g e | 127
 a) 4 V b) 8 V c) 12 V d) 16 V

996. 62.5×1018 electrons per second are flowing through a wire of area of cross-section 0.1 m2, the value of
current flowing will be
a) 1 A b) 0.1 A c) 10 A d) 0.11 A

997. If three bulbs 60W, 100W and 200 W are connected in parallel, then

a) 200 W bulb will glow more b) 60 W bulb will glow more

c) 100 W bulb will glow more d) All the bulbs will glow equally

998. Four resistances of 100 Ω each are connected in the form of square. Then, the effective resistance along the
diagonal points is
a) 200 Ω b) 400 Ω c) 100 Ω d) 150 Ω

999. Water of volume 2 litre in a container is heated with a coil of 1 kW at 27℃. The lid of the container is open
and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27℃ to 77℃ [Given
specific heat of water is 4.2 kJ/kg]
a) 8min 20 s b) 6min 2 s c) 7 min d) 14 min

100 Flash light equipped with a new set of batteries, produces bright white light. As the batteries wear out
0.

a) The light intensity gets reduced with no change in its colour

b) Light colour changes first to yellow and then red with no change in intensity

c) It stops working suddenly while giving white light

d) Colour changes to red and also intensity gets reduced

100 A potentiometer has uniform potential gradient. The specific resistance of the material of the
1. potentiometer wire is 10-7 ohm - meter and the current passing through it is 0.1 ampere; cross-section of
-6 2
the wire is 10 m . The potential gradient along the potentiometer wire is
a) 10-4 V/m b) 10-6 V/m c) 10-2 V/m d) 10-8 V/m

100 A hot electric iron has a resistance of 80 Ω and is used on a 200 V source. The electrical energy spent, if it is
2. used for two hours, will be

a) 8000 Wh b) 2000 Wh c) 1000 Wh d) 800 Wh

100 The thermocouple is based on the principle of

3.

P a g e | 128
 a) Seebeck effect b) Thomson effect c) Peltier effect d) Joule effect

100 The measurement of voltmeter in the following circuit is

4. +
6V
–

60
V

40

a) 2.4 V b) 3.4 V c) 4.0 V d) 6.0 V

100 In the circuit given, the correct relation to a balanced Wheatstone bridge is
5.

a) P = R b) P = S c) P = S d) None of these
Q S Q R R Q
100 In the circuit as shown in figure the
6.

a) Resistance R = 46 Ω

b) Current through 20 Ω resistance is 0.1 A

c) Potential difference across the middle resistance is 2 V

d) All option are correct

100 Two electric bulbs, each designed to operate with a power of 500 W in 220 V line are connected in series in
7. a 110 V line. The power generated by each bulb will be

a) 31.25 W b) 40 W c) 60 W d) 3.125 W

100 A heater draws a current of 2A when connected to a 250 V source. The rate of energy dissipation is
8.

a) 500 W b) 1000 W c) 250 W d) 125 W

100 A 100 ohm galvanometer gives full scale deflection at 10 mA. How much shunt is required to read 100 mA
9.

a) 11.11 ohm b) 9.9 ohm c) 1.1 ohm d) 4.4 ohm

P a g e | 129
101 The figure shows a circuit diagram of a ‘Wheatstone Bridge’ to measure the resistance G of the
0. P R
galvanometer. The relation = will be satisfied only when
Q G

a) The galvanometer shows a deflection when switch S is closed

b) The galvanometer shows a deflection when switch S is open

c) The galvanometer shows no change in deflection whether S is open or closed

d) The galvanometer shows no deflection

101 When the key K is pressed at time t = 0, which of the following statements about the current I in the
1. resistor AB of the given circuit is true
A B
2V K 1000Ω
1000Ω

1F C

a) I = 2 mA at all t

b) I oscillates between 1 mA and 2mA

c) I = 1 mA at all t

d) At t = 0, I = 2 mA and with time it goes to 1 mA

101 The equivalent resistance between points A and B of an infinite network of resistances, each of 1 Ω,
2. connected as shown is

1Ω 1Ω 1Ω
A -----

1Ω 1Ω 1Ω

B -----

a) Infinite b) 2 Ω c) 1+ 5 Ω d) zero
2
101 If 100 kWh of energy is consumed at 66 V in a copper voltmeter, then the mass of copper liberated will be
3. (Given ,ECE of Cu = 0.33×10-6kg C-1)

a) 1.65 kg b) 1.8 kg c) 3.3 kg d) 3.6 kg

101 In the circuit shown in the figure, if the potential at point A is taken to be zero, the potential at point B is
4.

P a g e | 130
 a) -2V b) +1V c) -1V d) +2V

101 When a current is passed through water, acidified with a dilute sulphuric acid, the gases formed at the
5. platinum electrodes are

a) 1 vol. hydrogen (cathode) and 2 vol. oxygen (anode)

b) 2 vol. hydrogen (cathode) and 1 vol. oxygen (anode)

c) 1 vol. hydrogen (cathode) and 1 vol. oxygen (anode)

d) 1 vol. oxygen (cathode) and 2 vol. hydrogen (anode)

101 The temperature at which thermo emf is zero, is

6.

a) Temperature of inversion b) Temperature of cold junction

c) Neutral temperature d) None of the above

101 An emf of 0.9 V is generated when the temperature difference hot and cold junction of thermocouple is 75
7. K. Assuming that the thermo emf is directly proportional to the temperature difference, the extent to
which the thermo emf will change when the cold junction is heated up by 15 K is
a) 10% b) 20% c) 40% d) 60%

101 n identical cells, each of emf E and internal resistance r, are connected in series a cell A is joined with
8. reverse polarity. The potential difference across each cell, except A is

a) 2nE b) (n-2)E c) (n-1)E d) 2E

n-2 n n n
101 The ammeter A reads 2 A and the voltmeter V reads 20 V. The value of resistance R is (Assuming finite
9. resistance’s of ammeter and voltmeter)
R
A

a) Exactly 10 ohm b) Less than 10 ohm

c) More than 10 ohm d) We cannot definitely say

102 An electric heater kept in vacuum is heated continuously by passing electric current. Its temperature
0.

a) Will go on rising with time

b) Will stop after sometime as it will loose heat to the surroundings by conduction

c) Will rise for sometime and there after will start falling

P a g e | 131
 d) Will become constant after sometime because of loss of heat due to radiation

102 Two identical batteries each emf E =2V and internal resistance r = 1Ω are available to produce heat in an
1. external resistance by passing a current through it. The maximum Joulean power that can be developed
across R using these batteries is
a) 1.28 W b) 2.0 W c) 8 W d) 3.2 W
9
102 One junction of thermocouple is at 0℃ and the other is at T℃. The thermo emf (in volts) is given by
-6 -6 2
2. E = 20×10 T - 0.02×10 T
The maximum value of E is
a) 5 mV b) 1 m V c) 10 m V d) Zero

102 If t1 and t2 are the times taken by two different coils for producing same heat with same supply, then the
3. time taken by them to produce the same heat when connected in parallel will be

a) t + t b) t1t2 c) 2t1t2 d) t t
1 2 t1+t2 t1+t2 1 2

102 For what value of R in the circuit as shown in figure, current passing through 4Ω resistance will be zero.
4. B C D

2 4 R

6V

A F E
9V 3V

a) 1 Ω b) 2 Ω c) 3 Ω d) 4 Ω

102 In the adjoining circuit diagram each resistance is of 10 Ω. The current in the arm AD will be
5. E

B F i

i
A C

a) 2i b) 3i c) 4i d) i
5 5 5 5
102 An electric fan and a heater are marked as 100 watt, 220 volt and 1000 watt, 220 volt respectively. The
6. resistance of the heater is

a) Zero b) Greater than that of the fan

c) Less than that of the fan d) Equal to that of the fan

102 In the figure, current through the 3Ω resistor is 0.8 ampere, then potential drop through 4 Ω resistor is
7.

P a g e | 132
 3

4
6

+ –

a) 9.6 V b) 2.6 V c) 4.8 V d) 1.2 V

102 In a potentiometer experiment two cells of e.m.f.’ s E1 and E2 are used in series and in conjunction and the
8. balancing length is found to be 58 cm of the wire. If the polarity of E is reversed, then the balancing length
2
E1
becomes 29 cm. The ratio of the e.m.f. of the two cells is
E2
a) 1 : 1 b) 2 : 1 c) 3 : 1 d) 4 : 1

102 An energy source will supply a constant current into, the load, if its internal resistance is
9.

a) Equal to the resistance of the load

b) Very large as compared to the load resistance

c) Zero

d) Non-zero but less than the resistance of the load

103 The electric bulbs have tungsten filaments of same length. If one of then gives 60 watt and other 100 watt,
0. then

a) 100 watt bulb has thicker filament

b) 60 watt bulb has thicker filament

c) Both filaments are of same thickness

d) It is possible to get different wattage unless the lengths are different

103 The value of current I in figure is

1. 3A

1A 2A

a) 4A b) 6A c) 3A d) 5A

103 A constant current i is passed through a resistor. Taking the temperature coefficient of resistance into
2. account, indicate which of the plots shown in figure best represents the rate of production of thermal
energy in the resistor

P a g e | 133
 dU
dt d
c
a
b

Temp

a) a b) b c) c d) d

103 A torch battery consists of two cells of 1.45 volt and an internal resistance 0.15 Ω. If each cell sends current
3. through the filament of the lamps having resistance 1.5 ohm, the value of current will be

a) 16.11 amp b) 1.611 amp c) 0.1611 amp d) 2.6 amp

103 Thermoelectric constant of a thermocouple are α and β. Thermoelectric power at inversion temperature is
4.

a) α b) -α c) α d) - α
β β
103 A galvanometer whose resistance is 120Ω gives full scale deflection with a current of 0.005 A so that it can
5. read a maximum current of 10 A. A shunt resistance is added in parallel with it. The resistance of the
ammeter so formed is
a) 0.06 Ω b) 0.006 Ω c) 0.6 Ω d) 6 Ω

103 An electron revolves 6×1015times/sec in circular loop. The current in the loop is
6.

a) 0.96 mA b) 0.96 μ A c) 28.8 A d) None of these

103 A galvanometer of resistance 25 Ω measures 10-3A. shunt required to increase range upto 2A is
7.

a) 12.5 Ω b) 0.125 Ω c) 0.125 Ω d) 1.25 Ω

103 The maximum current that can be measured by a galvanometer of resistance 40Ω is 10mA. It is converted
8. into a voltmeter that can read upto 50V. The resistance to be connected is series with the galvanometer (in
ohm) is
a) 2010 b) 4050 c) 5040 d) 4960

103 A student has 10 resistors of resistance 'r'. The minimum resistance made by him from given resistors is
9.

a) 10 r b) r c) r d) r
10 100 5
104 n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is
0. connected in series to this combination. The current through R is

a) nE b) nE c) E d) nE
R+nr nR+r R+nr R+r
104 A thermoelectric refrigerator works on
1.

a) Joule effect b) Seeback effect c) Peltier effect d) Thermonic emission

104 A wire of length 100 cm is connected to a cell of emf 2 V and negligible internal resistance. The resistance
2. of the wire is 3 Ω. The additional resistance required to produce a potential drop of 1 milli volt per cm is
P a g e | 134
 a) 60 Ω b) 47 Ω c) 57 Ω d) 35 Ω

104 An electric bulb is rated 60W, 220V. The resistance of its filament is
3.

a) 708 Ω b) 870 Ω c) 807 Ω d) 780 Ω

104 In a Ag voltameter 2.68 g of silver is deposited in 10 min. The heat developed in 20Ω resistor during the
4. same period will be

20

a) 192 kJ b) 192 J c) 200 J d) 132 kJ

104 A railway compartment is lit up by thirteen lamps each taking 2.1 A at 15 V. The heat generated per
5. second in each lamp will be

a) 4.35 cal b) 5.73 cal c) 7.5 cal d) 2.5 cal

104 If R1 and R2 be the resistances of the filaments of 200 W and 100 W electric bulbs operation at 220 V, then
6.
()
R1
R2
is

a) 1 b) 2 c) 0.5 d) 4

104 In voltaic air cell if 5g zinc is consumed, how many ampere hours shall we get?
7.

a) 2.05 b) 8.2 c) 4.1 d) 5×5.38×10-3

104 A house, served by 220 V supply line, is protected by a 9 A fuse. The maximum number of 60 W bulbs in
8. parallel that can be turned on is

a) 11 b) 22 c) 33 d) 44

104 The temperature coefficient of resistance for a wire is 0.00125° C-1.At 300 K its resistance is 1 Ω.The
9. temperature at which the resistance becomes 1.5 Ω is?

a) 450 K b) 727 K c) 454 K d) 900 K

105 A 10 μF capacitor is charged to 500 V and then its plates are joined together through a resistance of 10 Ω.
0. The heat produced in the resistance is

a) 500 J b) 250 J c) 125 J d) 1.25 J

105 The ratio of the amounts of heat developed in the four arms of a balanced Wheatstone bridge, when the
1. arms have resistance P = 100 Ω; Q = 10 Ω;R = 300 Ω and S = 30 Ω respectively is

a) 3 : 30 : 1 : 10 b) 30 : 3 : 10 :1 c) 30 : 10 : 1 : 3 d) 30 : 1 : 3 : 10

105 If the resistivity of an alloy of ρ’ and that of constituent metals is ρ, then

2.
P a g e | 135
 a) ρ' > ρ b) ρ' < ρ

c) ρ' = ρ d) There is no simple relation between ρ and ρ’

105 The electrochemical equivalent of a material in an electrolyte depends on

3.

a) The nature of the material

b) The current though the electrolyte

c) The amount of charge passed through electrolyte

d) The amount of material present in electrolyte

105 A current passing through a copper voltmeter deposits 0.002 kg of copper on cathode plate in 100 min. If
4. there are 1025 copper atoms in one kg of copper, the electric charge delivered to cathode by Cu++ ions per
second will be
a) 0.53 C b) 0.71 C c) 1.06 C d) 10.06 C

105 Resistance of a wire at 20°C is 20Ω and at 500°C is 60Ω. At what temperature its resistance is 25Ω?
5.

a) 160°C b) 250°C c) 100°C d) 80°C

105 The equivalent resistance between points a and b of a network shown in the figure is given by
6.

a) 3 R b) 4 R c) 5 R d) 4 R
4 3 4 5
105 If the potential difference across the internal resistance r1 is equal to the emf E of the battery, then
7.
------- ---------- ------- ----------
- - - r -
- - - -
-E r - E- 2 -
- 1 - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - -- - - - - - - - - - - - - - - -- --- - - - - - - - - - - - - - - -
- - - -

a) R = r + r b) R = r1 c) R = r - r d) R = r2
1 2
r2 1 2
r1
105 In an electric heater 4 amp current passes for 1 minute at potential difference of 250 volt, the power of
8. heater and energy consumed will be respectively

a) 1 kW, 60 kJ b) 0.5 kW, 30 kJ c) 10 kW, 600 kJ d) None of these

105 In a metre bridge experiment, resistances are connected as shown in figure. The balancing length l1 is 55
9. cm. Now an unknown resistance x is connected in series with P and the new balancing length is found to be

P a g e | 136
 75 cm. The value of x is

a) 54 Ω b) 20 Ω c) 48 Ω d) 11 Ω
12 11 11 48
106 The plates of a charged condenser is connected to a voltmeter. If the plates are moved apart, the reading of
0. voltmeter will

a) Increase b) Decrease

c) Remain unchanged d) Information is insufficient

106 Three equal resistances, each of 10 Ω are connected as shown in figure. The maximum power consumed by
1. each resistance is 20 W. What is maximum power that can be consumed by the combination?
10 Ω
10 Ω
10 Ω

a) 5 W b) 15 W c) 30 W d) 60 W

106 6
Two resistance wires on joining in parallel the resultant resistance is ohms. One of the wire breaks, the
2. 5
effective resistance is 2 ohms. The resistance of the broken wire is
a) 3 ohm b) 2 ohm c) 6 ohm d) 3 ohm
5 5
106 For the circuit shown in the figure the potential difference between A and B will be (in volt)
3. 1Ω

2 2
V V
1Ω 3Ω

a) 2 b) 1.5 c) 1.0 d) Zero

106 Shown in the figure adjacent is a meter-bridge set up with null deflection in the galvanometer. The value of
4. the unknown resistor R is
55Ω R

20 cm

a) 13.75 Ω b) 220 Ω c) 110 Ω d) 55 Ω

P a g e | 137
106 The heat produced in 4 Ω resistance is 10 cal. The heat produced in 10 Ω resistance will be
5.

4Ω 6Ω

10Ω|

a) 25 cal b) 14 cal c) 10 cal d) 20 cal

106 To liberate two litres of hydrogen at 222.4 atmosphere from acidulated water the quantity of electricity
6. that must pass through is

a) 44.8 C b) 96500 C c) 193000 C d) 386000 C

106 In the circuit shown, the reading of ammeter when switch S is open and when switch S is closed
7. respectively are

a) 3 A and 4 A b) 4 A and 5 A c) 5 A and 6 A d) 6 A and 7 A

106 In the process of electrolysis, the current is carried out inside the electrolyte by
8.

a) Electrons b) Atoms

c) Positive and negative ions d) All the above

106 The resistance of an ammeter is 3 Ω and its scale is graduated for a current upto 100A. After an additional
9. shunt has been connected to this ammeter it becomes possible to measure currents upto 750A by this
meter. The value of shunt resistance is
a) 20Ω b) 2 Ω c) 0.2 Ω d) 2K Ω

107 The relation between Faraday’s constant F, electron charge e and avogadro number N is
0.

a) F = N/e b) F = Ne c) N = F2 d) F = N2e

107 The resistance of a wire is 10Ω. Its length is increased by 10% by stretching. The new resistance will now
1. be

a) 12Ω b) 1.2Ω c) 13Ω d) 11Ω

107 The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter
2. the change in the resistance of the wire will be

P a g e | 138
 a) 200 % b) 100 % c) 50 % d) 300 %

107 The power of heater is 750 W at 1000℃. What will be its power at 200℃ if a = 4×10-4 per℃ ?
3.

a) 400 W b) 990 W c) 250 W d) 1500 W

107 Masses of the three wires of same material are in the ratio of 1:2:3 and their lengths in the ratio of 3:2:1.
4. Electrical resistance of these wires will be in the ratio of

a) 1:1:1 b) 1:2:3 c) 9:4:1 d) 27:6:1

107 A wire of diameter 0.02 metre contains 1028 free electrons per cubic metre. For an electrical current of 100
5. A, the drift velocity of the free electrons in the wire is nearly

a) 1×10-19m/s b) 5×10-10m/s c) 2×10-4m/s d) 8×103m/s

107 Two uniform wires A and B are of the same metal and have equal masses. The radius of wire A is twice that
6. of wire B. The total resistance of A and B when connected in parallel is

a) 4 Ω when the resistance of wire A is 4.25 Ω

b) 5 Ω when the resistance of wire A is 4.25 Ω

c) 4 Ω when the resistance of wire B is 4.25 Ω

d) 5 Ω when the resistance of wire B is 4.25 Ω

107 A piece of fuse wire melts when a current of 15 ampere flows through it. With this current, if it dissipates
7. 22.5 W, the resistance of fuse wire will be

a) Zero b) 10 Ω c) 1 Ω d) 0.10 Ω

107 A current of 2.0 ampere passes through a cell of e.m.f. 1.5 volt having internal resistance of 0.15 ohm. The
8. potential difference measured, in volt, across both the ends of the cell will be

a) 1.35 b) 1.50 c) 1.00 d) 1.20

107 To draw maximum current from a combination of cells, how should the cells be grouped?
9.

a) Parallel b) Series

c) Mixed grouped d) Depends upon the relative values of internal and

external resistances
108 Watt-hour meter measures
0.

a) Electric energy b) Current c) Voltage d) Power

108 A 12 HP motor has to be operated 8 h/day. How much will it cost at the rate of 50 paise/kWh in 10 days?
1.

a) Rs 347 b) Rs 358 c) Rs 375 d) Rs 397

108 In the following star circuit diagram (figure), the equivalent resistance between the points A and H will be
2.

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 A i
r r 72°
B r C D r E

r r
F H J
r r
r r
G I

a) 1.944 r b) 0.973 r c) 0.486 r d) 0.243 r

108 A voltmeter has resistance of 2000 ohm and it can measure upto 2V. If we want to increase its range to 10
3. V, then the required resistance in series will be

a) 2000 Ω b) 4000 Ω c) 6000 Ω d) 8000 Ω

108 The thermocouple among the following that can produce maximum thermo-emf for the same temperature
4. difference between the junction is

a) Fe-Cu b) Ag-Au c) Sb-Bi d) Cu-Pb

108 In the circuit shown, the current though 8 ohm is same before and after connecting E. The value of E is
5.

a) 12 V b) 6 V c) 4 V d) 2 V

108 Heat produced in a wire of resistance R due to current flowing at constant potential difference is
6. proportional to

a) 12 b) 1 c) R d) R2
R R
108 In the circuit shown, the current through the 4 Ω resistor is 1 amp when the points P and M are connected
7. to a d.c. voltage source. The potential difference between the points M and N is

a) 0.5 V b) 3.2 V c) 1.5 V d) 1.0 V

108 The thermo emf of a thermo-couple is found to depend on temperature T(in degree Celsius) as
2
8. T
E = 4T - , where T℃ is the temperature of the hot junction. The neutral and inversion temperature of
200
the thermocouple are (in degree celsius)
a) 100, 200 b) 200, 400 c) 300, 600 d) 400, 800

108 In the above question, the resistance between the square faces is
9.

a) 3×10-9ohm b) 3×10-7ohm c) 3×10-5ohm d) 3×10-3ohm

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109 In the Wheatstone’s network given, P=10 Ω,Q = 20Ω, R=15 Ω, S=30 Ω, the current passing through the
0. battery (of negligible internal resistance) is

P S

Q R

+ -
6V

a) 0.36A b) Zero c) 0.18A d) 0.72A

109 When a current is passed in a conductor, 3℃ rise in temperature is observed. If the strength of current is
1. increased by two times, then rise in temperature will approximately be

a) 36℃ b) 27℃ c) 18℃ d) 9℃

109 To get a maximum current through a resistance of 2.5Ω, one can use m rows of cells each row having n
2. cells. The internal resistance of each cell is 0.5Ω. What are the values of m and n if the total number of cells
are 20?
a) m = 2, n = 10 b) m = 4, n = 5 c) m = 5, n = 4 d) n = 2, m = 10

109 If an ammeter is joined in parallel through a circuit, it can be damaged due to excess
3.

a) Resistance b) Current c) Voltage d) None of these

109 In the circuit shown, A and V are ideal ammeter and voltmeter respectively. Reading of the voltmeter will
4. be
2V

A V

1 1

a) 2 V b) 1 V c) 0.5 V d) Zero

109 Two identical batteries each of emf 2 V and internal resistance 1 Ω are available to produce heat in an
5. external resistance by passing current through it. The maximum Joulean power that can be developed
across the resistance using these batteries it.
a) 2W b) 3.2 W c) 1.28 W d) 8/9 W

109 When the temperature difference between hot and cold junctions of a thermo-couple is 100 K an emf of 1
6. V is generated. Assume the cold junction is heated by 20 K, the percentage change in thermo emf is

a) 20% b) 30% c) 40% d) 25%

109 There are n similar conductors each of resistance R. The resultant resistance comes out to be x when

P a g e | 141
7. connected in parallel. If they are connected in series, the resistance comes out to be

a) x/n2 b) n2x c) x/n d) nx

109 The total current supplied to the circuit by the battery as shown figure is
8.

a) 1A b) 6A c) 4A d) 2A

109 A galvanometer has a resistance of 25 ohm and a maximum of 0.01 A current can be passed through it. In
9. order to change it into an ammeter of range 10 A, the shunt resistance required is

a) 5/999 ohm b) 10/999 ohm c) 20/999 ohm d) 25/999 ohm

110 In cosmic rays 0.15 protons cm-2sec-1 are entering the earth’s atmosphere. If the radius of the earth is
0. 6400 km, the current received by the earth in the form of cosmic rays is nearly.

a) 0.12 A b) 1.2 A c) 12 A d) 120 A

110 The current i and voltage V graphs for a given metallic wire at two different temperatures T1and T2 are
1. shown in the figure. It is concluded that

a) T > T b) T < T c) T = T d) T = 2T
1 2 1 2 1 2 1 1

110 In a region 1019 α- particales and 1019 protons move to the left, while 1019 electrons move to the right per
2. second. The current is

a) 3.2 A towards left b) 3.2 A towards right c) 6.4 A towards left d) 6.4 A towards right

110 An electric heater rated 220 V and 550 W is connected to AC mains. The current drawn by it is
3.

a) 0.8 A b) 2.5 A c) 0.4 A d) 1.25 A

110 A galvanometer has 30 divisions and a sensitivity 16 μA/div. It can be converted into a voltmeter to read 3
4. V by connecting

a) Resistance nearly 6 k Ω in series b) 6k Ω in parallel

c) 500 Ω in series d) It cannot be converted

110 Three resistors are connected to form the sides of a triangle ABC, the resistance of the sides AB, BC and CA
5. are 40 ohm, 60 ohm and 100 ohm respectively. The effective resistance between the points A and B in ohm
will be
a) 32 b) 64 c) 50 d) 200

110 A potentiometer wire of length 1m and resistance 10 Ω is connected in series with a cell of emf 2V with
6. internal resistance 1 Ω and a resistance box including a resistance R. If potential difference between the
P a g e | 142
 ends of the wire is 1 mV, the value of R is
a) 20000 Ω b) 19989 Ω c) 10000 Ω d) 9989 Ω

110 An external resistance R is connected to a battery of e.m.f. V and internal resistance r. The joule heat
7. produced in resistor R is maximum when R is equal to

a) r b) r c) 2r d) Infinitely large
2
110 Two bulbs of 500 W and 200 W are manufactured to operate on 220 V line. The ratio of heat produced in
8. 500 W and 200 W, in two cases, when firstly they are connected in parallel and secondly in series will be

a) 5 : 2 b) 5 : 5 c) 2 : 5 d) 2 : 2
2 5 2 2 5 2 5 5
110 The current in a simple series circuit is 5.0.A. when an additional resistance of 2.0 Ω is inserted, the current
9. drops to 4.0 A. the original resistance of the circuit in ohm was

a) 1.25 b) 8 c) 10 d) 20

111 When connected across the terminals of a cell, a voltmeter measures 5V and a connected ammeter
0. measures 10 A of current. A resistance of 2 ohm is connected across the terminals of the cell. The current
flowing through this resistance will be
a) 2.5 A b) 2.0 A c) 5.0 A d) 7.5 A

111 In a potentiometer experiment for measuring the emf of a cell, the null point is at 480 cm when we have a
1. 400 Ω resistor in series with the cell and galvanometer. If the series resistances is reduced to half, the null
point will be at
a) 120 cm b) 240 cm c) 480 cm d) 600 cm

111 An aluminium (Al) rod with area of cross-section 4×10-6m2 has a current of 5 A flowing through it. Find
2. the drift velocity of electron in the rod. Density of Al= 2.7×103 kgm-3 and atomic wt.=27u. Assume that
each Al atom provides one electron.
a) 8.6×10-4ms-1 b) 1.3×10-4ms-1 c) 2.8×10-2ms-1 d) 3.8×10-3ms-1

111 Seven resistance are connected as shown in the figure. The equivalent resistance between A and B is
3. 10Ω

A 10Ω 3Ω B

5Ω 8Ω 6Ω 6Ω

a) 3 Ω b) 4 Ω c) 4.5 Ω d) 5 Ω

111 In a conductor if 3000 coulomb of charge enters and 3000 coulomb of charge exits in time 10 minutes,
4. then the current is

a) 5 ampere b) 10 ampere c) 2.5 ampere d) Zero

111 There are 8 equal resistance R. Two are connected in parallel, such four groups are connected in series, the
5. total resistance of the system will be

a) R/2 b) 2 R c) 4 R d) 8 R

111 Fifty electric bulbs, all identical, are connected in series across the mains of a 220 V supply. After one bulb
6. is fused, the remaining 49 bulbs connected in series across the same mains. The illumination will be
P a g e | 143
 a) More with 50 bulbs than with 48 bulbs b) More with 49 bulbs than with 50 bulbs

d) In the ratio (50) :(49) in the first and second case

2
c) Equal in both cases 2

respectively
111 If VAB = 4V in the given figure, then resistance X will be
7. 10 5V

A B

2V X

a) 5 Ω b) 10 Ω c) 15 Ω d) 20 Ω

111 A current i passes through a wire of length l, radius of cross-section r and resistivity ρ. The rate of heat
8. generation is

( )
2 2
a) i lρ b) i2 lρ c) i2l ρ/r d) il ρ/r
2 2
πr πr
111 Two wires that are made up of two different materials whose specific resistance are in the ratio 2 : 3,
9. length 3 : 4 and area 4 : 5. The ratio of their resistances is

a) 6 : 5 b) 6 : 8 c) 5 : 8 d) 1 : 2

112 Two electric bulbs (60W and 100W respectively) are connected in series. The current passing through
0. them is

a) More in 100W bulb b) More in 60W bulb c) Same in both d) None of these

112 Twelve wires of equal resistance R are connected to form a cube. The effective resistance between two
1. opposite diagonal ends will be

a) (5/6)R b) (6/5)R c) 3R d) 12R

112 E.C.E. of Cu and Ag are 7×10-6 and 1.2×10-6. A certain current deposits 14 gm of Cu. Amount of Ag
2. deposited is

a) 1.2 gm b) 1.6 gm c) 2.4 gm d) 1.8 gm

112 A current of 2A flows in an electric circuit as shown in figure. The potential difference(VR - VS), in
3. volts( V - V are potentials at R and S respectively) is
R S

3Ω 7Ω
P Q
2A 2A

a) -4 b) +2 c) +4 d) -2

112 Two wires of the same dimensions but resistivities ρ1 and ρ2are connected in series. The equivalent
4. resistivity of the combination is

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 a) ρ1+ρ2 b) ρ + ρ c) 2(ρ + ρ ) d) ρ ρ
2 1 2 1 2 1 2

112 If resistance of voltmeter is 10000Ω and resistance of galvanometer is 2Ω, then find R when voltmeter
5. reads 12V and galvanometer reads 0.1A.
R
G

a) 118Ω b) 120Ω c) 124Ω d) 114Ω

112 A 25 W, 220 V bulb and a 100 W, 220 V bulb are connected in parallel across a 440 V line
6.

a) Only 100 watt bulb will fuse b) Only 25 watt bulb will fuse

c) Both bulbs will fuse d) None of the bulbs will fuse

112 A battery is made by connecting 6 cells each having capacity 5 Ah at 1.5V. The battery will have capacity
7. equal to

a) 20 Ah at 9 V b) 30 Ah at 1.5 V c) 5 Ah at 9 V d) 5 Ah at 1.5 V

112 If a wire of resistance 20Ω is covered with ice and a voltage of 210 V is applied across the wire, then the
8. rate of melting of ice is

a) 0.85g/s b) 1.92g/s c) 6.56g/s d) All of these

112 A voltmeter of resistance 1000Ω gives full scale deflection when a current of 100 mA flows through it. The
9. shunt resistance required across it to enable it to be used as an ammeter reading 1 A at full scale deflection
is
a) 10000Ω b) 9000Ω c) 222Ω d) 111Ω

113 A certain wire has a resistance R. The resistance of another wire identical with the first except having
0. twice its diameter is

a) 2 R b) 0.25 R c) 4 R d) 0.5 R

113 As the temperature of hot junction increases, the thermo e.m.f

1.

a) Always increases b) Always decreases

c) May increases or de decreases d) Always remains constant

113 A uniform wire of 16 Ω is made into the form of square. Two opposite corners of the square are connected
2. by a wire of resistance 16Ω. The effective resistance between the other two opposite corners is

a) 32Ω b) 20Ω c) 8Ω d) 4Ω

113 A 100 W bulb B1, and two 60-W bulbs B2 and B3, are connected to a 250 V source, as shown in the figure.
3. Now W , W and W are the output powers of the bulbs B ,B and B , respectively. Then
1 2 3 1 2 3

P a g e | 145
 B1 B2

B3

250V

a) W > W = W b) W > W > W c) W < W = W d) W < W < W

1 2 3 1 2 3 1 2 3 1 2 3

113 What is the equivalent resistance of the circuit

4. 4V , 1  2
+ – 2
2 A

4

a) 6 Ω b) 7 Ω c) 8 Ω d) 9 Ω

113 An immersion heater is rated 836 watt. It should heat 1 litre of water from 10℃ to 40℃ in about
5.

a) 200 sec b) 150 sec c) 836 sec d) 418 sec

113 A resistance of 2 Ω is connected across one gap of a meter-bridge(the length of the wire is 100cm) and an
6. unknown resistance, greater than 2 Ω is connected across the other gap. When these resistances are
interchanged, the unknown resistance is
a) 3 Ω b) 2 Ω c) 4 Ω d) 6 Ω

113 A wire of resistance 12 ohms per meter is bent to form a complete circle of radius 10 cm. The resistance
7. between its two diametrically opposite points A and B as shown in the figure, is

a) 0.6 π Ω b) 3 Ω c) 6 π Ω d) 6 Ω

113 Metals have

8.

a) Zero resistivity b) High resistivity c) Low resistivity d) Infinite resistivity

113 A galvanometer acting as a voltmeter should have

9.

a) Low resistance in series with its coil b) Low resistance in parallel with its coil

c) High resistance in series in series with its coil d) High resistance in parallel with its coil

114 The electromotive force of a primary cell is 2 volt. When it is short-circuited it gives a current of 4 ampere.
0. Its internal resistance in ohm is

a) 0.5 b) 5.0 c) 2.0 d) 8.0

114 In the circuit given here, the points A, B and C are 70V, zero, 10 V respectively. Then
B
1.
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 20Ω

A
10Ω D
30Ω
C

a) The point D will be at a potential of 60V

b) The point D will be at a potential of 20V

c) Currents in the path AD, DB and DC are in the ratio of 1:2:3

d) Currents in the path AD, DB and DC are in the ratio of 3 : 2 : 1

114 A 500 W heating unit is designed to operate from a 115 volt line. If the line voltage drops to 110 volt, the
2. percentage drop in heat output will be

a) 10.20% b) 8.1% c) 8.6% d) 7.6%

114 1
Resistors P and Q are connected in the gaps of the meter bridge. The balancing point is obtained m from
3. 3
2
the zero end. If a 6 Ω resistance is connected in series with P the balance point shifts to m from the same
3
end. P and Q are
a) 4,2 b) 2,4

c) Both (a) and (b) d) Neither (a) nor (b)

114 4 cells each of emf 2 V and internal resistance of 1Ω are connected in parallel to a load resistor of 2Ω. Then
4. the current through the load resistor is

a) 2 A b) 1.5 A c) 1 A d) 0.888 A

114 A potential divider is used to give outputs of 4 V and 8 V from a 12 V source. Which combination of
5. resistances, (R1:R2:R3) gives the correct voltages?

a) 2 : 1 : 2 b) 1 : 1 : 1 c) 2 : 2 : 1 d) 1 : 1 : 2

114 Two identical heaters of 220V , 1000 W are placed in parallel with each other across 220V line, then the
6. combined power is

a) 1000 W b) 2000 W c) 500 W d) 4000 W

114 Length of a hollow tube is 5m, it’s outer diameter is 10 cm and thickness of it’s wall is 5 mm. If resistivity of
7. the material of the tube is 1.7×10-8Ω×m then resistance of tube will be
P a g e | 147
 a) 5.6×10-5Ω b) 2×10-5Ω c) 4×10-5Ω d) None of these

114 The mass of ions deposited during a given interval of time in the process of electrolysis depends on
8.

a) The current b) The resistance c) The temperature d) The electric power

114 A current of 0.01mA passes through the potentiometer wire of a resistivity of

9. 109 Ω-cm and area of cross-section 10-2cm2. The potential gradient is

a) 109Vm-1 b) 1011Vm-1 c) 1010Vm-1 d) 108Vm-1

115 A galvanometer of resistance 25Ω giving full scale deflection for a current of 10 milliampere, is to be
0. changed into a voltmeter of range 100 V by connecting a resistance of 'R' in series with galvanometer. The
value of resistance R in Ω is
a) 10000 b) 10025 c) 975 d) 9975

115 n conducting wires of same dimensions but having resistivites 1,2,3…..n are connected in series. The
1. equivalent resistivity of the combinations is

a) n(n+1) b) n+1 c) n+2 d) 2n

2 2 2n n+1
115 Arrange the order of power dissipated in the given circuits, if the same current is passing through all the
2. circuits. The resistance of each resistor is r.

a) P >P > P > P b) P >P > P > P

1 2 3 4 2 3 4 1

c) P >P > P > P d) P =P = P = P

4 3 2 1 1 2 3 4

115 Three electric bulbs of rating 60W each are joined in series and then connected to electric mains. The
3. power consumed by these three bulbs will be

a) 180 W b) 60 W c) 20 W d) 20 W
3
115 A 3 V battery with negligible internal resistance is connected in a circuit as shown in the figure. The
4. current I, in the circuit will be

3Ω 3Ω

3V

3Ω

P a g e | 148
 a) 1A b) 1.5A c) 2A d) 1 A
3
115 Two conductors of the same material have their diameters in the ratio 1 : 2 and their lengths in the ratio 2 :
5. 1. If the temperature difference between their ends is the same, then the ratio of amounts of heat
conducted per second through them will be
a) 4 :1 b) 1 :4 c) 8 :1 d) 1 :8

115 An electric kettle boils some water in 16 min. Due to some defect, it becomes necessary to remove 10%
6. turns of heating coil of the kettle. Now, how much time will it take to boil the same of water?

a) 17.7 min b) 14.4 min c) 20.9 min d) 13.7 min

115 A rod of a certain metal is 1.0 m long and 0.6 cm in diameter. Its resistance is 3.0×10-3Ω. Another disc
7. made of the same metal is 2.0 cm in diameter and 1.0 mm thick. What is the resistance between the round
faces of the disc?
a) 1.35×10-8Ω b) 2.70×10-7Ω c) 4.05×10-6Ω d) 8.10×10-5Ω

115 The equivalent resistance between the points A and B will be (each resistance is
8. 15 Ω)
15Ω
D C

15Ω
15Ω 15Ω

A B

a) 30 Ω b) 8 Ω c) 10 Ω d) 40 Ω

115 An ammeter, suspected to give inaccurate reading, is connected in series with a silver voltameter. The
9. ammeter indicates 0.54 A. A steady current passed for one hour deposits 2.0124 g of silver. If the E.C.E. of
-3 -1
silver is 1.118×10 g/C , then the error in ammeter reading is
a) + 0.04 A b) + 0.02 A c) -0.03 A d) -0.01 A

116 The cell has an emf of 2V and the internal resistance of 3.9 Ω, the voltage across the cell will be
0.

a) 1.95 V b) 1.5V c) 2V d) 1.8V

116 Two bulbs of 100 W and 200 W working at 220 V are joined in series with 220 V supply. Total power
1. consumed will be

a) 65 W b) 33 W c) 300 W d) 100 W

116 A current of 1 mA is flowing through a copper wire. How many electrons will pass a given point in one
2. second
-19
[e = 1.6×10 Coulomb]
a) 6.25×1019 b) 6.25×1015 c) 6.25×1031 d) 6.25×108

116 A wire of resistance R is elongated n- fold to make a new uniform wire. The resistance of new wire
3.

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 a) nR b) n2R c) 2nR d) 2n2R

116 Three unequal resistors in parallel are equivalent to a resistance 1 Ω. If two of them are in the ratio 1:2 and
4. if no resistance value is fractional, the largest of the three resistance in ohm is

a) 4 b) 6 c) 8 d) 12

116 An electric kettle has two heating coils. When one coil is used, water in the kettle boils in 5 minutes, while
5. when second coil is used, same water boils in 10 minutes. If the two coils, connected in parallel are used
simultaneously, the same water will boil in time
a) 3min 20 sec b) 5 min c) 7min 30 sec d) 2min 30 sec

116 An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from room
6. temperature 20℃? The temperature of boiling water is 100℃

a) 0.63 minutes b) 6.3 minutes c) 12.6 minutes d) 12.8 minutes

116 The three resistances of equal value are arranged in the different combinations shown below. Arrange
7. them in increasing order of power dissipation

II

IV
III

a) III < II < IV <I b) II < III < IV < I c) I < IV < III < II d) I < III < II < IV

116 When a metal conductor connected to the left gap of a meter bridge is heated, the balancing point
8.

a) Shifts towards right b) Shifts towards left

c) Remains unchanged d) Remains at zero

116 The electro chemical equivalent of metal is 3.3×10-7kgC-1. The mass of the metal liberated at the cathode
9. when a 3 A current is passed for 2 s, will be

a) 19.8×10-7 kg b) 9.9 ×10-7 kg c) 6.6 ×10-7 kg d) 1.1 ×10-7 kg

117 The resistance of a wire of iron is 10 ohm and temp. coefficient of resistance is 5×10-3/℃. At 20℃ it carries
0. 30 milliampere of current. Keeping constant potential difference between its ends, the temperature of the
wire is raised to 120℃. The current in milliampere that flows in the wire is
a) 20 b) 15 c) 10 d) 40

117 One kilowatt hour is equal to

1.

P a g e | 150
 a) 36×105 joules b) 36×103 joules c) 103 joules d) 105 joules

117 A battery of 6 volts is connected to the terminals of a three metre long wire of uniform thickness and
2. resistance of the order of 100Ω. The difference of potential between two points separated by 50 cm on the
wire will be
a) 1 V b) 1.5 V c) 2 V d) 3 V

117 Two resistances of 400 Ω and 800 Ω are connected in series with 6 volt battery of negligible internal
3. resistance. A voltmeter of resistance 10,000 Ω is used to measure the potential difference across 400 Ω.
The error in the measurement of potential difference in volt approximately is
a) 0.01 b) 0.02 c) 0.03 d) 0.05

117 If potential V = 100 ± 0.5 Volt and current I = 10 ± 0.2 amp are given to us, then what will be the value
4. of resistance

a) 10 ± 0.7 ohm b) 5 ± 2 ohm c) 0.1 ± 0.2 ohm d) None of these

117 An electric bulb rated for 500 W at 100V is used in a circuit having a 200 V supply. The resistance R that
5. must be put in series with the bulb, so that the bulb drawn 500 W is

a) 18Ω b) 20Ω c) 40Ω d) 700Ω

117 In the diagram shown, the reading of voltmeter is 20 V and that of ammeter is 4 A. The value of R should be
6. (Consider given ammeter and voltmeter are not ideal)
V
20V

R
A
4A

a) Equal to 5 Ω b) Greater than 5 Ω

c) Less than 5 Ω d) Greater or less than 5 Ω depending on the material

of R
117 A current of 2 A flows in a system of conductors as shown. The potential difference (VA - VB) will be
7. A
2 3
2A
D C
3 2

a) +2V b) +1V c) -1V d) -2V

117 A bulb has specification of one kilowatt and 250 volts, the resistance of bulb is
8.

a) 125 Ω b) 62.5 Ω c) 0.25 Ω d) 625 Ω

117 The direction of current in an iron-copper thermocouple is

9.

a) From copper to iron at the hot junction b) From iron to copper at the hot junction

c) From copper to iron at cold junction d) No current will flow

P a g e | 151
118 What is the total resistance of the circuit?
0.
2
4V 2
+ - A
2
1
4

a) 6 Ω b) 7 Ω c) 8 Ω d) 9 Ω

118 In the circuit shown, the point 'B' is earthed. The potential at the point 'A' is
1. 5 7 B
A
10

50V C

3
E D

a) 14 V b) 24 V c) 26 V d) 50 V

118 If in the circuit, power dissipation is 150 W, then R is

2.

a) 2 Ω b) 6 Ω c) 5 Ω d) 4 Ω

118 In the circuit shown below, the cell has an e.m.f. of 10 V and internal resistance of 1 ohm. The other
3. resistances are shown in the figure. The potential difference VA - VB is
E=10V
r=1

4 A 2 1

C
2 B 4

a) 6 V b) 4 V c) 2 V d) -2 V

118 A Daniel cell is balanced on 125cm length of a potentiometer wire. Now the cell is short-circuited by a
4. resistance 2 ohm and the balance is obtained at 100 cm. The internal resistance of the Daniel cell is

a) 0.5 ohm b) 1.5 ohm c) 1.25 ohm d) 4/5 ohm

118 Charge Q is divided into two parts which are then kept some distance apart. The force between them will
5. be maximum if the two parts are having the charge

a) Q/2 each b) Q/4 and 3Q/4

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 c) Q/3 and 2Q/3 d) e and (Q-e),where e =electronic charge

118 When 1 g hydrogen (ECE = 1.044×10-8 kg C-1) forms water, 34 kcal heat is liberated. The minimum voltage
6. required to decompose water is

a) 0.75 V b) 3 V c) 1.5 V d) 4.5 V

118 E denotes electric field in a uniform conductor, I corresponding current through it, vd drift velocity of
7. electrons and P denotes thermal power produced in the conductor, then which of the following graph is
incorrect
a) vd b) P c) P d) P

E E vd i

118 The two ends of a uniform conductor are joined to a cell of e.m.f. E and some internal resistance. Starting
8. from the midpoint P of the conductor, we move in the direction of current and return to P. The potential V
at every point on the path is plotted against the distance covered (x). Which of the following graphs best
represents the resulting curve
a) V b) V c) V d) V

E <E E
<E

X X X X

118 Three wires of copper, iron and nickel are joined to form three junctions as shown in Fig. When the
9. temperature of junction 1 is kept 50℃ with the other two junctions at 0℃, the sensitive galvanometer
gives a deflection of 14 divisions. When the temperature of junction 3 is kept 50℃, with the other two
junctions at 0℃, the galvanometer gives a deflection of 11 divisions. Then the deflection given by the
galvanometer, when temperature of the junction 2 is kept at 50℃, with the other two junctions at 0℃, will
be

G
Cu Cu

Fe Ni

1 2 3

a) 3 div b) 11 div c) 14 div d) 25 div

119 Six resistors, each of value 3 Ω are connected as shown in the figure. A cell of emf 3V is connected across
0. AB. The effective resistance across AB and the current through the arm AB will be

3 D 3Ω
3

C D
3 F 3

A B
3

a) 0.6 Ω, 1 A b) 1.5 Ω , 2 A c) 0.6Ω,2 A d) 1.5 Ω, 1 A

119 In Seebeck series Sb appears before Bi. In a Sb - Bi thermocouple current flows from
P a g e | 153
1.

a) Sb to Bi at the hot junction b) Sb to Bi at the cold junction

c) Bi to Sb at the cold junction d) None of the above

119 What is the ratio of heat generated in R and 2R

2.

a) 2 :1 b) 1 :2 c) 4 :1 d) 1 :4

119 Which of the following statement is correct

3.

a) Electric field is zero on the surface of current carrying wire

b) Electric field is non-zero on the axis of hollow current carrying wire

c) Surface integral of magnetic field for any closed surface is equal to μ0 times of total algebraic sum of
current which are crossing through the closed surface
d) None

119 In the arrangement of resistance shown below, the effective resistance between points A and B is
4. 5 10 15
P

A 10 10 B

Q
10 20 30

a) 20 Ω b) 30 Ω c) 90 Ω d) 110 Ω

119 If 10 A deposits 10.8 g of silver in 25 min, how much copper would deposit when 9 A current flows for 20
5. min.?

a) 3.81 g b) 6.35 g c) 10.1 g d) 12.7 g

119 The relation between Faraday constant (F), chemical equivalent (E) and electrochemical equivalent (Z) is
6.

a) F = EZ b) F = Z c) F = E d) F = E
2
E Z Z
119 Two resistances are joined in parallel whose resistance is 3/5Ω. One of the resistance wire is broken and
7. the effective resistance become 3Ω.The resistance in ohm of the wire that got broken was

a) 4/3 b) 2 c) 6/5 d) 3/4

119 A moving coil galvanometer has a resistance of 10Ω and full scale deflection of 0.01A. It can be converted
8. into voltmeter of 10V full scale by connecting into resistance of

a) 9.90Ω is series b) 10Ω in series c) 990Ω in series d) 0.10Ω in series

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119 If a high power heater is connected to electric mains, then the bulbs in the house become dim, because
9. there is a

a) Current drop b) Potential drop c) No current drop d) No potential drop

120 A 10 m long wire of 20Ω resistance is connected with a battery of 3 volt e.m.f. (negligible internal
0. resistance) and a 10 Ω resistance is joined to it is series. Potential gradient along wire in volt per meter is

a) 0.02 b) 0.3 c) 0.2 d) 1.3

120 A certain charge liberates 0.8 gm of O2. The same charge will liberate how many gm of silver
1.

a) 108 gm b) 10.8 gm c) 0.8 gm d) 108 gm

0.8
120 A student measures the terminal potential difference (V) of a cell (of emf E and internal resistance r) as a
2. function of the current (I) flowing through it. The slope, and intercept, of the graph between V and I, then,
respectively, equal
a) E and –r b) -r and E c) r and –E d) -E and r

120 The charge on the capacitor of capacitance C shown in the figure below will be
3. E 2Ω

R2
I
R1

a) CE b) CER1 c) CER2 d) CER2

R1+r R2+r R1+r
120 A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point
4. length l. If the wire is replaced by another wire of same material but with double the length and half the
thickness, the balancing point is expected to be
a) 1 b) 1 c) 8l d) 16l
8l 4l
120 The equivalent resistance between the terminals A and B in the following circuit is
5.
5Ω 10 Ω 5Ω
A B

10 1

5 10 5
C D

a) 10 Ω b) 20 Ω c) 5 Ω d) 30 Ω

120 A block has dimensions 1 cm, 2 cm, 3 cm. Ratio of the maximum resistance to minimum resistance
6. between any point of opposite faces of this block is

a) 9 : 1 b) 1 : 9 c) 18 : 1 d) 1 : 6

120 Two rods of same material and length have their electric resistances in ratio 1 : 2. When both rods are
7. dipped in water, the correct statement will be

P a g e | 155
 a) A has more loss of weight b) B has more loss of weight

c) Both have same loss of weight d) Loss of weight will be in the ratio 1 : 2

120 Three resistances each of 1 ohm, are joined in parallel. Three such combinations are put in series, then the
8. resultant resistance will be

a) 9 ohm b) 3 ohm c) 1 ohm d) 1 ohm

3
120 As the temperature of hot junction of a thermo-couple is increased (while cold junction is at constant
9. temperature), the thermo e.m.f

a) Increases uniformly at constant rate

b) Increases slowly in the beginning and more rapidly at higher temperatures

c) Increases more rapidly in the beginning but less rapidly at higher temperatures

d) Is minimum at neutral temperature

121 Pick out the wrong statement

0.

a) In a simple battery circuit, the point of lowest potential is the negative terminal of the battery

b) The resistance of an incandescent lamp is greater when the lamp is switched off

c) An ordinary 100 W lamp has less resistance than a 60 W lamp

d) At constant voltage, the heat developed in a uniform wire varies inversely as the length of the wire used

121 A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of
1. resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of resistance r is connected
across it. What is the maximum current which can be sent through this galvanometer if no shunt is used
a) 0.01 A b) 0.02 A c) 0.03 A d) 0.04 A

121 In a Wheatstone’s bridge all the four arms have equal resistance R. If the resistance of the galvanometer
2. arm is also R, the equivalent resistance of the combination as seen by the battery is

a) R/2 b) R c) 2 R d) R/4

121 A torch bulb rated at 4.5 W, 1.5 V is connected as shown in figure. The emf of the cell needed to make the
3. bulb glow at full intensity if

a) 4.5 V b) 1.5 V c) 2.67 V d) 13.5 V

121 A 60 watt bulb operates on 220V supply. The current flowing through the bulb is
4.

a) 11/3 amp b) 3/11 amp c) 3 amp d) 6 amp

121 A battery of emf E produces currents I1 and I2 when connected to external resistances R1 and R2

P a g e | 156
5. respectively. The internal resistance of the battery is

a) I1R2-I2R1 b) I1R2+I2R1 c) I1R1+I2R2 d) I1R1-I2R2

I2-I1 I1-I2 I1-I2 I2-I1
121 The potential difference between A and B in the following figure is
6.

a) 24 V b) 14 V c) 32 V d) 48 V

121 The value of i1 in the circuit diagram will be

7. E

8V B
i i
2 2
5
4 i1 4
A C
5
2 2

a) 1A b) 1 A c) 3 A d) 3 A
2 4 2
121 In the circuit shown here, E1 = E2 = E3 = 2V and R1 = R2 = 4 ohm. The current flowing between points
8. A and B through battery E is
2
E1 R1

E2
A B

E3 R2

a) Zero b) 2 amp from A to B c) 2 amp from B to A d) None of the above

121 A battery of emf 10 V and internal resistance 3Ω is connected to a resistor as shown in the figure. If the
9. current in the circuit is 0.5 A, then the resistance of the resistor will be

a) 19 Ω b) 17 Ω c) 10 Ω d) 12 Ω

122 The equivalent resistance between A and B in the given circuit is

0.

P a g e | 157
 C

3Ω 3Ω
3Ω
D
3Ω 3Ω
A B
3Ω

a) 3Ω b) 6Ω c) 12Ω d) 1.5Ω

122 The material of fuse wire should have

1.

a) A high specific resistance and high melting point

b) A low specific resistance and low melting point

c) A high specific resistance and low melting point

d) A low specific resistance and a high melting point

122 Resistance of a voltameter is 2Ω, it is connected in series to a battery of 10 V through a resistance of 3Ω. In
2. a certain time mass deposited on cathode is 1g. Now the voltameter and the 3Ω resistance are connected in
parallel with the battery. Increase in the deposited mass on cathode in the same time will be
a) 0 b) 1.5 g c) 2.5 g d) 2 g

122 The length of a wire of a potentiometer is 100cm, and the emf of its stand and cell is E volt. It is employed
3. to measure the emf of a battery whose internal resistance is 0.5Ω. If the balance point is obtained at
l = 30cm from the positive end, the emf of the battery is
a) 30E b) 30E
100.5 100-0.5
30(E-0.5i)
c) , Where i is the current in the d) 30E
100
100
potentiometer wire.
122 Resistance of tungsten wire at 150℃ is 133Ω. Its resistance temperature coefficient is 0.0045/℃. The
4. resistance of this wire at 500℃ will be

a) 180Ω b) 225Ω c) 258Ω d) 317Ω

122 Resistors of resistance 20Ω and 30Ω are joined in series with a battery of emf 3V. It is desired to measure
5. current and voltage across the 20Ω resistor with the help of an ammeter and voltmeter. Identify the
correct arrangement of ammeter (A) and voltmeter (V) out of four possible arrangements shown in figure.
Given below
A V

V
20 30 20 30
a) b) A
+ - + -
3V 3V

P a g e | 158
 V
A
A 20 30
20 30 V
c) d) + -
+ -
3V
3V

122 Which arrangement of four identical resistance should be used to draw maximum energy from a cell of
6. voltage V

a) b)

c) d)

122 A capacitor is connected to a cell of emf E having some internal resistane r. The potential difference across
7. the

a) Cell is < E b) Cell is E c) Capacitor is > E d) Capacitor is < E

122 A wire of resistance 5.5 ohm is drawn out uniformly so that its length is increased twice. Then its new
8. resistance is

a) 44Ω b) 42Ω c) 40Ω d) 22Ω

122 Silver and copper voltameters are connected in parallel with a battery of emf 12 V. In 30 min 1 g of silver
9. and 1.8 g of copper are liberated. The energy supplied by the battery is

a) 720 J b) 2.41 J c) 24.12 J d) 4.34×104 J

123 Two bulbs of 250 V and 100 W are first connected in series and then in parallel with a supply of 250 V.
0. Total power in each of the case will be respectively

a) 100 W, 50 W b) 50 W, 100 W c) 200 W, 150 W d) 50 W, 200 W

123 An ammeter gives full scale deflection when a current of 2A flows through it. The resistance of ammeter is
1. 12 Ω. If the same ammeter is to be used for measuring a maximum current of 5A, then ammeter must be
connected with a resistance of
a) 18 Ωin parallel b) 8 Ωin parallel c) 18 Ω in series d) 8 Ω in series

123 A conductor with rectangular cross-section has dimensions (a×2a×4a) as shown in figure. Resistance
2. across AB is R1, across CD is R2 and across EF is R3.Then

a) R = R = R b) R > R > R c) R > R > R d) R > R > R

1 2 3 1 2 3 2 3 1 1 3 2

P a g e | 159
123 The current flowing in a coil of resistance 90 Ω is to be reduced by 90%. What value of resistance should
3. be connected in parallel with it

a) 9 Ω b) 90 Ω c) 1000 Ω d) 10 Ω

123 In the absence of applied potential, the electric current flowing through a metallic wire is zero because
4.

a) The electrons remain stationary

b) The electrons are drifted in random direction with a speed of the order of 10-2 cm s-1

c) The electrons move in random direction with a speed of the order close to that of velocity of light

d) Electrons and ions move in opposite direction

123 The resistors P,Q and R in the circuit have equal resistance. The battery, of negligible internal resistance,
5. supplies a total power of 12 W. What is the power dissipated by heating in resistor R?

a) 2 W b) 4 W c) 3 W d) 6 W

123 The temperature coefficient of resistance of a wire is 0.00125 K-1. At 300K, its resistance is 1Ω. The
6. resistance of the wire will be 2Ω at

a) 1154 K b) 1100 K c) 1400 K d) 1127 K

123 The resistance of a heater coil is 110 ohm. A resistance R is connected in parallel with it and the
7. combination is joined in series with a resistance of 11 ohm to a 220 volt main line. The heater operates
with a power of 110 watt. The value of R in ohm is
a) 12.22 b) 24.42

c) Negative d) That the given values are not correct

123 The resistance of a wire at room temperature 36℃ is found to be 10Ω. Now to increase the resistance by
8. 10%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the
wire is 0.002 per ℃]
a) 36℃ b) 83℃ c) 63℃ d) 33℃

123 A current of 2 A passing through conductor produces 80 J of heat in 10 seconds. The resistance of the
9. conductor is

a) 0.5 Ω b) 2 Ω c) 4 Ω d) 20 Ω

124 What is immaterial for an electric fuse?

0.

a) Its specific resistance b) Its length

c) Its radius d) Current flowing through it

124 A primary cell has an e.m.f. of 1.5 volt, when short-circuited it gives a current of 3 ampere. The internal
1. resistance of the cell is

a) 4.5 ohm b) 2 ohm c) 0.5 ohm d) 1/4.5 ohm

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124 The heating coils rated at 220 volt and producing 50 cal/sec heat are available with the resistance 55 Ω,
2. 110 Ω, 220 Ω and 440 Ω. The heater of maximum power will be of

a) 440 Ω b) 220 Ω c) 110 Ω d) 55 Ω

124 Which of the following graphs shows the variation of thermoelectric power with temperature difference
3. between hot and cold junction in thermocouples

dE dE dE dE
d d d d

a) b) c) d)

   

124 Five equal resistors when connected in series dissipated 5 W power. If they are connected in parallel, the
4. power dissipated will be

a) 25 W b) 50 W c) 100 W d) 125 W

124 On passing 96500 coulomb of charge through a solution CuSO4 the amount of copper liberated is
5.

a) 64 gm b) 32 gm c) 32 kg d) 64 kg
+
124 In a neon discharge tube 2.9×1018Ne ions move to the right each second while 1.2×1018 electrons move to
6. the left per second. Electron charge is 1.6×10-19C. The current in the discharge tube

a) 1 A towards right b) 0.66 A towards right c) 0.66 A towards left d) Zero

124 V
For a metallic wire, the ratio (V =applied potential difference and i=current flowing ) is
7. i

a) Independent of temperature

b) Increases as the temperature rises

c) Decreases as the temperature rises

d) Increases or decreases as temperature rises depending upon the metal

124 The resistance of the series combination of two resistance is S. When they are joined in parallel, the total
8. resistance is P. If S = nP, then the minimum possible value of n is

a) 4 b) 3 c) 2 d) 1

124 In a wire of circular cross-section with radius r, free electrons travel with a drift velocity V when a current
9. I flows through the wire. What is the current in another wire of half the radius and of the same material
when the drift velocity is 2V
a) 2I b) I c) I/2 d) I/4

125 Two electric bulbs, one of 200 volt 40 watt and the other 200 volt 100 watt are connected in a house
0. wiring circuit

a) They have equal currents through them

b) The resistance of the filaments in both the bulbs is same

P a g e | 161
 c) The resistance of the filament in 40 watt bulb is more than the resistance in 100 watt bulb

d) The resistance of the filament in 100 watt bulb is more than the resistance in 40 watt bulb

125 In which of the following substances does resistance decrease with increase in temperature?
1.

a) Copper b) Carbon c) Constantan d) Silver

125 Thirteen resistances each of resistance RΩ are connected in the circuit as shown in the figure. The effective
2. resistance between points A and B is
R R
A

R
R R R

R
R R R

R R

a) 4R Ω b) 2RΩ c) RΩ d) 2 RΩ
3 3
125 The resistance of a 10 m long wire is 10 Ω. Its length is increased by 25% by stretching the wire uniformly.
3. The resistance of wire will change to (approximately)

a) 12.5 Ω b) 14.5 Ω c) 15.6 Ω d) 16.6 Ω

125 The masses of the three wires of copper are in the ratio 5 : 3 : 1 and their lengths are in the ratio 1 : 3 : 5
4. the ratio of their electrical resistance is

a) 5 : 3 : 1 b) 125 :15 :1 c) 1 : 15 : 125 d) 1 : 3 : 5

125 A milliammeter of range 0-30mA has internal resistance of 20 Ω. The resistance to be connected in series
5. to convert it into a voltmeter of maximum reading 3V is

a) 49 Ω b) 80 Ω c) 40 Ω d) 30 Ω

125 Given figure shows a rectangular block with dimensions x, 2x and 4x. Electrical contacts can be made to
6. the block between opposite pairs of faces (for example, between the faces labelled A - A, B - B and C - C).
Between which two faces would the maximum electrical resistance be obtained (A - A : Top and bottom
faces, B - B : Left and right faces, C - C : Front and rear faces)

a) A - A

b) B - B
P a g e | 162
 c) C - C

d) Same for all three pairs

125 Resistance as shown in figure is negative at

7.
I A C

a) A b) B c) C d) None of these

125 In the circuit shown in the figure, the current through

8. 3Ω 2Ω 2Ω

9V 8Ω 8Ω 4Ω

2Ω 2Ω 2Ω

a) The 3Ω resistor is 0.50 A b) The 3Ω resistor is 0.25 A

c) The 4Ω resistor is 0.50 A d) The 4Ω resistor is 0.25 A

125 The potential gradient along the length of a uniform wire is 10 volt/metre. B and C are the two points at 30
9. cm and 60 cm point on a meter scale fitted along the wire. The potential difference between B and C will be

a) 3 volt b) 0.4 volt c) 7 volt d) 4 volt

126 Current is flowing with a current density J = 480Acm-2 in a copper wire. Assuming that each copper atom
0. contributes one free electron and given that
23 -1
Avogadro number=6.0×10 atoms mol
-3
Density of copper=9.0g cm
-1
Atomic weight of copper =64 g mol
The drift velocity of electrons is
a) 1 mm s-1 b) 2 mm s-1 c) 0.5 mm s-1 d) 0.36 mm s-1

126 Two wires of the same material but of different diameters carry the same current i. If ratio of their
1. diameters is 1:2, then the corresponding ratio of their mean drift velocities will be

a) 4:1 b) 1:1 c) 1:2 d) 1:4

126 How much work is required to carry a 6 μC charge from the negative terminal to the positive terminal of a
2. 9 V battery

a) 54×10-3J b) 54×10-6J c) 54×10-9J d) 54×10-12J

126 An electric bulb rated 220 V, 100 W is connected in series with another bulb rated 220 V, 60 W. If the
3. voltage across the combination is 220 V, the power consumed by the 100 W bulb will be about

a) 25 W b) 14 W c) 60 W d) 100 W

126 When a current of 1 ampere is passed through a conductor whose ends are maintained at temperature
4. difference of 1℃, the amount of heat evolved or absorbed is called

P a g e | 163
 a) Peltier coefficient b) Thomson coefficient

c) Thermoelectric power d) Thermo e.m.f.

126 In the figure given below, the current passing through 6Ω resistor is
5. 6

1.2 A

4

a) 0.40 ampere b) 0.48 ampere c) 0.72 ampere d) 0.80 ampere

126 For a given temperature difference which of the following pairs will generate maximum thermo-emf?
6.

a) Lead-nickel b) Copper-iron c) Gold-silver d) Antimony-bismuth

126 The resistance of ideal voltmeter is

7.

a) Zero b) Greater than zero but finite value

c) Infinite d) 5000 Ω

126 A 100 watt bulb working on 200 volt and a 200 watt bulb working on 100 volt have
8.

a) Resistances in the ratio of 4 :1

b) Maximum current ratings in the ratio of 1 :4

c) Resistances in the ratio of 2 :1

d) Maximum current ratings in the ratio of 1 :2

126 The graph between resistivity and temperature, for a limited range of temperatures, is a straight line for a
9. material like

a) Copper b) Nichrome c) Silicon d) Mercury

127 The emf of a generator is 6 V and internal resistance is 0.5 kΩ. The reading of a voltmeter having an
0. internal resistance of 2.5 kΩ is

a) 10-3V b) 10 V c) 5 V d) 0.5 V

127 A metallic block has no potential difference applied across it, then the mean velocity of free electrons at
1. absolute temperature T is

a) Proportional to T b) Proportional to T

c) Zero d) Finite but independent of T

127 An ammeter reads 0.90 A when connected in series with a silver voltmeter that deposits 2.60 g of silver in
2. 40 min. By what percentage is the ammeter reading is correct? Atomic weight of silver = 108 and 1
F=96500 C?
a) 5% b) 7% c) -5% d) -7%
P a g e | 164
127 A thermo-emf V appears across a conductor maintained at a temperature difference T. The thomson
3. coefficient is then given by
2 2
a) -T2 d V b) T2 dV c) -T d V d) - 1 dV
2 2 2
dT dT dT T dT
127 The amount of heat generated in 500Ω resistance, when the key is thrown over from contact 1 to 2, as
4. shown in figure is
5 F

330
500
2
1

E = 200 V

a) 10℃ b) 7.5℃ c) 5.0℃ d) 2.5℃

127 A circuit consists of five identical conductors as shown in figure. The two similar conductors are added as
5. indicated by the dotted lines. The ratio of resistances before and after addition will be

a) 7/5 b) 3/5 c) 5/3 d) 6/5

127 Two bulbs X and Y having same voltage rating and of power 40 W and 60 W respectively are connected in
6. series across a potential difference of 300 V, then

a) X will glow brighter b) Resistance of Y will be greater than X

c) Heat produced in Y will be greater than X d) Voltage drop in X will be greater than Y

127 Two voltameters, one of copper and another of silver, are joined in paralleled. When a total charge q flows
7. through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of
copper and silver are z1 and z2 respectively, the charge which flows through the silver voltameter is
q q
a) z b) z c) q z1 d) q z /z
1+ 1 1+ 2 z 2 1
z2 z1 2

127 In the circuit shown P ≠ R, the reading of the galvanometer is same with switch S open or closed. Then
8.

P a g e | 165
 a) I = I b) I = I c) I = I d) I = I
R G P G Q G Q R

127 The density of copper is 9×103 kg/m3and its atomic mass is 63.5 u. Each copper atom provides one free
9. electron. Estimate the number of free electrons per cubic metre in copper.

a) 1019 b) 1023 c) 1025 d) 1029

128 Find the true statements

0.

a) Ohm’s law is applicable to all conductors of electricity

b) In an electrolyte solution, the electric current is mainly due to the movement of electrons

c) The resistance of an incandescent lamp is lesser when the lamp is switched on

d) Specific resistance of a wire depends upon its dimension

128 A galvanometer can be converted into a voltmeter by connecting

1.

a) Low residence in parallel b) Low residence in series

c) High residence in parallel d) High residence in series

128 By mistake a voltmeter is connected in series and an ammeter is connected in parallel with a resistance in
2. an electrical circuit. What will happen to the instrument?

a) Voltmeter is damaged b) Ammeter is damaged

c) Both are damaged d) None is damaged

128 In a thermocouple, the neutral temperature is 270℃ and the temperature of inversion is
3. 525℃. The temperature of cold junction would be

a) 30℃ b) 255℃ c) 15℃ d) 25℃

128 Two electric lamps of 40 watt each are connected in parallel. The power consumed by the combination will
4. be

a) 20 watt b) 60 watt c) 80 watt d) 100 watt

128 Consider the following two statements A and B, and identify the correct choice out of given answers
5. A. Thermo e.m.f. is minimum at neutral temperature of a thermocouple
B. When two junctions made of two different metallic wires are maintained at different temperatures, an
electric current is generated in the circuit
a) A is false and B is true b) A is true and B is false

c) Both A and B are false d) Both A and B are true

128 Variation of current passing through a conductor as the voltage applied across its ends is varied as shown
6. in the adjoining diagram. If the resistance (R) is determined at the points A,B,C and D, we will find that

P a g e | 166
 a) R = R b) R > R c) R > R d) None of these
C D B A C B

128 There resistances of 4 Ω each are connected as shown in figure. If the point D divides the resistance into
7. two equal halves, the resistance between points A and D will be
A

4Ω 4Ω

B C
4Ω

a) 12 Ω b) 6 Ω c) 3 Ω d) 1 Ω
3
2 -8
128 A copper wire of cross-sectional area 2.0 mm , resistivity =1.7×10 Ωm, carries a current of 1A. The
8. electric field in the copper wire is

a) 8.5×10-5Vm-1 b) 8.5×10-4Vm-1 c) 8.5×10-3Vm-1 d) 8.5×10-2Vm-1

128 The thermo-emf of a thermocouple varies with the temperature θ of the hot junction as E = aθ + bθ2 in
9. volts where the ratio a/b is 700℃. If the cold junction is kept at 0℃, then the neutral temperature is

a) 700℃. b) 350℃.

c) 1400℃. d) No neutral temperature is possible for this

thermocouple
129 2, 4 and 6 S are the conductance of three conductors. When they are joined in parallel, their equivalent
0. conductance will be

a) 12 S b) (1/12)S c) (12/11)S d) (11/12)S

129 Two resistances are connected in two gaps of a meter bridge. The balance point is 20cm from the zero end.
1. A resistance of 15Ω is connected is series with the smaller of the two. The null point shifts to 40cm. The
value of the smaller resistance in ohm is
a) 3 b) 6 c) 9 d) 12

129 Two batteries, one of emf 18 volt and internal resistance 2Ω and the other of emf 12 volt and internal
2. resistance 1Ω, are connected as shown. The voltmeter V will record a reading of
V

18V 2

12V 1

a) 15 volt b) 30 volt c) 14 volt d) 18 volt

129 The drift velocity does not depend upon

3.
P a g e | 167
 a) Cross-section of the wire b) Length of the wire

c) Number of free electrons d) Magnitude of the current

129 The emf of a thermocouple, one junction of which is kept at 0℃, is given by e = at + bt2. The Peltier co-
4. efficient will be

a) (t + 273)(a + 2bt) b) (t + 273)(a - 2bt) c) (t - 273)(a - 2bt) d) (t - 273)(a + 2bt)

129 For electroplating a spoon, it is placed in the voltmeter at

5.

a) The position of anode

b) The position of cathode

c) Exactly in the middle of anode and the cathode

d) Anywhere in the electrolyte

129 Two resistors of 6 Ω and 9Ω are connected in series to a 120 volt source. The power consumed by the 6 Ω
6. resistor is

a) 384 W b) 576 W c) 1500 W d) 1200 W

129 The V - i graph for a conductor at temperatures T1 and T2 are as shown in the figure. (T2 - T1) is
7. proportional to
T2
V

T1


i

a) cos 2θ b) sin θ c) cot 2θ d) tan θ

129 For a thermocouple, the inversion temperature is 600℃ and the neutral temperature is 320℃. Find the
8. temperature of the cold junction?

a) 40℃ b) 20℃ c) 80℃ d) 60℃

129 A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. Now
9. W ,W and W are the output powers of the bulbs B ,B and B respectively, then
1 2 3 1 2 3

a) W > W = W b) W > W > W c) W < W = W d) W < W < W

1 2 3 1 2 3 1 2 3 1 2 3

130 A resistor is constructed as hollow cylinder of dimensions ra = 0.5 cm and rb = 1.0cm and
0. ρ = 3.5×10-5Ωm. The resistance of the configuration for the length of 5 cm cylinder is ….×10-3Ω.

a) 7.42 b) 10.56 c) 14.38 d) 16.48

130 The V - i graph for a conductor makes an angle θ with V-axis. Here V denotes the voltage and i denotes

P a g e | 168
1. current. The resistance of conductor is given by

a) sin θ b) cos θ c) tan θ d) cot θ

130 If a wire is stretched to make it 0.1% longer, its resistance will

2.

a) Increase by 0.2% b) Decrease by 0.2% c) Decrease 0.05% d) Increase by 0.05%

130 Two filaments of same length are connected first in series and then in parallel. For the same amount of
3. main current flowing the ratio of the heat produced is

a) 2 : 1 b) 1 : 2 c) 4 : 1 d) 1 : 4

130 Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is
4. decreased to 2/3 of its initial value, then the same amount of water will boil with the same supply voltage
in
a) 15 minutes b) 12 minutes c) 10 minutes d) 8 minutes

130 Antimony and bismuth are usually used in a thermocouple, because

5.

a) Negative thermal e.m.f. is produced b) Constant thermal e.m.f. is produced

c) Lower thermal e.m.f. is produced d) Higher thermal e.m.f. is produced

130 Specific resistance of copper, constantan and silver are 1.78×10-8, 39.1×10-8 and 10-8 Ω-m respectively.
6. Which of these is the best conductor of heat and electricity?

a) Copper b) Constantan c) Silver d) All of them

130 Two bars of radius r and 2r are kept in contact as shown. An electric current I is passed through the bars.
7. Which one of following is correct?
I/2

I /2
2r

A
B

a) Heat produced in bar BC is 4 times the heat b) Electric field in both halves is equal
produced in bar AB
c) Current density across AB is double that of across d) Potential difference across AB is 4 times that of
BC across BC
130 In producing chlorine through electrolysis 100 watt power at 125 V is being consumed. How much
8. chlorine per minute is liberated? E.C.E. chlorine is 0.367×10-6kg/coulomb

a) 24.3 mg b) 16.6 mg c) 17.6 mg d) 21.3 mg

130 The amount of heat produced in a resistor when a current is passed through it can be found using
9.

a) Faraday’s Law b) Kirchhoff’s Law c) Laplace’s Law d) Joule’s Law

P a g e | 169
131 A beam contains 2×108 doubly charged positive ions per cubic centimeter, all of which are moving with a
0. speed of 105 m/s. The current density is

a) 6.4 A/m2 b) 3.2 A/m2 c) 1.6 A/m2 d) None of these

131 Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt. If the same
1. resistors are connected in parallel across the same e.m.f., then the power dissipated will be

a) 10 watt b) 30 watt c) 10/3 watt d) 90 watt

131 When a current passes through a wire whose different parts are maintained at different temperatures,
2. evolution or absorption of heat all along the length of wire is known as

a) Joule effect b) Seebeck effect c) Peltier effect d) Thomson effect

131 To get the maximum current from a parallel combination of n identical cells each of internal resistance r
3. and external resistance R, when

a) R ≫ r b) R ≪ r c) R = r d) None of these

131 Which of the following is vector quantity

4.

a) Current density b) Current c) Wattless current d) Power

131 The resistance Rt of a conductor varies with temperature t as shown in the figure. If the variation is
5. represented by R = R [1+αt+βt2], then
t 0
Rt

a) α and β are both negative b) α and β are both positive

c) α is positive and β is negative d) α is negative and β are positive

131 In the circuit shown in the figure reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is
6. V when only S is closed and reading of voltmeter is V when both S and S are closed. Then
2 2 3 1 2
3R
R S1
6R
S2
V

a) V > V > V b) V > V > V c) V > V > V d) V > V > V

3 2 1 2 1 3 3 1 2 1 2 3

131 Two batteries of e.m.f. 4 V and 8 V with internal resistances 1 Ω and 2 Ω are connected in a circuit with a
7. resistance of 9 Ω as shown in figure. The current and potential difference between the points P and Q are

P a g e | 170
 1 4V 8V 2
P Q
r1 r2

9

a) 1 A and 3V b) 1 A and 4V c) 1 A and 9V d) 1 A and 12V

3 6 9 2
131 Assume that each atom of copper contributes one free electron. What is the average drift velocity of
8. conduction electrons in a copper wire of cross-sectional area 10-7 m2, carrying a current of 1.5 A? (Given
-3 -3 23
density of copper =9×10 kgm ; atomic mass of copper =63.5; Avogadro’s number= 6.023×10 per
gram atom)
a) 1.1×10-2ms-1 b) 1.1×10-3ms-1 c) 2.2×10-2ms-1 d) 2.2×10-3ms-1

131 The internal resistance of a primary cell is 4 Ω . It generates a current of 0.2 A in an external resistance of
9. 21 Ω. The rate at which chemical energy is consumed in providing the current is

a) 0.42 J s-1 b) 0.84 J s-1 c) 1 J s-1 d) 5 J s-1

132 The current in the primary circuit of a potentiometer is 0.2A. the specific resistance and cross-section of
0. the potentiometer wire are 4×10-7Ωm and 8×10-7m2 respectively. Potential gradient will be equal to

a) 0.2 V/m b) 1V/m c) 0.3V/m d) 0.1V/m

132 A cell of e.m.f. E connected with an external resistance R, then p.d. across cell is V. The internal resistance
1. of cell will be

a) (E-V)R b) (E-V)R c) (V-E)R d) (V-E)R

E V V E
++
132 The current flowing in a copper voltmeter is 1.6 A. The number of Cu ions deposited at the cathode per
2. minute are

a) 1.5×1020 b) 3×1020 c) 6×1020 d) 1×1019

132 There are three resistance coils of equal resistance. The maximum number of resistances you can obtain
3. by connecting them in any manner you choose, being free to use any number of the coils in any way is

a) 3 b) 4 c) 6 d) 5

132 A galvanometer of resistance 22.8 Ω measures 1A. How much shunt should be used, so that it can be used
4. to measure 20A?

a) 1Ω b) 2Ω c) 1.2Ω d) 2.2Ω

132 If two electric bulbs have 40 W and 60 W rating at 220 V, then the ratio of their resistances will be
5.

a) 9 :4 b) 4 :3 c) 3 :8 d) 3 :2

132 The negative Zn pole of Daniell cell, sending a constant current through a circuit, decreases in mass by 0.13
6. g in 30 min. If the electrochemical equivalent of Zn and Cu are 32.5 and 31.5 respectively, the increase in
the mass of the positive Cu pole in this time is
a) 0.180 g b) 0.141 g c) 0.126 g d) 0.242 g

132 A lead-acid battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.5 Ω, the
P a g e | 171
7. maximum current that can be drawn from the battery will be

a) 30 A b) 20 A c) 6 A d) 24 A

132 When an electrical appliance is switched on, it responds almost immediately, because
8.

a) The electrons in the connecting wires move with the speed of light

b) The electrical signal is carried by electromagnetic waves moving with the speed of light

c) The electrons move with speed which is close to but less than speed of light

d) The electron are stagnant

132 A wire has resistance of 24 Ω is bent in the following shape. The effective resistance between A and B is
9.

a) 24 Ω b) 10 Ω c) 16 Ω d) None of these
3

P a g e | 172
Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

: ANSWER KEY :

1) d 2) b 3) a 4) a 85) a 86) c 87) c 88) a

5) a 6) c 7) b 8) b 89) b 90) a 91) c 92) b

9) b 10) c 11) b 12) c 93) a 94) b 95) b 96) a

13) d 14) b 15) a 16) c 97) a 98) b 99) a 100) b

17) b 18) d 19) a 20) d 101) a 102) d 103) a 104) d

21) a 22) a 23) a 24) c 105) d 106) b 107) c 108) b

25) b 26) a 27) d 28) d 109) b 110) a 111) d 112) c

29) b 30) d 31) a 32) a 113) c 114) b 115) d 116) c

33) a 34) c 35) b 36) a 117) c 118) b 119) b 120) b

37) a 38) d 39) b 40) a 121) a 122) d 123) b 124) b

41) c 42) b 43) d 44) a 125) b 126) a 127) a 128) b

45) c 46) d 47) d 48) a 129) b 130) c 131) c 132) c

49) d 50) c 51) d 52) c 133) b 134) d 135) d 136) d

53) b 54) c 55) b 56) c 137) a 138) b 139) b 140) c

57) d 58) a 59) a 60) b 141) b 142) a 143) a 144) b

61) a 62) d 63) d 64) d 145) a 146) d 147) b 148) a

65) c 66) a 67) c 68) b 149) c 150) c 151) c 152) b

69) b 70) d 71) d 72) d 153) a 154) c 155) c 156) c

73) d 74) c 75) c 76) d 157) c 158) b 159) c 160) b

77) b 78) d 79) a 80) a 161) c 162) c 163) b 164) c

81) a 82) d 83) a 84) c 165) d 166) a 167) b 168) a

P a g e | 173
169) c 170) c 171) b 172) d 289) c 290) d 291) c 292) b

173) d 174) b 175) c 176) a 293) d 294) d 295) b 296) c

177) c 178) c 179) d 180) b 297) b 298) c 299) c 300) a

181) d 182) b 183) d 184) c 301) d 302) a 303) b 304) d

185) a 186) c 187) b 188) b 305) b 306) c 307) b 308) b

189) d 190) c 191) c 192) a 309) a 310) a 311) a 312) c

193) b 194) d 195) d 196) c 313) a 314) a 315) c 316) a

197) b 198) b 199) c 200) d 317) b 318) b 319) b 320) b

201) c 202) a 203) b 204) a 321) a 322) c 323) d 324) a

205) a 206) c 207) a 208) b 325) a 326) a 327) d 328) d

209) b 210) a 211) b 212) b 329) a 330) c 331) a 332) a

213) c 214) a 215) d 216) a 333) b 334) b 335) a 336) c

217) a 218) b 219) a 220) b 337) a 338) d 339) c 340) c

221) b 222) d 223) b 224) d 341) d 342) c 343) b 344) a

225) c 226) d 227) c 228) a 345) a 346) c 347) a 348) b

229) c 230) b 231) b 232) c 349) d 350) c 351) d 352) b

233) a 234) b 235) c 236) b 353) b 354) c 355) d 356) a

237) d 238) b 239) d 240) a 357) b 358) a 359) b 360) c

241) a 242) c 243) b 244) c 361) b 362) c 363) b 364) a

245) b 246) b 247) b 248) d 365) d 366) b 367) c 368) d

249) b 250) d 251) d 252) d 369) c 370) c 371) c 372) a

253) d 254) d 255) d 256) c 373) a 374) b 375) c 376) c

257) c 258) c 259) d 260) d 377) a 378) a 379) b 380) a

261) c 262) a 263) a 264) d 381) a 382) b 383) b 384) a

265) c 266) a 267) d 268) a 385) c 386) d 387) c 388) c

269) a 270) d 271) c 272) a 389) c 390) a 391) d 392) b

273) d 274) a 275) b 276) c 393) c 394) b 395) d 396) d

277) a 278) a 279) a 280) b 397) b 398) a 399) c 400) c

281) c 282) b 283) c 284) d 401) c 402) a 403) c 404) d

285) a 286) d 287) c 288) c 405) b 406) d 407) c 408) b

P a g e | 174
409) c 410) a 411) a 412) a 529) b 530) b 531) b 532) c

413) c 414) c 415) d 416) c 533) c 534) c 535) b 536) a

417) c 418) a 419) c 420) a 537) c 538) d 539) d 540) b

421) a 422) c 423) c 424) d 541) b 542) d 543) b 544) c

425) a 426) c 427) c 428) d 545) c 546) d 547) c 548) b

429) a 430) b 431) c 432) a 549) c 550) d 551) c 552) d

433) d 434) b 435) d 436) b 553) a 554) c 555) d 556) d

437) c 438) d 439) a 440) a 557) d 558) a 559) c 560) c

441) b 442) d 443) d 444) b 561) d 562) b 563) d 564) a

445) b 446) c 447) a 448) d 565) d 566) a 567) c 568) b

449) d 450) b 451) c 452) c 569) b 570) c 571) c 572) b

453) d 454) c 455) c 456) c 573) c 574) b 575) b 576) c

457) a 458) a 459) c 460) c 577) d 578) a 579) d 580) a

461) c 462) d 463) b 464) b 581) c 582) a 583) d 584) c

465) b 466) a 467) b 468) d 585) a 586) c 587) b 588) c

469) a 470) d 471) a 472) c 589) d 590) d 591) a 592) a

473) b 474) c 475) c 476) c 593) b 594) b 595) c 596) c

477) c 478) a 479) a 480) b 597) a 598) b 599) d 600) c

481) d 482) d 483) c 484) d 601) b 602) a 603) a 604) d

485) b 486) c 487) a 488) d 605) d 606) a 607) c 608) d

489) b 490) a 491) c 492) c 609) b 610) a 611) c 612) c

493) c 494) a 495) c 496) a 613) d 614) c 615) d 616) a

497) b 498) c 499) b 500) b 617) d 618) a 619) a 620) a

501) b 502) c 503) c 504) c 621) c 622) b 623) b 624) a

505) c 506) a 507) b 508) a 625) d 626) c 627) b 628) a

509) b 510) d 511) c 512) a 629) b 630) d 631) a 632) b

513) d 514) a 515) a 516) c 633) a 634) c 635) d 636) c

517) d 518) a 519) b 520) c 637) c 638) b 639) a 640) d

521) a 522) c 523) d 524) c 641) a 642) c 643) b 644) d

525) d 526) d 527) c 528) a 645) d 646) b 647) a 648) d

P a g e | 175
649) b 650) b 651) c 652) c 769) c 770) d 771) d 772) c

653) b 654) b 655) d 656) b 773) b 774) b 775) c 776) c

657) b 658) c 659) a 660) b 777) c 778) a 779) a 780) d

661) c 662) d 663) a 664) a 781) b 782) b 783) d 784) d

665) a 666) c 667) b 668) b 785) c 786) b 787) d 788) a

669) d 670) b 671) d 672) c 789) a 790) a 791) b 792) d

673) c 674) d 675) a 676) d 793) d 794) a 795) c 796) a

677) a 678) b 679) b 680) b 797) b 798) c 799) d 800) d

681) c 682) b 683) b 684) b 801) d 802) d 803) a 804) b

685) b 686) c 687) b 688) b 805) c 806) c 807) d 808) d

689) c 690) c 691) c 692) c 809) a 810) d 811) c 812) d

693) a 694) a 695) d 696) c 813) c 814) b 815) d 816) a

697) c 698) d 699) b 700) c 817) c 818) a 819) a 820) a

701) a 702) c 703) d 704) c 821) c 822) a 823) c 824) a

705) c 706) d 707) b 708) d 825) b 826) b 827) c 828) b

709) b 710) c 711) c 712) c 829) d 830) b 831) c 832) a

713) b 714) d 715) c 716) b 833) b 834) a 835) b 836) a

717) a 718) a 719) a 720) a 837) d 838) b 839) c 840) a

721) b 722) d 723) a 724) a 841) a 842) b 843) d 844) a

725) a 726) c 727) c 728) c 845) c 846) b 847) b 848) d

729) a 730) a 731) a 732) b 849) d 850) a 851) c 852) b

733) c 734) c 735) d 736) a 853) c 854) d 855) a 856) a

737) a 738) c 739) d 740) c 857) b 858) d 859) d 860) d

741) a 742) b 743) b 744) c 861) a 862) c 863) a 864) a

745) a 746) c 747) c 748) b 865) b 866) b 867) d 868) a

749) c 750) d 751) d 752) b 869) b 870) d 871) d 872) a

753) b 754) b 755) a 756) b 873) d 874) c 875) a 876) a

757) d 758) c 759) d 760) d 877) c 878) a 879) a 880) b

761) b 762) b 763) a 764) a 881) a 882) b 883) d 884) c

765) a 766) d 767) b 768) d 885) c 886) c 887) c 888) b

P a g e | 176
889) c 890) c 891) d 892) d 1009) a 1010) c 1011) d 1012) c

893) b 894) b 895) d 896) b 1013) b 1014) b 1015) b 1016) a

897) d 898) a 899) b 900) d 1017) b 1018) d 1019) c 1020) d

901) a 902) c 903) d 904) a 1021) b 1022) a 1023) b 1024) b

905) c 906) d 907) b 908) d 1025) a 1026) c 1027) c 1028) c

909) a 910) d 911) c 912) d 1029) c 1030) a 1031) a 1032) d

913) b 914) a 915) c 916) a 1033) b 1034) b 1035) c 1036) a

917) b 918) c 919) d 920) c 1037) a 1038) d 1039) b 1040) a

921) c 922) b 923) a 924) b 1041) c 1042) c 1043) c 1044) a

925) d 926) d 927) a 928) d 1045) c 1046) c 1047) c 1048) c

929) b 930) b 931) a 932) d 1049) b 1050) d 1051) b 1052) a

933) c 934) a 935) b 936) a 1053) a 1054) c 1055) d 1056) b

937) c 938) d 939) c 940) a 1057) c 1058) a 1059) c 1060) a

941) d 942) c 943) b 944) d 1061) c 1062) d 1063) d 1064) b

945) a 946) a 947) a 948) b 1065) a 1066) d 1067) b 1068) c

949) a 950) a 951) a 952) d 1069) b 1070) b 1071) a 1072) d

953) a 954) a 955) b 956) d 1073) b 1074) d 1075) c 1076) a

957) b 958) a 959) c 960) d 1077) d 1078) d 1079) d 1080) a

961) c 962) b 963) b 964) b 1081) a 1082) b 1083) d 1084) c

965) c 966) a 967) a 968) b 1085) c 1086) b 1087) b 1088) d

969) a 970) b 971) b 972) b 1089) d 1090) a 1091) b 1092) a

973) a 974) c 975) d 976) c 1093) b 1094) d 1095) a 1096) a

977) b 978) b 979) a 980) b 1097) b 1098) c 1099) d 1100) a

981) c 982) a 983) b 984) b 1101) a 1102) c 1103) b 1104) a

985) c 986) c 987) b 988) b 1105) a 1106) b 1107) a 1108) a

989) b 990) b 991) a 992) c 1109) b 1110) b 1111) c 1112) b

993) c 994) c 995) b 996) c 1113) b 1114) a 1115) b 1116) d

997) a 998) c 999) a 1000) d 1117) d 1118) a 1119) c 1120) c

1001) c 1002) c 1003) a 1004) d 1121) a 1122) c 1123) c 1124) a

1005) c 1006) d 1007) a 1008) a 1125) a 1126) c 1127) c 1128) c

P a g e | 177
1129) d 1130) b 1131) c 1132) d 1249) c 1250) c 1251) b 1252) d

1133) d 1134) c 1135) b 1136) a 1253) c 1254) c 1255) b 1256) c

1137) a 1138) c 1139) c 1140) a 1257) a 1258) d 1259) a 1260) d

1141) d 1142) c 1143) b 1144) d 1261) d 1262) b 1263) b 1264) b

1145) b 1146) b 1147) a 1148) a 1265) b 1266) d 1267) c 1268) b

1149) d 1150) d 1151) a 1152) b 1269) b 1270) c 1271) c 1272) d

1153) c 1154) b 1155) d 1156) b 1273) c 1274) c 1275) c 1276) a

1157) b 1158) b 1159) a 1160) a 1277) b 1278) a 1279) d 1280) c

1161) a 1162) b 1163) b 1164) b 1281) d 1282) d 1283) c 1284) c

1165) a 1166) b 1167) a 1168) a 1285) a 1286) d 1287) c 1288) c

1169) a 1170) a 1171) a 1172) a 1289) d 1290) a 1291) c 1292) c

1173) d 1174) d 1175) b 1176) c 1293) b 1294) a 1295) b 1296) a

1177) b 1178) b 1179) a 1180) c 1297) c 1298) a 1299) d 1300) a

1181) b 1182) b 1183) d 1184) a 1301) d 1302) a 1303) c 1304) c

1185) a 1186) c 1187) c 1188) b 1305) d 1306) c 1307) a 1308) c

1189) d 1190) d 1191) b 1192) a 1309) d 1310) a 1311) d 1312) d

1193) c 1194) a 1195) a 1196) c 1313) b 1314) a 1315) b 1316) b

1197) d 1198) c 1199) b 1200) c 1317) a 1318) b 1319) c 1320) d

1201) b 1202) b 1203) c 1204) c 1321) b 1322) b 1323) b 1324) b

1205) b 1206) a 1207) a 1208) c 1325) d 1326) c 1327) d 1328) b

1209) c 1210) b 1211) b 1212) b 1329) b

1213) d 1214) b 1215) d 1216) d

1217) a 1218) b 1219) b 1220) d

1221) c 1222) b 1223) d 1224) c

1225) c 1226) b 1227) b 1228) d

1229) d 1230) d 1231) b 1232) d

1233) d 1234) c 1235) a 1236) c

1237) a 1238) b 1239) b 1240) b

1241) c 1242) d 1243) a 1244) d

1245) b 1246) b 1247) b 1248) a

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 ACTIVE SITE TUTORIALS

Date : 28-07-2019 TEST ID: 417

Time : 22:09:00 PHYSICS

Marks : 5316

3.CURRENT ELECTRICITY

: HINTS AND SOLUTIONS :

1 (d) 30 l
=
P+Q (100-l )
net emf
Current =
net resistance 30 37.5
=
P+Q (100-37.5)
2+2+2 6
or I = = = 1.2A
1+1+1+2 5 30 37.5
=
P+Q 62.5
2 (b)
30×62.5
The bridge will be balanced when the shunted P+Q=
37.5
3×S
resistance of value 2Ω ie, 2 = . On solving
3+S P + Q = 50 …(i)
S = 6Ω
IInd case
4 (a)
30 l
=
P R R PQ (100-l )
Using Wheatstone principle = =
Q S 100-l P+Q

35 35 7 30(P+Q) 71.4
= = = =
100-35 65 13 PQ (100-71.4)

5 (a) 30×50 71.4

=
PQ 28.6
If resistances of bulbs are R1 and R2 respectively
then in parallel 30×50×28.6
PQ =
71.4
1 1 1 1 1 1

() () ()
= + ⇒ 2 = 2 + 2 P ≈ 600 ….(ii)
RP R1 R2 V V V
Pp P1 P2
So, from Eqs. (i) and (ii)
⇒PP = P1 + P2 P = 30Ω and Q = 20Ω

6 (c) 7 (b)

Ist case

P a g e | 179
 r= ( )
l1-l2
l2
'
( )
l -2
×R = 1 ×5
2
…(i)
For given information

ρB = 2ρA

and r = ( ) l1-3
3
×10 …(ii) rB = 2rA

on solving (i) and (ii), r = 10 Ω And RA = RB

8 (b) ρAlA ρBlB

∴ 2 = 2
πrA πrB
I×t
No. of ions liberated (n) =
e×Valency(p) ρAlA 2ρA×lB
⟹ 2 =
π(2rA)
2
πrA
1
ie, n ∝
P lB 2
⟹ = = 2:1
nAg P lA 1
3
∴ = Al =
nAl PAg 1
14 (b)
9 (b) 2
V P1 R 6 4 2
P= ⇒ = 2 ⇒ = = ⇒P2 = 9 W
2 R P2 R 1 P2 6 3
Vt
Heat produced = = mL
4.2 R
15 (a)
2
V (210)2×1 An ideal cell has zero resistance
or m = = = 2.62 g
4.2 R L 4.2×50×80
16 (c)
10 (c)
i = q/t = ne/t
iG 1×0.018 0.018
S= g = = = 0.002Ω
(i-ig) 10-1 9 2×1 19
or n = it e = -19 = 1.25×10
1.6×10
11 (b)
17 (b)
-4 -2
Order of drift velocity = 10 m/sec = 10 cm/sec
1
In parallel Pconsumed ∝ Brightness ∝
12 (c) R

Ammeter is always connected in series with PA > PB [Given] ∴ RA < RB

circuit
18 (d)
13 (d)
5
P 10
Let (ρA,lA,rA,AA) and (ρB,lB,rB,AB)are specific I= = = 500 A
V 200
resistances, lengths, radii and areas of wires A and
-3
B respectively. and W = zlt = 0.367×10 ×500×1 = 0.1835 g

Resistance of 19 (a)

ρAlA ρl Equivalent resistance of the combination

A = RA = = A A2
AA πRA
(2+2)×2 8 4
= = = Ω
2+2+2 6 3
Resistance of

ρBlB ρl
B = RB = = B 2B
AB πrB
P a g e | 180
 2 2

P Q
2

20 (d)
6 6
Resistance of a conductor varies linearly with Current i = = A
6+4+1 11
temperature as 6 60
P.D. between A and B, V = ×10 = V
11 11
Rt = R0(1+αt)
25 (b)
For the first conductor
1 division = 1μA
Tt = R0(1+α1t1)
1

40μV
Current for 1℃ = = 4μA
Rt -R 10
or α1 = 1 0 ….(i)
t
1
1μA = ℃ = 0.25℃
Similarly, for second conductor 4

Rt2-R0 26 (a)
α2 = ……..(ii)
t2
Two resistances of each side of triangle are
From Eq. (i) and Eq. (ii), we get connected in parallel. Therefore, the effective
resistance of each arm of the triangle would be
α1 t r×r r
= 2 = = . The two arms AB
α2 t1 r+r 2

21 (a) and AC are in series and they together are in

parallel with third one.
Ammeter is always connected in series and
'
voltmeter in parallel. ∴ R (r/2) + (r/2) = r

22 (a) Total resistance

G 25 25 25 1 1 2 3
S= = = 5 = 5 = + =
i 5 10 -1 10 R r r r
-1 -6 -1
ig 50×10
-4 R = r/3
= 2.5×10 Ω
27 (d)
23 (a)
I = neAvd
V iR iρL iρ
Potential gradient = = = =
L L AL A
1
or vd =
0.2×40×10
-8
-2
neA
= -6 = 10 V/m
8×10 I
or vd ∝
A
24 (c)

The given circuit can be redrawn as follows

P a g e | 181
 ' '
vd I /A' 2I/2A dE
∴ = = =1 = a + bt
vd I/A I/A dt
'
or v d = vd = v When t = tn, ie, neural temperature, then

28 (d) dE
=0
dt
Let the resistance of the wire be R, then we know
that resistance is proportional to the length of the a
∴ 0 = a + btn or tn = -
wire. So each of the four wires will have R/4 b
resistance and they are connected in parallel. So
The temperature of inversion
the effective resistance will be
ti = 2tn = t0
1
R1
=
4
R ()
4⇒R1 =
R
16
= 2tn - 0 = -
2a
b
29 (b)
Thermoelectric power
By Faraday’s law, m ∝ it
dE
m it m 4×120 m P= = a + bt
∴ 1 = 1 1⇒ = ⇒m2 = dt
m2 i2t2 m2 6×40 2
34 (c)
30 (d)
Since, charge (q)=current (i) × times (t)
1 coulomb ×1volt = 1 joule
Therefore, charge is equal to area under the curve.
Hence, option (d) is incorrect.
∴ Ist rectangle =q=lb=2
31 (a)
IInd rectangle =q=lb=2
-3
i G i.G G 100×10 ×40
=1+ ⇒ =1+ ⇒
ig S Vg S 800×10
-3
1
IIIrd triangle = q = lb = 2
40 2
=1+
S
Hence, ratio is 1:1:1.
⇒S = 10Ω
35 (b)
32 (a)
The internal resistance of battery is given by

( ) ( )
This is a balanced Wheatstone bridge. Therefore
E 40 9×10
no current will flow from the diagonal resistance r= -1 R = -1 ×9 = = 3Ω
V 30 30
10Ω
36 (a)
∴ Equivalent resistance
(10+10)×(10+10) 1
= = 10Ω Conductivity σ = …(i)
(10+10)+(10+10) ρ

33 (a) 1
and conductance G =
R
1 2
E = at + bt …(i)
2 ⇒GR = 1 ….(ii)

Differentiating Eq. (i), w.r.t., t GR

From equation (i) and (ii) σ =
ρ
We have

P a g e | 182
38 (d) Ammeter is always connected in series and
Voltmeter is always connected in parallel
Let the current in 12Ω resistance is i
45 (c)
Applying loop theorem in closed mesh AEFCA
-3
m 5×10
12i = -E + E = 0 It = = -7
z 3.387×10
∴i = 0 4
5×10
= Ah = 4.1 Ah
39 (b) 3.387×60×60

() ()
2 2 46 (d)
2 P V V
P∝V⇒ = ⇒P = P
P0 V0 V0 0 R1 (1+αt1) 5 (1+α×50)
= ⇒ = ⇒α
R2 (1+αt2) 6 (1+α×100)
40 (a)
1
= per ℃
P=
2
V PP R
⇒ = S =
(R1+R2) (R +R )
= 1 2
2
200
R PS RP R1R2/(R1+R2) R1R2
Again by Rt = R0(1 + αt)
(R +R ) R 1
2

( )
100
⇒ = 1 2 ⇒ 1 = 1
25 R1R2 R2 1 ⇒5 = R0 1+ ×50 ⇒R0 = 4Ω
200
41 (c) 47 (d)
For semiconductors, resistance decreases on Given, the resistance of wire R=12Ω. The wire is
increasing the temperature bent in square form
42 (b) B
3
Given circuit is equivalent to
3 3
3

R1 = 3 + 3 = 6Ω

R2 = 3 + 3 = 6Ω
So the equivalent resistance between points A and
R1 = 6 Ω
B is equal to

6×3 A B
R= = 2Ω
6+3
R2 = 6 Ω
43 (d)

watt×hour 1 1 1
Energy consumed in kWh = ' = +
1000 R 6 6

10×50×10 1 2
⇒ For 30 days, P = ×30 = 150kWh or ' =
1000 R 6
'
44 (a) or R = 3Ω

P a g e | 183
48 (a) Let R be the resistance of each lamp and V be the
voltage supplied to the circuit. Current in the
197.1 circuit is
Chemical equivalent of gold = = 65.7
3
V 2V
200×5 I1 = =
Gold to be deposited = =10g R×R 3R
100 R+
R+R
Electrochemical equivalent of gold Current flowing through B or C,

( )
W2 65.7 -4 -1 I1
z2 = z z = ×0.1044×10 gC 1 2V V
W1 1 2 1.008 I2 = = =
2 2 3R 3R
m When C is fused, the whole current flows through
Also m = zlt, t =
zl A and B.
10 '
Then , I2 = V/2R
⇒ =
( 65.7
1.008
-4
×0.1044×10 ×2 ) So current through A decreases and current
through B increases. Therefore brilliance of A
= 7347.9s
decreases and that of B increase.
49 (d)
56 (c)
2
I ×6 = 60 or I = 10 A 2
V
As for an electric appliance R = .
Current through upper branch = 2 10A. Heat P
produced per second 3Ω =
For first bulb, its resistance
2
(2 10) ×3 cal = 120 cal. 2
V 250×250
R2 = = = 625 Ω
50 (c) P1 100

R R
For second bulb, its resistance

R 2R / 2R / 2
R R R V 200×200
R2 = 2 =
A R B  A R B P2 100
R R R
R 2R / 2R /
= 400 Ω
R R
Now, in series potential divides in proportion to
resistance.
2R
Hence Req = [Since it’s a balanced Wheatstone
3 R2
So, V2 = V
bridge] (R1+ R2)
51 (d) Where V is supply voltage.
Because cell is in open circuit ∴ Potential drop across bulb B2.
54 (c) 400
V2 = ×250
(625+400)
I 20
vd = = 29 -6 -19
nAe 10 ×10 ×1.6×10 = 97.56 V
-3
= 1.25×10 m/s
= 98 V
55 (b)

P a g e | 184
57 (d) 60 (b)
2
27 V
Equivalent weight of aluminium = =9 Power, P =
3 R
2
So 1 faraday = 96500 C are required to liberate 9 V (60)2
R= = = 22.5Ω
gm of Al P 160

Now, according to Ohm’s law

V=IR
1A 4 A 16
60
∴ I=
22.5
V
2 C
⟹I = 2.6A
1A 16 B 4

58 (a) Here, t = 60s

In the following circuit potential difference ne

As I=
t
between
I×t
C and A is VC - VA = 1×4 = 4 …(i) ⟹ n=
e

26×60 21
= -19 ≈ 10
1.6×10

61 (a)
-6
e -7 (30×10 )×θ
i= ⇒3×10 = ⇒θ = 0.5°
R 50

62 (d)

Since E1(10V) > E2(4V)

C and B is VC - VB = 1×16 = 16 …(ii)
So current in the circuit will be clockwise
On solving equations (i) and (ii) we get
1 E1 e E2 2
a b
VA - VB = 12V
10V 4V
i
59 (a) 3

9
Equivalent resistance R = = 1Ω
9
Applying Kirchhoff’s voltage law
1A 1A 1A 1A 1A 1A 1A 1A 1A
+ -1×i + 10 - 4 - 2×i - 3i = 0⇒i = 1A(a to b via e)
9V 9 9 9 9 9 9 9 9 9
–
V 10-4
∴ Current = = = 1.0 ampere
A R 6

63 (d)
9
Current i = = 9A
1 -6 1 -6 2
Given, V = 10×10 t - ×10 t
40
Current passing through the ammeter = 5A

P a g e | 185
 At neutral temperature 1 5
R = ×2 = 5Ω
2
dV
=0
dt 25 25 5
∴x= = =
(10+5)L 15.L 3L
-6 1 -6
∴ 10×10 - ×10 tn = 0
20 E = x.l2

or tn = 200℃ 5 3L
⇒1 = .l2⇒l2 = = 0.6L
3L 2
Also at neutral temperature, thermo-emf is
maximum. 66 (a)

Thus, Q
Voltage sensitivity =
V
-6 1 -6
Vmax = 10×10 (200) - ×10 (200)2
40 Q
Current sensitivity =
I
-3 -3
= 2×10 - 1×10 = 1 mv
Also, potential difference
64 (d)
V = IG
For conversion of galvanometer (of resistance)
into voltmeter, a resistance R is connected in VS Q/V I I
Hence, = = =
series Is Q/I V IG

V1 V2 Vs 1
∴ ig = and ig = ∴ =
R+G 2R+G Is G

V1 V2 V 2R+G 2(R+G)-G 67 (c)

⇒ = ⇒ 2 = =
R+G 2R+G V1 R+G (R+G)
In steady state the branch containing capacitors,
G VG can be neglected. So reduced circuit is as follows
= 2-
(R+G) ⇒V2 = 2V1 - (R+G) ⇒V2 < 2V1
1

65 (c)

E 2.5 2.5×10 25
I= = and V = I.R = =
RT 10+R 10+R 10+R

V 25
x= = 2
L (10+R)L V (2)2
Power P = = = 1W
R 4
E = x.l1
68 (b)
25 L
⇒1 = × ⇒25 = 20 + 2R
(10+R)L 2 According to Kirchhoff’s first law

5 (5A)+(4A)+(-3A)+(-5A)+I=0
⇒2R = 5⇒R =
2
Or I=-1A
∴ Now the resistance is doubled

P a g e | 186
 20V, 1.5Ω
4A

i 3Ω 2Ω i
X 3A
i/2 P
5A

l
X 2Ω 3Ω
5A
i/2 Q

69 (b) Potential difference between X and P,

When the heating coil is cut into two equal parts

and these parts are joined in parallel, the
VX - VP = ()
5
2
×3 = 7.5V ….(i)

resistance of coil is reduced to one fourth, so 5

power consumed will become 4 times i.e. 400 Js
-1 VX - VQ = ×2 = 5V ….(ii)
2
70 (d) On solving (i) and (ii) VP - VQ = -2.5 volt;VQ > VP
The emf of the standard cell must be greater than i
that of experimental cells, otherwise balance Short Trick : (VP-VQ) = (R -R ) = 52 (2-3) = -2.5
2 2 1
point is not obtained
⇒VQ > VP
71 (d)
74 (c)
1 2
E = αt + βt , graph between E and t will be a
2 Initially the inductance will oppose the current
parabola, such that first emf increases and then which tries to flow through the inductance. But
decreases 10Ω and 20Ω can conduct. The current will be

72 (d) 2V 1
= A
30Ω 15
Potential difference between A and B
75 (c)
VA - VB = 1×1.5
Given circuit can be reduced to a simple circuit as
⇒VA - 0 = 1.5V⇒VA = 1.5V shown in figures below

Potential difference between B and C

VB - VC = 1×2.5 = 2.5V

⇒0 - VC = 2.5V⇒VC = -2.5V

Potential difference between C and D

VC - VD = -2V⇒ - 2.5 - VD = -2⇒VD = -0.5V

73 (d)

5
Req = Ω
2 76 (d)

20 Let polarity of m cells in a 12 cells battery is

i= = 5A
5 reversed, then equivalent emf of the battery
+1.5
2 = (12 - 2m)E

Now the circuit can be drawn as

P a g e | 187
 (12-2m)E developed is
A 2
Vt
H1 = …(i)
2R

When two heater wires are connected in parallel,

the heat developed is
2 E 2 2
Vt 2V t
H2 = = …(ii)
R/2 R
R
Dividing Eq. (i) by Eq. (ii), we get

H1 1
= or H1 :H2 = 1 :4
H2 4
When 12-cell battery and 2-cell battery aid each
other, then current through the circuit 80 (a)
(12-2m)E+2E (P+Q)(R+S) 4
i1 = Reff =
R (P+Q+R+S) = 3 R
(14-2m)E 81 (a)
or 3 = ….(i)
R
In the first case, Zit = m
When they oppose each other, the current
through the circuit. i
In the second case, Z× ×4t = m
4
(12-2m)E-2E
i2 =
R 82 (d)

(10-2m)E For conductors, resistance ∝ Temperature and for

or 2 = …(ii)
R 1
semiconductor, resistance ∝
Temperature
Dividing Eq. (i) by Eq. (ii), we have
83 (a)
3 14-2m
=
2 10-2m Since resistance connected in arms CE, ED, CF and
FD will form a balanced Wheatstone bridge,
or 30 - 6m = 28 - 4m
therefore, the resistance of arm EF becomes
or 2m = 2 ineffective. Now resistance of arm CED or
CFD = 2 + 2 = 4Ω. Effective resistance of these
or m=1 4×4
two parallel arm= = 2Ω
4+4
77 (b)
-19 15 -4
Now resistance of arm ACDB = 2 + 2 + 2 = 6Ω,
i = qv = 1.6×10 ×6.6×10 = 10.56×10 A is in parallel with resistance arm AB = 2Ω. Thus,
= 1mA effective resistance between A and B
78 (d) 6×2 3
= = Ω
2 6+2 2
l 1 A d
R = ρ and P ∝ ⇒P ∝ ⇒P ∝ ⇒PA = 2PB
A R l l 84 (c)
79 (a) dQ d
= (5t +3t+1) = 10t + 3
2
i=
dt dt
Let the resistance of each heater wire is R. When
two wires are connected in series, the heat
P a g e | 188
 When t = 5s, i = 10×5 + 3 = 53 A 1 1 1 1 μ+2R+μ (R+r)
∴ = + + = =
Req 4R 2r 4R 4μR 2Rr
85 (a)
2Rr
The current taken by the silver voltameter ⇒ Req =
R+r
n 1 89 (b)
I1 = = -4 = 0.496 A
Zt 11.2×10 ×30×60
dT d 2 3
= (at -bt ) = 2at - 3bt
2
and by copper voltameter dt dt
1.8 dE
I2 = -4 = 1.515 A When t = tn(ie, neutral temperature), =0
6.6×10 ×30×60 dt
The current I = (I1+I2) = 2.011 A 2 2a
∴ 0 = 2atn - 3btn or tn = .
3b
Power P = IV = 2.011×12 = 24.132 J/sec
90 (a)
86 (c)
The circuit may be redrawn as shown in the
Average current
adjacent figure
50+100+50 200
i= = mA 2×2
3 3 Here Eeq = 12V,req = = 1Ω
2+2
m 3m m
z= = -3 = Eeq 12 12
it 200×10 ×30 2 i= = = = 2Ω
R+req 5+1 6
87 (c)

The equivalent current due to motion of electrons

5 + +
is given by 12 V 2Ω 12 V
2Ω
-19
e 1.6×10
I= = -18
t 1.594×10

= 1.0037×10
-1 91 (c)

= 100.37×10 A
-3 i = nAevd

= 100.37mA i
or vd =
nAe
88 (a)
Total number of free electrons in the unit length
In the circuit arrangement PSTQ is a balanced of conductor, N = nA×1.
Wheatstone bridge, hence resistance 2R joined in
arm AB be omitted. Similarly, resistance 2R joined Total linear momentum of all free electrons per
in arm BC may also be omitted. unit length

i i i
= (Nm)vd = nAm× = =
nAe (e/m) s

92 (b)

As m = z l t = z () V
R
t ie, m ∝ Vt

P a g e | 189
 m2 Vt The resistivity of metal increases when it is
∴ = 22
m1 V1t1 converted into an alloy
'
Vt 6×45×2 ∴ ρ >ρ
or m2 = 2 2 ×m1 = =1.5g.
V1t2 12×30
98 (b)
93 (a)
This is because of secondary ionisation which is
Neon bulb is filled with gas, so the resistance is possible in the gas filled in it
infinite; hence no current flows through it.
99 (a)

Using RT = RT [1 + α(T2 - T1)]

2 1

⇒R100 = R50[1 + α(100 - 50)]

(7-5)
⇒7 = 5[1+(α×50)]⇒α = = 0.008/℃
250

-t/RC
100 (b)
Now, Vc = E(1 - e )
Ammeter is made by connecting a low resistance
-t/RC
⇒ 120 = 200(1 - e ) shunt S in parallel with galvanometer G. since G
and S are in parallel, the potential difference
-t/RC 2
⇒e = across them is same.
5

⇒ t = RCln 2.5
i ig
t 5
⇒ R= = 6
i - ig
Cln 2.5 2.303×2×10 log 2.5
6
= 2.7×10 Ω S

94 (b) Ammeter

vd ∝ 1/l. Therefore, drift velocity is halued

95 (b) ig×G = (i-ig)×S

1 1 1 1 3 Given, G = R, i = 4ig
= + + =
RP R R R R
ig i R
S= ×R = g ×R =
4ig-ig 3ig 3

101 (a)

Maximum current flows through bulb (1)

R Therefore, it will lights brightly.
⇒RP = Ω
3
102 (d)
⇒RS = R + R = 2 RΩ
igG G i=ig 10-1 9
S= ⇒ = = =
R 7R (i-ig) S ig 1 1
⇒Rnet = RP + RS = 2R + = Ω
3 3
104 (d)
96 (a)
P a g e | 190
 Potentiometer works on null deflection method. drawn from the battery outside it
In balance condition no current flows in
secondary circuit. 110 (a)

106 (b) Slope of graph

The circuit can be simplified as follows I 1

= =
V R
B C
i 30 If experiment is performed at higher temperature
i i then resistance increase and hence slope
A D
i
40 40V decrease, choice (a) is wrong.
F E
40 80V Similarly in choice (b) and (c) resistance increase.

But for choice (d) resistance R increases, so slope

Applying KCL at junction A decreases

i3 = i1 + i2 …(i) 111 (d)

2
Applying Kirchhoff’s voltage law for the loop Vt
Heat produced, H = . When voltage is halved,
ABCDA R
the heat produced becomes one-fourth. Hence
-30i1 - 40i3 + 40 = 0 time taken to heat the water becomes four time.

⇒ - 30i1 - 40(i1+i2) + 40 = 0 112 (c)

2
⇒7i1 + 4i2 = 4 …(ii) V H R 4 2
H= t⇒ 1 = 2 = =
R H2 R1 2 1
Applying Kirchhoff’s voltage law for the loop
ADEFA 113 (c)

-40i2 - 40i3 + 80 + 40 = 0 Given that

⇒ - 40i2 - 40(i1+i2) = -120 emf EN = 1.5rN

⇒i1 + 2i2 = 3 …(iii) Where rN is the internal resistance of nth cell.

On solving equation (ii) and (iii) i1 = -0.4A Total emf E = E1 + E2 + E3 + … + En

107 (c) = 1.5[r1+r2+r3+…+rn]

From Faraday’s law, m/E = constant Total internal resistance

where m = mass of substance deposited, E = r = r1 + r2 + r3 + … + rn

chemical equivalent
Etotal
m2 E ∴ Current i =
108 rtotal
∴ = 2 ⇒m2 = ×1.6 = 5.4g
m1 E1 32
1.5[r1+r2+r3+…+rn]
i=
108 (b) [r1+r2+r3+…+rn]
Based on Peltier effect Hence, i = 1.5A
109 (b) 114 (b)
The current through the voltameter is same as The given network is a balanced Wheatstone
P a g e | 191
 18 Now, dividing Eq. (iii) by Eq. (ii), we get
bridge. It’s equivalent resistance will be R = Ω
5
E R+r r
= =1+
V V 5V V R R
So current from the battery i = = =
R 18/5 18
E r
-1 =
115 (d) V R

The resistance of 40 W bulb will be more and 60

W bulb will be less
or ( ) E-V
V
R=r

116 (c) Hence, internal resistance

E = aT + bT
2
r= ( ) E-V
V
R
At temperature of inversion, E = 0,
120 (b)
2
∴ aTi + bTi = 0
In the given circuit, resistors 4R and 2R are
a connected in parallel while resistance R is
⇒Ti = -
b connected in series to it.
-6
10×10 Hence, equivalent resistance is
⇒Ti = - -6 = 500℃
(0.02×10 )

117 (c) R 4
R
3 |

10k
i/2
A
i/2
30V 10k
10k

B 1 1 1
' = +
R 4R 2R
10
Equivalent resistance R = 10 + = 15 kΩ 6R
2 = 2
8R
30 -3
Current i = = 2×10 A ' 8
15 R = R
6
Hence, potential difference between A and B
4
= R
( ) 3
-3
2×10 3
V= ×10×10 = 10 Volt
2
'' 4 7R
⟹R = R + R=
118 (b) 3 3

Let the potential difference across battery is V and Given, emf is E volts, therefore
internal resistance of the cell is r, then
E 3E
i= =
E = V + ir ….(i) R 7R

V = iR ………(ii)

Now, from Eqs. (i) and (ii) we have

E = iR + ir = i(R + r) …..(iii)
P a g e | 192
 '2 2
7R/3
Erms = Erms×4 = (2.9)2×4
-1
[ ∵ Erms = 2.9Vm (Given)]
' -1 -1
or Erms = 2.9×2Vm = 5.8Vm

123 (b)

ti + tc
Neutral temperature, tn =
2

ti +10°
⇒ 285° =
2
Potential difference across R is
570° = ti + 10°
3R 3E
V = ir = ×R =
7R 7 or ti = 560°

Potential difference across 2R is 124 (b)

' 3E 4E Here, V<E

V = E- =
7 7
∴ E = V + Ir
121 (a)
For first case
Drift velocity, vd =
eτV
mL
∵E= [ ]
V
L E = 12 +
12
16
r ….(i)

Where the symbols have their usual meaning

For second case
If the temperature are not same, τ cannot be
same. Then none of the given options is correct 11
E = 11 + r ……(ii)
10
vd V1 1
If temperatures are same, then 1
= = From Eqs. (i) and (ii),
vd V2 2
2

12 11
122 (d) 12 + r = 11 + r
16 10

The light from bulb spread out uniformly in all 20

⟹r = Ω
directions. 7

For a 100 W bulb, intensity at a distance of 3 m is 125 (b)

Power 100 The amount of chlorine

I= =
Area 4π(3)2

As I = ε0cE
2
rms
⇒E
2
rms
=
I
…(i)
m = zlt = z ()
P
V
t [∵P=VI]
ε0c

For a 400 W bulb, intensity at the same point is

= 0.367×10 ×
-6
( 100×1000
125
×60 )
'
' 400 '2 I = 0.017616 kg = 17.616 g
I = ⇒Erms =
4π(3)2
ε0c
126 (a)
'2 '
E I 400 4π(3)2
rms
= = × RParallel < RSeries. From graph it is clear that slope of
E
'2
rms
I 4π(3)2 100
the line A is lower than the slope of the line B.

P a g e | 193
 Also slope = resistance, so line A represents the 2E
V1 = E - ir1 = E - r =0
graph for parallel combination (r1+r2+R) 1

127 (a)

Near room temperature, the electric resistance of

a typical metal conductor increases linearly with
temperature.

R = R0(1+αT)

Where α is the thermal resistance coefficient. 2Er1

⇒E = or r1 + r2 + R = 2r1 = R
128 (b) (r1+r2+R)
= (r1 - r2)
ρL ρ×1
R= ⇒0.7 =
A 22 ( 134 (d)
1×10 )
-3 2

7
When a circuit is made up on any two metals in
-6
ρ = 2.2×10 ohm - m thermoelectric series, the current flows across the
cold junction from the later occurring metal in the
129 (b) series to the one occurring earlier. In
thermoelectric series Bismuth comes earlier than
In the part c b d,
Antimony. So, at cold junction current flows from
Vc+Vd Antimony to Bismuth and at hot junction it flows
Vc - Vb = Vb - Vd⇒Vb =
2 from bismuth to Antimony.

In the part c a d 135 (d)

Vc+Vd Resistivity is the property of the material. It does

Vc - Va > Va - Vd⇒ > Va⇒Vb > Va
2 not depend upon size and shape

130 (c) 136 (d)

m = Zi⇒t =
m
Zi
=
m×F
E×i [ ]
∵Z=
E
F
As circuit is open, therefore no current flows
through circuit. Hence potential difference across
X and Y=EMF of battery =120V
27×96500 12062.5
t= = 12062.5sec = hr
108×2 3600 137 (a)
= 3.35 hr 2 2
I Rt I Rt
H= = cal
132 (c) J 4.2

P1 R 2 138 (b)
We know that = 2 =
P2 R1 1
Kirchhoff’s second law is ∑V = 0
133 (b)
It states that the algebric sum of the potential
As both cells are in series, the circuit current differences in any loop including those associated
emf’s and those of resistive elements, must equal
E+E 2E zero.
i= =
r1+r2+R r1+r2=R
This law represents ‘conservation of energy’.
As terminal potential drop across 1st cell is zero,
hence 139 (b)

P a g e | 194
 2
When length and radius both are doubled, in P = i R = (1)2×5 = 5W
ρL
accordance with relation R = the resistance of
A
wire is reduced to 1/2 of its initial value. As at
constant voltage the heat produced H ∝ 1/R,
hence heat produced is doubled

140 (c)

Approximate change in resistance=2×% change

in length by stretching
145 (a)
141 (b)
We know that thermoelectric power
By using Kirchhoff’s junction law as shown below
dE
S=
dT

Given, E = K(T-Tr) T0-[ 1

2
(T+Tr) ]
By differentiating the above equation w.r.t. T and
1 1
putting T = T0, we get S = kT0
2 2

We get i = 8A 146 (d)

142 (a) Three resistances are in parallel,

dq 1 1 1 1 3
I=
2
= 3t + 2t + 5 ∴ ' = + + =
dt R R R R R

2
∴ dq = (3t + 2t + 5)dt The equivalent resistance

t=2 R 6
R’ = Ω = = 2Ω
∫(3t +2t+5)dt 3 3
2
∴q=
t=0
148 (a)

| |
3 2
3t 2t
+ 5t 2=t +t +5t 2 = 22 C
3 2
= + Internal resistance of the cell
3 2 0 0

143 (a) r= [ ]
E
V
-1 R

[ ]
i G+S ig S 2.5 1 1.5
= ⇒ = = = r= -1 14 = 1Ω
ig S i G+S 27.5 11 1.4
144 (b) 149 (c)
Current flowing through 2Ω resistance is 3A, so From Kirchhoff’s second law
P.D. across it is 3×2 = 6V
V = ∑ir (for closed mesh)
6
Current through the bottom line = = 1A
1+5 Where V is potential difference, i the current and r
the resistance.
∴ Power dissipated in 5Ω resistance is
∴ E + E = Ir + Ir = 2Ir
P a g e | 195
 E 153 (a)
or I= ….(i)
r
Equivalent resistance of the given network
Vx - Vy = E - Ir
Req = 75Ω
Putting the value of I from Eq (i), we get
∴ Total current through battery,
E
Vx - Vy = E - ×V = 0 3
r i=
75
150 (c)
3 3
i1 = i2 = =
P 2 75×2 150
Here, =
Q 3

we know i i2
R1
R3 R 4 30
P l
= i1
Q 100-l
3V R2
2 l
⟹ = R5
3 100-l

⟹l = 40 cm

151 (c) i2
R1
I 20
vd = = 29 -6 -19
i1
nAe 10 ×10 ×1.6×10
-3 R2
= 1.25×10 m/s

152 (b)

Total resistance

Or R = 20 + 40

R = 60Ω Current through resistance

Given G=15V 3 60
R4 = ×
150 (30+60)
V 15
Current I = = 3 60
R 60 = ×
150 90
I = 0.25A
2
= A
V 150
Potential gradient =
l
V4 = i4×R4
20×0.25 -1
= = 0.5Vm 2
10 = ×30
150
PD across 240 cm
2
= = 0.4 volt
E = 0.5× 2.4 5

E = 1.2V 154 (c)

P a g e | 196
 Current through a conductor is constant at even 161 (c)
cross-section of the conductor
θ0 = 2θn - θi = 2×210 - (600 - 273)
155 (c)
= 420 - 327 = 93℃
Mass deposited = density × volume of the metal
162 (c)
m = p×A×X …(i)
Let the circuit be as shown
Hence from Faraday’s first law m = Zit …(ii)
2Ω 2Ω 2Ω
So from equation (i) and (ii)

Zit
Zit = ρ×Ax⇒x = A B
ρA
2Ω
-4 -3
3.04×10 ×10 ×1×3600 -6
= = 2.4×10 m
9000×0.05
Equivalent resistance between A and B is
= 2.4μm
1 1 1 2
157 (c) = + =
R 2 (2+2+2 ) 3
The given circuit can be simplified as follows 3
R= = 1.5Ω
2
10
3
Q
Therefore, 4 resistances are required.
3 R
163 (b)
10
3 -7
 P Q riron ρiron 1×10
= = -8 = 2.4
rCopper ρcopper 1.7×10
(3 + R)
164 (c)
2 3
10×(3+R) 30+10R For a fuse I ∝ r
R=3+ =3+
10+3+R 13+R 2 3
I1 r1
∴ 2 = 2
39+3R+30+10R 69+13R I2 r2
R= =
13+R 13+R

( )
2 3
3 0.02
2 2 =
13R + R = 69 + 13R⇒R = 69Ω I2 0.03

()
158 (b) 3
3/2
I2 = 3× A
2 2
R ∝ l ⇒ If l doubled then R becomes 4 times
165 (d)
159 (c)
Let n cells be in series and m in parallel, then
Ammeter is used to measure the current through
the circuit nE E
=
R+nr r
160 (b) R+
m
2
E 2 ER
i= ⇒P = i R⇒P =
r+R (r+R )2
2
Power is maximum when r = R⇒Pmax = E /4r
P a g e | 197
 ⟹ n R+[ ] r
m
= R + nr
Given, A = 16, B = 0.08

1 2
∴ E = 16T - ×0.08×T
⟹nRm + nr = Rm + mnr 2

⟹6 + 2r = 3 + 4r Since, at temperature of inversion emf is zero, we

have
⟹2r = 3
2
0 = 16T - 0.04 T
⟹ r = 1.5Ω
16
166 (a) ⇒T = = 400℃
0.04
AC 170 (c)
The ratio will remain unchanged.
CB
The equivalent circuit can be redrawn as
167 (b)
P = 4Ω R
V2 - V1 = E - ir = 5 - 2×0.5 = 4volt
4Ω

⇒V2 = 4 + V1 = 4 + 10 = 14 volt A 10 B
Ω
Q = 4Ω
168 (a) S = 4Ω

Power in electric bulb

P R
V
2 we have, =
P= Q S
R
4 4
So, resistance of electric bulb ie, =
4 4
2
V So, the given circuit is a balanced Wheatstone’s
R=
P bridge.
Given, P1 = 25 W, P2 = 100 W, Hence, the equivalent resistance

V1 = V2 = 220 volt (4+4)×(4+4)

RAB =
(4+4)+(4+4)
Therefore, for same potential difference V
8×8 64
1 = = = 4Ω
R∝ 8+8 16
P
171 (b)
Thus, we observe that for minimum power,
resistance will be maximum and vice - versa. Force =Electric intensity ×charge

Hence, resistance of 25 W bulb is maximum and Potential diffsrence

= ×charge
100 W bulb is minimum. distance

169 (c) -19 V -19

∴ 4.8×10 = ×1.6×10
5
Let temperature of cold junction be 0℃ and that
of hot junction be T℃. The relation for thermo- or V = 15 volt
emf is given by
172 (d)
1 2
E = AT - BT 1
2 In stretching of wire R ∝ 4 , where d = Diameter
d
P a g e | 198
 of wire E2 = V2I2t2 = 14×5×15 = 1050 Wh

173 (d) Hence, watt-hour efficiency of the battery is given

by
Total current through the circuit
10 3 E2
i= = A η= ×100 = 0.875×100 = 87.5%
1000 250 E1
+500
3
Now voltmeter reading 179 (d)
2 3
= iv×RV = × ×500 = 4V Total power spend across two resistors connected
3 250
2 2
174 (b) V V
in parallel to battery = +
R1 R2
V iR r R
E = xl = = ×l⇒E = × ×l
l L (R+Rh+r) L 3×3 3×3 36
= + = = 18
2 2/3 2
10 5
⇒E = × ×3 = 3V = 3×3×2 J
(5+4+1) 5

175 (c) 180 (b)

i = nAevd 1 A 1
Conductance C = = ⇒C ∝
R ρl l
i 1
or vd = ie, vd ∝ 181 (d)
nAe A
2 3
As A increases vd decreases, because i remains
i1
constant
i
176 (a)
i2 6 4
R = ρl A or R ∝ l/A. Thus, resistance is least in a
wire of length L/2 and area of cross-section 2A
Resistance of upper branch R1 = 2 + 3 = 5Ω
177 (c)
Resistance of lower branch R2 = 4 + 6 = 10Ω
i 5.4
Vd = = 28 -6 -19
nAe 8.4×10 ×10 ×1.6×10 i1 R 10
Hence = 2 = =2
-3
i2 R1 5
= 0.4×10 m/sec = 0.4 mm/sec
Heat generated across 3 Ω (H1)
2
i ×3 4
178 (c) = 12 = =2
Heat generated across 6 Ω (H2) i2×6 2
The power of the battery, when charged, is given
∴ Heat generated across 3 Ω = 120 cal/sec
by
182 (b)
P1 = V1I1
At resonance both bulbs will glow with same
The electrical energy dissipated is given by
brightness. At resonance,
E1 = P1t1
XL = XC
i.e., E1 = V1I1t1 = 15×10×8 = 1200 Wh
1
Similarly, the electrical energy dissipated during Or 2πfL =
2πfC
the discharge a battery is given by,

P a g e | 199
 1
Or f=
2π LC

183 (d)

()
2 2
1 R1 l 1 r2 1 5 2
R∝ 2 ⇒ = × ⇒ = × ⇒l2 = 20m
r R2 l2 r21 1 l2 1

184 (c)
i1 3
If an identical battery is connected in opposition, = ⇒i1 = i2 ∴ i2 = 0.5A = i1
i2 3
net emf = E - E = 0 and the current through
circuit will be zero, although each one of them has 189 (d)
constant emf.
Current through the galvanometer
185 (a)
3 -3
I= = 10 A
The circuit given in figure can be redrawn as (50+2950)
shown here. Here two resistances are joined in
-3
series and the combination is joined in parallel Current for 30 divisions = 10 A
with the third resistance. Since in parallel
Current for 20 divisions
grouping effective resistance is even less than the
smallest individual resistance, hence net 10
-3
2 -3
resistance will be maximum between the points P = ×20 = ×10 A
30 3
and Q
50 Ω

2950Ω
186 (c)
3V
iS 50×12
ig = ⇒10 = ⇒12 + G = 60⇒G
S+G 12+G
= 48Ω
For the same deflection to obtain for 20 divisions,
187 (b) let resistance added be R
igS = (i-ig)G⇒ig(S+G) = iG 2 -3 3
∴ ×10 =
3 (50+1R)
ig G 8
⇒ = = = 0.8
i S+G 2+8 or R = 4450Ω

188 (b) 190 (c)

1.5 Suppose resistance R is corrected in series with

i1 + i2 = = 1 amp
3/2 bulb

90
Current through the bulb i = = 3A
30

P a g e | 200
 R i 30V, 90W P R
= '
Q S
90 V 30 V
2 2
= '
2 S
120 V
'
∴ S = 2Ω
Hence for resistance Now,
V = iR⇒90 = 3×R⇒R = 30Ω
1 1 1
191 (c) ' = +
S S r
Rmax 2
1 1 1 1 1 3-2
Rmax = nR and Rmin = R/n⇒ =n or = '- = - =
Rmin r S S 2 3 6

192 (a) 1 1
=
r 6
The voltage per unit light of the metre wire PQ is

(6.00 mV
0.600 m )
i.e. 10 mV/m. Hence potential r = 6Ω

difference across the metre wire is 195 (d)

10mV/m×1m = 10 mV. The current drawn from
Let the value of shunt be r. Hence the equivalent
10 mV
the driver cell is i = = 2 mA Sr
5Ω resistance of branch containing S will be
S+r
The resistance
(2V-10mV) 1990 mV P Sr/(S+r)
R= = = 995 Ω In balance condition, = . This gives
2 mA 2 mA Q R
r = 8Ω
193 (b)
197 (b)
To make range n times, the galvanometer
resistance should be G/n, where G is initial 2.2×10
3
1
resistance P = Vi⇒i = = A
22000 10
194 (d)
( )
2
2 1
Now loss of power = i R = ×100 = 1 W
10
Let a resistance r ohm be shunted with resistance
S, so that the bridge is balanced. 198 (b)

If S’ is the resultant resistance of S and r, then Let resistance for bulb filament at o°C be R0 and at
a temperature θ°C its value be 200 Ω. Then,
In balanced position
100 = R0(1+α×100) = R0(1 + 0.005×100)

P
= R0(1.5) …(i)
Q

and 200 = R0(1+α×θ) = R0(1 + 0.005×θ)

S
= R0(1.005θ) …(ii)
R
1+0.005θ
Dividing Eq. (ii) by Eq.(i), we get 2 =
r
1.5

P a g e | 201
 3 = 1 + 0.005θ Current through resistance P and Q,

2 4 1
⇒θ = = 400°C i1 = = A
0.005 90+110 50

199 (c) 1
VA - VB = Pi1 = 90× = 1.8 V
50
Since the current coming out from the positive
terminal is equal to the current entering the Current through resistance R and S,
negative terminal, therefore, current in the
4 1
respective loop will remain confined in the loop i2 = = A
40+60 25
itself
1
∴ Current through 2Ω resistor = 0 VA - VD = Ri2 = 40× = 1.6 V
25
200 (d)
VB - VD = (VA-VD) - (VA - VB)
Graph (d) represents the thermal energy
= 1.6 - 1.8 = -0.2V
produced in a resistor.
203 (b)
201 (c)
100 32 70 100
Potential difference across 1 MΩ resistor is By using e0 = e0 + e32 + e70
100 100
18V×1×10 Ω
6
18V×1×10 Ω
6 ⇒200 = 64 + 76 + e70 ⇒e70 = 60 μV
VP - VB = 6 = 6 = 15V
(0.2+1)×10 Ω 1.2×10 Ω
204 (a)
VB = -15V [Given]
Given, E1 = 1.5V, l1 = 27 cm,
∴ VP - VB = 15V or VP = 15V + VB
l2 = 54 cm, E2 = ?
= 15V - 15V = 0V
E1 l
= 1
E2 l2

E1l2
or E2 =
l1

1.5×54
or E2 =
27

E2 = 3V
Potential difference across 200 kΩ resistor is
205 (a)
6
18V×0.2×10 Ω
VA - VP = 6 Joule effect is not reversible
(0.2+1)×10 Ω

18V×0.2×10 Ω
6 206 (c)
= 6 = 3V
1.2×10 Ω
The given circuit can be redrawn as follows
VA = +3V [Given]
2
2 2 2 1 2
∴ VA - VP = 3V or VP = VA - 3V 

= + 3V - 3 V = 0 V 2

202 (a)
P a g e | 202
 I’ 4Ω
4Ω
⇒Req = 5Ω
4Ω
207 (a) I 4Ω I 6Ω
4Ω
E 5 i
i= = = 1A 4V 4V
R+r 4.5+0.5

V = E - ir = 5 - 1×0.5 = 4.5 Volt

208 (b) i 4V

ne
Current I =
l
-19
n×1.6×10
3.2 = So, net resistance,
1

3.2 19
R=2.4+1.6=4.0Ω
n= -19 = 2×10
1.6×10
Therefore, current from the battery.
209 (b)
V 4
i= = = 1A
As the current and the other factors are same for R 4
both the galvanometers
Now, from the circuit (b),
N ∝ tan θ
4I’ =6I
N1 tan 60°
= = 3 ⟹ I =
' 3
I
N2 tan 30° 1/ 3 2

N1 But i=I+I’
⟹ =3
N2
3 5
=I+ I= I
210 (a) 2 2

Potential difference 5
Resistance = ∴ 1= I
Current 2

211 (b) 2
⟹ I= = 0.4A
5
S 0.01 S 50
ig = i ⇒ = ⇒S = = 0.05Ω
G+S 10 50+S 999 214 (a)

212 (b) Vab = ig.G = (i-ig)S

The sensitivity of potentiometer can be increased S

by decreasing the potential gradient i.e. by
increasing the length of potentiometer wire

1 ( i - ig )
[Sensitivity ∝ ∝ Length]
P.G.
g
i
213 (c) G
i a ig b
We can simplify the network as shown

P a g e | 203
 ( ) ()
2 2 3 3
G P PR 2.2×10 ×2.2×10 ×10
∴ i = 1+ i PL = R= 2 =
S g V V 22000×22000
= 0.1W
Substituting the values we get,
220 (b)
i = 100.1mA
As 5Ω resister is joined in parallel to series
215 (d)
combination of 4Ω and 6Ω (ie, total resistance
2E 10Ω), V =constant.
In series i =
2+R
i1 R 10
and = 2 = =2
( ) i2 R1 5
2
2 2E
∴ J1 = i R = .R
2+R
i1
or i2 =
In parallel 2

E Now heat produced per second in5Ω resistor

i=
0.5+R
2 2 -1
H1 = i1R1 = i1×5 = 100Js …(i)

( )
2
2 E
J2 = i R = .R
0.5+R and for 4Ω resistor

()
2
J1 4(0.5 +R)2 2 i 2
= 2.25 = H2 = i R2 = 1 ×4 = i1 …(ii)
J2 (2 +R)2 2
2

Solving we get, R = 4 Ω Simplifying Eqs. (i) and (ii), we get

216 (a) H2 1 1 -1
= or H2 = ×100 = 20Js
100 5 5
With rise in temperature the thermal velocity of
the electron increases. Relaxation time and hence 221 (b)
drift velocity will decrease.
The resistance of one wire
217 (a)
l
R1 = ρ 1
2 A
I1 = ×6 = 2A 1
2+4
and the resistance of second wire
I2 = 4 amp
l
R2 = ρ 2
218 (b) A2

vd = i/nAe Ratio of their resistance

28 -6 -19
= 0.21/(8.4×10 ×10 ×1.6×10 ) R1 l A
= 1× 2
-5 -1
R2 A1 l2
= 1.56×10 ms
∵Mass=density×volume
219 (a)
∵Mass=density ×area ×length
For power transmission power loss in line

()
2
PL = i R R1 l ρA l
2

Or = 1 × 22
R2 l2 ρA l
1 1
If power of electricity is P and it is transmitted at
P
voltage V, then P = Vi⇒i =
V
P a g e | 204
 () t×(R1+ R2)
2
R1 l m t
Or = 1 × 2 ⇒ = 1 …(iii)
R2 l2 m1 R1R2 R1

R1 9 3 R1 27 On using Eqs. (i) and (iii), we get

Or = × or =
R2 16 2 R2 32
t1t2
t=
R1:R2 = 27:32 t1+t2

222 (d) 225 (c)

We know that the current in the circuit Error in measurement = Actual value – Measured
value
E
I=
R+r Actual value = 2V

and power delivered to the resistance R is 998 

V
2
2 ER
P=IR= i
(R+r)2

dP + –
It is maximum when =0 2V 2
dR

dP
dR
=E
2
[
2 (r+R) -2R(r+R)
(r+R)4
=0 ] i=
2
998+2
Or (r+R)2 = 2R(r+R) 1
= A
500
Or R = r
Since E = V + ir
223 (b)
1 998
T +T 10+530 ⇒V = E - ir = 2 - ×2 = V
Tn = i c = = 270℃ 500 500
2 2
998
224 (d) ∴ Measured value = V
500
Let R1 and R2 be the resistances of the coils
998 -3
⇒ Error = 2 - = 4×10 volt
2 2 500
V t1 V t2
H= and H =
R1 R2 226 (d)

t1 t R t The circuit shown is a balanced Wheatstone

⇒ = 2 , ie, 2 = 2 …(i)
R1 R2 R1 t1 bridge. So, there is no current flowing through
10Ω resistance. Therefore, by replacing 10Ω
Now in parallel resistance by 20Ω resistance, current in the circuit
will be as such ie., 4A.
' R1R2
R = = R1
R1+R2 227 (c)

( ) ( )
2
Vt l1 75
∴H = ' …(ii) r=R -1 = 10 -1
R l2 65
2 2
Vt V t1 = 10×0.0154 = 1.54Ω
Now, ' =
R R1
228 (a)

P a g e | 205
 Suppose m rows are connected in parallel and V iR
Potential gradient x = =
each row contains n identical cells (each cell L L
having E = 15 V and r = 2Ω)
2 15 3
⇒x = × = volt/cm
For maximum current in the external resistance R, (15+5) 10 2000
nr
the necessary condition is R = 232 (c)
m
l
n×2 Resistance, R = ρ
⇒12 = ⇒n = 6m …(i) A
m
1
Total cells = 24 = n×m …(ii) R∝l∝
A
On solving equations (i) and (ii) n = 12 and
∴R is maximum when
m = 2 i.e. 2 rows of 12 cells are connected in
parallel A
length = 2L and area = .
2
229 (c)
233 (a)
Current drawn from the cell in resistance R1 will
be I = E/(R1 + r) 22.4 litre H2 = 1 mole of H2 = N molecules of H2

Therefore, heat produced in R1 ie, = 2N atoms of H

2
E R1t So charge required to liberate 22.4 litre of
H1 =
(R1+r)2 H2 = 2Ne = 2F

2
E R2t Hence charge required to liberate 11.2 litre of
Heat produced in R2 ie, H2 = H2 = F
(R2+r)2
As per question H1 = H2 234 (b)

2 2 Current through external resistance,

E R1t E R2t
or 2 =
(R1+r) (R2+r)2 i=
nE
=
5×2
= 2A
nr+R 5×0.2+4
On solving we get;
235 (c)
r = R1R2
2 2
V V
When bulbs are connected in series, P = ' =
= 100×40=63.25Ω R 3R

230 (b) When bulbs are connected in parallel

2 2 2
V ' V V ×3
R= P = = = 3×3P = 9P
P R" R

1 236 (b)
or R∝
P
11.2 L of H2 is liberated by 96500 C 22.4 L of H2 is
1 1 1 liberated by
∴ > >
R100 R60 R70
96500×2 = 193000 C.
Hence, the correct option is (d)
237 (d)
231 (b)

P a g e | 206
 Case (i) E + E = (r + r + 5)×1.0 241 (a)

or 2E = 2r + 5 …(i) Each part will have a resistance r = R/10

(
Case (ii) E =
r×r
r+r )
+5 ×0.8
Let equivalent resistance be rR, then

1 1 1 1

( )
= + + ……. 10 times
r rR r r r
or E = +5 0.8 or E = 0.4r + 4.0 …(ii)
2
1 10 10 100 R
Multiplying Eq.(ii) by 2 and equating with Eq.(i), ∴ = = = ⇒rR = = 0.01R
rR r R/10 R 100
we get
242 (c)
2r + 5 = 00.8r + 8
X 6
3
or 1.2 = 3 or r = = 2.5Ω
1.2 G
4 6
239 (d) A B
C

6Ω and 6Ω are in series, so effective resistance is

12Ω which is in parallel with 3Ω, so 5V

1 1 1 15
= + =
R 3 12 36 Resistance of the part AC

36 RAC = 0.1×40 = 4Ω and RCB = 0.1×60 = 6Ω

⇒R =
15
X 4
In balanced condition = ⇒X = 4Ω
V 4.8×15 6 6
∴I= = = 2A
R 36
Equivalent resistance Req = 5Ω so current drawn
240 (a) from battery
Let R0 = resistance of filament at room 5
i= = 1A
temperature 5

Rt = resistance of filament at 2500℃ 244 (c)

-19 15 -3
Similarly powers, P0 and Pt. i = ev = 1.6×10 ×6.8×10 = 1.1×10 amp

Here, voltage remains the same. 245 (b)

()
2 2
V i i v i1 r2 ' v
P0 = vd = 2 ⇒vd ∝ 2 ⇒ ' = × ⇒v =
R0 neπr r v i2 r1 2
2 2
V t V 246 (b)
or R0 = , R =
P0 Pt
E (M/P)
m= z= It = It
also Rt = R0[1+α(2500-20)] F F
3
and P0 = Pt[1 + α(2500-20)] mFP (0.254×10 )×96500×2
or t = =
MI 63.5×100
-3
= 50[1 + 4.5×10 (2500-20)]
= 7720 s.
= 608W
247 (b)

P a g e | 207
 In series, current, 100×540×4.2
or t = 3 = 210 s
1.08×10
2E
i1 =
2+2r 250 (d)

In parallel, current, Heat developed by 210 W electric bulb in 5 min is

given by
E 2E
i2 = =
r 4+r W 210×5×60
2+ H= = = 15000 cal
2 J 4.2

According to the question 251 (d)

i1 = i2 4×4 6×6
Equivalent resistance = + = 5ohm So
4+4 6+6
2E 2E 20
⟹ = the current in the circuit = = 4 ampere Hence
4+r 2+2r 5
the current flowing through each resistance = 2
⟹ r = 2Ω
ampere
248 (d)
252 (d)
In the given circuit 4Ω resistors are connected in
Here, 2Ω and 2Ω are in parallel
parallel, this combination is connected in series
with 1Ω resistance. 1 1 1
∴ = +
R 2 2
2 1Ω
2×2
R= = 1Ω
2+2

6V
Now, internal resistance (1Ω), 2Ω, 4Ω and
resistance R in series.

∴ Rnet = 1Ω + 2Ω + 4Ω + 1Ω = 8Ω
1 1 1 2 1
∴ ' = + = =
R 4 4 4 2
Hence, current
'
⟹ R = 2Ω V 4
I= = = 0.5A
R 8
Also, R’’=2 Ω +1Ω =3Ω
253 (d)
From Ohm’s law, V = iR
The given circuit is a balanced wheatstone bridge
V 6
∴ i= = = 2A circuit. Hence potential difference between A and
R 3
B is zero
249 (b)
254 (d)
3
Heat produced by heater per second = 1.08×10 J 2 2 2 2
V ρl V AV AV
P= but R = ⇒P = = . Since
Heat taken by water to form steam mL R A ρl / A ρl l
1
is constant as per given condition so P ∝
= 100×540 cal ρ

= 100×540×4.2 J 255 (d)

3
∴ 1.08×10 ×t = 100×540×4.2 Current in 9Ω is 2A, so that in 6Ω is 3A. Total
current is 2 + 3 = 5A. Potential drop 5×2 = 10
P a g e | 208
 V Resistor’s of

256 (c) 6
Ω and 2.8Ω are connected in series.
5
2
V
H=
R
6/5Ω
2
V
⇒R =
H

(25)2
= = 25 Ω
25
6V 2.8Ω
257 (c)

If a cell is connected between points A and C, no

current will flow in arms BE and ED. Therefore, 6
Hence, R’’ = Ω + 2.8Ω
the resistance of arms BE and ED an be removed. 5
Now resistance between points A and C will be = 1.2Ω + 2.8Ω = 4.0Ω
the resistance of three parallel arms, each of
resistance= R + R = 2R From Ohm’s law,

∴ Total resistance Rpwill be V 6

Current i = = = 1.5A
R 4.0
1 1 1 1 3 2R
= + + = or RP =
RP 2R 2R 2R 2R 3 261 (c)

258 (c) The given circuit can be simplified as follows

I∝Q

Ig 1
= A B C D
I 2

S 1
=
G+S 2
5R
40 1 ∴ RAD =
= ⟹G = 40Ω 6
G+40 2
262 (a)
259 (d)
Specific resistance is independent of dimensions
The resistance of the cell is independent of e.m.f
of conductor but depends on nature of conductor.
260 (d)
263 (a)
In the given circuit the resistors of 2Ω and 3Ω are
Here three resistance of 4Ω each are connected in
connected in parallel hence, equivalent resistance
4
is parallel so that their combined resistance = Ω. It
3
1 1 1 5 is in series with ammeter, battery and last 4Ω
' = + = resistance.
R 2 3 6

6 4 16
∴R =
'
Ω ∴ Net resistances R = +4= Ω
5 3 3

Also in steady state, the circuit is shown as. ∴ Current in main circuit =ammeter reading
P a g e | 209
 E 2V 3 The equivalent circuits are as shown below
i= = = A
R 16 8
Ω C C
3 C
2 2
264 (d) A B A B
A
 
Slope of V - i curve = resistance. Hence 2 2
1 D B D D
R = = 1Ω
1

265 (c) Clearly, the circuit is a balanced Wheatstone

bridge. So effective resistance between A and B is
When rod is bent in the form of square, then each 2Ω
side has resistance of
268 (a)
1
Ω. As shown R1, R2 and R3 are connected in
4 Ti+Tc
Tn =
series, so their equivalent resistance 2

1 Ti = 2Tn - Tc = 540℃
R2 = Ω
4
1 270 (d)
R1 =
4
1
R3 = Ω
1 4 The last two resistances are out of circuit. Now 8Ω
R = 1Ω
4
is in parallel with (1+1+4+1+1)Ω

8
∴ R = 8Ω||8Ω = = 4Ω⇒RAB = 4 + 2 + 2
2
= 8Ω

'
R = R1 + R2 + R3 271 (c)

1 1 1 3 As 3Ω and 6Ω resistances are

= + + = Ω
4 4 4 4
In parallel their equivalent resistance will be 2Ω.
Now, R’ and R4 are connected in parallel, so Here 2Ω and 4Ω are in series, their equivalent
equivalent resistance of the circuit is resistance will be 6Ω. From current distribution
law
'
R ×R4
R= '
R +R4 i 6Ω
1

=
( )( )
3 1
4 4
18 A

3
4()()
+
1
4
i
12 Ω

3
( )
16 3
= = Ω
1 16
12×18
i1 = = 12A
266 (a) 18

F = Ne = 6×10 ×1.6×10
23 -19
6×18
i2 = = 6A
18
267 (d)

P a g e | 210
 Dividing eq. (i) by eq. (ii)
3Ω
i 1 xl1

( )
'
8 l1

( ) ( ) ( )
12 +1 2
A 3 l2 l1 l1 8 l
6Ω = = 2 ⇒ +1 × = 1
12 l l l2 36 l2
xl2 1 +1 l2 1 +1
i 1 l2 l2
2
'

8 l1
⇒(y +1+2y)×
2
=y [Where y = ]
Now, 12A current is entering in parallel 36 l2
combination of 3Ω and 6Ω again from current 2
⇒8y + 8 + 16y = 36y
distribution law
2
⇒8y - 20y + 8 = 0
' 6×12
i1 = = 8A
9 1
On solving, we get y = or 2
2
' 3×12
i2 = = 4A.
9 l1 1
∴y = = or 2
∴ Potential difference across 3Ω resistance l2 2

=8×3=24V 274 (a)

272 (a) Internal resistance,

R=ρ
l
A
m l
= 2 .
ne τ A
r=R ( ) (
l1-l2
l2
=1
540-500
500
=
40
500
= 0.08 Ω )
273 (d) 275 (b)

Let x is the resistance per unit length then 2Ω 3Ω 2Ω

A
4Ω 2Ω
Equivalent resistance 1Ω
10Ω

1Ω
1.8Ω 5Ω
2.2Ω

2Ω 3Ω
A
 B

1Ω
4Ω 2Ω

R1R2
R=
R1+R2 1.8Ω 5Ω
Short circuited 2.2Ω

8
⇒ =
(xl1)(xl2) ⇒ 8 = x l1l2 2Ω

3Ω
B
4Ω
3 xl1+xl2 3 l1+l2 A
1Ω
4Ω 2Ω
8 l1
⇒ =x …(i)
3 l1
+1 5Ω
l2 
B
Also R0 = xl1 + xl2⇒12 = x(l1+l2) 2Ω 1Ω 5Ω
 A B

⇒12 = xl2 ( ) l1
l2
+1 …(ii)

P a g e | 211
 RAB = 8Ω temperature of inversion is

276 (c) ti = 2tn - tc⇒ti = 2×270 - 40 = 500℃

P×t = ms ∆t 282 (b)

3
⇒ 2.1×10 ×t i1 15 3
= = …(i)
i2 5 1
= 1.5×4200×[(100+273)-(20+273)]
i2 9  6
1.5×4200×80
⇒ = 3 = 240s
2.1×10 12 

i
277 (a)
i1 5
In potentiometer experiment in which we find
internal resistance of a cell, let E be the emf of the
cell and V the terminal potential difference, then Also
H
= i R⇒45 = (i1) ×5
2 2

t
E l
= 1 ⇒i1 = 3 A and from equation (i) i2 = 1 A
V l2

Where l1 and l2 are lengths of potentiometer wire So i = i1 + i2 = 4 A

with and without short circuited through a
Hence power developed in 12 Ω resistance
resistance.
2
P = i R = (4)2×12 = 192W
Since,
283 (c)
E R+r
= [∵E=I(R+r)and V=IR]
V R I 2/ 3 2
K= = = A
tan θ tan 60° 3
R+r l1
∴ -
R l2 284 (d)

r 110 r 10 2E
or 1 + = or = i=
R 100 R 100 R+R1+R2

1 From cell (2) E = V + iR2 = 0 + iR2

or r = ×10 = 1Ω
10
R
279 (a)
i
Resistivity of a material is its intrinsic property
and is constant at particular temperature.
Resistivity does not depend upon shape E, R 1 E, R 2
(1) (2)
280 (b)

Here internal resistance is given by the slope of 2E

⇒E = ×R ⇒R = R2 - R1
graph R+R1+R2 2

x 1 y 285 (a)
i.e. . But conductance = =
y Resistance x
Temperature coefficient of a semiconductor is
281 (c) negative
No charge in neutral temperature but
P a g e | 212
286 (d) Potential difference across length l = Kl

E = i1(R1+r) = i2(R2 + r) = 10×0.5 = 5 V

i2R2-i1R1 293 (d)

On solving, r =
(i1-i2)
The given circuit can be simplified as follows
288 (c)

The circuit consists of three resistances (2R,2R

and R) connected in parallel

289 (c)

Here resistances 4 Ω, 6Ω, 12Ω and 24 Ω are in

parallel. Their effective resistances, RP will be

1 1 1 1 1
= + + +
RP 4 6 12 24

6+4+2+1 13 24
= = or RP =
24 24 13

Total resistance between A and B

24 128 294 (d)

=3+ +5= = 9.85 Ω
13 13
6×1.5
290 (d) Rtotal = 2 + = 3.2kΩ
6+1.5
The current density of electrons in a metallic 24V
22 -3
conductor = 10 cm = 10 m
28 -3 (a)I = = 7.5mA = IR
3.2kΩ 1

291 (c)

The given circuit can be redrawn as follows

IR =
2 ( ) RL
RL+R2
I

A B 1.5
5 5 5 C I= ×7.5 = 1.5mA
2/3V 2/3V 2/3V
7.5

IR = 6mA
L
2V

A C
(b)VR = (IR )(RL) = 9V
5 5 D 5 L L

(c)
PR
1
=
(I )R
2
R1 1
=
(7.5)2(2) 25
=
For identical resistance, potential difference PR
2
(I )R
2
R2 2
(1.5)2(6) 3
distributes equally among all. Hence potential
2 (d) When R1and R2are interchanged, then
difference across each resistance is V, and
3
4 R2RL 2×1.5 6
potential difference between A and B is V = = kΩ
3 R2+RL 3.5 7

292 (b) Now potential difference across RL will be

-1
Here, potential gradient, K = 10Vm ;

P a g e | 213
 []
6 V 4
On solving, V = 4 Volt, i3 = = = 2A
7 2 2
VL = 24 3V
6
6+
7 2
10 V 4 C 5V
Earlier it was 9V A i1 i3 i2 B
2
V 2
2
Since, P = or P ∝ V
R

In new situation potential difference has been S

decreased three times. Therefore, power
dissipated will decrease by a factor of 9.

295 (b)
299 (c)
Mass of copper deposited, 5 5 -1
m = zI t;so 9 = z×10 or z = 9/10 g C
m = volume×density
m = zI t = (9×10 )×50×(20×60) = 5.4 g.
5

= (area×thickness)×density
300 (a)
= [2×(12×3)×0.002]×8.9 g
Potential difference
Potential gradient =
m [2(12×3)×0.002×8.91] Length
t= = -5 = 776s.
zI 33×10 ×5
301 (d)
296 (c)
Charge supplied per minute = 3.2×60 = 192 C
Suppose n resistors are used for the required job. +2
Suppose equivalent resistance of the combination Charge 2e liberates one Cu ion
is R' and according to energy conservation it’s +2
'
∴ No of Cu ion liberate by 192 C
current rating is i
192 192 20
Energy consumed by the combination = n× = = -19 = 6×10
2e 2×1.6×10
(Energy consumed by each resistance)
302 (a)
() ( ) () ( )
' 2 ' 2
'2 ' i2 R 4 5
⇒i R = n×i R⇒n = × = × -2
i R 1 10 ρl -8 50×10 -6
R= = 50×10 × -2 2 = 10 Ω
=8 A (50×10 )
297 (b) 303 (b)

24-12 360 The equivalent circuit of the given circuit is as

i= = 4 A, Time of charging t =
3 V.i shown

360
⇒t = = 7.5 hours 3Ω
12×4

298 (c)
6Ω
Let V be the potential at C 1.5
2Ω

Using Kirchhoff’s first law i1 + i2 = i3 _ V = 9V

10-V 5-V V-0

+ =
4 2 2
P a g e | 214
 Resistances 6Ω and 2Ω are in parallel For series grouping

' 6×2 3 C
∴ R = = Ω Ce =
6+2 2 2

3 RC
Resistances Ω and 1.5Ω are in series t1 = In 2
2 2

'' 3 t2 1
∴ R = + 1.5 = 3Ω = t2 = 2.5s
2 t1 4

Resistances 3Ω and 3Ω are in parallel 305 (b)

3×3 3 Resistance of combination Re = 4R
∴ R= =
3+3 2
∆Re ∆R
V =
The current, I = Re R
R
5×100
9 = = 5%
= = 6A 100
3/2
306 (c)
304 (d)
2
Since H ∝ i , so on doubling the current, the heat
If Ce be the effective capacitance, then
C produced and hence the rise in temperature
becomes four times

307 (b)
C
The given circuit can be redrawn as
R
i1 9 1
C C = =
i2 18 2

and i = i1 + i2
R
i2 6 3

10 
i
1 i1 18 
VC = V0
2

q q i
= 0 i
Ce 2Ce ⇒ =1+ 2 =1+2=3
i1 i1

( ) = q2
()
t
-
⟹q0 1-e RC 0 2 P10Ω i
2
10
From P = i R⇒ = × ⇒P10Ω = 10W
e

P18Ω i1 18
⟹ t = RCe In2
308 (b)

( )
For parallel grouping
α 500
tn = = = 100℃
2C β 5
Ce =
2
ti+tc t +0
Also tn = ⇒100 = i ⇒ti = 200℃
∴ t2 = 2RC In2 2 2
P a g e | 215
309 (a) 313 (a)

According to Seebeck effect The emf of the circuit is

310 (a) E = E1 + E2

l = 4V + 2V = 6V
R = ρ and mass m = volume
A
(V)×density(d) = (A l)d In the given circuit, 3Ω and 6Ω are connected in
parallel, hence equivalent resistance is
Since wires have same material so ρ and d is same
for both 1 1 1 3 1
' = + = =
R 3 6 6 2
Also they have same mass
'
1 ⟹R = 2Ω
⇒Al = constant⇒l ∝
A
Total resistance of circuit is

() ()
2 4
R1 l A A r2
⇒ = 1× 2 = 2 = R = 1Ω + 1Ω + 2Ω = 6Ω
R2 l2 A1 A1 r1
From Ohm’s law V = iR
( )
4
34 r
⇒ = ⇒R2 = 544 Ω
R2 2r V 6
⟹i = = = 1A
R 6
311 (a)
The 3Ω and 6Ω resistors are in parallel, hence
(Rated voltage)2
Resistance of a bulb =
Rated power i1R1 = i2R2 = V

1 ∴ i1×3 = i2×6
For a given voltage, R ∝
P
⟹i1 = 2i2 and i1 + i2 = 1
∴ R40 > R60 > R100
2
2i2 + i2 = 1
Rate of heat produced, H = I R
1
When the bulbs are connected in series, the 3i2 = 1⟹i2 = A
3
current flowing through each bulb is same
314 (a)
∴ H ∝ R . As R40 > R60 > R100
Kirchhoff’s first law is based on the law of
∴ H40 > H60 > H100⇒B1 > B2 > B3 conservation of charge

312 (c) 315 (c)

L Let the resultant resistance be R. If we add one

Resistance of the slab = ρ where
A more branch, then the resultant resistance would
be the same because this is an infinite sequence

R1 = 1 X
The potential across R is I×R = V A

The length of the slab is doubled. Therefore the R2 = 2 R

resistance is 2R. Assuming that the same current

B
is passed, the potential across the new resistance Y

is l×2R = 2V

P a g e | 216
 2 2
RR2 2 H V 1 1 lρ
∴ + R1 = R⇒2R + R + 2 = R + 2R = P = ⇒P ∝ also R ∝ ∝
R+R2 t R R A A.lρ
2
2
⇒R - R - 2 = 0⇒R = -1 or R = 2ohm l 2
⇒R ∝ ⇒R ∝ l [for same mass]
m
2×R 2
R=2+2+ ⇒2R + R = 8 + 4R + 2R 2
2+R PA l 4
So = 2B = ⇒PA = 20W
PB lA 1
2 4± 16+32
⇒R - 4R - 8 = 0⇒R = =2±2 3
2 323 (d)

R cannot be negative, hence mAg = mCu×EAg/ECu

R = 2 + 2 3 = 5.46Ω
= 2×108/(63.6/2) = 6.8 mg.
316 (a)
324 (a)
Here, all resistance are in parallel.
dE
2 At neutral temperature, =0
V dT
∴ ∆H = t
R
325 (a)
1
∴ ∆H ∝
R VA - VB = emf of the cell = 2 V

Hence, (a) is correct. 1 2 1 -6 -6

∴U = CV = ×2×10 ×(2)2 = 4×10 J
2 2
318 (b)
326 (a)
E E
Current, i = when R decreases to 0, i = .
R+r r The temperature of the wire increases to such a
value at which, the heat produced per second
Similarly, potential difference V = iR when R
equals heat lost per second due to radiation ie,
decreases to 0,

V = 0.
2
I ( )
ρl
πr
2 = H×2πrl, where H is heat lost per

319 (b) second per unit area due to radiation.

2 3
Kirchhoff’s second law is based on the law of Hence, I ∝ r
conservation of energy 2 3
I1 r1
2 = 3 or r2 = r1(I2/I1)
2/3
So
320 (b) I2 r2
1/3
Thermo-electric power P =
dE
; at tn, E→ = 1×(3.0/1.5)2/3 = 4 mm.
dθ
maximum 327 (d)

So P→ zero Charge = Current × Time = 5×60 = 300 C

321 (a) 328 (d)

2
Vt 1 V
Heat produced H = = H ∝ Hence Resistance, R = = cot 40°
4.2R R i
H1 R
= 2 329 (a)
H2 R1

322 (c)
P a g e | 217
 proportional to ECE of the substance.

335 (a)

3×2
In series, i = …(i)

( )
3r+2
GS
G= + S''
G+S 2 2×3
In parallel, i = = …(ii)
2 r/3+2 r+2×3
GS '' '' G
G- = S ∴S =
G+S G+S From Eqs. (i) and (ii), 3r + 2 = r + 6 or r = 2Ω
330 (c) 3×2
From eqs. (i), i = = 0.75A
3×2+2
E - V = ir or r = (E-V)i
336 (c)
and V = iR = 4.5×10 = 45V

(50-45) 5
∴ r= =
4.5 4.5

= 1.10 Ω

331 (a)

R1 P 100 5
= 2 = = . Resistance of 40 W bulb is
R2 P1 40 2
5 2
times than 100 W. In series, P = i R and in
2
2
V
parallel, P = . So 40 W in series and 100 W in
R R×16
∴ + 10 = 18, on solving we get, R = 16Ω
parallel will glow brighter R+16

332 (a) 337 (a)

A particular temperature, the resistance of a When wire is stretched to doubled its length, its
superconductor is zero resistance becomes four times

1 1 338 (d)
⇒G = = =∞
R 0
Current in wire i = Anevd
333 (b)
28 -3
Here, i = 1A, n = 8×10 electron/m
2E
In series, i1 =
2+2r A = 5×10 m
-7 2

E 2E 28
∴ 1 = 8×10 ×1.6×10 ×5×10 ×vd
-19 -7
In parallel, i2 = =
r 4+r
2+
2 1
or vd = 28 -19 -7
8×10 ×1.6×10 ×5×10
2E 2E
Since i1 = i2⇒ = ⇒r = 2Ω
4+r 2+2r Now,

334 (b) 1 28 -19 -7

t= = 8×10 ×1.6×10 ×5×10
vd
The amount of decomposition (ie, mass of the
substance liberated during electrolysis) is 2
= 64×10 = 6.4×10 s
3

P a g e | 218
339 (c) A shunt connected in parallel is given by

Voltmeter is an instrument which measure the IgG

S=
potential difference between two points. A high I-Ig
resistance is connected in series with coil of the
galvanometer to convert it into voltmeter. This If I < Ig, then the value of S is negative. Hence, a
resistance is either 2000Ω or more than that. galvanometer cannot be converted into an
ammeter of range I < Ig.
Resistance connected in series is given by
342 (c)
V
R= -G ..(i)
Ig When corrent is passed through a junction of two
different metals, the heat is either evolved or
Eq. (i) is the value of the resistance required to absorbed at the junction. This effect is known as
convert the galvanometer to voltmeter of range 0 Peltier effect.
to V.
343 (b)
From the relation
P1 = P2 = 60 W; when bulbs are connected in
V
R = -G series then total power
Ig
P1P2 60×60
nV Ps = = =30 W
R= -G P1+P2 60+60
()
V
G 344 (a)
or R = nG - G = G(n-1) The current in the circuit are assumed as shown
Hence, R = (n - 1) G in the fig.

340 (c)

We will require a voltmeter, an ammeter, a test

resistor and a variable battery to verify Ohm’s
law.

Voltmeter which is made by connecting a high

resistance with a galvanometer is connected in
parallel with the test resistor.

Further, an ammeter which is formed by Applying KVL along the loop ABDA, we get
connecting a low resistance in parallel with -6i1 - 3 i2 + 15 = 0 or 2i1 + i2 = 5 ….(i)
galvanometer is required to measure the current Applying KVL along the loop BCDB, we get
through test resistor. -3(i1-i2) - 30 + 3i2 = 0 or -i1 + 2i2 = 10 ….(ii)
Solving equation (i) and (ii) for i2, we get i2 = 5 A
The correct option is (c).

341 (d) 345 (a)

2
A galvanometer can be converted into an E = αt + bt . At temperature of inversion E is
ammeter by using a low resistance wire in parallel minimum
with the galvanometer. The range of ammeter can i.e., E = 0
be increased but cannot be decreased. The reason
is that a series resistor cannot change the current 2 a
∴ αti + bti = 0, i.e., t1 = -
that will produce full deflection. b

P a g e | 219
346 (c) From Eqs. (i) and (ii)

1 V 2V
R∝ ;where τ = Relaxation time =
τ 100+R 100+R

When lamp is switched on, temperature of R = 900Ω

filament increases, hence τ decreases so R
increases 351 (d)

347 (a) The rate at which heat is developed

2
Slope of the V - i curve at any point equals to V (110)2
H= = = 1210 W
resistance at that point. From the curve slope for R 10
T1 > slope for T2⇒RT > TT . Also at higher 352 (b)
1 2

temperature resistance will be higher so T1 > T2

Shunt of an ammeter,
348 (b)
Ig×G

[∫ ]
3 3 S=
t=3 I-Ig
∫ ∫t dt
2
dQ = Idt⇒Q = Idt= 2 tdt+3
t=2
2 2 5×G G
= =
100-5 19
= [t ]2 + [t3]32 = (9-4) + (27-8) = 5 + 19
2 3

= 24C 353 (b)

A = πr = ρl Ror r = (ρl πR)

2 1/2
349 (d)

( )
-8 1/2
R 1.7×10 ×0.5
i= r= = 0.367 mm
r+R 3.14×2
2
P=iR 354 (c)
2
ER If a charged particle of charge q resolves in a
⟹ P=
(r+R)2 circular orbit of radius r with frequency v, then
the orbital current is given by
Power is maximum when r=R
2 I = qv
E
∴ Pmax =
4R

E
2
or I=q
ω
2π (
∵=2πv⟹v=
ω
2π )
or Pmax =
4r ev
I= (∵v=rω)
2πr
350 (c)
355 (d)
When a resistance of 100Ω is connected in series
current, For maximum power r = R

V 356 (a)
i= ……….(i)
100+R
Initially : Resistance of given cable
When a resistance of 1000Ω is connected in
series, the its range double l
R=ρ …(i)
π×(9×10 )
-3 2

2V
Current, i = ……(ii)
1100+R Finally : Resistance of each insulated copper wire

P a g e | 220
 is 1 1 1
'' = +
R 6 6
' l
R =ρ -3 2 . Hence equivalent resistance
π×(3×10 )
''
R = 3Ω
of
Now between points A and B 1Ω, 3 Ω and 1Ω are

( )
'
R 1 l in series.
cable Req = = × ρ …(ii)
π×(3×10 )
-3 2
6 6
Therefore, resultant resistance is
l l
R=1+3+1=5Ω


9 mm 359 (b)

(vd)e ie nh 7 5 5
on solving equation (i) and (ii), we get Req = 7.5 i = neAvd⇒ = × = × =
(vd)h ih ne 4 7 4
Ω
360 (c)
357 (b)
Electroplating only provides a thin deposition of a
From the Ohm’s law
metal on the surface which in no way can give
2E hardness to the metal
I=
2r+R
361 (b)
In parallel combination of two cells, the current
Yellow, Violet and Gold
through the external resistance r will be
362 (c)
' E 2E
I = =
r r+2R When there is no deflection, then this
+R
2 temperature is called inversion temperature. It is
given by the relation
If I = I’ then 2r + R = r + 2R
θi+θc
⟹ r = R = 3Ω θn =
2
358 (a)
Where θc is temperature of cold junction = 20℃
Equivalent circuit of the given circuit is and neutral temperature θn = 270℃

1Ω C 2Ω ∴ θi = 2θn - θc = 540 - 20 = 520℃

A

363 (b)
6Ω 2Ω
E1 l 1.08 400
= 1⟹ =
B
E2 l2 E2 440
1Ω D 2Ω

440×1.08
⟹E2 = = 1.188V.
400
Between points C and D resistors 2Ω, 2Ω and 2Ω
364 (a)
are in series, therefore, their equivalent
resistance, The relative position of metals in the electro
' chemical series determines the emf between the
R = 2 + 2 + 2 = 6Ω
two metals placed in an electrolyte.
Resistors R’ and 6Ω are in parallel, therefore their
365 (d)
equivalent resistance is given by
P a g e | 221
 e R 2 10 E 2
E= ×l = × ×0.4 i= = = 0.5;
(R+Rh+r) L (10+40+0) 1 R+r 3.9+0.1
= 0.16V
V = E - ir = 2 - 0.5×0.1 = 1.95V
366 (b)
368 (d)
I = 4 - 0.08t A
For greater sensitivity of meter bridge the
dq resistance (R) taken in the resistance box should
Or = 4 - 0.08t A
dt be such that the null point is nearly in the middle
of the wire. In this position all resistance P, Q, R
50
or q =
∫ 0
(4-0.08t)dt C and S become nearly equal. The emf of cell
depends upon the size and area of electrodes.

[ ]
2 50
0.08t 369 (c)
or Ne = 4t- = 100 C
2 0
I 1 V
vd = = ×
Where N is number of electrons. nAl nAe R

100 100 1 V V
or N = = -19 = × =
e 1.6×10 nAe (ρl/A) neρl
20
= 6.25×10 As vd is independent of area of cross-section
hence drift velocity will not change, when
367 (c) diameter is doubled

370 (c)

Given circuit can be redrawn as follows

R
⇒Req =
2

P a g e | 222
371 (c) 12×4
R= + 2 = 5Ω
12+4
R AC 20
By Wheatstone bridge, = = ⇒R = 20
80 BC 80 E 12
I= = = 2A
Ω R+r 6

372 (a) I1 + I1 = 2A
2
V 1 1
P= ⇒P ∝ and R ∝ l I∝
R R R

1 P1 l 2 ∴ I1 = 0.5A, I2 = 1.5A
∴P∝ ⇒ = 2 =
l P2 l1 1
377 (a)
373 (a)
Let the temperature of molten metal is t℃
It is found that temperature of inversion (Ti) is as -6
The thermo-emf e = 10×10 t volt
much above the neutral temperature (Tn) as
neutral temperature is above the temperature of e 10 t
-5
10 t
-5

the cold junction (T), ie, Current in the circuit = = amp

R+RG 8+1.6 9.6

Ti - Tn = Tn - T or Ti = 2Tn - T V 8×10
-3
But i = =
RG 8
But, here the cold junction is kept at 0℃, hence
T = 0. -5
10 t 8×10
-3
9.6×10
-3
∴ = or t = -5 = 960℃
9.6 8 10
Ti
Thus, Ti = 2Tn or Tn =
2 378 (a)
374 (b) V
2
P=
R
2
Resistance across XY = Ω
3 V
2
∴ Rhot =
P

200×200
2 2V = = 400 Ω
A 1000
2
X Y
400
2 Rcold = = 40 Ω
10
2
379 (b)

Given, E = 2.2V, R = 5Ω
Total resistance
V = 1.8V, r = ?
2 8
=2+ = Ω Now current in the circuit
3 3

Current through ammeter E

i=
R+r
2 6 3
= = = A
8/3 8 4

375 (c)

P a g e | 223
 V E circuit, hence it can be redrawn as follows
or =
R R+r
12 12
4 8
1.8 2.2
= a
5 5+r b

a b

10 2 4
or r = Ω 6
9 6

380 (a)
12×6
E1E2 = l1/l2. As, E1/E2, therefore l1 > l2. Therefore RAB = = 4Ω
(12+6)
the null point for thr cell of emf E2must be at
383 (b)
shorter length than that of cell E1.Thus null point
on potentiometer wire should shift towards left of In general, ammeter always reads less than the
C. actual value because of its resistance

381 (a) 384 (a)

Current given by cell In the circuit shown total external resistance

R = 2Ω+ parallel combination of two 4 Ω
E
I= resistors + parallel combination of three 15 Ω
R+r
resistors
Power delivered in first case
4 15
=2+ + = 2 + 2 + 5 = 9Ω

( ) 2 3
2
2 E
P1 = I R 1 = R
R1+r 1
As E = 10V and i = 1A, hence internal resistance
Power delivered in second case r of the cell should have a value given by

( )
E
2 E 10
P2 = I R 2 =
2
R E = i(R+r) or r = -R = - 9 = 1Ω
R2+r 2 i 1

If 4Ω resistors are replaced by 2Ω resistors , then

Power delivered is same in the both the cases.
as before

( ) ( )
2 2
E E
R1 = R ' 2 15
R1+r R2+r 2 R =2+ + = 2 + 1 + 5 = 8Ω
2 3
R1 R2
2 = ' E 10
(R1+r) (R2+r)2 ∴ New circuit current i = '
R +r
=
8+1
= 1.11A

R1(R2+r +2R2r) = R2(R1+r +2R1r)

2 2 2 2
385 (c)
2 2 2 2
R1R2 + R1r + 2R1R2r = R2R1 + R2r + 2R1R2r Let G be resistance of galvanometer and ig the
2 2 2 2
current which on passing through the
R1R2 - R2R1 = R2r - R1r galvanometer produces full scale deflection. If i is
the maximum current, and since, G and S are in
R1R2(R2-R1) = r (R2-R1)
2

parallel.
r = R1R2

382 (b)

Given circuit is a balanced Wheatstone bridge

P a g e | 224
 5
H 34×10 ×4.2
V= = = 1.48 = 1.5V.
i ig G i q 96500,000

390 (a)

The reciprocal of resistance is called conductance

S
391 (d)
Ammeter 2
In case of stretching of wire R ∝ l

⇒ If length becomes 3 times so Resistance

'
becomes 9 times i.e. R = 9×20 = 180Ω

392 (b)
ig×G = (i-ig)×S

ig S
⟹ =
i S+G

Given, G=36Ω, S=4Ω

ig 4 4 Since deflection in galvanometer is zero so

∴ = =
i 36+4 40 current will flow as shown in the above diagram
i VA
⟹ig = 12 12
10 Current I = = =
R1+R 500+100 600
ig 1
∴ 100× = ×100 So VB = IR
i 10
12
i = ×100 = 2V
⟹ e % = 10% 600
i
393 (c)
386 (d)
nE
R = 91×10 ≈ 9.1kΩ
2
In series combination of cells current, i =
nr+R
387 (c) E
'
In parallel combination of cell, i =

( )
(l1-l2) ' (r n)+R
60-50
r= ×R = ×6 = 1.2 Ω
l2 50 ' nE E nE
If i = i then = =
nr+R (r n)+R r+nR
388 (c)
It will be so if r = R
ρ(L) ρL ρ
R= = =
A tL t 394 (b)
ie, R is independent of L. 240 ×240
Resistance of 40 W bulb =
40
Hence the correct option is (c).
= 1440 Ω
389 (c)
240
m 1 It’s safe current = = 0.167 A
q= = = 96500,000 C 1440
z (1/96500,00)

P a g e | 225
 240 ×240 400 (c)
Resistance of 60 W bulb =
60
R1 and R2 are in series
= 960 Ω
∴ R12 = R1 + R2 = 4Ω
240
It’s safe current = = 0.25 A
960

When connected in series to 420 V supply, then C 2Ω

the current D
R2
2Ω
420 420 R
I= = 4
1440+960 2400 R3 E
R1 2Ω
4Ω R 5
= 0.175 A 4Ω R6 2Ω
R7
Thus, current is greater for 40 W bulb, so it will B
A 8Ω
fuse.

395 (d)
R12and R3 are in parallel
The sensitivity of the thermocouple will be

= 500μV/℃ - (-72μV/℃) = 572μV/℃ R3×R12

R123 =
R3+R12
Therefore for a 100℃ temperature difference, the
thermo e.m.f. will be 4×4
=
4+4
-6 -3
E = 572×10 ×100(volt) = 57.2×10
= 57.2 mV = 2Ω

396 (d) R123and R4 are in series

The resistance of an ideal voltmeter is considered ∴ R1234 = R123 + R4

as infinite
= 2Ω + 2Ω
397 (b)
= 4Ω
For power to be maximum
R1234 and R5 are in parallel
External resistance = Equivalent internal
resistance of the circuit ∴ R12345 = 2Ω

398 (a) R12345 and R6 are in series

vd =
eE
m
e V
τ= τ
m l ()
or Vd∞V
2Ω + 2Ω = 4Ω

= R123456
Therefore, drift velocity is doubled
Now, R123456and R7are in parallel
399 (c)
4×8
The specific resistance (ρ) is the characteristic of ∴ Rcomb =
4+8
the material of conductor. Its value depends only
on the material of conductor and its temperature. 32 8
= = Ω
Its value does not depend on the length and area 12 3
of cross-section of the conductor.
P a g e | 226
401 (c) 2 3
Volta
-6
l 64×10 ×198
R = ρ ⇒7 = ⇒r = 0.024cm
A 22 2 i
×r
7 V

402 (a)

Drift velocity is defined as the average velocity V V

Finally main current i = =
with which free electrons get defined towards the 3+1 4
positive end of the conductor under the influence V
of an external electric field. Hence current through voltameter i2 =
8
Drift velocity is given by 2

eEτ
vd = Volta 2
m
i
V V
But E=
l

(if l is length of the conductor and V is constant

potential difference applied across the ends of the m
∵ Rate of deposition (R) = = Zi⇒R ∝ i
conductor) t

eVτ R2-R1 i -i
∴ vd = ⟹vd ∝ V ∴ % drop in rate = ×100 = 2 1 ×100
ml R1 i1

So, when the potential difference is doubled the

drift velocity will be doubled. =
V V
-
8 5( )
×100 = -37.5%
V
403 (c) 5

Net voltage =12-8=4V 405 (b)

Net resistance =9+2+1=12Ω Pt = mSθ

Current through the circuit 1×4200×15

P= W= 1050 W
60
4 1
= = A
12 3 406 (d)

Terminal voltage across E2 is Heating effect of current

1 407 (c)
= E2 - Ir2 = 12 - ×2
3
230×230
R= = 529 Ω
2 34 100
= 12 - = = 11.34 V
3 3
2
V 115×115
∴H = ×t = ×10×60 = 15 kJ
404 (d) R 529

Initially current through the voltmeter 408 (b)

V V
i1 = = Potentiometer is based on null deflection
(3+2) 5
409 (c)

P a g e | 227
 Due to the negligible temperature co-efficient of Or P1 = 30 W, P2 = 20 W
resistance of constantan wire, there is no change
in it’s resistance value with change in temperature P1 3
∴ =
P2 2
410 (a)
414 (c)
Neutral temperature is defined as temperature of
a hot junction of a thermocouple at which the Lowest resistance will be in the case when all the
electromotive force of the thermocouple attains resistors are connected in parallel
its maximum value when cold junction is
1 1 1
maintained at a constant temperature of 0℃. = + …. 10 times
R 0.1 0.1
Hence, for a given thermocouple neutral
temperature is a constant. 1
= 10 + 10…… 10 times
R
411 (a)
1 1
For one wire cable, = 100 i.e. R = Ω
R 100
Resistance, R = ρl/π(9×10 )
' -3 2
= 5Ω 415 (d)
For other wire of cable, e e ev
i= = =
t 2πr/v 2πr
Resistance, R = ρl/π(3×10 )
' -3 2

2
2 2 e 2 2
= 9 ×5/3 = 45Ω Here, v = and r = h /me
h
When six wires each of resistance R’ are 2 3 2 5
e(e h) e ×me me
connected in parallel, their effective resistance ∴i = 2 2 = 3 = 3
2π(h /me ) 2πh 2πh
will be
2 5
4π me
R' 45 i=
Rp = = = 7.5Ω h
3
6 6
416 (c)
412 (a)
For the same length and same material,
After short circuiting, R2 becomes meaningless
R2 A 3
413 (c) = 1 = or R2 = 3R1
R1 A2 1
In series,
The resistance of thick wire, R1 = 10 Ω
P ×P
Ps = 1 2 = 12 W The resistance of thin wire
P1+P2
= 3R1 = 3×10 = 30 Ω
In parallel, Pp = P1 + PP2 = 50 W
Total resistance = 10 + 30 = 40 Ω
∴ P1P2 = 12×50 = 600
417 (c)
Now, (P1-P2) = (P1+P2) - 4P1P2
2 2

Resistance of copper part of wire

= (50)2 - 4×600 ρ .L ρ .L
Rc = c = c 2 and
= 2500 - 2400 = 100 Ac πr

∴ P1 - P2 = 10 Resistance of nickel portion of wire

P a g e | 228
 ρc.L ρc.L
Rn = = 2 2
An π(R -r )
ie., i2×1.5R = 3R×i3
As these two resistances are in parallel, hence
conductance And i2 + i3 = i

of the nickelled wire 2i i

⟹ i2 = and i3 =
1 1 1 πr
2 2 2
π(R -r ) 3 3
C= = + = +
R Rc Rn ρc.L ρn.L
Now, Vi = iR

[ ]
2 2 2
π r R -r
= + 2i
L ρc ρn V2 = ×1.5R = iR
3
418 (a)
i
V3 = ×3R = iR
All the conductors have equal lengths. Area of 3
cross-section of A is {( 3a) -( 2a) } = a
2 2 2
ie, V1 = V2 = V3
Similarly area of cross-section of B = Area of
2
421 (a)
cross-section of C = a
3×6
l Equivalent resistance R = 4 + = 6Ω and
Hence according to formula R = ρ ; resistances 3+6
A E 3
of all the conductors are equal i.e. RA = RB = RC main current i = = = 0.5A
R 6

419 (c) Now potential difference across the combination

R 20×10 
V
3
of 3Ω and 6Ω, V = 0.5×
3×6
3+6 ( )
= 1Volt

5V The same potential difference also develops

i
110V across 3Ω resistance

422 (c)
110
Here i = 3
20×10 +R I 20
vd = = 29 -6 -19
nAe 10 ×10 ×1.6×10
∵ V = iR⇒5 = ( 110
3
20×10 +R
×20×10 )
3 -3
= 1.25×10 m/s

5
423 (c)
5 10 5
⇒10 + 5R = 22×10 ⇒R = 21× = 420 KΩ
5 In series combination, the net resistance

420 (a) Rs = R + 2R = 3R

Here, V2 = V3 Heat produced in Rs,

1.5 R 2 2
V V
Hs = = …(i)
Rs 3R
R V2
i I parallel combination, the net resistance
2
V1 V3 R×2R 2R 2
RP = = = R
R+2R 3R 3

3R
Heat produced in RP,

P a g e | 229
 2 2 2
V V 3V =10 - 0.5 × 3
HP = = = …(ii)
RP 2R/3 2R
=10 – 1.5= 8.5 V
Dividing Eq. (i) by Eq. (ii), we obtain
429 (a)
2
Hs V /3R 2
= 2 = θ θ σ
Hp 3V /2R 9 σi = = .G = σVG⟹ i = σv
i iG G
424 (d)
430 (b)
E = 2.2 V, R = 4 Ω, V = 2V 2
V
Energy liberated = t
r= ( ) ( )
E
V
-1 R =
2.2
2
-1 ×4 = 0.1×4 = 0.4Ω
(120)2 ( 5
R

= × 10×60) = 14.4×10 J
425 (a) 6

Thermo electric power, S ∝ θ 431 (c)

S100 l
100 100 5 R=ρ
∴ = or S100 = S80× = S80 A
S80 80 80 4
432 (a)
Therefore % change in thermo electric power

( )
As the current in heater filament increases, it gets
S100-S80
= ×100 more heated, hence its temperature increases and
S80
thereby its resistance increases. Due to which the

( )
5 current will decrease. Hence the variation of V
S -S and i for heater filament will as shown in Fig.(a)
4 80 80
= ×100
S80
433 (d)
= 25%
By using H = σQ∆θ
426 (c)
⇒H = (10×10 )×10×(60-50) = 10 J = 1 mJ
-6 -3

The bulbs are in series, hence they will have the

434 (b)
same current through them
2
ENi = at + bt
427 (c) -Cu

= (16.3×10 )(100) + (-0.021×10 )×(100)2

-6 -6
E
By using i =
R+r -3
= 1.42×10 V
E
⇒0.5 = ⇒E = 5.5 + 0.5r …(i) 435 (d)
11+r
V
E From the relation, current i =
and 0.9 = ⇒E = 4.5 + 0.9r …(ii) r
5+r
2
On solving these equations, we have r = 2.5Ω or 4 =
r
428 (d)
1
or r = Ω
Terminal voltage of the battery after closing the 2
circuit is
436 (b)
V = E - ir
P a g e | 230
 P R R1R2 2×8 8
For balanced Wheatstone bridge = Hence, Req = = = Ω
Q S R1+R2 (2+8) 5

12 x+6 446 (c)

⇒ = ⇒x = 6Ω
(1/2) (1/2)
In the steady state, no current flows through the
437 (c) branch containing the capacitor. So, the
equivalent circuit will be of the form as shown
On doubling the length of wire its resistance is
below:
doubled and slope of V - I graph is doubled

438 (d) 2Ω

Heat generated in both the cases will be same

3Ω
because the capacitor has the same energy
initially
2.8
1 2 1 -6
= CV = ×200×10 ×(200)2 = 4J E=6V
2 2

439 (a)
The effective resistance of the circuit is
E ∝ l (balancing length)

440 (a)
2×3
The first two bands indicate the first two R= + 2.8 = 1.2 + 2.8 = 4Ω
2+3
significant figures of the resistance in ohm. The
third band indicates the decimal multiplier and The current through the circuit is
the last band stands for the tolerance in percent i1 2Ω
about the indicated value
i i
443 (d)

Effective resistance between the points A and B is (i - i1) 3Ω

32 8
R= = Ω
12 3
E 6
i= = = 1.5A
444 (b) R 4

R1R2 Let current i1 flows through 2Ω resistance.

R1 + R2 = 9 and = 2⇒R1R2 = 18
R1+R2
∴ 2×i1 = (i-i1)×3
R1 - R2 = (R1+R2) -4R1R2 =
2
81-72 = 3
⟹ 2i1 = (1.5-i1)×3
R1 = 6Ω, R2 = 3Ω
⟹2i1 = 4.5 - 3i1
445 (b)
⟹5i1 = 4.5

ρ- same, l- same, A2 =
1
4 [ r
A1 as r2= 1
2 ] ⟹i1 = 0.9A

447 (a)
l R A R 1
By using R = ρ ⇒ 1 = 2 ⇒ 1 = ⇒R1 = 2Ω
A R2 A1 8 4 The equivalent resistance between C and D is

P a g e | 231
 1 1 1 1 2 ' 3 450 (b)
= + + = or R = = 1.5Ω
R' 6 6 3 3 2
4 24
8
Now the equivalent resistance between A and B as 20
' 16
R = 1.5Ω and 2.5Ω are connected in series, so 4 20
A 16 B A B

R" = 1.5 + 2.5 = 4Ω
6 6
9
Now by ohm’s law, potential difference between A 6 12
and B is given by VA - VB = iR = 2×4.0 = 8 volt 6 18

448 (d)
24×12
RAB =
To find equivalent resistance across (24+12) = 8Ω

BC, AB and AC is in series 451 (c)

'
R = 6 + 6 = 12Ω Let resistors A, B and C have equal resistance R.

12 Ω and 6Ω is in parallel Let I be the total current then the current in

resistor A is I and in resistor B and C are I/2.
A

So, heat produced in resistor A is

6 6
2
HA = I R …(i)

B C
and heat produced in resistor B is
6

()
2 2
I IR
HB = R= …(ii)
2 4
Total resistance,
and heat produced in resistor C is
12×6
R= 2
12+6 IR
HC = …(iii)
4
72
R= = 4Ω
18 Hence, it is clear that the heat produced will be
maximum in A.
449 (d)
452 (c)
R
R200 = 800
1+αt q 4
i= = = 2 ampere
t 2
R800 R
R200 = -4 = 800 453 (d)
1+4×10 ×600 1.24

V
2 120×120
=P ……(i) R60 = = 240Ω
R800 60

2
V
and = P' ……(ii)
R200

P' R R800
∴ = 800 =
P R200
( ) R800
1.24
120 120
Current = A= A
'
P = 1.24P = 1.24×500 = 620 W 240+6 246
P a g e | 232
 Voltage across bulb ∆R ∆D ∆R
= -4 or = -4×0.25 = 1.0%
R D R
120
= ×240 volt = 117.1 volt
246 456 (c)

Here, 2Ω, 3Ω and 6Ω are in parallel. So potential

drop across them will be the same. As heat
2
V 1
produced, = t ie, H ∝ , so maximum heat will
R R
be generated across 2Ω resistance. Similarly 4Ω
and 5Ω are also in parallel, so more heat will e
120×120 generated across 4Ω. Now the effective circuit will
R240 = 60Ω
240 become

Resistance of parallel combination 1Ω 20/9 Ω

60×240 Vi Ve
= = 48Ω.
60+240
I
Total resistance = (48+6)Ω = 54Ω.
V
120
Current I = A
54
20 29
Voltage across parallel combination Total resistance, = 1 + = Ω
9 9
120
= ×48 volt = 106.7 volt V 9V
54 Current, I = = A
29/9 29
Change in voltage = (117.1-106.7) = 10.4 V.
9V 9
V1 = ×1 = V
454 (c) 29 29

V
2
R (200)2 4 9V 20 20V
R=
2
or R ∝ V ∴ 1 = = . and V2 = × =
P R2 (300)2 9 29 9 29

Power spent across 2 Ω,

When bulbs are connected in series, the current I

( )
2
is same through each. As P = I R or P ∝ R( as I is 9V
2
2
same in series), V1 29 40.5V
2
P1 = = =
2 2 (29)2
P1 R 4
so = 1 = .
P2 R2 9 Power spent across 4 Ω,

( )
2
455 (c) 20V
2 2
V2 29 50V
On stretching, volume (V) remains constant. So P2 = = =
4 4 (29)2
V = Al or l = V/A. ∴ P2 > P1. Hence maximum heat is produced in
ρl ρV ρV 16ρV 4Ω resistance.
Now, R = = 2 = 2 4 = 2 4
A A π D /16 πD 457 (a)
Taking logarithm of both the side and When ammeter is connected in parallel to the
differentiating it we get circuit, net resistance of the circuit decreases.
Hence more current is drawn from the battery,

P a g e | 233
 which damages the ammeter Potential gradient

458 (a) e R
x= .
(R+Rh+r) L
According to kirchhoff’s voltage law only option
(a) is correct 2 10
⟹ x=
(990+10) × 2
459 (c)
-1
= 0.01 Vm
80
Total resistance of the circuit = + 20 = 60 Ω
2 462 (d)

2 1 ig S 4 4 1
⇒ Main current i = = A = = = =
60 30 i G+S 36+4 40 10
Combination of voltmeter and 80Ω resistance is 463 (b)
connected in series with 20Ω, so current through
1 Let the voltage across any one cell is V, then
20Ω and this combination will be same = A
30
V
Since the resistance of voltmeter is also 80Ω, so
r1 r2
this current is equally distributed in 80Ω
1 E E
resistance and voltmeter [i.e. A through each]
60
R
1
P.D. across 80Ω resistance = ×80 = 1.33V
60

460 (c) V = E - ir = E - r1 ( 2E
r1+r2+R )
But V = 0

2Er1
⇒E - =0
r1+r2+R

⇒r1 + r2 + R = 2r1
Here, emf of each cell, ε = 0.2V
⇒R = r1 - r2
Internal resistance of each cell, r = 1Ω
464 (b)
External resistance, R = 10Ω
The given circuit can be redrawn as
The total emf of 5 cells = 5ε = 5(0.2)V = 1V

Total internal resistance of 5 cells

= 5r = 5(1)Ω = 5Ω

Total resistance of the circuit

= R + 5r = 10 + 5 = 15Ω

The current in the external circuit,

Which is a balanced Wheatstone’s bridge and
5ε 1V 1
I= = = A hence, no current flows in the centre resistor, so
R+5r 15Ω 15
equivalent circuit would be as shown below.
461 (c)
P a g e | 234
 R4 = 10 + 6 = 16Ω
10Ω 20Ω
30Ω

Now, R4and 16 Ω are in parallel

1 1 1
I 5Ω 10Ω
15Ω ∴ = +
I R 16 16
5V 5V
⇒ R = 3Ω
30 Ω || 15 Ω

470 (d)
I
(220)2 (220×0.8)2
P1 = and P2 =
5V R1 R2

P2 (220×0.8)2 R1 P2 ( )2 R1
= × ⇒ = 0.8 ×
P1 (220)2 R 2 P1 R2
V 5 Here, R2 < R1 (because voltage decreases from
So, I = = = 0.5A
R 10
220 V→220×0.8 V
465 (b)
It means heat produced → decreases)
q = it = n×2e
R1
So > 1⇒P2 > (0.8)2P1⇒P2 > (0.8)2×100 W
it 2×32 20 R2
n= = -19 = 2×10
2e 2×1.6×10
P2 (220×0.8)i2
Also = , Since i2 < i1 [we expect]
466 (a) P1 220i1

Rate of flow of electrons in a conductor is low but P2

number density of free electrons in a conductor is So < 0.8⇒P2(100×0.8)
P1
very high. The drifting of electrons over the entire
length of the conductor contributes to the current Hence the actual power would be between
throughout the conductor 100×(0.8)2W and

467 (b) (100×0.8) W

m = Z i t ⇒20×10 =
-3
( 32
96500)×0.15×t
471 (a)

vd = i n Ae;where n = Nρ/M
= 6.7 min = 6 min.42 sec
26 3 29
= 6.023×10 ×9×10 /63 = 0.860×10
469 (a)
28
= 8.6×10
1 1 1 5 1
= + = = ⇒R1 = 2Ω
R1 10 2.5 10 2 2 22 ( -3)2 2
and A = πD 4 = × 10 4m ;
7
Now 2Ω and 10 Ω are in series
11 -6 2
= ×10 m
R2 = 10 + 2 = 12Ω 14

R2 and 12Ω are in parallel 1.1

vd =
1128 -6 -19
8.6×10 × ×10 ×1.6 ×10
1 1 1 14
= + ⇒R3 = 6Ω
R3 12 12

Now R3and 6Ω are in series

P a g e | 235
 -4
1 100×10 -4 n
= +3 = = 1.0×10 m/s So main current i = A
9.6×10 96 2

= 0.1 mms
-1
Also E = V + ir⇒120 = 100 + () n
2
×10⇒n = 4
472 (c)
476 (c)
Charge flowing in 30 min,
' R 1
R = = = 0.1Ω
q = area under graph n 10

= [ 0.1×10
2
+0.1×10+
0.1×10
2
×60 ] 477 (c)

igG 10×99
= 120C Shunt resistances S = = = 11Ω
(i-ig) (100-10)

M 31.5 478 (a)

∴ m = zq = q= ×120 = 0.039 g
F 96500

473 (b) V
Potential gradient x = =
iR
=
( )
i
ρL
A
=
iρ
L L L A
When wire is cut into two equal parts then power
dissipated by each part is 2P1 479 (a)

So their parallel combination will dissipate power The given circuit can be simplified as follows

P2 = 2P1 + 2P1 = 4P1 2 18 2

4.5
7
P2 15V 15V
Which gives =4 6 1  6 18
P1
0.5 0.5
10
474 (c)
8 8
t2 5
dQ
i=
dt
⇒dQ = idt⇒Q =
∫ idt=∫ (1.2t+3)dt
t1 0 On further solving equivalent resistance R = 15Ω

[ ]
2 5
1.2t 15
= +3t = 30C Hence current from the battery i = = 1A
2 0
15

475 (c) 480 (b)

When each bulb is glowing at full power, Let R be the resistance of each bulb.
2
50 1 V
Current from each bulb = i =
'
= A We have P =
100 2 R
2 2
V V
1 or R = =
P 100
2
When the bulbs are connected in series, the
n
i voltage across each of them is V/2. Hence, the
100 V total power consumed is

() ()
2 2
120 V, 10  V V
2
' 2 2 V
P = + =
R R 2R

P a g e | 236
 ' P 100 485 (b)
∴P = = =50 W
2 2
W
W = qV also P = i×V =
481 (d) t

ρl 486 (c)
R=
A
According to Faraday’s first law of electrolysis is
or R ∝ l
m = zIt
R l L
∴ 1 = 1 = =4 where z is the electrochemical equivalent of the
R2 l2 L/4
substance
R
Or R2 =
4
(∵R1=R) As voltameters are connected in series, same
current will pass through them for same time
In parallel combination of such four resistances.
mCu z z m 0.475
∴ = Cu or Cu = Cu = = 3.65
1 1 1 1 1 mCr zCr zCr mCr 0.130
' = + + +
R R2 R2 R2 R2
487 (a)
1 1 1 1 1
or ' = + + + Given circuit is a balanced Wheatstone bridge. So,
R R/4 R/4 R/4 R/4
diagonal resistance of 2Ω will be ineffective.
1 4 4 4 4
or ' = + + +
R R R R R
2Ω 2

1 16
or ' =
A B
R R 2Ω

2 2
R '
or R =
16

482 (d) Equivalent resistance of upper arms

ig S 5 S G =2+2=4Ω
= ⇒ = ⇒S =
i G+S 100 G+S 19
Equivalent resistance of lower arms
483 (c)
=2+2=4Ω
60×8×30
Total kWh consumed= = 14.4 4×4
1000 RAB = = 2Ω
4+4
Hence cost = 14.4×1.25 = 18 Rs
488 (d)
484 (d)
In series, effective resistance,

()
2
l1 l R l A l r
R1 = ρ and R2 = ρ 2 ⇒ 1 = 1 . 2 = 1 2 1 1 1 1
A1 A2 R2 l2 A1 l 2 r1 Reff = R1 + R2 + R3⇒ = + +
σeff σ1 σ2 σ3
l1 1 r 2 r 1 R 1
Given = and 1 = or 2 = ⇒ 1 = σ2σ3+σ1σ3+σ1σ2
l2 2 r2 1 r1 2 R2 8 =
σ1σ2σ3
2
H V /R R 8 σ1σ2σ3
∴ Ratio of heats 1 = 2 1 = 2 = ∴ σeff =
H2 V /R2 R1 1 σ2σ3+σ1σ3+σ1σ2

P a g e | 237
489 (b)

The resistance of a metal increases with

increasing temperature this is because, with
increase in temperature the ions of the conductor
vibrate with greater amplitude and the collision
between ions and electrons becomes more R'' is given by
frequent.
1 1 1 50+R
490 (a) '' = + =
R R 50 50R

The power drawn by the bulb is

V
2
⇒R =
''
( )
50R
50+R
P=
R According to ohm’s law
2
V ''
V = IR''
⇒R =
P

or R ∝
1 ⇒
100
3
= I. ( )
50R
50+R
P

(as V is same in parallel) ⇒

3 50R ( )
100 50+R
= I …(ii)

It means that greater power will have less Now, total resistance of circuit
resistance and therefore, draws more current.
Hence, current flowing in bulb B will be more. '''' 50R
R = 50 +
50+R
491 (c)
(2500+100R)
⇒R" =
Human body, though has a large resistance of the (50+R)
order, of KΩ(say 10kΩ), is very sensitive to
minute currents even as low as a few mA. Now, V" = IR"

( )
Electrons, excites and disorders the nervous
100 50+R 2500+100R
system of the body and hence one fails to control ⇒100 =
3 50R (50+R)
the activity of the body
⇒150R = 2500 + 100R
492 (c)
2
⇒R = 50 kΩ
V 220×220
Resistance of bulb R = = = 484Ω
P 100 494 (a)

Power when the bulb of operated on a voltage Equivalent resistance in the second case
'2
= R1 + R2 = R
' ' V 110×110
V = 110V will be P = = = 25W
R 484 1
Now, we know that P ∝
R
493 (c)
Since in the second case the resistance (R1+R2) is
Internal resistance of voltmeter is R
higher than that in the first case (R1)
Therefore effective resistance across B and C,
Therefore power dissipation in the second case
will be decreased

495 (c)

P a g e | 238
 2
Vt 1 Here, I = I1 + I2 ….(i)
Heat H = ⇒H ∝ [If V, t constant]
R R
I1 1
( )
R×2R and = =2
I2 2
HS RP 3R 2
⇒ = = =
HP RS (R+2R) 9
1Ω
496 (a) I I1 3Ω
I
W W 1000
P= = Vi⇒V = = = 1.38 V 2Ω I2
t it 2×6×60

497 (b)
∴ I = I1 + I2 = 2I2 + I2
2
V
Heat produced by the heater H = ×t
R I 2I
∴ I2 = and I1
3 3
For 220V heater heat produced
2 2
∴ P = I ×3 = 3I
(220) 2
H1 = ×5
R
( )
2 2
2 2I 4I
P1 = I1×1 = ×1 =
3 9
For 110V heater heat produced
2
2 I 22
(110)2 P2 = I2×2 = ×2 = I
H2 = ×L 9 9
R
2 2
P3 = I ×3 = 3I
Now, H1 = H2
4 2
110×110 220×220×5 ∴ P1:P2:P3 = : = 4 :2 :27
t= 9 9
R R
501 (b)
t = 20 min
The figure can be drawn as follows
498 (c)

Equivalent resistance of the circuit Req = 100Ω

D 7 C C

2.4 10
Current through the circuit i = A 3  5
100 10 5
10
P.D. across combination of voltmeter and 100Ω A B
A B
10 10

2.4 C
Resistance = ×50 = 1.2V
100 10
5

Since the voltmeter and 100 Ω resistance are in 5 
parallel, so the voltmeter reads the same value i.e. B
A B A 10
1.2V 10

499 (b)

12 ⇒RAB = 5Ω
i= = 5A
(1+1)+0.4
502 (c)
500 (b)
The ratio of the weights deposited on cathodes

P a g e | 239
 will be in the ratio of their chemical equivalents. By balanced Wheatstone bridge condition
The chemical equivalent of copper = At. Wt/2 and 16 4
=
that of silver = At. Wt/1. X 0.5

503 (c) 8
⇒X = = 2Ω
4
A fully charged capacitor draws no current.
Therefore, no current flows in arm GHF. So the R 509 (b)
of arm HF is ineffective. The total resistance of the
From the given circuit
resistors in circuit
VA - (6×2) - 12 - (9×2) + 4 - (5×2) = VB
' (R+R)×R
is R = +R
(R+R)×R
Or VA - 12 - 12 - 18 + 4 - 10 = VB
(2+2)×2 10
= +2= Ω Or VA - VB =48 volt
(2+2)+2 3

E 10 510 (d)
Total current, i = ' = = 3A
R (10/3)
Let R0 be the initial resistance of both conductors.
In parallel circuit, the current divides in the
∴ At temperature θ their resistances will be,
inverse ratio of resistance, so current in arm
ABGD = 1A and current in arm AD = 2A R1 = R0(1 + α1θ)

Potential difference between G and D and R2 = R0(1 + α2θ)

= VG - VD = 1×2 = 2V For series combination, Rs = R1 + R2
Potential difference between D and F Rs(1+αsθ) = R0(1+α1θ) + R0(1 + α2θ)
= VD - VF = 3×2 = 6V
Where, Rs0 = R0 + R0 = 2R0
∴ VG - VF = (VG-VD) + (VD - VF)
∴ 2R0(1+α2θ) = 2R0 + R0θ(α1 + α2)
= 2 + 6 = 8V
α1+α2
or αs =
Potential difference across capacitor 2
= VG - VF = 8V
For parallel combination,
504 (c)
R1R2
Rp =
Amount of metallic sodium appearing, R1+R2

m = Zit = ( ) (
A
VF
it =
23
1×96500 )
×16×10×60 Rp0(1+αpθ) =
R0(1+α1θ)R0(1+α2θ)
R0(1+α1θ)+R0(1+α2θ)
= 2.3 gm
R0R0 R
505 (c) Where, Rp0 = = 0
R0+R0 2

V = E - IR = 15 - 10×0.05 = 14.5V R0
2
R (1+α1θ+α2θ+α1α2θ )
2

∴
2
(1+αpθ) = 0
R0(2+α1θ+α2θ)
506 (a)

Resistance of voltmeter should be high as α1 and α2 are small quantities.

507 (b) ∴ α1α2 is negligible.

P a g e | 240
 or αp =
α1+α2
2+(α1+α2)θ
α +α α +α
= 1 2 1- 1 2 θ
2 2 [ ( )] 518 (a)

Because with rise in temperature resistance of

as (α1+α2) is negligible conductor increases, so graph between V and i
2

becomes non linear

α1+α2
∴ αp = 519 (b)
2

511 (c) In VI graph, we will not get a straight line in case

of liquids
Current through resistance R will be zero if
520 (c)
E E1 E(R1+R2)
= or E1 = Here, V = 10V, G = 500Ω
R2 R1+R2 R2
V 10 -2
512 (a) ig = = = 2×10 A
G 500
m
m = Zit⇒ = 1[constant] Now,
Zit
igG -2
2×10 ×500
513 (d) S= = -2 ≃ 1Ω in parallel.
i-ig 10-2×10
Drift velocity=mobility× intensity of electric field
521 (a)
vd
or vd = μE or μ =
() ( )
4
4
E R1 r R nr R
= 2 ⇒ = ⇒R2 = 4
R2 r1 R2 r n
or μ=
vd
V/l ( )
∵E=
V
l 522 (c)

0.5×2 -3
Slope is zero at neutral temperature
∴μ = = 4.5×10
220
523 (d)
-3 2 -1 -1
≈ 5×10 m V s
4.2×3.14×(0.2×10 )
2 -3 2
Rπr
l= = -8 = 1.1m
515 (a) ρ 48×10

Heat produced by heater is given by 524 (c)

H = Pt According to Kirchhoff’s first rule

Given, P = 100 W, t = 2 min = 2×60s = 12s I + 4 + 2-3-5 = 0

∴ H = 100×120 = 12×10 J
3 I + 6-8 = 0

516 (c) ⟹ I = 2A

i 1.344 525 (d)

vd = = -2 -19 22
nAe 10 ×1.6×10 ×8.4×10
Wire AB is uniform so current through wire AB at
1.344 every across section will be same. Hence current
= = 0.01cm/s = 0.1mm/s
10×1.6×8.4 density, J( = i/A) at every point of the wire will
be same
517 (d)
526 (d)
ti = 2tn - tc⇒ti = 2×350 - 30 = 670℃
After connecting a resistance R in parallel with

P a g e | 241
 voltmeter its effective resistance decreases. Hence E.m.f. is the value of voltage, when no current is
less voltage appears across it i.e. V will decreases. drawn from the circuit so E = 2V.
Since overall resistance decreases so more
current will flow i.e. A will increase 2
Also r = slope = = 0.4Ω
5
527 (c)
532 (c)
J
vd = ⇒v ∝ J [current density] They are the resistors made up of semiconductors
ne d
whose resistance decreases with the increase in
i 2i i temperature. This implies that they have negative
J1 = and J2 = = = J1;
A 2A A and high temperature coefficient of resistivity.
∴ (vd) = (vd) = v
1 2
They are usually made of metal oxides with high
528 (a) temperature coefficient of resistivity.

For a balance Wheatstone bridge 533 (c)

A D 10 4 Strength = 5×18 = 90AH

= ⇒ ≠ [Unbalanced]
B C 5 4
534 (c)
' '
A D A 4 '
= ⇒ = ⇒A = 5Ω -2
V = E - ir = 12 - 60×5×10 = 9V
B C 5 4

A'(5Ω) is obtained by connecting a 10 Ω 535 (b)

resistance in parallel with A Effective emf of circuit = 10 - 3 = 7 V
529 (b) Total resistance of
Let the e.m.f. of cell be E and internal resistance be circuit= 2 + 5 + 3 + 4 = 14Ω
r Current, i = 7/14 = 0.5 A
E E Potential difference between A and D
Then 0.5 = and 0.25 =
(r+2) (r+5)
= 0.5×10 = 5V
5+r
On dividing, 2 = ⇒r = 1Ω Potential at D = 10 - 5 = 5V
2+r
Hence, E cannot be at zero potential, as there is
530 (b)
potential drop at E
Mass of water = volume×density
536 (a)
= 1000×1 = 1000 g.
When a constant current is passed through a wire
Heat taken by water = mc ∆θ of uniform cross-section, the potential difference
across any portion of the wire is directly
= 1000×1(37 - 22) cal proportional to the length of that portion.

= 1000×15×4.2 J Potential gradient=0.2 mVcm

-1

energy spent ∴ Potential difference across potentiometer wire

Power of geyser=
time
=0.02 × 1= 0.02V
1000×15×4.2
= 1050 W.
60 Total resistance = r + R = 490 + R

531 (b)

P a g e | 242
 V1 R The galvanometer shows zero deflection ie,
∴ = 1
V2 R2 current through XY is zero.

0.02 R 500
⟹ = X Y
2 490+R G

⟹R = 4.9Ω
12 V R 2V
537 (c)

Given V1 = 50 volt, i1 = 11A;V2 = 60 volt,

i2 = 1A

If e.m.f. and internal resistance of battery are E As a result potential drop across R is 2V, circuit
and r respectively then P.D. across terminals of can be redrawn as
battery,
12
V = E - ir I=
500+R

We have 50 = E - 11r …(i)

500
And 60 = E - 1r …(ii)
I
From (i) and (ii),
12 V R 2V
E = 61V and r = 1Ω

538 (d)

R1 + R2 = R1(1+αt) + R2(1 - βt)

Voltage across R, V = IR
R β
⇒R1 + R2 = R1 + R2 + R1αt - R2βt⇒ 1 =
R2 α 12
⟹ 2= ×R
500+R
539 (d)
⟹1000 + 2R = 12R
22.4×1
V2 = = 22.4 litre at NTP
1 ⟹R = 100Ω

∵ 11.2 litre of H2 is liberated by 96,500 C 543 (b)

∴ 22.4 litre of H2 is liberated by Production of e.m.f. by temperature difference is

96500×2 = 1,93,000 C known as Seebeck effect

540 (b) 545 (c)

The moving coil galvanometer have their coil From Joule’s law,
wound on a metallic (copper or aluminium) 2
Vt
frame, so as to make the motion dead beat due to H= cal
RJ
the production of eddy currents. In the ballistic
galvanometer, on the other hand, the damping is Where V is potential difference, t the time and R
to be reduced to the minimum and hence the the resistance.
frame is of a non-conducting material eg, paper or
bamboo. R = R1 + R2

541 (b)
P a g e | 243
 t HJ '' V
Let = 2 = constant = k Current through CBD, I = amp
R V 4

⇒R1 = kt1 and R2 = kt2 Potential difference between C and B = VC - VB

∴ kt = kt1 + kt2 V V
= ×1 = volt
4 4
Or t = 15 + 20
Potential between A and B = VA - VB
t = 35 min
V V V
546 (d) ∴ VA - VB = VC - VB - (VC-VA) = - =-
4 2 4

In a conductor, the electron number density ie, ⇒VA - VB < 0 or, VA < VB
number of electrons per unit volume of a
28 -3
conductor is very large ( ≈ 10 m ), so large as VA < VB, so direction of current will be B to A
current in a conductor is obtained irrespective of
their small drift speed. 549 (c)

547 (c) Here points B and D are common. So 2R in arm DC

and 2 R in arm CB are in parallel between C and B.
As resistances are in parallel Their effective resistances

1 1 1 2R×2R
= + + …5000 times = =R
Req 220 220 2R+2R

1 5000 The modified and simpler circuit will be shown in

=
Req 220 figure. The effective resistance between A and B is

220
Req =
5000

220
I=
220
5000

I = 5000 A

548 (b) R×(R+R) 2

Reff = = R
R+(R+R) 3

550 (d)

Full scale deflection current

150
= mA = 15mA
10

Ig R G
V
Current though arm CAD, I = amp
8 | |
V

Potential difference between C and A = VC - VA

V V Full scale deflection voltage

= ×4 = volt
8 2

P a g e | 244
 150 2R×2R
= mV = 75mV ∴ Equivalent resistance= = RΩ
2 2R+2R

Galvanometer resistance 553 (a)

75mV The tolerance level of resistance is mostly 1%,

G= = 5Ω
15mA 2%, 5% and 10%. In old days 20% was also
common, but these are now rare. Now a days 5%
Required full scale deflection voltage.
tolerance in treands.
V = 1 × 150 = 150 volt
554 (c)
Let resistance to be connected in series is R.
Resistance of wire given by
⟹ V = Ig(R+G)
l
R=ρ
-3 A
∴ 150 = 15×10 (R+5)
4 Also, volume (V)=Length(l)×Area (A)
⟹10 = R + 5
2
Where, A = πr (r is radius)
⟹R = 10000 - 5 = 9995
When the wire is stretched its volume (V)
551 (c)
remains constant.
1 2
Total energy stored in capacitor, Etotal = CV Hence,
2
ρV
1 -6 2 -4 R= ….(i)
= ×3×10 ×10 = 1.5×10 J 2 4
πr
2
When radius is halved
2
Energy dissipated in 2Ω = ×E
(2+4) total ' pV
R = ……………(ii)
()
4
2 r
2 -4 -4 πr
= ×1.5×10 = 0.5×10 J = 0.05 mJ 2
6
' 2 4
552 (d) R 16ρV π r
∴ = 2 4× = 16
R πr ρV
The equivalent circuit of these network is as '
shown in figure, which is a balanced Wheatstone ⟹R = 16R
bridge. Therefore no current will flow in the
Hence, new resistance increases to sixteen times
resistance of arm PQ. When cell is connected to
its original value.
points A and B
555 (d)
Therefore effective resistance of arm
APS = (R + R = 2R) will be in parallel to the Comparing the given equation with standard
total resistance of arm AQS( = R + R = 2R) equation

P 1 2
E = αt + βt
R R 2

R 1 1 1
A S B α = 40 and β = - ⇒β = -
2 20 10
R R
Q α -40
Hence neutral temperature tn = - =
β -1/10

P a g e | 245
 ⇒tn = 400℃ ρl
For wire P, 20 = 2 …(i)
πr
Alternate solution : We know at neutral
temperature Similarly for wire Q,

dE ρ(2l)
=0 8= ….(ii)
π(r )
' 2
dθ
2
θ 2θ Dividing Eq.(i) by Eq. (ii), we have
E = 40θ - ⇒40 - = 0⇒θ = 400℃
20 20
ρl π(r )
' 2
20
= 2×
556 (d) 8 πr ρ(2l)

Only current through the conductor of non-

()
' 2
r
⟹5 =
uniform area of cross-section is constant. Drift r
velocity or drift speed vary inversely with the
'
area of cross-section of the conductor ⟹r = 5r

557 (d) 561 (d)

Equivalent resistance between P and Q Voltage across B3 is greatest hence B3 will show
maximum brightness. In series combination of
1 1 1 1 48 bulbs, the bulb of lesser wattage will glow more
= + + ⇒RPQ = Ω
RPQ (6+2) 3 (4+12) 25 bright. Hence W2 > W1.
Current between P and Q;i = 1.5A So , W1 < W2 < W3.
So, potential difference between P and Q
562 (b)
48
VPQ = 1.5× = 2.88V E
25 Specific resistance k =
j
559 (c)
563 (d)
Given that the resistance of the total wire is 4Ω.
Kirchhoff’s Ist law or KCL states that the algebraic
C sum of current meeting at any junction is equal to
zero. In other words we can say that “the sum of
A B all the currents directed towards a junction in a
circuit is equal to the sum of all the currents
D directed away from that junction.” Thus, no
charge has been accumulated at any junction i.e.,
charge is conserved, and hence, we can say that
Here, ACB(2Ω) and ADB (2Ω) are in parallel. KCL (∑i = 0) is based on conservation of charge.

So, the resistance across any diameter is Kirchhoff’s IInd law or KVL states that algebraic
sum of changes in potential around any closed
1 1 1 2 resistor loop must be zero. In other words
⟹ = + = =1
R 2 2 2 “around any closed loop, voltage drops are equal
⟹ R = 1Ω to voltage rises”. No energy is gained or lost in
circulating a charge around a loop, thus, we can
560 (c) say that KVL is based on conservation of energy.

ρl 564 (a)
Resistance, R=
A
The circuit diagram may be redrawn as shown
P a g e | 246
 here.
Vp - Vq = ( 6 12×6
+
3 12+6 )
(0.5) = (2+4)(0.5) = 3V
2
Obviously, ICAD = ICBD = A
12 572 (b)

2 2 mH WH 1
∴ VC - VA = A×5Ω = V = =
( )
2 2
15 3 MCu WCu 63.5
2
2 4
and Vc - VB = A×10Ω = V
15 3 2
⇒mH = ×0.3175 = 0.01g
2 63.5
4 2 2
∴ VA - VB = (VC-VB) - (Vc-VA) = V- V = V
3 3 3 2g of H2 occupies volume at NTP = 22.4 L
566 (a) ⇒0.01 g of H2 occupies volume at NTP given by:
V 10
R= -G = -3 - 1 = 999Ω 22.4
ig 10×10 V= ×0.01L
2
567 (c) ⇒V = 11.2×0.01×1000 cc = 112 cc
Rt = R1(1+α1t) and Rt = R2(1+α2t) 573 (c)
1 2

Also Vt
2
H V
2
H= or =
Req = Rt + Rt ⇒Req = R1 + R2 + (R1α1 + R2α2)t 4.2 R t 4.2 R
1 2

⇒Req = (R1+R2) 1+ [ ( R1α1+R2α2

R1+R2
.t )] ⇒800 =
20×20
4.2×R
⇒R =
5
42
= 0.119 ≈ 0.12Ω

R1α1+R2α2 574 (b)

So αeff =
R1+R2 The circuit will be as shown

569 (b) 10V

m = zit
5
m zit A
If V is the volume, then V= =
ρ ρ
-7 10
zit 3.3×10 ×1.5×20×60 i= = 2A
Thickness = = -4 5
Aρ 50×10 ×9000
-5 575 (b)
Thickness = 1.3×10 m
ρl ' ρl R
570 (c) R= for first wire and R = = for second
a 4a 4
If resistance does not vary with temperature P wire
consumed
576 (c)

() ( )
2 2
VA 110
= ×PR = ×100 = 25W. But in Regarding Kirchhoff’s junction rule, the circuit can
VR 220
be redrawn as
second case resistance decreases so consumed
power will be more than 25 W

571 (c)

P a g e | 247
 Current density of drinking electrons j = nev
7 -3 7 6 -3
1A n = 5×10 cm = 5×10 ×10 m
B
-1 -19 -6 -2
v = 0.4 ms , e = 1.6×10 C⇒j = 3.2×10 Am

Current density of ions

10 A I
-6 -6 A
C = (4-3.2)×10 = 0.8×10 2
A m
6A
-1
This gives v for ions = 0.1 ms
D
580 (a)
2A

VA + VB + VC ∝ 740

VA + VB ∝ 440

VB + VC ∝ 540

Current in arm, AB = 10 - 6 = 4A Hence VA:VB:VC = 1:1.2:3.5

Current in arm, DC = 6 + 2 = 8A
581 (c)
Current in arm, BC = 4 + 1 = 5A
The number density (n) of conduction electrons
Hence, I = 5 + 8 = 13A in the copper is a characteristic of the copper and
29
is about 10 at room temperature for both the
577 (d) copper rod X and the thin copper wire Y.

V 100 Both X and Y carry the same current I since they

By using R = - G⇒R = -3 - 5 = 19,995Ω
ig 5×10 are joined in series

578 (a) From I = nAvq

-19
μV Where q is the electron charge of 1.6×10 C,
Thermo-emf of thermocouple = 25 .
℃
v is the drift velocity in the conductor and A is the
Let θ be the smallest temperature difference. cross-sectional area of the conductor.

Therefore, after connecting the thermocouple We may conclude that rod X has a lower drift
with the galvanometer, thermo-emf velocity of electrons compared to wire Y since rod
X has a larger cross-sectional area. This is so
E = 25 ( ) μV
℃
×θ(℃) because the electrons in X collide more often with
one another and with the copper ions when
= 25θ×10 V
-6 drifting towards the positive end. Thus, the mean
time between collisions of the electrons is more in
Potential drop developed across the galvanometer X than Y.
-5 -4
= iR = 10 ×40 = 4×10 V 582 (a)
-4 -6
∴ 4×10 = 25θ×10 As steady current is flowing through the
conductor, hence the number of electrons
4 2
∴θ= ×10 = 16℃ entering from one end and outgoing from the
25
other end of any segment is equal. Hence charge
579 (d) will be zero

P a g e | 248
583 (d) 2E
I=
R+r1+r2
The given circuit can be simplified as shown
below in circuit R2, R5 and R3,R4 are in series and Potential difference across first cell
then their resultant is connected parallel.
V = E - Ir1 = 0
Similarly R7,R8 and R6,R9 are in series and their
resultant is connected parallel on simplifying this E E
we get their equivalent circuit
r1 r2

2Er1
E- =0
R+r1+r2

[ R+r1+r2-2r1
R+r1+r2
=0 ]
Now current in circuit
⟹ R + r2 - r1 = 0
V 6 3
I= = 3 ×10 = 2mA
R 3×10 Ω ⟹ R = r1 - r2

584 (c) 589 (d)

For a galvanometer Resistivity depends only on the material of the

conductor
NIAB = C θ
590 (d)
θ NAB
⟹ =
I C

θ
Here, is called the sensitivity of galvanometer
I
so to increase the sensitivity of galvanometer, C
should be decreased and N, A and B should be
increased
Applying Kirchhoff’s second law for closed loop
586 (c)
AEFBA,
E = 2.2volt, V = 1.8 volt, R = 5R
We get

r= ( ) ( )
E
V
-1 R =
2.2
1.8
-1 ×5 = 1.1Ω -(I1+I2)×5 - I1×2 + 2 = 0 or 7I1 + 5I2 = 2 …(i)

587 (b) Again, applying Kirchhoff’s second law for a

closed loop DEFCD, we get
z = m/It.From graph, It = Area OABC
-(I1+I2)×5 - I2×2 + 2 = 0
1
= (10+30)× 100 = 2
2 1000 or 5I1 + 7I2 = 2 …(ii)

∴ z = m/2 = -0.5 m Multiplying (i) by 5 and (ii) by 7, we get

588 (c) 35I1 + 25I2 = 10 …(iii)

P a g e | 249
 35I1 + 49I2 = 14 …(iv) Thermo-electric power is
dE T 3
Subtracting (iv) from (iii) we get, = KT0 + K 0 = K T0.
dT 2 2
1 595 (c)
-24 I2 = -4⇒I2 = A
6

Substituting the value of I2 in equation (i), we get Cost = ( 60×8×30

1000 )
×1.25 =Rs 18.

1 7 1 596 (c)
7I1 = 2 - 5× ⇒7I1 = ⇒I1 = A
6 6 6
Let potential of P1 is 0 V and potential of P2 is V0.
The current through the 5Ω is P
Now apply KCL at P2.
1 1 1
= I1 + I2 = A+ A= A 2
6 6 3
2Ω 1Ω
10 Ω
591 (a) 5V -2
5V 2V
Chemical energy reduced P1

= VIt
V0-5 V -0 V -(-2)
= 6×5×6×60 = 10800 + 0 + 0 =0
2 10 1
4
= 1.08×10 J
5
⟹ V0 =
592 (a) 16

V
2 So, current through 10Ω resistor is
H1 = t
R
V0
from P2to P1.
V
2
10
H2 = t
R/2
597 (a)
H2
∴ =2 Given current I = 2A
H1
6Ω
⇒H2 = 2H1

593 (b) +
E 20V

∵ ig = 10% of
i G 90
i= ⇒S = = = 10Ω
10 (n-1) (10-1)
(i) If current is in clockwise direction then from
594 (b) Kirchhoff’s second law
1
E = K(T-Tr)T0 +
2 2
K(T - Tr ) 6×2 + 20 - E = 0
2
⟹ E = 32V
dE 1
= KT0 + K×2T = KT0 + KT
dT 2 (ii) If the current is in anticlockwise direction,
then
At temperature T = T0/2,

P a g e | 250
 ( ) ( )
1/2 1/2
6×2 + E - 20 = 0 RV 3×3 3
or l = = =
ρ ρ ρ
⟹ E-8 = 0
602 (a)
E=8
An emf of the order of a few microvolt is
598 (b) generated which is proportional to (t2 - t1).
According to Kirchhoff’s law iCD = i2 + i3
603 (a)
599 (d) If m = Number of rows
nE E And n = Number of cells in a row
Short circuited current i = = i.e. i doesn’t
nr r
depend upon n Then m×n = 100 …(i)

600 (c) nr
Also condition of maximum current is R =
m
When A is area of cross-section of wire, and n be
number of free electrons per unit volume, then 1×n
⇒25 = ⇒n = 25 m …(ii)
relation between electric current (i) and drift m
velocity (vd) is
On solving (i) and (ii), m = 2
i = ne Avd
604 (d)
Number of atoms in 63 g of copper is equal to
23
According to Faraday’s law of electrolysis
Avogadro’s number ie, 6×10
m = zit
Volume of 63 g copper
Here, i = 1.5 A, t = 10 min = 10×60 s
63 63 3
= = = 7cm
density 9 z = 30×10 gC
-5 -1

23
6.02×10 3 Hence, mass of copper deposited on the electrode
n= per cm
7
-5 -2
m = 30×10 ×1.5×10×60 = 27×10 = 0.27 g

( )
29
6.02×10 3
= per m
7 605 (d)

Area A = πr
2
Let d be the density of the material of copper wire.
Let l1,l2 be the lengths of copper wires of diameter
= π(0.5×10 ) m2
-3 2
1 mm and 2 mm respectively. As
Hence, drift velocity Mass=volume×density =(πD /4)l.d
2

1.1 π(10
-3 2
) l ×d = π× (2×10-3)2 l ×d
vd =
( ) So, 1 =
29
6.02×10
×1.6×10 ×π×(0.5×10 )
-19 -3 2
4 1
4 2
7
-3 -1
or l1 = 4l2
vd = 0.1×10 ms
ρl l
vd = 0.1 mms
-1 Now, R = 2 ie,R ∝ 2
πD /4 D

601 (b) R1 l D
2
2
∴ = 1 × 22 = 4×2 = 16
2 R2 l2 D 1
ρl ρl ρl
R= = =
A V/l V
P a g e | 251
606 (a) From Eqs. (i) and (ii)

e R i2 = 2.87A
Potential gradient = .
(R+Rh+r) L
612 (c)
2 5 V V
= × = 0.5 = 0.005 V
2
(15+5+0) 1 m cm Applying P=
R
608 (d)
R1 = 1 Ω, R2 = 0.5 Ω and R3 = 2 Ω, V1 = V2 = V3
e R 5 5 = 3 volt
E= . ×l⇒0.4 = × ×l
(R+Rh+r) L (5+45+0) 10
(3)2
∴ P1 = = 9W
⇒l = 8 m 1

609 (b) (3)2

P2 = = 18 W and
0.5
m
Charge →q =
z (3)2
P3 = = 4.5 W
2
Where, z = electro chemical equivalent
∴ P2 > P1 > P3
m 0.067
∴ z= = = 0.001117
It 1×60 613 (d)

108 4 Assuming current I flows through the circuit

q= = 9.67×10 C/geq
0.001117

610 (a)

Work done in delivering q coulomb of charge

from clouds to ground.

W = Vq 2
Energy dissipated in load = I R
6 6
= 4×10 ×4 = 16×10 J
Energy dissipated in the compete circuit
2
The power of lighting strike is = I (r + R)
6 2
W 16×10 6 IR R
P= = = 160×10 W ∴ The efficiency = 2 =
t 0.1 I (R+r) R+r

= 160 MW Relation between resistance, mass and cross-

sectional area
611 (c)
l×α V
In closed loop ABGFEHA R=p or R = ρ 2
α×α α

[]
10 - i2×1 + i1×0.5 - 6 = 0
m ρ m
R=ρ 2 =
dα d α2
0.5i1 - i2 = -4
m
In closed loop BCDEB I∝ 2
α
(i1+i2)×12 + i2×1 - 10 = 0 614 (c)
12ii + 13i2 = 10 From

P a g e | 252
 Rt = R0(1+αt) 2
* (EI - I r) represents the power output of the
source of emf.
5 = Ro(1+50α) …(i)
617 (d)
and 6 = Ro(1+100α) …(ii)
As resistance ∝ Lengt
5 1+50α
∴ =
6 1+100α 12
∴ Resistance of each arm = = 4Ω
3
1
⟹α =
200 4×8 8
∴ Reffective = = Ω
4+8 3
Putting value of αin Eq. (i), we get
8

(
5 = R0 1+50×
1
200 ) 4 4

∴ R0 = 4Ω 4

615 (d)
618 (a)
To convert a moving coil galvanometer (MCG)
In the steady state, no current flows through
into a voltmeter a high resistance R is connected
capacitor branch.
in series with (MCG) as shown below.
r P E
R
G

V g = ig G | (V - V g)

Q E

C
V

R 2E

616 (a) 2r

(i) Rate of chemical energy consumption

= 1.5×2 =3 W
Current in the circuit
(ii) Rate of energy dissipation inside the cell
net emf
i=
= 2×2×0.1 = 0.4 W net resistance

(iii) Rate of energy dissipation inside the resistor 2E-E E

= =
r+2r 3r
= (3 - 0.4) W = 2.6 W
So, potential drop across capacitor
(iv) Power output of source = (3 - 0.4) W = 2.6 W
E E
* EI represents rate of chemical energy V = ir = ×r =
3r 3
consumption of the cell.
619 (a)
2
* I r represents the rate of energy dissipation
inside the cell. Given that, the resultant voltage across the
1.2V,0.2Ω
P a g e | 253
 2
battery terminal=1.5V ⇒R - R - 2 = 0⇒R = -1 or R = 2ohm

622 (b)

When two similar bulbs of different powers are

connected in series, then

E,0.4Ω 1 1 1
= +
P P1 P2

Given, P1 = 200 W, P2 = 100 W

Let I be the current in the circuit then total
resistance =0.6Ω 1 1+2
⇒ =
P 200
Hence, V = IR
200
⟹1.5 = I×0.6 ⇒P = = 66.7 W
3
1.5 5
⟹I = ⟹I = A 623 (b)
0.6 2
This is a balanced Wheatstone bridge circuit. So
Now, applying Kirchhoff’s second law in the
potential at B and D will be same and no current
circuit
flows through 4R resistance
0.4I + 0.2I + 1.2 - E=0
624 (a)
5
0.6 × + 1.2 = E 3
2 Equivalent resistance of the circuit R = Ω
2
⟹ E = 2.7V
V 3
∴ Current through the circuit i = = = 2A
620 (a) R 3/2

Effective resistance of n resistance each of the 625 (d)

resistance r in series Rs = r×n = R(as per
question); so r = R/n. When these resistances are
connected in parallel, the effective resistance

R/n 2
Rp = r/n = = R/n
n
Here, V = 10 V, R = 10kΩ, G = 100 Ω = 0.1 kΩ
621 (c)
V = Ig(G+R)
Let the resultant resistance be R. If we add one
more branch, then the resultant resistance would V 10V 10V
be the same because this is an infinite sequence Ig = = =
G+R (10+0.1)kΩ 10.1kΩ
R1 = 1
= 0.99mA ≈ 1 mA
X
A

R2 = 2 R

B
Y

RR2 2
∴
R+R2
+ R1 = R⇒2R + R + 2 = R + 2R Here, I = 1A, (I-Ig)S = IgG

P a g e | 254
 -3
IgG 1×10 ×100
S= = -3 ig
I-Ig 1-1×10
-3 i- i g
100×10 100 1
≈ = Ω= Ω = 0.1Ω
1 1000 10
S
626 (c)

ρλ
Resistance,R =
A ∴ ig×G = (i-ig)×S
For given problem,
ig S 5 1
⟹ = = =
R∝λ
2 i-ig G 50 10

R1 λ
2 ⟹10ig = i - ig
∴ = 12
R2 λ2
⟹11ig = i

()
2
λ2
R2 = ×R1 = 15.6 Ω ig 1
λ1 ⟹ =
i 11
627 (b)
630 (d)
e R
E= . ×l If T is the smallest temperature difference that
(R+Rh+r) L can be detected, then
-3 2 10 -6 -6
⇒10×10 = × ×0.4⇒R = 790Ω 40×10 T = 100×10
(10+R+0) 1
⇒T = 2.5℃
628 (a)
631 (a)
Let l be the original length of wire and x be its
length stretched uniformly such that final length The circuit diagram is follows:
is 1.5 l
1
l
1.5

(l – x 0.5l 1
1.5l i

(l-x) (0.5l+x)
Then 4R = ρ +ρ ' where
A A 10 V
' x
A = A
(0.5l+x)

l l-x (0.5l+x)2 For 1Ω resistors in parallel, the resultant

∴ 4ρ =ρ +ρ
A A xA resistance is
2 2
1l x lx x 1 1 1 1
or 4l = l - x + + + or = = +
4x x x l 8 R R1 R2

629 (b) Where, R1 = R2 = 1Ω

The galvanometer G and shunt S is connected in
parallel, hence potential difference is the same.

P a g e | 255
 1 1 1 2 1 1 1 1
∴ = + = ' = 2 + 3 + 4
R 1 1 1 R R R R

⟹ R = 0.5Ω 1 1 1
= + +
50 50 75
This 0.5Ω resistor is connected in series with1.5Ω
resistor. 30+30+20
=
1500
Hence, equivalent resistance is
80 4
R’=0.5+1.5=2Ω = =
1500 75

From Ohm’s law, current flowing in the circuit is ' 75

∴ R = Ω
given by 4

V = iR ' 75
R = R1 + R = 100 +
4
V 10
⟹i = = = 5A
R 2 475
= Ω = 118.75Ω
4
632 (b)
-2
635 (d)
Length l = 1 cm = 10 m
Two resistances in series are connected in parallel
with the third. Hence
1 1 1 3 8
= + = ⇒Rp = Ω
Rp 4 8 8 3
1 cm

100 cm
636 (c)
1 cm
Manganin or constantan are used for making the
potentiometer wire
Area of cross-section A = 1 cm ×100 cm
2 -2 2 637 (c)
= 100 cm = 10 m

() ()
2 2 2
-2 J1 I/πr1 r2 3 9
10 -7 -7 = 2 = = =
Resistance R = 3×10 × -2 = 3×10 Ω J2 I/πr2 r1 1 1
10

633 (a) 638 (b)

When temperature of the hot junction of a

A A
2 thermocouple rises, the thermo-emf increases and
2
2 2 becomes maximum at a particular temperature.
2 
2 2 This temperature is called neutral temperature.
B B
1 2
Given, e = at + bt
2
2×2
RAB = = 1Ω Differentiating with respect to t, we get
2+2

634 (c) de
= a + bt
dt
R2, R3and R4 are in parallel order, so their
equivalent resistance For maximum value of e,

P a g e | 256
 de 4
=0 VA - VD = ×90 = 1.8V
dt 200

∴ a + bt = 0 ∴ VB - VD = (VA-VD) - (VA - VB)

a 1.8 - 1.6 = 0.2 V

⇒t = -
b
646 (b)
639 (a)
Given that,
4
R = 32×10 ± 5%
l1 4 r 2
640 (d) = and 1 =
l2 3 r2 3

()
2
2 2 ρl i ρVt Here, l1and l2are the length of the wires while
H = i RT = i t= 2 [V = volume = Al]
A A
r1and r2 are the radii of the wires.

() ()
4 4
1 H r 2 16 Now, we know that
⇒H ∝ 4 ⇒ 1 = 2 = =
r H2 r1 1 1
V = IR⟹IR = constant
642 (c)
⟹I1R1 = I2R2
When key K is opened, bulb B2 will not draw any
current from the source, so that terminal voltage I1 R
or = 2 ….(i)
of source increases. Hence, power consumed by I2 R1
bulb increases, so light of the bulb becomes more.
The brightness of bulb B1 decreases. But we know that the resistance of the wire is

643 (b) ρl
R=
A
Let the internal resistance of cell be r, then
Hence, from Eq (i)
E 1.5
i= ⇒15 = ⇒r = 0.06Ω I1 ρl /A
R+r 0.04+r = 2 2
I2 ρl1/A1
644 (d)
Here,
Given circuit is a balanced Wheatstone bridge
circuit. So there will be no change in equivalent ρl1 ρl
R1 = ,R2 = 2 and ρ1 = ρ2 = ρ
resistance. Hence no further current will be A1 A2
drawn
Because both wires are of same material.
645 (d)
I1 lA
∴ = 2 1
Current through arm ABC, I2 l1A2

= 4/(40+60) = 0.04 A I1 l πr
2

⟹ = 2 12
I2 l1πr2
Potential difference across A and B,
2
VA - VB = 0.04×40 = 1.6V I lr
⟹ 1 = 2 12
I2 l2r2
Current through arm ADC,
l2 3 r 2
= 4/(90+110) = 4/200A Here, = and 1 =
l1 4 r2 3

Potential difference between A and D,

P a g e | 257
 ()
2
I1 3 2 200 W
⟹ = ×
I2 4 3

I1 1
Or =
I2 3 200 W

647 (a)

When temperature is raised, the ions/atoms of

( )
the conductor start vibrating with increased
amplitude of vibration and greater frequency. Due
to which the electrons moving towards the
positive end of conductor will suffermore rapid '
P = PA + PB = 200 + 200
collisions and hence time of relaxation
(τ)decreaseds. As vd ∝ τ, thus drift velocity = 400 W
decrease. Therefore vd ∝ 1/T
Now P' and bulb C are in serried. So, the resultant
648 (d) power of the combination is

R 2 8 2 '' 400×400
RAB = +R= +2= =2 Ω P = = 200 W
3 3 3 3 400+400

649 (b) 652 (c)

dQ Resistances at C and B are not in the circuit. Use

i= = 10 + 4t or dQ = (10+4)dt
dt laws of resistances in series and parallel excluding
the two resistance
2
4t 2
Integrating it, we get Q = 10t + = 10t + 2t
2 653 (b)
2
When t = 10s, Q = 10×10 + 2×10 = 300 C Potential gradient along wire

650 (b) potential difference along wire

=
length of wire
Let q be the charge flowing through copper
voltmeter. -3 I×40 -1
∴ 0.1×10 = Vcm
1000
The charge flowing through silver voltmeter
= (2000 - q) Current in wire.

Now, m = zCuq = zAg×(2000 - q); 1 E

I= A or I = '
400 R+R
ZAg
∴q= ×(2000 - q) 2 I
ZCu ∴ =
40+R 400
-6
1.008×10 Or R = 800 - 40 = 760Ω
= -7 ×(2000 - q)
3.36×10
654 (b)
q = 1500 C
Just for your knowledge remember, voltaic cell
651 (c) uses dil. H2SO4 ; Dry cell uses NH4Cl + ZnCl2
Bulbs A and B are in parallel. paste; Daniel cell uses dil. H2SO4 ; Lead
Accumulator uses dil. H2SO4 and Ni-Fe cell or
There, effective power is Alkaline Accumulator uses KOH solution

P a g e | 258
655 (d) ⟹ 10I = 1

Temperature of cold junction Tc, temperature of 1

∴ I= = 0.1A
inversion Ti and neutral temperature Tn are 10
related as follows
660 (b)
Ti+Tc
Tn = ⇒Ti = 2Tn - Tc Equivalent circuit of this combination of
2
resistances is as shown in figure. The effective
As Tn is constant for a given thermocouple Ti resistance of arm
decreases with increase in Tc 4×4
EG = = 2Ω
4+4
656 (b)

2 ∆P 2∆i 4
P = i R⇒ = [R→Constant] A B
P i
4 4
⇒ % change in power = 2×% change in current
E G
= 2×1 = 2% 4 4

658 (c) 4
D C
2
PR 1
Power loss in transmission PL = 2 ⇒PL ∝ 2
V V

659 (a) Total resistances between A and B will be

All the resistances are in parallel order, so voltage 1 1 1 1 3 4

= + + = or R = Ω
across them will be equal. R 4 4 4 4 3

661 (c)
60 Ω

I The equivalent circuit is as shown in figure (a)

15 Ω 5 Ω and (b)
1A I1 1A
1 - I - I1 10 Ω R R
R 2R
R
A B
6R

∴ 60I = (15+5)I1 R
4R 2R 4R
R
⟹ 60I = 20I1

⟹ I1 = 3I E,4 Ω E,4 Ω
(a) (b)

Again (15+5)I1 = 10(1-I-I1)

⟹ 2I1 = 1 - I - I1
Since, the network of resistances is a balanced
⟹ 2(3I) = 1 - I - 3I Wheat stone bridge, so resistance between points
A and B of network figure (b) is given by
⟹ 6I + 4I = 1
1 1 1 2+1 1
= + = = or R’ = 2R
R' 3R 6R 6R 2R

For maximum power to the network, R’ should be

P a g e | 259
 equal to internal resistance of the battery. So i×R1 i×4 i
Current through R2 is i2 = = =
'
R1+R2 12 3
R = 2R = 4 or R = 4/2 = 2Ω
2
Power dissipated in 3Ω resistor is P1 = i1×3 …(i)
662 (d)
Power dissipated in 8Ω resistor is
Let x watt be the power loss in transmission line 2
P2 = i2×8 …(ii)
in the form of heat

() ( )
2
P
2
10×1000
2 P1 i ×3 P (2i/3)2×3 12 3
∴x= R= ×0.2 ∴ = 12 or, 1 = = =
V 200 P2 i2×8 P2 (i/3)2×8 8 2

= 500W = 0.5 kW 3 3
P1 = ×P = ×2 = 3watt
2 2 2
Efficiency of transmission
∴ Power dissipated across 3Ω resistor is 3 watt
power delivered by line 10 kW
= =
power supplied to line 10 kW+x kW 668 (b)

10 10 The filament of the heater reaches its steady

= = = 0.95 = 95%
10+0.5 10.5 resistance when the heater reaches its steady
temperature, which is much higher than the room
663 (a)
temperature. The resistance at room temperature
v v
2 2 is thus much lower than the resistance at its
H1 = H2⇒ t1 = t ⇒t = 2t1⇒t1 = 30 min steady state. When the heater is switched on, it
R 2R 2 2
draws a larger current than its steady state
∴ t2 = 60 min current. As the filament heats up, its resistance
increases and current falls to steady state value
664 (a)
669 (d)
Significant figures Multiplier
Brown Black Brown When cells are in series, emf of the combination of
1 0 10
1 cells increases
1
∴ R = 10×10 = 100 Ω
670 (b)
665 (a)
Give that,
V 6 6
i= ⇒2 = = ⇒R = 1 Ω l1 d ρ 1
R 6×3 2+R = 1 = 1 =
+R l2 d2 ρ2 2
6+3

666 (c) and R1 = 10Ω

m = zq, z = atomic mass / valence We know that, the resistance of the wire

[ ( )]
2
667 (b) ρl ρl 4ρl d
R= = = ∵A=π
()
2 2
A d πd 2
π
Resistance of series combination of 3 Ω and 1 Ω is 2
R1 = 3 + 1 = 4Ω, R2 = 8Ω
So, the resistance of first wire is
Let i be the total current in the circuit
4ρ1l1
R1 = 2 ..(i)
i×R2 i×8 2i πd1
Current through R1 is i1 = = =
R1+R2 12 3
and the resistance of the second wire is

P a g e | 260
 4ρ2l2 Given, R=3Ω, r = 0.5Ω
R2 = 2 ..(ii)
πd2
m
∴ 3= ×0.5
On dividing Eq. (ii) by Eq (i) n

R2 ρ l d
2 m
⟹ =6
= 2 × 2 × 12 n
R1 ρ1 l1 d2
⟹m = 6n
()
2
R 2 2 1
⟹ 2 = × ×
10 1 1 2 Total number of cells= m×n = 24 …(i)

R2 From Eqs. (i) and (ii), we get …(ii)

⟹ = 1⟹R2 = 10Ω
10
6n×n = 24
671 (d) 2
⟹6n = 24
Net resistance of 3Ω and 7Ω resistors (in series)
2
⟹n = 4
R’=3+7=10Ω
⟹n = 2, m = 12
R’ and 10Ω are in parallel, so
673 (c)
' 10×10
R = = 5Ω
10+10 ρl ρ ρ×2l ρ R
RA = = and RB = = i.e. A = 1 :1
l×t t 2l×t t RB
'
R and 5Ω are in series, so
674 (d)
R’=5+5=10Ω
The network can be redrawn as follows
Now, R’ and 10Ω are in parallel, so
3 3 3
10×10 A B
R= = 5Ω
10+10
⇒Req = 9Ω
672 (c)
675 (a)
In mixed grouping the current in the external
circuit will be maximum when the internal Current flowing through both the bars is equal.
resistance of the battery is equal to the external Now, the heat produced is given by
resistance, 2
H = I Rt
mr
R= HAB R
n or H ∝ R or = AB
HBc RBC
m
=
(1 2r)2
(1 r)2 ( 1 1
∵R∝ ∝ 2
A r )
n 1
=
4

or HBC = 4HAB
3Ω
677 (a)

Applying Kirchhoff’s law

P a g e | 261
 (2+2) = (0.1+0.3+0.2)i⇒i = 20 A Drift velocity vd =
i 1 1
⇒vd ∝ or vd ∝ 2
3 neA A d

() ( )
2
Hence potential difference across A vP d d/2
2
1 1
⇒ = Q = = ⇒v = vQ
vQ dP d 4 P 4
20 4
= 2 - 0.1× = V [less than 2V]
3 3 682 (b)
20 Because as temperature increases, the resistivity
Potential difference across B = 2 - 0.3× =0
3
increases and hence the relaxation time decreases
678 (b) for conductors τ∝
1
ρ ( )
Here S consist of S1and S2 arranged in parallel,
683 (b)
hence
Given, that, the two resistances R and 2R are
S1S2
S= connected in parallel so the potential drop is
S1+S2
equal for both.
So, the balance condition will be
The thermal energy developed in a circuit is
P R R(S1+S2)
= = 2
Q S S1S2 Vt 1
H= ⇒H ∝
R R
679 (b)
H1 R H 2R
i1 R 10 2 ⇒ = 2⇒ 1 =
= 2 = = H2 R1 H2 R
i2 R1 5 1
[Here, R1=R, R2=2R]
Or H1:H2 = 2 :1

684 (b)

When switch Sis pressed, the resistance of circuit

decreases. Hence, the current in bulb A will
increase but the current in bulb B will decrease.
H 2
=P=iR
Also heat produced per sec i.e. Hence, the brightness of bulb A will increase and
t

()
that of bulb B will decrease.
()
2
P i 5 2 2
5 5 10
⇒ 5 = 1 × = × = ⇒P4 =
P4 i2 4 1 4 1 5 685 (b)
= 2cal/s
The equivalent resistance between two corners of
680 (b) equilateral triangle having resistance R in each
arm = 2R/3 = 2×4/3 = 8/3Ω
l1-l2
Here, r = ×2Ω
l2 686 (c)

()
2
Where l1 = 240 cm, l2 = 120 cm 2i 42
Thermal power in A = PA = 3R = i R
3 3
240-120

()
∴ r= ×2 2
120 i 22
Thermal power in B = PB = 6R = i R
3 3
120
= ×2×2Ω
120 Thermal power in
2
681 (c) C = PC = i R
P a g e | 262
 3 V 6
R= -G = -3 - 25 = 975Ω [In series]
A ig 6×10
R

6R
i C 690 (c)
B
∴ [εo] = [M L T A ] and [E] = [MLT -3A ]
-1 -3 4 2 -1

⇒PA:PB:PC ∴ [ ]
1 2
2
εoE = [M L T A ]×[MLT A ]
-1 -3 4 2 -3 -1 2

4 2
= : :1 = 4 :2 :3 -1 -2
= [ML T ]
3 3

687 (b) 691 (c)

Let n be the number of wrongly connected cells For figure (i) i1 = 7A

For figure (ii) i2 = 4 + 3 = 7A
Number of cells helping one another = (12 - n) For figure (iii) i3 = 5 + 2 = 7A
Total e.m.f. of such cells = (12 - n)E For figure (iv) i4 = 6 - 1 = 5A

Total e.m.f. of cells opposing = nE 692 (c)

Resultant e.m.f. of battery Length of the wire is

= (12-n)E - nE = (12 - 2n)E
RA
l=
Total resistance of cells = 12r ρ

4×π×(0.7×10 )
-3 2
(∵ resistance remains same irrespective of ∴l = = 280m
-8
connections of cells) 2.2×10

With additional cells 693 (a)

(a) Total e.m.f. of cells when additional cells help Resistance of V1 = 80×200
battery = (12-2n)E + 2E
=16000Ω
Total resistance = 12r + 2r = 14r
Resistance of V2=32000Ω
(12-2n)E+2E
∴ =3 …(i) The current flowing in the circuit is given by
14r

(b) Similarly when additional cells oppose the 80 1

i= = A
battery 16000 200

(12-2n)E-2E Total resistance of the circuit

=2 …(ii)
14r
=16000+32000=48000Ω
Solving (i) and (ii), n = 1
Line voltage= iR
688 (b)
48000
= = 240V
P R 2 2 200
By using = ⇒ = ⇒S = 3Ω
Q S 2 6S
694 (a)
(6+S)
According to ohm’s law V = iR
689 (c)
⇒loge V=loge i+loge R⇒loge i=loge V-loge R

P a g e | 263
 The graph between loge I and loge V will be a substance
straight line which cuts loge V axis and it’s
m1 I t
gradient will be positive ∴ = 1× 1
m2 I2 t2
695 (d)
xg 4A 60 min 2
Or = × or m = xg
Current for 50 divisions, m2 6A 40 min

1×50 698 (d)

Ig = = 2mA
10

∴
Ig
=
S
S= ( )
100-l
l
.R

I S+G

⟹I = ( )
S+G
I
Initially, 30 = ( )
100-l
l
×10⇒l = 25cm

S g

Or
Finally, 10 = ( )
100-l
l
×30⇒l = 75cm

I= ( )4+20
4
5mA = 30mA
So, shift = 50cm

699 (b)
696 (c)
Applying Kirchhoff’s law for the loops (1) and (2)
As voltage across the resistors R2 and R3 is same as shown in figure
and they show same dissipation of energy, so
2 R1 = 2Ω i1
V
using the relation for energy, H = t, we have
R E1 = 4V
i1
R2 = R3. Thus, the current in each resistor R2 and 1 (i1 – i2)

R3 will be I/2 R3 = 2Ω
i2 2 i2
R2 = 4Ω
ie,I1 = I/2 and I2 = I/2
E2 = 6V

For loop (1)

-2i1 - 2(i1-i2) + 4 = 0
Since the energy dissipation is same in all the
⇒2i1 - i2 = 2 …(i)
three resistors, so
2 2 For loop (2)
I R1t = I1R2t
-2(i1-i2) + 4i2 - 6 = 0
or I R1t = (I/2)2R2t
2

⇒ - i1 + 3i2 = 3 …(ii)
or R1 = R2/4
On solving equation (i) and (ii), i1 = 1.8A
697 (c)
700 (c)
According to Faraday’s first law of electrolysis,
mass deposited dQ d
= (5t +3t+1) = 10t + 3
2
Current, i =
dt dt
m = ZIt
When t=5s
Where Z = electrochemical equivalent of
P a g e | 264
701 (a)
i = 10×5 + 3 = 53A
=
(220)2
40
+
(220)2
60 [
= (220)2
1 1
+
40 60 ]
Because H2O is used as electrolyte = (220)2 [ 60+40
60×40 ]
= (220)2
100
2400( )
=
(220)2
24

702 (c) VS
2
24
∴ P1 = = (220)2× = 24W
Rs (220)2
In a potentiometer there is no current drawn from
the cell whose emf is to be measured whereas a When the bulbs are connected in parallel
voltmeter always draws some current from the
cell. Hence, the emf of a cell can be measured 1 1 1 1 40 60
= + ⇒ = +
accurately using a potentiometer. Rp RB RB RP (220)2 (220)2
1 2

703 (d) 1 100 (220)2

= 2 or Rp =
Rp (200) 100
For balanced Wheatstone bridge
2
VS 100
P S ∴ P2 = = (220)2× = 100 W
= Rp (200)2
R Q

704 (c) P1 24 W
∴ = = 0.24
P2 100 W
Resistance of the wire is given by
706 (d)
2 2
l l ρl
R=ρ =ρ = (∵Al=V)
A Al V The resistance AB, BC and CD in series. The total
resistance is
So,
R1 = 2 + 2 + 2 = 6Ω
2
R∝l (if density remains same)
The resistance AE, EF and FD in series. The total
'
R (2l)2 resistance is
or = =4
R (l)2
'
R2 = 2 + 2 + 2 = 6Ω
R = 4R
The resistance BE and CF are in effective
Hence, change in resistance
∵ R1and R2 are in parallel
=4R-R=3R
∴ The total resistance
change in resistance 3R
Therefore, = = 3:1
original resistance 1R 6×6
R= = 3Ω
6+6
705 (c)
The current in the circuit
(Rated voltage)2
Resistance of a bulb =
(Rated power) V 3
I= = = 1.0A
(220)2 (220)2 R 3
RB = Ω and RB = Ω
1
40 2
60 707 (b)
When the bulbs are connected in series, 1 1
P∝ and R ∝ l⇒P ∝
RS = RB + RB R l
1 2

P1 l P (100-10) 90
⇒ = 2⇒ 1 = = ⇒P = 1.11P1
P2 l 1 P2 100 100 2

P a g e | 265
 ()()
2
P2-P1 l R l r 2 1
2
1
% change in power = ×100 = 11% R ∝ 2 ⇒ 2 = 2 × 12 = × =
P1 r R 1
l 1
r2 1 2 2

708 (d) R1
⇒R2 = , specific resistance doesn’t depend
2
Potential difference across PQ i.e. p.d. across the
upon length, and radius
resistance of 20Ω is V = i×20
714 (d)
48
and i =
(100+100+80+20) = 0.16A The circuit reduces to
∴ V = 0.16×20 = 3.2V 3

709 (b)
3 3
I
Drift velocity, vd = A B
neA
6

5
⟹vd =
(5×10 )×(1.6×10-19)×(4×10-6)
26
9×6 9×6 18
RAB = = = = 3.6Ω
9+6 15 5
1 -2 -1
= = 1.56×10 ms
64 715 (c)

710 (c) For semiconductor the temperature coefficient of

resistance (α) is negative. Hence, resistivity will
Let current flow from b to a as shown
decrease with the temperature rise.

716 (b)
2
R1 m a
= 21 × 2
R2 a1 m2

( )( ) ( )( )
2
m1 a2 1 1
= = 1 2
=
( ) ( ) m2 a1 2 2
2 2
2 1 2
Ratio of thermal power is I 3R : I 6R :I R
3 3
2
Q1 IRt 1
= 2 12 =
4 2 Q2 IRt 2
or : :1 or 4 : 2 : 3.
3 3
717 (a)
711 (c)
R/2 R
Both plates have same thickness, RAB = =
2 4
ρl ρ2l R/
RR = and Rs =
ld 2ld

R1
∴ =1
R2 A R/ B

712 (c)
718 (a)
R = 56×10 ± 10% = 560 + 10%

() ( )
2 2
R1 l 10 5 1
= 1 ⇒ = = = R2 = 160Ω
713 (b) R2 l2 R2 20 16

P a g e | 266
719 (a) Case III When zero current, then

In closed loop EFGDE i=o

A
E
1
B
V=E

i 1 721 (b)

R
If B2 or B3 is disconnected, resistance is increased.
1
E
2 E 3 Due to which current in the circuit is decreased.
D
E C Therefore bulb B, will become dimmer.
i 2
722 (d)

F
R
G Specific resistance doesn’t depend upon length
2
and area

723 (a)

i2R2 = E2 m 4.572
m = Zit⇒Z = =
it 5×45×60
i2×30 = 3 -4
= 3.387×10 gm/C

i2 = 0.1A 724 (a)

In closed loop ABCEA Red, brown, orange, silver red and brown
represents the first two significant figures
-i1R1 - E1 + E2 + E3 = 0
Significant figures
Multiplier Tolerance
-i1×10 - 3 + 3 + 2 = 0 Red Brown
Orange Silver
3
2 1 10 ±10%
i1 = 0.2A 3
∴ R = 21×10 ± 10%
720 (a)
725 (a)
If E is the emf of the battery, r the internal
resistance, i the current drawn and V the PD
across the plates of battery, then

V = E - ir

Case I When the direction of current in the battery

is from –ve to +ve or outside +ve to –ve, then
VAB = I.RAB =
I.ρ.LAB
=
I.ρ()
L
2
=
I ρ.()
L
2
A1 π(2r)2 π4r
2
i = +ve (during discharging)
I ρ.L
V<E VAB = 2
8πr
Case II When the direction of current in the
I.ρ.L
battery is form +ve to –ve or outside –ve to +ve, VBC = I.RBC =
A2
then

i = -ve (during discharging)

V = E + ir

⟹V>E
P a g e | 267
 -6 2
L I.ρ.L = 2.97×10 m
I.ρ. 2
2 I.ρ.L VAB 8πr 2 1
= 2 = 2⇒ = = = i = 5A
π(r ) 2πr VBC I.ρ.L 8 4
2
2πr -7
ρ = 5×10 Ωm
VBC
VAB = Required resistance to convert the galvanometer
4
into ammeter.
Now for power loss
igG 0.05×50 2.5
R= = =
PAB = VA.B.I i-ig 5-0.05 4.95

PBC = VBC.I l 50
ρ =
A 99
PAB V 1 P
= AB = ⇒VAB = BC 50 A 50 2.97×10 50 29.7
-6
PBC VBC 4 4 l= × = × -7 = ×
99 ρ 99 5×10 99 5
726 (c) = 10×0.3 = 3m

P R 728 (c)
= ' [For balancing bridge]
Q S
Parallel
3
B 2 2 2

P = 9 Q = 11 6
A C

6
R = 4 Req = 4Ω
r
D
S' 729 (a)

Given, resistance of uniform wire=24Ω.

4×11
' 44
⇒S = =
9 9 When the wire is bent in the form of a circle, then
resistance will divide the wire in two equal at
1 1 1 opposite point in parallel
⇒ ' = +
S r 6

9 1 1
⇒ - =
44 6 r

132
⇒r = = 26.4 Ω
5

727 (c) The effective resistance between the two end

points on any diameter of the circle.
Resistance of galvanometer
12×12
R=
G = 50Ω 12+12

Full scale current ig = 0.05 144

or R =
-2 2
24
A = 2.97×10 cm
or R = 6Ω
-2 -4 2
= 2.97×10 ×10 m
731 (a)

To convert a galvanometer into an ammeter, a

P a g e | 268
 shunt resistance in parallel is connected to 10 2.5
=
galvanometer. 11 l2

ig 10l2 = 2.5×11
i
G
2.5×11
l2 = = 2.75M
i-i 10
g

735 (d)

R = ρl A = 10
S
Now length l1 = l + l/10 = 11 l/10

∴ New area A1 = Al/l1

Since, galvanometer G and shunt S are in parallel,
hence ∴ New resistance,

igG = (i-ig) R1 = ρl1 A1 = ρ(11/10)/(10/11)A

-3
igG 4×10 ×15 121 ρl 121
⟹ S= = -3 = = ×10 = 12.1 Ω
i-ig 6-4×10 100 A 100
-3
60×10 -3 736 (a)
= = 10×10 Ω = 10mΩ
5.996
In the following figure
732 (b)
Resistance of part PNQ;
-3
Here , m = 1 g = p kg;
M
i2
-8 -1 i
z = 1.044×10 kg C P

i1
H = 34 k cal. = 34×1000×4.2 J N Q
3V, 1Ω
-3 5
m 10 10
q= = -8 = C
z 1.044×10 1.044

H 34×1000×4.2 10
and V = = 5 = 1.5 V. R1 = = 2.5Ω and
q (10 /1.044) 4

733 (c) Resistance of part PMQ;

ε 3
2= R2 = ×10 = 7.5Ω
2+r 4

ε 2 9+r 1 R1R2 2.5×7.5 15

0.5 = or = ∴r = Ω Req = = = Ω
9+r 0.5 2+r 3 R1+R2 (2.5+7.5) 8
734 (c)
3 24
Main Current i = = A
L∝l 15 23
+1
8
L1 l
L2
= 1
l2 So, i1 = i× ( )
R2
R1+R2
=
24
×
7.5
23 2.5+7.5
= (
18
23
A )
P a g e | 269
 24 18 6 where k be restoring torque per unit twist, n be
and i2 = i - i1 = - = A
23 23 23 number of turns in the coil, B is strength of
magnetic field in which coil is suspended, A be
737 (a)
area of coil.
Wheatstone bridge is balanced, therefore
Since, restoring per unit twist (torsional constant)
P R 10 is minimum for galvanometer A, hence more
= or 1 = ⇒S = 10 ohm sensitive.
Q S S

738 (c) 742 (b)

I 20 Let the resistance of voltmeter is GΩ.

vd = = 29 -6 -19
nAe 10 ×10 ×1.6×10
-3 ∴ Total resistance of the circuit
= 1.25×10 m/s

739 (d) R= ( G×100

G+100
+50 Ω )
The resistance are connected in parallel hence, Total current
voltage will remain constant
V 10
i= =
( )
2
Vt G G×100
E= +50
R G+100
1 Voltage across 100Ω resistance
⟹E ∝
R

E1 R
= 2 =
2R
=
2
=i ( G×100
G+100
=
10
G×100 ) (
+50
×
G×100
G+100
)( )
E2 R1 R 1 G+100

740 (c) Reading of voltmeter =5V

V ∴ Voltage across 100Ω =5V

I=
R
R C R
∴5 =
(
10
G×100
+50
×
G×100
G+100
) ( )
A
F
R
E
B G+100
R R

On solving G = 100Ω
V

743 (b)

1 V The distribution of current is as shown in figure.

∴ Current in FC = = As per question,
2 2R

741 (a)
A 2 B 2 C 1 D
Current sensitivity of a galvanometer is defined as i i - i1 i1 i1/2 i1/2
the deflection produced in the galvanometer
when a unit current flows through it. E 4 2 1
1A
If θ is the deflection in the galvanometer when
H i G i1 F i 1/2 E1
current I is passed through it, then

θ nBA
Current Sensitivity Is = =
I k

P a g e | 270
 i1 The given circuit is
= 1 or i1 = 2A
2

In a closed circuit ACFG

i
2i + 2× 1 - 4(i-i1) = 0
2

7i1 = 4i 1 1 1 1 1 4
= + + ⇒ = ⇒Req. = 1Ω
Req. 4 2 4 Req. 4
7 7 7
or i = i1 = ×2 = A 749 (c)
4 4 2
l 2l l
Total resistance of the circuit between A and H is R1 ∝ ⇒R2 ∝ i.e.R2 ∝
A 2A A
4×3 26
=2+ = ∴ R1 = R2
4+3 7

7 26 751 (d)
EMF of cell is E = × = 13V
2 7
R∝l
744 (c)
R
Hence every new piece will have a resistance .
i = q/T = qv 10
If two pieces are connected in series, then their
-2
= 2×10 ×30 = 0.6 A 2R R
resistance = =
10 5
745 (a)
If 5 such combinations are joined in parallel, then
-6
q CV (10×10 )×40 R R
Current i = = = net resistance = =
t t 0.2 5×5 25
-3
= 2×10 A = 2 mA 752 (b)

746 (c) The current in a conductor depends on the drift

2 velocity of electrons
V
Resistance of electric bulb R = where
P 753 (b)
subscripts denote for rated parameters.
The circuit arrangement shown in figure (b) is the
(220)2 correct arrangement for verification of Ohm’s law.
R=
100 For convenience the same figure has been
V
2 redrawn here. In the figure, R is the resistance, for
Power consumed at 110 V, Pconsumed = which Ohm’s law is to be verified. Voltmeter V is
R
connected to its parallel and ammeter, cell and
(110)2 rheostat arrangement in the series.
∴ Pconsumed = = 25 W
(220)2/100

747 (c)

2+2 4
i= = A
1+1.9+0.9 3.8

4
For cell A E = V + ir⇒V = 2 - ×1.9 = 0
3.8
754 (b)
748 (b)
P a g e | 271
 me t0
Resistivity, ρ = 2R
∫(a-2bt) dt
2
ne t = 2
9
2 0
1 ne τ

[ ]
Conductivity, σ = = t0
p me
2R
∫(a -4b t -4dt)dt
2 2 2
=
755 (a) 9
0

110
[{ }]
2 2 2 to
Voltage across each bulb V =
'
= 55W so, 2R 2 4b t 4bat
2 = a t+ -
9 3 2 0
power consumed by each bulb will be

[ ]
2 3
2R 2 4b t0 2
500W 500W = a t0+ -2bat0
220V 220V 9 3

55V 55V a
but t0 = [from Eqn.(i)]
2b

[ ]
110 V 2 3 2
2R 2 a 4b a a
H= a× + 3 -2ab 2
9 2b 3 8b 4b
( )
2
' 55
P = ×500
220
[ ]
3 3 3 3
2R a a a aR
= + - =
9 2b 6b 2b 27b
125
= W
4 757 (d)

756 (b) 1 2
The energy stored in the capacitor = CV ;
2
2
Q = at - bt
This energy will be converted into heat in the
dQ
I= = a - 2bt …(i) resistor.
dt
1 -6
H= ×4×10 ×400×400
2
-2
= 32×10

= 0.32 J
Where, t = t0,I = 0
758 (c)
In loop BCDEB I1(2R) - (I-I1)R = 0
Potential difference between B and D is zero, it
or 3I1 = I means Wheatstone bridge is in balanced condition

I a-2bt B
∴ I1 = = X
3 3 21 6 8X
3 
8 (8  X)
t0 15 3

∫I (2R)dt
2
∴H = 1
A C
0 18 15 4 6

6 4
6 4
D

P a g e | 272
 P R 21 18
So = ⇒ = ⇒X = 8Ω
Q S 8X 6
3+
(8+X)

759 (d)

Resistance of heater 762 (b)

2
V (100)2 ρL ρ2L
RH = = = 10 Ω R= ; R1 = = 2R
P 1000 A A
Total resistance of circuit m = A L d;m1 = A 2 L d = 2m
' 10×R 100+20R (3V)2t
R = 10 + =
10+R 10+R Now, = mc∆T …(i)
R
Current in heater (NV)2t
and = 2mc∆T …(ii)
R 100 R 5×R 2R
IH = I. = × =
10+R 100+20R (10+R) 5+R
Solving (i) and (ii), we get N = 6.
10+R
763 (a)
225R
∴ Power P = I RH⇒ ×10 = 62.5
(5+R)2
H

∴ R = 5Ω

760 (d)

Let m1 = m, then m2 = 3m and m3 = 5m. Again -3

GS V 25×10
let L3 = l, then L2 = 3l and L1 = 5l. If σ be the = G =
G+S I 25
density of copper, then
GS
m = 0.001Ω
m 3m 5m G+S
A1 = 1 = ,A = and A3 =
L1σ 5lσ 2 3lσ lσ
Here S << G So
5A1
Hence A2 = and A3 = 25A1 S = 0.001 Ω
3

ρL1 ρL2 764 (a)

ρ.5l ρ.3l 3
∴ R1 = = ,R2 = = = R,
A1 A1 A2 5A1 25 1 2
VRated 1
PRated = ⇒R ∝ [V - constant]
R PRated
ρL3 ρ.l R
and R3 = = = 1
A3 25A1 125 So bulb of high power will have less resistance

3 R 765 (a)
∴ R1:R2:R3 = R1: R1: 1 = 125:15:1
25 125
m 0.99
m = Zi t ⇒i = = = 2.5A
761 (b) Zt 0.00033×1200

Using Kirchhoffs first law we can find current Hence heat generated in the coil is
distribution in the given part of electric circuit as
H = i Rt = (2.5)2×0.1×1200 = 750 J
2
shown in the adjoining figure. From figure
i = 1.7A
766 (d)

P a g e | 273
 The three resistances between A and B are straight line passing through origin
parallel,
771 (d)
1 1 1 1
= + + Since it’s a balanced Wheatstone bridge, the
Rcomb R1 R2 R3
circuit can be redrawn as
1 1 1
= + +
9 9 9

1 3
=
Rcomb 9

⟹ Rcomb = 3Ω
12I = 30(1.4 I)
768 (d)
12I = 42 - 30I
The temperature difference is 20℃ = 20 K. So
∴ I = 1A
that thermo emf developed
μV 772 (c)
E = αθ = 40 ×20K = 800 μV
K
We know that when current flow is same then
4
Hence total emf = 150×800 = 12×10 μV = 120 resistors are connected in series, hence resultant
mV resistance is
769 (c) '
R = R1 + R2 = 10Ω + 20Ω = 30Ω
Given, Thomson’s coefficient, σ = 3μV/℃ Also since, cell are connected in opposite
dV directions, the resultant emf is
σ=
dT
E = E1 - E2 = 5V - 2V = 3V
or d V = σdT
From Ohm’s law E = iR
so V = σ(T2 - T1)
E 3
∴ i= = = 0.1A
-4 -6
or 3×10 = 3×10 (T2 - 20) R 30

773 (b)
or 100 + 20 = T2
In parallel,
or T2 = 120° C
R1R2
770 (d) Rp =
R1+R2
Terminal voltage V = E - Ir. Hence the graph
5.0×10.0
between V and i will be a straight line having =
5.0+10.0
negative slope and positive intercept.
50
Thermal power generated in the external circuit = = 3.3Ω
2
15
P = EI - I r. Hence graph between P and I will be a
parabola passing through origin. Also,

Also at an instant, thermal power generated in the

2
cell = i r and total electrical power generated in

()
2
Ir r
the cell = Ei. Hence the fraction η = = I;
EI E
so η ∝ I. It means graph between η and I will be a
P a g e | 274
 ∆Rp Dividing Eq. (ii) by Eq. (i), we get
×100
Rp -3
0.6 1+1.5×10 t
∆R1 ∆R2 = -3
= ×100 + ×100 0.3 1+1.5×10 ×27
R1 R2
∆(R1+R2) ⟹2(1+1.5×10 ×27) = 1 + 1.5×10 t
-3 -3
+ ×100
R1+R2
-3 -3
⟹2 + 81×10 = 1 + 1.5×10 t
0.2 0.1 0.3
= ×100 + ×100 + ×100 ⟹2 + 0.081 = 1 + 1.5×10 t
-3
5.0 10.0 15

= 7% 1.081
⟹t = -3 = 720°C = 993K
1.5×10
Rp = 3.3Ω ± 7%
776 (c)
774 (b)
1 dR
Temperature coefficient of resistance=
The given circuit can be simplifies as follows Rt dt

1 d 2
r r = R (1 + αt + βt )
Ro(1+αt+βt ) dt o
2
r
r r
P Q
r α+2βt
r r = 2
1+αt+βt


r r
778 (a)
r r

( )
P Q
r r
r=
l1-l2
l2
'
×R ⇒r =
50 (
55-50
)
×10 = 1Ω


2r

2r 779 (a)
P Q
2r
ρ1l1 ρl
R1 = and R2 = 2 2 . In series Req = R1 + R2
A A

ρeq.(l1+l2) ρl ρl ρ l +ρ l
= 1 1 + 2 2 ⇒ρeq = 1 1 2 2
' 2r 2 3 A A A l1+l2
R = = × = 1Ω
3 3 2
780 (d)
775 (c)
In parallel combination Eeq = E = 6V
Given, R300 = 0.3Ω, Rt = 0.6Ω
781 (b)
T = 300K = 27°C
From the figure
Temperature coefficient of resistance,
-3 -1 4Ω 6Ω
α = 1.5×10 K
I1
∴ R300 = R0(1+α×27) 5Ω
I2
0.3 = R0(1+1.5×10 ×27) …(i)
-3

Again, Rt = R0(1+αt)
10I1 = 5I2 I2 = 2I1 ….(i)
0.6 = R0(1+×10 ×t) …(ii)
-3

Heat produced in resistance

P a g e | 275
 2 2 2
I2R = I2×5 V (30)2
R0 = = = 10 Ω
P 90
-1
∴×5 = 10 cals ….(ii)
Current in the lamp
From Eqs. (i) and (ii) ,we have
30
i= = 3Ω
I2 1 10
I1 = = 2 = A
2 2 2
When lamp is operated on a 120 V, then
Hence, heat produced in resistance of 4Ω resistance
'
2 1 -1 ' V 120
I2×4 = ×4 = 2 cals R = = = 40 Ω
2 i 30

782 (b) Thus, for proper glow, the resistance required to

the put in series will be
No current flows through the capacitor branch in
'
steady state. Total current supplied by the battery R = R - R0 = 40 - 10 = 30 Ω

6 3 789 (a)
i= =
2.8+1.2 2
When r is internal resistance of the battery and i
3 3 the charging current, then
Current through 2 Ω resistor = × = 0.9A
2 5
V = E + ir
783 (d)
V = 2 + 5×0.1 = 2.5V
96500 coulombs of charge is needed to deposit
one gram equivalent of an element at an electrode 791 (b)

784 (d) Current flowing in the circuit

E 10-4 1
Pressing the key does not disturb current in all i= = = A
R 20+10 5
resistances as the bridge is balanced. Therefore,
deflection in the galvanometer in whatever 1
P.D. across AC = ×20 = 4V
direction it was, will stay 5

785 (c) P.D. across AN = 4 + 4 = 8V

Same mass, same material i.e. volume is same or 792 (d)

Al = constant -6
A V A 2 10 -6
ρ=R = × = × = 10 Ωm

() () l i l 4 0.5
2 4
l R l A A d2
Also, R = ρ ⇒ 1 = 1 × 2 = 2 =
A R2 l2 A1 A1 d1
793 (d)

( )
4
24 d Given, V = 1V
⇒ = = 16⇒R2 = 1.5Ω
R2 d/2
Ig = 1mA
786 (b)
-3
Ig = 1×10 A
π = T dE/dT.
Resistance of galvanometer
787 (d)

Resistance of the lamp

P a g e | 276
 Rg = 50Ω 2 16
= I ×2 = ×8.4×2
9
V
Rg = -R -1
Ig g ≈ 30 Js

1 796 (a)
or Rs = -3 - 50
10
The resistance of a conductor
or R1 = 950Ω
1
∴R =ρ 2
' 10 πr
R2 = -3 - 50 = 9950Ω
10
1
⟹ R∝ 2
or R2 = R'2 - R1 = 9950 - 950 r

50 2
or R2 = 9000Ω (a)R1 = = 800×10 Ω
(0.25×10-1)2
794 (a)
100 2
(b)R2 = = 400×10 Ω
Ti+ Tc (0.5×10-1)2
Neutral temperature, Tn =
2
200 2
(c)R3 = = 200×10 Ω
or Ti = 2Tn - Tc = 540℃. (1×10-1)2

795 (c) 300 2

(d)R4 = = 133.3×10 Ω
(1.5×10-1)2
The simplified circuit is shown below
Hence, electrical resistance of first wire is
maximum.
I1 6Ω 9Ω
I 2Ω I 797 (b)
I2 5Ω
Using the concept of balanced wheat stone bridge,
we have

From figure, P R
=
Q S
15I2
15 I1 = 5I2 or I2 = = 3I1 x 10
5 =
52+1 48+2
I2 4I
∴ I = I1 + I2 = + I2 = 2 …(i) 10×53
3 3 x=
50
2
But I2 ×5 = 42
= 10.6Ω
2 42
Or I2 = = 8.4 798 (c)
5
Let equivalent resistance between A and B be R,
Putting value of I2 in Eq. (i), we get
then equivalent resistance between C and D will
4 also be R
I= × 8.4
3

Therefore, heat dissipated across 2 Ω

P a g e | 277
 A 1
C
90% of earlier value, because fall in temperature
is small. Hence, option (d) is correct.
R 1 R
803 (a)
1
B D Electric fuse is a type of over current protection
device. They are engineered to contribute a
negligible amount of extra resistance to the
' R
R = +2=R circuits they protect. This is largely accomplished
R+1
by making the fuse wire as short as possible.
2
⇒R - 2R - 2 = 0 Fuses are primarily rated as current amperes. A
fuse wire of certain material and gauge will blow
2± 4+8 at a certain current no matter how long it is.
∴R = = 3 +1
2 Since, length is not a factor in current rating the
shorter it can be made the less resistance it will
799 (d)
have end to end.
Change q = it = 0.5 A×3600 sec = 1800
804 (b)
coulomb
Given, I = 1 A, t = 10 s, q = It, q = 10 C
800 (d)
2+ -19
2 Charge of Cu = 2e = 2×1.6×10 C
V 220 ×220
Power P = ⇒300 =
R R The number of copper atoms deposited at the
cathode
22 ×22
R=
3 10 19
= -19 = 3.1×10
2×1.6×10
110 ×110 ×3
Again P =
22 ×22 805 (c)
P = 75 Ohm’s Law is not obeyed by semiconductors
75 ×100 806 (c)
P% = = 25%
300
i G 1 20 20
The percentage reduction in power = 1 + ⇒ -3 = 1 + ⇒S = ≈ 0.02Ω
ig S 10 S 900
P = 100 - 25 = 75%
807 (d)
801 (d)
The resistance of voltmeter is too high, so that it
Neutral temperature is independent of draws negligible current from the circuit, hence
temperature of cold junction. potential drop in the external circuit is also
negligible
802 (d)
808 (d)
Let the resistance of the lamp filament be R. Then
100 = (220)2/R. When the voltage drops, Full deflection current
-4 -4
expected power is ig = 25×4×10 = 100×10 A

P = (220 ×0.3)2/R' V 25
Using R = -G = -4 - 50 = 2450Ω in
Ig 100×10
Here, R' will be less than R, because now the rise series
in temperature will be less. Therefore, P is more
than (220 ×0.9)2 R = 81 W. But it will not be 809 (a)

P a g e | 278
 The resistance of 25 W bulb is greater than 100 W 6i×R = E …...(ii)
bulb. So for the same current, heat produced will
be more in 25 W bulb. So it will glow more From Eqs. (i) and (ii), we have
brightly 6iR = 5ir
810 (d) 5
Or R = r
E i
6
i F
2i i Hence, r=6Ω
B
A
2i 2i 5
∴R = ×6 or R = 5Ω
i 6
2i
2i G
6i H 811 (c)
2
2i V (18)2
D i C 6i
P= = = 54W
R 6

812 (d)
E
Rseries = R1 + R2 + R3 + …

813 (c)

Let ABCDEFGH be the skeleton cube formed by Since current i is independent of the value of R6, it
joining twelve equal wires each of resistance r. Let is clear that the circuit is of a balanced
the current enters the cube at corner A and after Wheatstone bridge. As per condition of balance,
passing through all twelve wires, let the current we have
leaves at G, a corner diagonally opposite to corner
R1 R
A. = 2 ⇒ R1R4 = R2R3
R3 R4
For the sake of convenience, let us suppose
814 (b)
that the total current is 6i. At A, this current is
divided into three equal parts each (2i) along AE, If we take R1 = 4Ω, R2 = 12Ω, then in series
AB and AD as the resistance along these paths are resistance
equal and their end points are equidistant from
exit point G. At the points E, B and D, each part is R = R1 + R2 = 4 + 12 = 16Ω
further divided into two equal parts each part
equal to i. The distribution of current in the In parallel, resistance
various arms of skeleton cube is shown according
4×12
to Kirchhoff’s first law. The current leaving the R= = 3Ω
4+12
cube at G is again 6i.
So, R14Ω and R2 = 12Ω
Applying Kirchhoff’s second law to the closed
circuit ADCGA, we get 815 (d)
2ir + ir + 2ir = E Discharging energy
Watt hour efficiency =
Charging energy
Or 5ir = E ……(i)
14×5×15
Where E is the emf of the cell of neglegible = = 0.875 = 87.5%
15×8×10
internal resistance. If R is the resistance of the
cube between the diagonally opposite corners A 816 (a)
and G, then according to Ohm’s law, we have
P a g e | 279
 Effective emf in circuit = 100 - 12 = 88V V m zlt
Thickness d = = =
A ρA ρA
Current in circuit
=
(3.3×107)×(1.5)×(20×60) -6
= 6.6×10 m.
effective emf 88 9000×(50×10×2)×10
-4
i= =
resistance R
826 (b)
88 88
Or R = = = 88Ω
i 1 Total resistance between points P and Q,

817 (c) If m gram of the ice mean m given time t, then As

per question,
These questions are done by hit and trial method
only. You check all the options one by one till you
(10)2× 20 (10×60) = m×80×4.2
get the final desired result 3

819 (a) 100×20×10×60 3

or m = = 1.19×10 g
3×80×4.2
Total external resistance
= 1.19 kg.
'
R = R/3
827 (c)
For maximum heat generation
For a two cell battery
Rext = rint
2E E
I= =
R 2r r
= 0.1
3
Similarly, for a n cell battery
R = 0.3 Ω
nE E
I= -
820 (a) nr r

Neutral temperature remains same. So, current in the circuit does not depend on
number of cells in the battery.
Inversion temperature, Ti = 2Tn - Tc
Hence, the correct graph will be (c).
= 2×285 - 10 = 560℃
828 (b)
821 (c)
m = Zi t and i t =Area of given curve
( )
2 5 2
V V V R+5 4
+ = 4× or =
R 5 R+5 5R R+5 = Area of triangle + Area of rectangle

On solving, we get R = 5Ω. 1

⇒i t = ×(2×60)×1 + (6-2)×60×1 = 300
2
822 (a)
m m
3/2 ∴Z= =
It is called safe current and is proportional to r it 300

823 (c) 829 (d)

By using Rt = R0(1 + αt) As circuit is open, therefore no current flows

through circuit.
3×R0 = R0(1+4×10 t)⇒t = 500℃
-3

Hence, potential difference across X and Y

825 (b)
=emf of battery =120V.
P a g e | 280
830 (b) fuse.

When bulb glows with full intensity, then voltage 835 (b)
across it will be 1.5 V and voltage across 3 Ω
resistance will be 4.5 V e.R 10×3
Potential gradient = =
(R+r).L (3+3)×5

= 1V/m = 10 mV/cm

836 (a)
2
V
Power dissipated P = ;
Reff
4.5 Reff is least in case of figure (a). Hence power
Current through 3 Ω resistance i = = 1.5A
3
dissipated in circuit (a) is maximum
Same current will flow between X and Y
So VXY = iRXY⇒1.5 = 1.5RXY⇒RXY = 1Ω 838 (b)

831 (c) Required arrangement is shown in figure.

ρ1l ρl l B
R = R1 + R2 = + 2 = (ρ1+ρ2) …(i)
A A A 1 2

ρ(2l)
R= …(ii) 3
A A C

ρ1+ρ2 3V
From Eqs.(i) and (ii), 2ρ = ρ1 + ρ2 or ρ =
2

833 (b)
The equivalent circuit will look like (since the two
Because H has positive charge resistances of 1Ω and 2Ω are in series, which from
3Ω which is in parallel with 3Ω resistance).
834 (a)
3Ω
220 ×220
Resistance of 25 W bulb = = 1936
25
Ω 3
A C
220
It’s safe current = = 0.11 A
1936

220 × 220 3V
Resistance of 100 W bulb = = 484 Ω
100

220
It’s safe current = = 0.48 A
484 Therefore, the effective resistance is

when connected in series to 440 V supply, then (1+2)×3 3

= Ω
the current (1+2)+3 2

440
i= (3/2) Ω
(1936 +484)
= 0.18 A

Thus, current is greater for 25 W bulb, so it will

P a g e | 281
 4X l
A C
=
I Y 100-l
3V
B
X l
1 ⟹ = …(ii)
2 Y 4(100-l)
I2 3Ω I 2 = 1A
Therefore, form Eqs. (i) and (ii)
A C
I 1 = 1A I 1+ I 2= 1A
l 20
3V =
4(100-l) 80

∴ Current in the circuit, l 1

⟹ =
4(100-l) 4
3
I= = 2A
()3
2
⟹ l = 100 - l

⟹ 2l = 100
I
∴ Current in 3Ω resistor= = 1A Hence, l=50cm
2
841 (a)
839 (c)
Given, galvanometer resistance G=240Ω
mAg/mZn = EAg/EZn = 108/31
I IG I
or mAg = mZn×108/31= w×108/31 G

3.48 w = 3.5 w I- I
G

840 (a)

Meter bridge is an arrangement which works on

S
Wheatstone’s principle, so the balancing condition
is

R l
= 1
S l2
Shunt resistance S = ?
Where l2 = 100 - l1
4
Ist case R = X, S = Y, l1 = 20 cm, IG = I
100
l2 = 100 - 20 = 80cm
From figure voltage through the circuit.
X 20
∴ = …(i) (I-IG)S = IGG
Y 80

IInd Case Let the position null point is obtained at

a distance l from same end.
or ( )
I-
4I
400
S=
4I
100
×240

∴ R = 4X, S = Y, l1 = l, l2 = 100 - l 4×240

or S= = 10Ω
96
So, from Eq. (i)
842 (b)

In series, the current I, is same in two bulbs.

2
V
ResistanceR =
P
P a g e | 282
 And potential drop (V) = IR 850 (a)

∴ Potential difference across 60 W bulb is greater

than the potential difference across 200 W bulb. ( 1 2 1 2
Thermo electric power E = k TT0-T0Tr- T0+ Tr
2 2 )
844 (a)
-3
S=
dE
dT [ 1
= k T-0- ×2T+0
2 ]
E 2.4×10 -4
E = xl = iρl⇒i = = = 4×10 A
ρl 1.2×5 = k[T0 - T]

845 (c) At T = T0/2

Since the unit of electrical energy is kilowatt hour 1

S= kT
(kWh), 2 0

So total number of units consumed is 851 (c)

N = (0.1×8 + 0.3×4)×30 Power of the combination

P 1000
(Because June has 30 days) Ps = = = 500W
n 2
∴ N = 60 units 852 (b)
Total cost = 60×0.5 = 30 Rs. For no current through galvanometer, we have

( ) ( )
846 (b) E1 12
X = E⇒ X = 2⇒X = 100Ω
To shift the balance point on higher length, the 500+X 500+X
potential gradient of the wire is to be decreased. 853 (c)
The same can be obtained by decreasing the
current of the main circuit, which is possible by E1 = ?, l1 = 60cm;E2 = 3V, l2 = 45cm
increasing the resistance in series of
potentiometer wire. In balance condition

847 (b) E1 l E 60
= 1⇒ 1 = ⇒E1 = 4 volt
E2 l2 3 45
7 1 1
= + ⇒R = 3Ω
12 4 R 854 (d)

848 (d) E1 l +l (6+2) 2

= 1 2 = =
E2 l1-l2 (6-2) 1
P 4.5
Current in the bulb = = = 3A
V 1.5 855 (a)
1.5 According to Seebeck effect
Current in 1Ω resistance = = 1.5A
1
856 (a)
Hence total current from the cell
i = 3 + 1.5 = 4.5A From Kirchhoff’s first law at junction P,

By using I1 + I2 = 6 …(i)
E = V + ir⇒E = 1.5 + 4.5×(2.67) = 13.5V
From Kirchhoff’s second law to the closed circuit
849 (d) PQRP,
Colliding electrons lose their kinetic energy as
heat
P a g e | 283
 -2I1 - 2I1 + 2I2 = 0 859 (d)

⟹ - 4I1 + 2I2 = 0 Suppose current though different paths of the

circuit is allows :
⟹2I1 - I2 = 0 ……(ii)
28 54

Adding Eqs. (i) and (ii), we get

6V
1 2
3I1 = 6⟹I1 = 2A i3

From Eq. (i), 8V 12 V

I2 = 6 - 2 = 4A
After applying KVL for loop (1) and loop (2)
857 (b)
1
We get 28i1 = -6 - 8⇒i1 = - A
The given circuit can be redrawn as follows 2

1
A
R and 54i2 = -6 - 12⇒i2 = - A
F C
3
R
R R 5
hence i3 = i1 + i2 = - A
6
D
R B 860 (d)
2
i Rt = ms ∆t
Equivalent resistance between A and B is R and
V ⇒ (4)2×7×3×60
current i =
R 3
= 0.1×4.2×10 ×[T-(20+273)]
858 (d)
∴ T = 341 K = 68℃
Resistance of a conductor
861 (a)
ρl
R= ………………..(i) Current through R is maximum when total
A
internal resistance of the circuit is equal to
As ρ depends on the material, external resistance

so R depends on the material. 862 (c)

According to the given formula in Eq.(i), it A voltmeter is a high resistance device and is
depends on length. Moreover resistance always connected in parallel with the circuit.
∝temperature. While an ammeter is a low resistance device and
is always connected is series with the circuit. So,
If R0= resistance of conductor at 0° C, to use voltmeter in place of ammeter a high
resistance must be connected in series with the
Rt= resistance of conductor at t°C, circuit.
and a,β =temperature coefficient of resistance, 863 (a)
then
-2
V l 2 50×10 -6
Rt = R0(1+αt+βt )
2 R= =ρ ⇒ =ρ -3 2 ⇒ρ = 1×10 Ωm
i A 4 (1×10 )
The resistance of a straight conductor does not 864 (a)
depend on shape of cross-section.
P a g e | 284
 2
V
P=
R
| V |
2
V a b
∴ Rhot =
R
i g

200×200
=
100
R
= 400Ω G

400 Voltmeter
Rcold = = 40Ω
100

865 (b)
18 18 -19 Hence, a voltmeter is made by connecting a high
Current, i = (2.9×10 + 1.2×10 )×1.6×10
resistance in series with a pivoted type moving
= 0.66 A towards right coil galvanometer G. The value of R depends upon
the range of the required voltmeter.
866 (b)
869 (b)
3 = 1.5(1+r)⇒r = 1Ω

867 (d)

When a single heater (resistance R1 = R) is

connected to 220 V, then it will consume a power
P1 =1000 W. If two such identical heaters are
connected in parallel (total resistance R2 = R1/2
When switch S is open, the corresponding
= R/2) to some source, then it will consume
equivalent circuit diagram is as shown in the
power P2.
figure
P2 R
= 1
P1 R2

⇒ P2 = 2P1

P2 = 2000 W

868 (a)
The equivalent resistance between A and B is
A voltmeter is an instrument used to measure the
potential difference between two points in an 12×6 6×12
Req = + = 4 + 4 = 8Ω
electrical circuit directly in volts. Voltmeter is 12+6 6+12
essentially a galvanometer which is connected in
When switch S is closed, the corresponding
parallel across two points in the circuit between
equivalent circuit diagram is as shown in the
which the potential difference is to be measured.
figure below
The potential difference read by the voltmeter is
slightly less than the actual value to be measured.
Hence, the resistance of the voltmeter should be
as high as possible so, that on connecting it in a
circuit across two points the potential difference
may not fall appreciably.

P a g e | 285
 ' 1
P = P
2

1
So, power becomes of initial value.
2

873 (d)

P 40 2
Bulb (I):Rated current I1 = = = amp
V 220 11
2
The equivalent resistance A and B is V (220)2
Resistance R1 = = = 1210Ω
'''''''''''''''''''''''''''''''' 8×8 P 40
Req = = 4Ω
8+8
100 5
Bulb (II):Rated current I2 = = amp
870 (d) 220 11

The bridge ABCD is balanced if (220)2

Resistance R2 = = 484 Ω
100
10 30
= ⇒R1 = 3Ω When both are connected in series across 40 V
R1 9
supply
When the bridge is balanced, no current flows in
R1 R2
the arm BD. Therefore, R2 can have any finite
value

871 (d)
40 V
R
Resistance of each part will be ; such n parts are
n
R Total current through supply
joined in parallel so Req = 2
n
40 40 40
I= = = = 0.03A
872 (a) P1+P2 1210+484 1254

The rate of dissipation of electric energy is called This current is less than the rated current of each
electric power bulb. So neither bulb will fuse

W = Vit Short Trick : Since VApplied < VRated' neither bulb

will fuse
The electric power dissipated will be is given by
874 (c)
W Vit
P= = = Vi
t t m = zit
2
V 5 9 -1
= …(i) 9 = z×10 ⇒z = 5g C
R 10
-5
When resistance is doubled, then let electric ∴ m = zit = 9×10 ×50×(20×60) =5.4 g.
'
power is P .
875 (a)
2
' V
∴ P = …(ii) For maximum energy, we have
2R

From Eqs, (i) and (ii), we get External resistance of the circuit

= Equivalent internal resistance of the circuit i.e.

P a g e | 286
 r Heat absorbed by water = mass×specific heaat
R=
2 capacity

876 (a) ×rise in temperature

Given problem is the case of mixed grouping of = 1×4200×(100 - 15)

cells
-3
= 4.2×85×10
nE
So total current produced i =
nr heat absorbed
R+ Efficiency = ×100
m heat produced
3
Here m = 100, n = 5000, R = 500Ω 85×4.2×10
= 4 ×100 = 79%
45×10
E = 0.15 V and r = 0.25Ω
880 (b)
5000×0.15 750
⇒i = = ≈ 1.5 A
5000×0.25 512.5 If number of collisions of the free electrons with
500+
100 the lattice is decreased, the time of relaxation of
electrons will increase. Due to which drift velocity
877 (c) of electrons will increase and hence current will
Let time taken in boiling the water by the heater is increase
t s. Then 881 (a)
Pt Power is the amount of work done or energy (Q)
Q = ms∆T ⇒ = ms∆T
J
transferred per unit of time (t).
836
∴ t = 1×1000(40-10) H
4.2 ∴ P=
t
836
t = 1000×30 H = mc ∆θ
4.2
Where m is mass, c the specific heat and ∆θ the
1000×30×4.2
t= = 150s temperature difference.
836
mc ∆θ
878 (a) ∴P =
t
Effective resistance of three resistances between
∆θ = (35-20)℃ = 15℃
C and
C = 1000, 4.2 J = 1 cal
R×2R 2
D= = R
R+2R 3 t = 1min =60s

Total resistance between A and B 1×1000×15×4.2 -1

∴P = Js = 1050 W
60
2 8 8
=R+ R + R = R = ×3 = 8Ω
3 3 3 882 (b)

879 (a) Drift velocity,

Heat produced H = Vit = Pt J i

vd =
neA
Where, P = Vi watt
4 Where I is current, n the number of electrons, e
∴ H = 500×15×60 = 45×10 J
the electron charge and A the area of cross-

P a g e | 287
 section of wire. VA - VB = iR + V - V = iR

Number of electrons per unit volume V

⇒ Potential drop across C =
21
3
2×10
n=
A×100 886 (c)

So, current in the wire If l is the balancing length for R1, and l' for R2,

i = neAvd V0 ER1 V ER1

.l = ⇒ 0 .l = as R = 8R1
21 L0 (r+R1+R2) L0 ( 1+9R1) 2
r
2×10 -19
i= ×1.6×10 ×A×0.25
A×100 V0 ' E.R2 E.8R1
l = =
L0 (r1+9R1) (r1+9R1)
=0.8A

883 (d) l
'
R 8R1 '
∴ = 2 = ⇒l = 8 l
l R1 R1
m l
Resistance of a conductor, R = 2
ne τ A 887 (c)

Where the symbols have their usual meaning Production of heat at junctions due to current is
known as Peltier effect
As the temperature increases, the relaxation time
τ decreases 889 (c)

884 (c) If volume and density remains same, then

resistance of wire
2 2
V 220
In India, power, P1 = = 2
R R R∝l
'2
V (110)2 Where lis the length of the wire
In USA, power, P2 = ' = '
R R
( ) ( )
2 2
' 10 11
∴ R = 1+ l = R
220 (110)22
100 10
As P1 = P2, so = '
R R
Hence,
' (110) R
2
or R = R= R -R
'
21
(220)2 4 = = 21%
R 100
885 (c)
Therefore, change in resistance of wire =21%
Moving anticlockwise from A
890 (c)
V R i
nE n×1.5
In series i = ⇒0.6 = ⇒n = 10
V
nr+R n×0.5×20
A
B C 891 (d)
2V

2R i The temperature coefficient of resistivity for

thermister is negative. Therefore by increase in
temperature the resistivity of the thermister
-iR - V + 2V - 2iR = 0 decreases.
V 892 (d)
or 3iR = V or i =
3R
In steady state capacitor is fully charged and no
P a g e | 288
 current flows through it Similarly effective value of parallel combination of
24Ω and 8Ω resistance is given by

24×8
R2 = = 6Ω
24+8

48V

3Ω 10 6Ω 1Ω
A B

∴ No current passes through 4Ω

Hence, the circuit may be redrawn as shown in
1 1 1 1 the adjacent figure, where total resistance across
= + +
Reff 1 2 3 A and B, R = 3 + 10 + 6 + 1 = 20Ω. As
potential across R2( = 66Ω) is 48 V, hence
6+3+2 11
= =
6 6 R 48×20
VAB = 48× = = 160V
R2 6
6
⇒Reff = Ω
11 896 (b)

Current =
6×11
6
= 11A m = zIt = z () P
V
t

( )
-6 -6
Q = CV = 0.5×10 ×6 = 3.0×10 C = 3μC -6 100×1000
= 0.367×10 ×60 kg
125
893 (b)
= 0.017616 kg = 17.616 g.
Let R = 100Ω
897 (d)
' 10
∴ R = 100 + 100×
100 X 20 1
= ⇒X = Ω = 0.25Ω
1 80 4
= 110Ω
'
898 (a)
∴ ∆R = R - R
As we know that,
= 110 - 99 = 11Ω
2
V
894 (b) P=
R

() ( )
2 2
l l R l d L 2d V
2
R ∝ ∝ 2⇒ 1 = 1× 2 = =1 ⇒R =
A d R2 l2 d1 4L d P

⇒R2 = R1 = R 1
⇒ R∝
P
895 (d)
Hence, the resistance of 25 W bulb is greater than
Effective value of resistance of parallel the resistance of 100 W bulb.
combination of 20Ω, 30 Ω, 60Ω is R1, where
Now, both the bulbs are joined in series so the
1 1 1 1 3+2+1 6 1 current will be same.
= + + = = =
R1 20 30 60 60 60 10
So, heat produced by the bulbs
R1 = 10Ω
2
H = i Rt ⇒ H ∝ R
P a g e | 289
 So, the heat produced by the 25 W bulb is greater
than the bulb of 100 W, because its resistance is
more than that of 100 W bulb. Hence, 25 W bulb
will glow brighter.

899 (b)

Temperature of inversion is

Ti = 2Tn - T0
1000 2500
∴ Ti = 2×270 - 10 = 530℃ Req = 500 + =
3 3

900 (d) ∴ Current drawn from the cell

i G 5 60 10 3
=1+ ⇒ =1+ ⇒S = 15Ω i= = A
ig S 1 S (2500/3) 250

901 (a) Reading of voltmeter i.e.

3 5
R = G(n-1) = 50×10 (3-1) = 10 Ω 3 1000
Potential difference across AB = × =4
250 3
902 (c) V
2
220 904 (a)
Rbulb = = 484Ω;
100
i 9i
220
2 In figure (b) current through R2 = i - =
Rgeyser = = 48.4Ω 10 10
1000
Potential difference across R2 = Potential
When only bulb is on, difference across
220×484 9 i R 11
Vbulb = = 217.4 volt R⇒R2× i = R× i.e. R2 = = Ω
484+6 10 10 9 9
When geyser is also switched on, effective 11 11
resistance of bulb and geyser ×
R2×R 9 1 11
Req = = = Ω
(R2+R) 11 11 10
484×48.4 +
= = 44 Ω 9 1
484+48.4
Total circuit resistance
220×44
Vbulb = = 193.6 V 11
(44+) = + R1 = R = 11⇒R1 = 9.9Ω
10
Hence, the potential drop = 217.4×193.6
905 (c)
= 23.8 V≈ 24 V.
Number attached for brown, black, green and
903 (d) silver are 0, 5, ± 10% Therefore the resistance of
given resistor
Resistance between A and
5 6
1000×500 1000 = 10×10 Ω ± 10% = 1.0×10 Ω ± 10%
B= =
(1500) 3
906 (d)
So, equivalent resistance of the circuit
Let R be the equivalent resistance. Then addition/
subtraction of one more seat of resistors R1,

P a g e | 290
 R2and R3will not affect the total resistance. Thus, Here, R = R1, S = R2

R1 l1
∴ =
R2 (100-l1)
Ist case

R1+10 50
= ………(i)
R2 50
R = R1+ (parallel combination of R and R2)+R3
⟹R1 + 10 = R2

R = R1 + ( )
RR3
R+R3
+ R2 IInd case

R1 40
2
⇒ R + RR3 = RR1 + R1R3 + RR3 + RR2 + R2R3 =
R2 60

⇒ R - R(R1+R2) = (R2+R1)R3 = 0
2
60
⟹ R2 = R …..(ii)
40 1
(R1+R2)± (R1+R2) +4(R1+R2)R3
2

⇒R = So, Eqs. (i) and (ii) give

2

As R cannot be negative, hence 60

R1 + 10 = R
40 1
R=
1
[(R +R + (R1+R2) +4(R1+R2)R3
2 1 2
2
] ⟹
60
R - R = 10
40 1 1
=
1
[
(R +R )+ (R1+R2)(R1+R2+4R3)
2 1 2
] ⟹
20
R = 10
40 1
907 (b)
10×40
⟹R1 =
1 P1 R 200 R 20
P∝ ⇒ = 2⇒ = 2 ⇒R2 = 2R1
R P2 R1 100 R1
∴ R1 = 20Ω
908 (d)
910 (d)
64
Equivalent weight of copper = = 32 Resistance of original wire is
2

Equivalent weight of Cu l
R=ρ
Equivalent weight of Ag A

Weight of Cu deposited ρ ,being the specific resistance of wire. When the

=
Weight of Ag deposited wire is cut in two equal halves then resistance
becomes
10.8×32
Weight of copper deposited = = 3.2gm
108 ' ρl/2 R
R = =
A 2
909 (a)
Thus, the net resistance of parallel combination of
The balance condition of a meter bridge
two halves is given by
experiment

R l1
=
S (100-l 1
)
P a g e | 291
 ' '
R ×R Ti+TC
Rnet = ' ' Tn = ⇒Ti = 2Tn - TC
R +R 2
'
R R 6 917 (b)
= = = = 1.5Ω
2 2×2 4 2
E = 16T - 0.04T
911 (c)
At temperature of inversion, E = 0
Charge (q) is given by
2
∴ 16Ti - 0.04Ti = 0
q=
∫Idt ⇒T =
16
= 400℃
0.04
Given, I = 1.2t + 3
918 (c)
Integrating the expression using
For portion CD slope of the curve is negative i.e.
n+1
x
∫ resistance is negative
n
x dx=
n+1
919 (d)
We have
In the normal condition current flows from X to Y
q=
∫Idt=1.2 ∫t dt+3∫dt through cold junction. After increasing the
temperature of hot junction beyond temperature

[]
2 5 of inversion the current is reversed i.e. X to Y
t
q = 1.2 + 3[t]50 through hot junction or Y to X through cold
2 0
junction
1.2
q= ×25 + 3×5 920 (c)
2
R = kl1 and R + X = kl2
q = 15 + 15 = 30C
921 (c)
912 (d)
Before connecting the voltmeter, potential
Equivalent resistance of parallel resistors is
difference across 100Ω resistance
always less than any of the member of the
resistance system 10 100

913 (b) Vi

Supercurrent always flows on the surface of the

superconductor. V

914 (a)
100 10
0.9(2+r) = 0.3(7+r)⇒6 + 3r = 7 + r⇒r V1 = ×V = V
(100+10) 11
= 0.5 Ω
Finally after connecting voltmeter across 100Ω
915 (c) equivalent resistance
m 0.972 100×900
i= = = 0.5 A = 90Ω
Zt 0.00018×3×3600 (100+900)
916 (a) Final potential difference

P a g e | 292
 900 R

10 100
9
Vf A
A
1
V A
10 A
0 81 Ω

90 9
Vf = ×V = V Voltage across R= Voltage across ammeter
(90+10) 10
⟹ 9R = 0.81 × 1
V -V
% error = i f ×100
Vi 0.81
⟹ R= = 0.09Ω
9
10 9
V- V
11 10 926 (d)
= ×100 = 1.0
10
V Consider a concentric spherical shell of radius x
11
and thickness dx as showing in figure. Its
923 (a) resistance, dR is

From Joule’s law, the heat produced is given by ρdx

dR = 2
2
4πx
V
P=
R

Where V is potential and R the resistance.

When one bulb is fused, the total resistance of the

circuit decreases. Hence, (P) illumination
increases.

924 (b)

i ×G
S= g =
-3
10×10 ×50
-3 =
50
Ω in parallel
∴ Total resistance, R =
ρ b dx
∫
4π a x2
= [ ]
ρ 1 1
-
4π a b
i-ig 1-10 ×10 99
927 (a)
925 (d)
Cu voltameter with soluble electrodes obeys
To increase the range of ammeter we have to ohm’s law. In water voltameter, in the beginning
connect a small resistance in parallel when V is small (< 1.7 volt), very little current
flows, the voltameter does not obey ohm’s law. As
soon as V exceeds 1.7 volt (back e.m.f.) the
current increases steadily according to ohm’s law
0 81 Ω
928 (d)
1
A
The resistance R of a particular conductor is
related to the resistivity ρ of its material by

ρl
(shunt), let its value be R. R=
A
Apply KCL at junction to divide the current.
Or

P a g e | 293
 RA 1+x 1+2x
ρ = resistivity = x=1+ =
l 1+x 1+x
2
Given, R=0.072Ω or x + x = 1 + 2x
-6 2
A = 2mm×2mm = 4×10 m , l=12
1
-6
A' A
0.072×4×10
∴ ρ=
12
1 x
-8
= 2.4×10 Ωm
B' B
929 (b)

Shunt

IgG 2 1± 5 1+ 5
S= or x - x - 1 = 0 or x = =
I-Ig 2 2

Ig Since negative value of R is not possible

1
Here, =
I 34 933 (c)
1 S Current capacity of a fuse wire should be slightly
∴ =
34 S+G
greater than the total rated load current
G 3663
∴ S= = = 111Ω 934 (a)
33 33

930 (b)

R2R3 4×4
RAB = R1 + + R4 = 2 + + 2 = 6Ω
R2+R3 4+4

931 (a)

Effective resistance RP of 4 Ω, 6Ω and 12 Ω in

parallel will be

1 1 1 1 6 1
= + + = = or Rp = 2Ω
Rp 4 6 12 12 2
RAB = 3.12Ω
Total resistance of circuit = 2 + 2 = 4 Ω
935 (b)
The battery current, i = 4/4 = 1A 2
i Rt = Cθ = 3C;C = Thermal capacity
932 (d) 2
when i1 = 2i⇒Cθ1 = 4i Rt = 4×3C⇒θ1 = 12℃
The x be the total resistance of infinite network of
resistance connected to points A and B. Therefore 936 (a)
the addition of one step of resistances in the
The given circuit is a balanced Wheatstone bridge,
infinite network of resistances will not change the
hence it can be redrawn as follows
total resistance x of the network. Therefore
equivalent circuit will be as shown in figure. Then
total resistances between A’ and B’ is x given by

P a g e | 294
 7 therefore, total internal resistance (rp) of all the
3 4 cells is given by
A B 1 1 1
= + + …upto m terms
6 8 rp nr nr
14
m
=
nr
7×14 14
⇒Req = = Ω nr
(7+14) 3 or rp =
m
938 (d)
Total resistance in their circuit = R + nr/m
Battery is short circuited so potential difference is
zero Effective emf of all the cells= nE

939 (c) The current in the external resistance is given by

When we measure the emf of a cell by the nE mnE

I= =
R+nr/m mR+nr
potentiometer then no current flows in the circuit
in zero-deflection condition ie, cell is in open The current I will be maximum, if mR + nr is
circuit. Thus, in this condition the actual value of a minimum.
cell is found. In this way, potentiometer is
equivalent to an ideal voltmeter of infinite Mathematically, it can be shown that mR + nr is
resistance. minimum, if

940 (a) mR = nr

If the cells are connected as shown in figure, they nr

or R =
are said to be connected in mixed grouping. Let m
there be n cells in series in one row and m rows of
Here, mn = 45 (given )….(i)
cells in parallel. Suppose all the cells are identical.
Let each cell be of emf E and internal resistance r. and m×2.5 = n×0.5
| nE | Or n = 5m ………..(ii)

From Eqs. (i) and (ii), we have

A B
2
5m = 45
2
or m = 9
I
or m=3
R
∴ n=15

941 (d)

The thermo-couple works in a closed circuit. Since

the two pipes are isolated, only b and a are closed-
In each row, there are n cells in series, therefore circuits.
their total internal resistance=nr.
In circuit b, the two constantans wires to the cold
Their total emf = nE. copper
Since, there are m rows of cells in parallel
P a g e | 295
 pipe produce 2 opposing currents. A

In circuit a, the copper-constantan junction is 3 3

maintained across the cold pipe. A steady current i1

can flow. B C
i2 3
i
942 (c)
2V
1 2
E = 40t + t at
10
3×(3+3)
Inversion, E will be minimum Equivalent resistance R = = 2Ω
3+(3+3)
dE
dt
=0
Current i =
2
2
= 1A. So, i1 = 1×( )
3
3+6
1
= A
3
d
dt ( 1 2
40t+ t = 0
10 ) Potential difference between A and
1
B = ×3 = 1volt
1 3
40 + t=0
5
946 (a)
t = -40×5 2 4
V (200)2 4×10
P= ⇒100 = ⇒R = 2 = 400 Ω
t = -200℃ R R 10

943 (b) V 100 1

Now, i = = = amp
R 400 4
0.224 -1
Hydrogen liberated per sec = Ls .
100 947 (a)

In order to liberate 11.2 L of hydrogen per sec The potential difference across 300
charge passed = 96500 Cs
-1
Ω = 60 - 30 = 30V Therefore the effective
resistance of voltmeter resistance R and 400 Ω in
0.224 -1 parallel will be equal to 300 Ω, as 60 V is equally
To liberate = Ls of hydrogen
100 R×400
divided between two parts. So 300=
R+400
96500 0.224
= ×
11.2 100 or 300R + 120000 = 400R or R = 1200 Ω
-1
= 19.30 Cs = 19.3A 948 (b)
944 (d) When 1 bulb fuses, the total resistance of the
circuit decreases hence the current increases.
The speed at which current travels through the 2
Since P = i R, therefore illumination increases
conductors means the speed of electric effect
travelling through conductor which is at a speed 949 (a)
of light

945 (a) Here RXWY =

R
2πr
×(rα) =
Rα
2π [ ]
∴α=
l
r
The circuit can be drawn as follows R R
and RXZY = ×r(2π-α) = (2π - α)
2πr 2π

P a g e | 296
 Rα R 956 (d)
× (2π-α)
RXWYRXZY 2π 2π Rα
Req = = = 2 (2π - α) V = xl⇒iR = xl
RXWY+RXZY Rα R(2π-α) 4π
+
2π 2π
( )
-3
2×10 -2
⇒i×10 = -2 ×50×10 = 0.1
950 (a) 10
-3
q = it = current × time ⇒i = 10×10 A = 10 mA

951 (a) 957 (b)

As galvanometer deflection remains unaffected The power of 1st bulb

with switch S open or closed, hence the bridge 2
circuit is balanced. Hence, Ip = IQand IR = IG V (220)2
F1 = =
R1 R1
However, as P ≠ R, hence IP ≠ IR
(220)2
100 =
952 (d) R1

Resistance of bulb, 200 ×200

or R1 = = 400 Ω
100
2
V 220×220 484
R= = = Ω The power of IInd bulb
P 300 3

(110)2 110×110
(200)2
' P2 =
New power, P = = = 75 watt R2
R 484/3
reduction (200)2
or 200 =
R2
300-75
Of power = ×100 = 75%
300
200 ×200
or R2 = = 200 Ω
953 (a) 200

Peltier coefficient is directly proportional to The bulbs are joined in series.

absolute temperature T. So, R = R1 + R2
954 (a)
= 400 + 200 = 600 Ω
ρL 4ρL
Resistance of a wire R = = 2 where D is The total power
A πD
diameter of wire V
2
(200)2
P= =
R 600
1
As R ∝ L and R ∝ 2 , hence it is clear that
D P = 66.7 W
L
resistance will be maximum if 2 is maximum. On
D 958 (a)
calculation we find
First colour gives first digit, second colour gives
L second digit and third colour gives the multiplier
2 maximum when, L = 50 cm and D = 0.5 mm
D and fourth colour gives the tolerance

955 (b)

1
i∝
R

P a g e | 297
 () ()
5 2 2 2 2
= 36×10 Ω ± 10% R r -r r d
⇒ A = 221 = 2 -1 = 2 -1
RB r1 r 1
d1
959 (c)

()
2
2
According to loop rule, = -1 = 3
1
2.5 - 0.5I - 2I = 0
962 (b)
⇒I = 1A
Shunt is connected to the galvanometer
qo
VA - VB = = 2I = 2V iS
C ig =
S+G
-6
qo = C×2 = 2×10 ×2 = 4μC
100×1
1=
(1+G)
l 2
⟹ G = 99Ω
10
A B 964 (b)
+ -
2.5V
l l Energy = P×t = 2×1×30 = 60 kWh = 60 unit
0.5 Ω
965 (c)

960 (d) BC, CD and BA are known resistance.

() The unknown resistance is connected between A

2
R l 2
In Stretching of wire R ∝ l ⇒ 1 = 1 and D.
R2 l2

( )
R1 2 Hence, the correct option is (c).
100
If l1 = 100, then l2 = 110 ⇒ =
R2 110 966 (a)
⇒R2 = 1.21 R1 5 4 3
RPQ = r, RQR = r and RPR = r
11 11 11
Resistivity doesn’t change with stretching
∴ RPQ is maximum.
961 (c)
967 (a)
ρl
For conductor A, RA = 2,
πr1 For ohmic resistance V ∝ i⇒V = Ri (here R is
ρl constant)
For conductor B, RB = 2 2
π(r2-r1)
968 (b)

During charging of lead-acid accumulator, the

specific gravity of H2SO4 increases.

969 (a)

m = ZIt

200 108
= ×I×60×60
100 96500

I = 50mA

970 (b)
P a g e | 298
 Let the length of various edges in increase order 400×10,000 5000
RP = = Ω
be l,x,2l respectively 400+10,000 13

ρ2l 2ρ ρl ρ Total resistance of circuit

Rmax = = ;Rmin = =
xl x 2lx 2x
5000 15400
= + 800 = Ω
Rmax 13 13
∴ =4
Rmin
6 39
Current in the circuit, i = = A
971 (b) 15400/13 7700

E 4 Potential difference across

i= ⇒1 = = r = 2Ω 39 5000
R+r 2+r voltmeter=iRP = ×
7700 13
Short circuit, is when terminals of battery are
connected directly, then current which flows is = 1.95 A
E 4
iSC = = = 2A 978 (b)
r 2
Emf’s E1 and E2 are opposing each other. Since
972 (b)
E2 > E1 so, current will move from right to left.
E 6
i= = = 12 amp
r 0.5 E1 E2
r1 = 1Ω r2 = 2Ω
A
C
973 (a) 2 4
V V

1
Internal resistance ∝
Temperature
i R = 5Ω
974 (c)

Resistance are in parallel

Current in circuit
R
∴ Req =
3 E2-E1
i=
R+r1+r2
975 (d)
4-2 2
V = = = 0.25A
E = ;E is constant (volt. gradient) 5+1+2 8
l
The potential drop between points A and C is
V V 1.1 V 180×1.1
⇒ 1 = 2⇒ = ⇒V = = 1.41 V
l1 l2 140 180 140 VA - VC = E1 + ir1

976 (c) = 2 + (0.25×1)

8 4 = 2.25V
The current in the circuit = =
5+1 3
979 (a)
4 4
Now VC - VE = ×1⇒VE = - V
3 3 Magnetic field due to a long straight wire of radius
a carrying current I at a point distant r from the
977 (b) centre of the wire is given as follows

Here, the resistance of 400 Ω and 10000 Ω are in μ0Ir

parallel, their effective resistance Rp will be B= 2 for r < a
2πa

P a g e | 299
 2
μ0I v ∆P 2∆V ∆R
B= for r = a P= ⇒ = +
2πr R P V R

μ0I ∆P
B= for r > a = 2×2.5 + 0 = 5%
2πr P

The variation of magnetic field B with distance r 986 (c)

from the centre of wire is shown in the figure
1 1 1 1 1
A B

1 1
RAB = 2 + =2 Ω
3 3

980 (b) 987 (b)

'
Let resistivity at a distance 'x from left end be The temperature of the wire increase to such a
ρ = (ρ0+ax). Then electric field intensity at a value at which the heat produced per second
'
distance 'x from left and will be equal to equals heat lost per second due to radiation.
i(ρ0+ax)
( )
iρ
E= = 2 ρl
A A i 2 = H×2πrl
πr
Where i is the current flowing through the
Where H is heat lost per second per unit area due
conductor. It means E ∝ ρ or E varies linearly
' to radiation.
with distance 'x . But at x = 0,E has non-zero
3
value. Hence (b) is correct 2
Hence, i ∝ r

981 (c) i1
2
r1
3

So, 2 = 3
2 i2 r2
E (2)2
Pmax = = = 2W
4r 4×0.5

()
2
i 3
Or r2 = r1 2
982 (a) i1
2 2
V
= P⇒R =
V
=
220×220
= 484Ω [Here :r1=1 mm, i1=1.5 A, i2=3 A]
R P 100

( )
2/3
3 1/3
984 (b) r2 = 1× =4 mm
1.5
Let i be the current through arm ADC. Then
988 (b)
current through arm ABC = (2.1-i). As there is no
deflection in the galvanometer, hence Here, Ig = 1A

(20+5)i = (8+2)(2.1 - i) I = 10A

or 25i = 21 - 10i or 35i = 21

or i = 21/35 = 3/5 = 0.6 A

985 (c)

Resistance of bulb is constant

From figure

P a g e | 300
 IgG = (I - Ig)S Switch S2 is open so capacitor is not in circuit

G I-I
= g
S Ig

Substituting the given values, we get

G 10A-1A 9
= =
S 1A 1

989 (b)

V Q Vt 20×2×60
i= = ⇒Q = = = 240C 24
R t R 10 Current through 3Ω resistor = = 4A
3+3
991 (a) Let potential of point 'O' shown in fig. is VO
Then using ohm’s law
e V e El VO - Va = 3×4 = 12V …(i)
vd = × τ or vd = . τ [ ∴ V = El]
m l m l
24
Now current through 5Ω resistor = 4A
∴ vd ∝ E 5+1
So V0 - Vb = 4×1 = 4V …(ii)
992 (c) From equation (i) and (ii) Vb - Va = 12 - 4 = 8V

418t = 1×4180×20 or t = 200 s 996 (c)

-1 -1
Note that in this case, C = 418 J/kg [℃ ] ne
18
62.5×10 ×1.6×10
-19
i= = = 10 ampere
-1
t 1
J = 4.18 J cal
997 (a)
993 (c)
In parallel PConsumed ∝ PRated
1Ω X 2Ω

I1
1 Ixy 2 998 (c)

I2 3Ω Y 4Ω The figure can be drawn as follows

A B 100  C
D

100 100
-i1 + 0×ixy + 3i2 = 0 i.e. i1 = 3i2 …(i)


Also -2(i1-ixy) + 4(i2+ixy) = 0 A B

100 

i.e. 2i1 - 4i2 = 6ixy …(ii)

Also VAB - 1×i1 - 2(i1-ixy) = 0⇒50 = i1 + 2(i1-ixy) 200×200

RAC = = 100Ω
200+200
= 3i1 - 2ixy …(iii)
999 (a)
Solving (i), (ii) and (iii), ixy = 2A Heat gained by water = Heat supplied by
container heat lost
994 (c)

By Kirchhoff’s current law

995 (b)
P a g e | 301
 ⇒mS∆θ = 1000t - 160t S -3 S -3
ig = i ⇒10×10 = ×100×10
G+S 100+S
2×4.2×1000×50
⇒t = = 8min 20 sec
840 1000
90 S = 1000⇒S = = 11.11 Ω
90
100 (d)
0 101 (c)
As batteries wear out, temperature of filament of 0
flash light attains lesser value, therefore intensity In balance condition, no current will flow through
of radiation reduces. Also dominating wavelength the branch containing S
(λm) in spectrum, which is the red colour,
101 (d)
increases
1
At time t = 0 i.e. when capacitor is charging,
100 (c)
current
1
Potential gradient
-7 2
iρ 0.1×10 -2 i= = 2mA
(x) = = -6 = 10 V/m 1000
A 10
When capacitor is full charged, no current will
100 (c)
pass through it, hence current through the circuit
2 2
V 200×200×2
Energy = t= = 1000 Wh 2
R 80 i= = 1mA
2000
100 (d)
101 (c)
4
If the voltmeter is ideal then given circuit is an 2
Let x be the equivalent resistance of entire
open circuit, so reading of voltmeter is equal to
network between A and B. Hence, we have
the e.m.f. of cell i.e., 6V

100 (a) i 1Ω
7 2
Resistance of each bulb R = V /P. A
i i - i1
When connected in series total resistance of bulbs V
1Ω X
= 2R
'
Current in each bulb , I = V /2R; B

2
Power generated by each bulb = I R

( )
' 2
V V' V'2 RAB = 1+ resistance of parallel combination of
= ×R = = 2
2R 4R (V /P) 1Ω and xΩ

(110)2×500 x
= =31.25 W ∴ RAB = 1 +
4×(220)2 1+x

100 (a) x
∴ x=1+
8 1+x
P = Vi = 250×2 = 500 W
2
⟹ x+x =1+x+x
100 (a)
2
9 ⟹x - x - 1 = 0

P a g e | 302
 1+ 1+4 (n-2)E
⟹ x= Current in the circuit is i = ×r
2 nr

1+ 5 Potential difference across each cell is

= Ω
2
(n-2)E 2E
V = E - Ir = E - ×r =
101 (b) nr n
3 6
Energy, E = 1 kWh = 3.6×10 J = QV 101 (c)
9
or Q = E/V 2R > 20⇒R > 10Ω

E 102 (d)
m = zQ = z
V 0
After some time, thermal equilibrium will reach

=
(0.33×10-6)(3.6×106) = 1.8 kg.
66 102 (b)
1
101 (b) For maximum power Rext = Rint
4
Current from D to C = 1A When batteries are connected in parallel.

∴ VD - VC = 2×1 = 2V 1
Rint = = 0.5Ω
2
VA = 0 ∴ VC = 1V, ∴ VD - VC = 2
E 2
Current I = = = 2Ω
⇒VD - 1 = 2 ∴ VD = 3V Rext+Rint 0.5+0.5

∴ VD - VB = 2 ∴ 3 - VB = 2 ∴ VB = 1V Maximum power is given by

2
101 (b) Pmax = i Rext
7
Here, E1 = 0.9V,θ - θ0 = 75E;E2 = ?when ⇒Pmax = (2)2×0.5 =2 W
'
θ0 = θ0 + 15 If we assume batteries to be connected in series,
' then
∴ temperature difference θ - θ0 = θ - (θ0 + 15)
Rint = 1 + 1 = 2Ω
= (θ-θ0) - 15 = 75 - 15 = 60 K
2E 2×2
E θ-θ0 60
'
4 Current I = = = 1A
As 2 = = = Rext+Rint 2+2
E1 θ-θ0 75 5
So, maximum power is now given by
or % decrease in thermo emf
2
Pmax = i Rext = 1×2 = 2 W
=
E1-E2
E1
×100 = ( )
1-
E2
E1
×100
In either case Pmax = 2 W

= ( )
1-
5
4
×100 = 20%
102 (a)
2
For E to be maximum
101 (d)
8 dE -6 -6
= 20×10 - 0.02×10 ×2T = 0
When one call is wrongly connected in series, the dT
emf of cells decrease by 2 E, but internal
resistance of cells remains the same for all the ⇒ Tn = 500℃
cells.
P a g e | 303
 -6 -6
∴ Emax = 20×10 (500) - 0.02×10 (500)2 =5 mV From equation (i) and (ii)

102 (b) 2i i 2i
i1 = ,i = ⇒iAD =
3 5 2 5 5
1 1 1
In parallel = +
tp t1 t2 102 (c)
6
t1t2 1
tp = For constant voltage, we know that P ∝
t1+ t2 R

So higher the power, lower will be the resistance

102 (b)
4 102 (c)
Since no current is to flow in the 4Ω resistance,
7
hence resistance 4Ω becomes ineffective in Current through 6Ω resistance in parallel with 3Ω
current. resistance = 0.4 A

B C D So total current = 0.8 + 0.4 = 1.2 A

4 Potential drop across 4Ω = 1.2×4 = 4.8V

2 R
102 (c)
6V
8
A F E E1 l +l 58+29 3
9V 3V = 1 2 = =
E2 l1-l2 58-29 1

102 (c)
Current through resistances 2 Ω is 9
E
I=
9-6 3 R+r
i= = A
2 2
E
I= = constant
3 R
In circuit ABCDEFA, 9 - 3 = (2 + R)×
2
where, R=external resistance
or 12 = 6 + 3R or R = 2Ω
r= internal resistance =0
102 (a)
103 (a)
5
(i – i1 – i2) E 0
(i – i1 – i2) 1 1
PRated ∝ and R ∝
B F i¬ R [Thickness of filament]2
i2
(i – i1)
i (i1 + i2) So PRated ∝ (Thickness of filament)2
A C

103 (a)
D i1
1
From Kirchhoff’s first law, in an electric circuit the
Applying Kirchhoff’s law in mesh ABCDA algebraic sum of the currents meeting at any
junction is zero,
-10(i-i1) - 10i2 + 20i1 = 0 ⇒3i1 - i2 = i …(i)

And in mesh BEFCB

-20(i-i1-i2) + 10(i1+i2) + 10i2 = 0

⇒3i1 + 4i2 = 2i …(ii)

P a g e | 304
 3A dE 1
P= = α + β.(2T)⇒α + β.T
dT 2

1A 2A
α+β ( )
-2α
β
= α - 2α = -α

I 103 (c)
5
GS
Resistance of shunted ammeter =
G+S

i G GS i .G
Also =1+ ⇒ = g
ie., ∑i = 0 ig S G+S i

∴ Taking inward direction of current as positive GS 0.05×120

⇒ = = 0.6 Ω
and outward as negative, we have G+S 10

1A - 3A - 2A + I = 0 103 (a)
6 15 -19
⟹ I = 4A i = 6×10 ×1.6×10 = 0.96mA

103 (d) 103 (d)

2 2
8
Thermal energy is resistor is U = i Rt To convert galvanometer into voltmeter, the
necessary value of resistance to be connected in
2
R = R0(1+αθ)⇒U = i R0(1+αθ)t [where θ series with the galvanometer is
= temp]
V
R= -G
dU 2 Ig
So = i R0(1 + αθ)
dt
50
With time temperature increases, hence dU/dt = -3 - 40
10×10
increases. This is best shown by curve (d)
= 5000 - 40 = 4960Ω
103 (b)
3 103 (b)
Here two cells are in series 9
To obtain minimum resistance, all resistors must
Therefore total emf = 2E be connected in parallel.
Total resistance = R + 2r Hence equivalent resistance of combination of
r
2E 2×1.45 2.9 29 combination =
∴i = = = = 10
R+2r 1.5+2×0.15 1.8 18
= 1.611 amp 104 (a)
0
103 (b) Total e.m.f. = nE, Total resistance R + nr
4 nE
1 2 ⇒i =
E = αT + βT R+nr
2
104 (c)
For inversion temperature
2
e R
-2α Potential gradient x = .
E = 0, T = (R+Rh+r) L
β

Thermo electric power

P a g e | 305
 -3
10 2 3 3600 C,
⇒ -2 = × ⇒Rh = 57Ω
10 (3+Rh+0) 1
∴ 1 A h = 3600 C
104 (c) 4
1.476×10
3 ⇒q = Ah = 4.1 Ah
V
2
(220)2 3600
R= = = 807 Ω
P 60
104 (c)
104 (a) 8
(220)2
4 Resistance of each bulb =
m 2.68 2.68 965 60
I= = = × ≈ 4A
Zt 108 108 6
×10×60 If the number of bulbs is n then effective
96500
resistance
2 2
Energy = I Rt = 4 ×20×600 = 192 kJ
(220)2
=
104 (c) 60n
5
From Joule’s law, for a current carrying conductor V 220×60n
∴i = 9= = ⇒ n = 33
at a definite temperature the rate of production of Reff 220×220
heat is given by
104 (b)
H = Vit 9
R2 = Ro(1+αt2) and Rt = Ro(1 + αt1)
1

Where i is current, V the potential difference and t

the time. Rt 1+αt2
∴ 2
=
Rt 1+αt1
1
Given, V = 1.5 volt, i = 2.1 A, t = 1s
1.5 1+0.00125×t2
∴ H = 15×2.1×1 = 31.5 J or =
1 1+0.00125×27
Also, 1 cal = 4.2 J
On solving we get;
31.5 t2 = 454°C = 454 + 273 = 727K
∴H = = 7.5 cal
4.2
105 (d)
104 (c) 0
Heat produced = energy stored in capacitor
6 2
V
Power of bulb, P = 1 2 1
CV = ×(10×10 )×(500)2 = 1.25 J
-6
R =
2 2
P2 R
∴ = 1 105 (b)
P1 R2
1
Let I be the total current passing through
R1 100
Or = = 0.5 balanced wheat
R2 200
stone bridge. Current through arms of resistances
104 (c) P and
7
m = zq Q in series is
-3
m 5×10 I×300 3
⇒q = = -7 I1 = = I and current through arms of
z 3.387×10 330+110 4
4
q = 1.476×10 C Resistances R and S in series is

Since a current of 1 A for 1 h gives a charge of

P a g e | 306
 I2 =
I×110
330+110
1
= I
4
or (
20 = R0 1+
1
220 )
×20 …(iii)

∴ Ratio of heat developed per sec Rt = R0(1+αt)

HP :HQ :HR :HS

(
25 = R0 1+
1
220
×t ) ..(iv)

( ) ( ) ( ) ( )
2 2 2 2
3 3 1 1
= I ×100 : I ×10 : I ×300 : I ×30
4 4 4 4 Dividing Eq. (iv) by Eq (iii), we get

= 30 :3 :10 :1.
25
=
(1+
1
220
×t )
⟹4 +
4t
=5+
100
105 (a)
2
The resistivity of metal increases when it is
20
(1+
1
220
×20 )220 220

converted into an alloy or t = 80°C

ρ' > ρ 105 (b)

6
105 (c) RN = R + R/3
4
Charge delivered to cathode per second = 4R/3
25 -19
0.002×10 ×2×1.6×10
=
100×60

= 1.06 C

105 (d)
105 (c)
5
Here, R20 = 20, R500 = 60Ω,Rt = 25Ω, 7
The total emf = E + E = 2E
∵ Rt = R0(1+αt) Total resistance =R + r1 + r2
Where α is the temperature coefficient of ∴ Current flowing through the circuit
resistance.
2E
∴ R20 = R0(1+α×20) i=
R+r1+r2

or 20 = R0(1+α×20) ..(i) According to question E = ir1

R500 = R0(1+α×500) E
⟹ i=
r1
or 60 = R0(1+α×500) …(ii)
E 2E
Dividing Eq. (ii) by Eq. (i), we get ∴ =
r1 R+r1+r2
60 1+α×500
= R = r1 - r2
20 1+α×20
105 (a)
or 3 + 60α = 1 + 500α
8
Power P = VI = 250×4 = 1000W = 1kW
2 1 -1
or α= = °C
440 200 Energy = P×t = 1 kW×60 sec = 60 kJ
Again, R20 = R0(1+α×20) 105 (c)
9
P a g e | 307
 2
For the given meter bridge ∵ P = I R⇒I ∝ R

P l1 P R ' 15
= ∴ = ⇒P = ×20 = 30Ω
Q 100-l1 P' R' 10

l1 = 55cm⇒100 - l1 = 45cm 106 (d)

2
∵ P = 3Q Suppose resistance of wires are R1 and R2 then

45 9 27 6 RR
⇒Q = 3× = 3× = Ω = 1 2 . If R2 breaks then R1 = 2Ω
55 11 11 5 R1+R2

When x is connected in series with P, l1 = 75cm 6 2×R2

Hence, = ⇒R2 = 3Ω
5 2+R2
P+x 75 cm 27
⇒ = ⇒3 + x = 3×
Q 25 cm 11 106 (d)
3
81 48 The equivalent circuit is given by
⇒x = - 3⇒x = Ω
11 11

106 (a) 3Ω 2V
0
The capacitance (C) of a capacity is defined as the
3Ω
ratio of the charge (q) given to the rise in the 2V
A B
potential (V) of the conductor. When the plates
are moved apart, the charge remains constant,
hence
Since, VA = VB
q = CV = constant ………..(i)
Potential difference is zero.
Capacitance (C) of a parallel plate capacitor is
106 (b)
given by
4
From balanced Wheatstone bridge concept,
ε0 A
C= ………..(ii)
d 550Ω 20
=
R 80
Where A is area of plates and d the separation
between them. ⟹ R = 220Ω

From Eqs. (i) and (ii), we have 106 (a)

5
ε0 A Voltage across parallel combination is same. The
q= .V = constant
d equivalent resistance of upper branch

When plates are moved apart, d increases, hence R = 4 + 6 = 10 Ω

value of C decreases and in order that charge (q)
remains constant V increases. So, I1×10 = I2×10

106 (c) Or I1 = I2 = I (say)

1
Resistance of the combination ie, Same current is flowing through both branches.

' 10×10 Now, it is given that

R = + 10 = 15Ω
10+10
H4Ω = 10 cal

P a g e | 308
 2
Or I t ×4 = 10 ie., ia×R = (i-ia)×S

2 10 iaR
Or It= or S = …………(i)
4 i-ia

Therefore, heat produced in 10 Ω resistance is Given, ia = 100A, i = 750A, R = 13Ω

2 10×10 100×13
H10Ω = 10×I t = = 25 cal Hence, S = = 2Ω
4 750-100
106 (d) 107 (b)
6 0
P1V1 22.4×2 Faraday constant = 1 mole electron charge = Ne
V2 = = = 44.8 L
P2 1
23 -19
= 6.02×10 ×1.6×10 = 96500
11.2 L of H2 is liberated by charge = 96500 C
107 (a)
44.8 L of H2 is liberated by charge 1 2
Since R ∝ l ⇒ If length is increased by 10%,
96500 resistance increases by almost 20%
= ×44.8 = 386000C
11.2 '
Hence new resistance R = 10 + 20% of 10
106 (b)
7 20
= 10 + ×10 = 12 Ω
When switch S is open total current trough 100
ammeter
20 107 (d)
i= = 4A 2
(3+2) '
Given l = l + 100% of l = 2l
20
When switch is closed i = = 5A
3+(2||2) Initial volume = final volume

106 (b) 2
ie, πr l = πr l
'2 '

9
Let ia be the current flowing through ammeter and '2 rl
2
2 l
⟹ r = ' = r ×
i the total current. So, a current l 2l

i - ia will flow through shunt resistance. '2 r

2
⟹ r =
2

( )
'
' l 2l ρl
∴R =ρ ' = ρ '2 ∵R=
i ia i A πr A
Ammeter
ρ4l
( i - ia ) = 2
πr
'
Thus, ∆R = R - R = 4R - R = 3R
S
Shunt 3R
∴ %∆R = ×100%
R
High reading ammeter

= 300%

107 (b)
Potential difference across ammeter and shunt 3 2 2
resistance is same, R1000 = V /750 and R200 = V /P;

P a g e | 309
 Now, R1000 = R200(1 + α×800) 9 To draw maximum current from a combination of
nr
2 2 cells, external resistance R = .Therefore,
V V -4 m
So, = (1 + 4×10 ×800)
750 P grouping of cells depends upon the relative values
of internal and external resistance.
or P = 750(1 + 0.32) = 990 W
108 (a)
107 (d)
0
4 Watt-hour meter measures electric energy
Mass, M = volume×density = Al×d
108 (a)
or A = M/ld
1
If a motor of 12 HP works for 10 days at the rate
Resistance R = ρl/A = ρl/(M/ld)
of 8 h/day then
2
ρl d
= Energy consumption = Power ×time
M
2 J
So R ∝ l /M = 12×746 ×(80×60×60)s
s
2 2 2
l1 l2 l3 9
Thus, R1:R2:R3 = : : = 12×746×80×60×60J = 2.5×10 J
M1 M2 M3
paise
3 2 1
2 2 2 Rate of energy = 50
= : : = 27:6:1 kWh
1 2 3
6
∴ 3.6×10 J energy cost = Rs 0.5
107 (c)
9
5 9 2.5×10
i 100 ∴ 2.5×10 J energy cost = 6 = Rs 347
By using vd = = 2×3.6×10
neA 28 -19 π
10 ×1.6×10 × ×(0.02)2
4 108 (b)
-4
2
= 2×10 m/sec Resistance of CD arm = 2rcos 72°=0.62r

107 (a) Resistance of CBFC branch

6

() ()
4 4
RA rB R 1 1 A
= ⇒ A = = ⇒RB = 16RA i
RB rA RB 2 16 r r 72°
B r C D r E
When RA and RB are connected in parallel then r r
equivalent resistance F H J
r r
r r
RARB 16 G I
Req = = R
(RA+RB) 17 A

If RA = 4.25Ω then Req = 4Ω i.e. option (a) is

correct
1
R
=
1
2r
+
1
0.62r
=
1 2.62
r 2×0.62 ( )
107 (d) 1 2.62 1.24r
= ∴R =
7 R 1.24r 2.62
P = i R⇒22.5 = (15)2×R⇒R = 0.10 Ω
2

' 1.24r
Equivalent R = 2R + r = 2× +r
107 (d) 2.62
8
V = E - ir = 1.5 - 2×0.15 = 1.20 Volt =r ( 2.48
2.62 )
+1 = 1.946r
107 (d)
P a g e | 310
 Because the star circuit is symmetrical about the After connecting E, the current through 8 Ω is also
line AH
1
I= A
∴ Equivalent resistance between A and H 2

1 1 1 1
= ' + ' ∴E = A×8Ω = 4V
Req R R 2

R
'
1.946 108 (b)
⇒Req = = r = 0.973r 6
2 2 2
V 1
At constant p.d., heat produced = i.e. H ∝
108 (d) R R
3
10 108 (b)
Here n = =5
2 7
Given current through 4Ω resistance
∴ R = (n-1)G = (5-1)2000 = 8000 Ω
Is 1A, so P.D. across upper Branch
108 (c)
i.e. P.D. between P and M is 4V
4
Seebeck arranged the metals in a certain sequence
Hence P.D. between M & N is
are called the thermoelectric series. The direction
of current at the hot junction is from metal 1
occurring earlier in series to one occurring later ×4 = 3.2V
1+0.25
in series. Some of the substances in series are Bi,
Ni, Co, Pd, Pt, Cu, Mn, Hg, Pb, Sn, Au, Ag, Zn, Cd, Fe,
Sb, Te.

The more separated the metals are in the series,

the greater is the value of thermo-emf generated.
Thus, emf developed in Sb-Bi thermocouple is
higher than other thermocouples.

108 (c) 108 (d)

5 8
Before connecting E, the circuit diagram is as dE 2T T
= 4- 4- ;
shown in the figure dT 200 100

At neutral point, T = Tn.

dT
= 0 = 4 - Tn/100
dT

or Tn = 400℃, Ti = 2Tn - T0

The equivalent resistance of the given circuit is = 2×400 - 0 = 800℃

Req = 6 Ω + 8 Ω + 10 Ω = 24 Ω 108 (d)
9
12V 1 In the above question for calculating equivalent
Current in the circuit, I = = A
24Ω 2 resistance between two opposite square faces

Before connecting E, the current through 8 Ω is 2

l = 100 cm = 1m,A = 1 cm = 10 m , so
-4 2

resistance
1
I= A
2

P a g e | 311
 2
-7 1 -3 From Eq.(i), m×5m = 20 or m = 4
R = 3×10 × -4 = 3×10 Ω
10
or m = 2. Therefore, n = 5×2 = 10
109 (a)
0 109 (b)
The balanced condition for Wheatstone bridge is 3
The resistance of ammeter is very low, so when it
P R is used in parallel through a circuit then excess
=
Q S current will flow through it thus, damaging it.

as is obvious from the given values. 109 (d)

4
No, current flows through galvanometer is zero. Zero (No potential difference across voltmeter)

Now, P and R are in series, so 109 (a)

5
Resistance R1 = P + R When two batteries are in series to the external
resistance
= 10 + 15 = 25Ω
R, total resistance of the circuit
Similarly, Q and S are in series, so
= R + 2r = R + 2×1 = R + 2Ω
Resistance R2 = R + S = 20 + 30 = 50Ω
Total emf of batteries = 2+2 = 4 V
Net resistance of the network as R1 and R2 are in
parallel. ∴ Joulean power across R

( )
2 2
1 1 1 4 4R
= + = ×R = .
R R1 R2 R+2 (R-2)2+8 R

25×50 50 Joulean power will be maximum if R - 2 = 0

∴R = = Ω
25+50 3
or R = 2Ω
Hence, 2
4 ×2
∴ Maximum joulean power = =2
V 6 (2 2)2+8×2
I= = = 0.36A watt
R 50
3
109 (a)
109 (b) 6
Assume thermo emf E is directly proportional to
1 2
H = I Rt = mc∆θ temperature difference T.

∴ ∆θ ∝ I
2 ie, E = aT

() ( )
1
2 2 2 2
∆θ I2 ∆θ 2I+I a=
-2
= 10 VK
-1
Hence = or = 100
∆θ2 I1 3 I
When cold junction is heated by 20 K, the
or ∆θ2 = 9×3 = 27℃.
temperature
109 (a)
difference T between junction becomes 80K, then
2
m n = 20 …(i) thermo

For maximum current R = n r m emf

-2
or 2.5 = n×0.5/m or n = 5m E = aT = 10 ×80 = 0.8 V

P a g e | 312
 -19 -2 -1
Percentage change of emf J = 0.15×1.6×10 Cm s
-19 4 -2 -1
E-E' = 0.15×1.6×10 ×10 Cm s
= ×100
E
2
∴ Current, i = JA = J4πr
1-0.8
= ×100 = 20%
= 0.15×1.6×10 ×10 ×4×3.14×(6.4×10 )
-19 4 6 2
1 on
solving
109 (b)
7 i = 0.12 A
R
In parallel, x = R = nx
n 110 (a)
1
In series, R + R + R…n times Slope of the graph will give us reciprocal of
2
= nR = n(nx) = n x resistance. Here resistance at temperature T1is
greater than that at T2. Since resistance of metallic
109 (c)
wire is more at higher temperature then at lower
8
The given circuit may be redrawn as shown in temperature, hence T1 > T2
adjacent figure. Resistance of parallel
combination of 2Ω and 6Ω, 110 (c)
2
- + An α- particle has a charge equal to 2 protons.
6V 2Ω Motion of α particle to the left, motion of proton
towards left and motion of electrons towards
A
1.5 B right, all will produce conventional current
6Ω towards left. The total current will be
3Ω 19
i = 10 ×(2×1.6×10 )
-19

+10 ×(1.6×10 ) + 1019×(1.6×10-19)

19 -19

2×6
R1 = = 1.5Ω
2+6 = 6.4 A

Now resistance of ABC arm= 1.5 + 1.5 = 3Ω 110 (b)

3
3×3 P 550
and total network resistance R = = 1.5Ω I= = = 2.5 A
3+3 V 220

∴ Total current supplied by the battery 110 (a)

6V 4
i= = 4A
1.5Ω (R+G)ig = V⇒(R+G) = V
ig
109 (d)
ig R
9 G
S S
ig = i ⇒0.01 = 10
G+S 25+S
∴ Value of R is nearly equal to 6kΩ
25
⇒1000S = 25 + S⇒S = Ω This is connected in series in a voltmeter
999
110 (a)
110 (a)
5
0 2 160×40
Surface area of earth, A = 4πr R= = 32Ω
160+40
Charge entering the earth per sec per unit area

P a g e | 313
 Series 5
C Current through the resistance i = = 2A
(2+0.5)
100+60 = 160
100 60
111 (c)
1
A B At null point, no current flows through
40
galvanometer. Therefore, the resistance
connected in series with galvanometer, at null
110 (b) point position will be in effective. Thus, null point
6 is obtained on potentiometer wire at the original
e -3 2
V = i.R. = .R⇒10 = ×10 position after the removal wire at the original
(R+Rh+r) (10+R+r)
position after the removal of series resistance
⇒R = 19,989Ω from galvanometer

110 (a) 111 (b)

7 3
For maximum joule heat produced in resistor The given circuit can be simplified as follows
external resistance = Internal resistance
10Ω

110 (a) 10Ω 3Ω

A
8 B
H Pt P 500 5 5Ω 8Ω 6Ω 6Ω
In parallel, 1 = 1 = 1 = =
H2 P2t P2 200 2
2 2 5Ω 3Ω
H1 IRt R V /P A
In series, = 2 1 = 1 = 2 1 B
H2 I R2t R 2
V /P2  5Ω 8Ω 3Ω

P2 200 2
= = =
P1 500 5
Now it is a balanced Wheatstone bridge
110 (b)
9 So,
Let the original resistance is R Ω.
8Ω
∴ V = IR
8Ω
V = 5×R = 5R …(i)

When 2 Ω resistance is inserted, then total

resistance 8×8 64
⇒RAB = = = 4Ω
8+8 16
=(R+2) Ω
111 (a)
∴ V = I’ (R + 2) = 4(R + 2) …(ii) 4
q 3000C
From Eqs. (i) and (ii), we get Current, I = = = 5A
t 10×60s
5R = 4(R + 2) 111 (b)
5
∴ R=8 Ω R
Resistance of parallel group =
2
111 (b)
0 R
5 ∴ Total equivalent resistance = 4× = 2R
Emf E = 5V, Internal resistance r = = 0.5Ω 2
10
111 (d)
P a g e | 314
6 If resistance of each bulb be R, then bias voltage V 112 (a)
2
v 4
is same in both cases, hence P = The effective resistance of combined wire
Rtotal

In the case Rtotal = 50R and in second case 49 R ρ( )l+l

A
ρl ρl
= 1 + 2
A A

( ) ( )
2 2
P1 49 50
∴ = ⇒P2 = .P1 (∵total length L=l + l)
P2 50 49
Or
or P2 > P1
ρ1+ρ2
111 (d) ρ=
2
7
VAB = 4 =
5X+2×10
X+10 [
E r +E r
⇒X = 20Ω, v= 2 1 1 2
r1+r2 ] 112 (a)
5
The circuit diagram with current variation can be
111 (a) drawn as
8 2
2 H 2 i ρl Let current Igflows through R and G.
H = i Rt⇒ =iR= 2
t πr
I Ig R I
G
111 (c)
9
l
Resistance = ρ V
A

R1 ρ l A 2 3 5 5
∴ = 1× 1× 2 = × × = For a voltmeter with full scale reading, we have
R2 ρ2 l2 A1 3 4 4 8
V = Ig(G+R)
112 (c)
0 V
Because in series current is same or R = -G
Ig
112 (a)
1 Given, G = 2Ω, V = 12volt, Ig = 0.1A
The effective resistance between two diagonally
opposite ends= 5 R/6 12
∴R = - 2 = 120 - 2 = 118Ω
0.1
112 (c)
2 112 (c)
-6
m1 Z mZ 14×1.2×10 6
= 1 ⇒m2 = 1 2 = -6 = 2.4g Because given voltage is very high
m2 Z2 Z1 7×10
112 (c)
112 (c)
7
3 When cells are connected in series: emf increase
Current through each arm
but current capacity remains unchanged. The emf
PRQ and PSQ=1A of 6 cells in series = 6×1.5 = 9 V and current
capacity = 5A h
Vp - VR = 3v
112 (c)
Vp - Vs = 7V 8 2
V (210)2
H= ×t = ×1 = mL
From Eqs. (i) and (ii), we get R 20

VR - Vs = +4V (210)2
∴ = m×80×4.2⇒m = 6.56 g/s
20
P a g e | 315
112 (d) J×m×s∆θ
⇒t = [For water 1 litre = 1 kg]
9 P
i G
By using = 1 +
ig S 4.2×1×1000×(40-10)
⇒t = = 150 sec
836
i 1000 1000
⇒ -3 = 1 + ⇒S = = 111Ω
100×10 S 9 4200×m×∆θ
Short Trick : use formula t =
P
113 (b)
0 113 (a)
1 1 1 6
R ∝ ⇒R ∝ 2 ∝ 2 [d = diameter of wire] R > 2Ω
A r d

113 (c) ∴ 100 - x > x

1
The graph between thermo emf and temperature P R
Applying, =
of hot junction is parabolic in shape Q S

113 (d)
2Ω R
2
According to the principle of Wheatstone’s bridge,
the effective resistance between the given points G
is 4Ω

B
| x | 100 - x |

4 4
A C F
16

4 4 R

D 2Ω
G

113 (d)
3 2 2 2 2 | x + 20 | |
V V V V 80 - x
P= so R = ⇒R1 = and R2 = R3 =
R P 100 60

(250)2 (250)2
Now W1 = 2 .R1, W2 = .R2
(R1+R2) (R1+R2)2
We have
(250)2
and W3 =
R3 2 x
= ….(i)
R 100-x
W1:W2:W3 = 15:25:64 or W1 < W2 < W3
R x+20
113 (c) = ……..(ii)
2 80-x
4
The voltmeter is assumed to have infinite Solving Eqs. (i) and (ii), we get
resistance. Hence (1+2+1) + 4 = 8Ω
R = 3Ω
113 (b)
5 113 (a)
W = JH⇒P×t = J×ms ∆θ 7
Length of wire = 2πr = 2π(0.1) = 0.2πm

Resistance of complete wire = 12×0.2π = 2.4πΩ

P a g e | 316
 ∴ Resistance of each semicircle = 1.2πΩ ⟹VD = 40V

1.2π 70-40
Hence equivalent resistance RAB = = 0.6πΩ ⟹i1 = = 3A
2 10

113 (c) 40-0

i2 = = 2A
8 20
RA
Resistivity ρ =
l 40-10
and i3 = = 1A
30
⟹ ρ∝R
114 (c)
∵ Metals have low resistance, therefore they have
2

() ( )
2
low resistivity. V 110
2
Pconsumed = A ×PR = ×500 = 457.46 W
VR 115
113 (c)
9 So, percentage drop in power output
Voltmeter has high resistance and is always
connected in parallel with the circuit. So, to (500-457.46)
convert a galvanometer into voltmeter, a high = ×100 = 8.6%
500
resistance must be connected in series with it so
that it draws negligible current from the circuit. 114 (b)
3
114 (a) P 1/3 1/3
= =
0 Q 1-1/3 2/3
E = V + ir
⟹P:Q = 1:2
E 2
After short-circuiting, V = 0⇒r = = = 0.5Ω P = k, Q = 2k
i 4

114 (d) 2
1 P+6 3
=
Applying Kirchhoff’s law at point D, we get Q 2
1-
3
I1 = I2 + I3
P+6 2
⟹ =
VA-VD V -0 V -V Q 1
= D + D C
10 20 30
k+6 2
=
VD V -10 2k 1
or 70 - VD = + D
2 3
⟹k + 6 = 4k
A ⟹k = 2
i1
10Ω ∴ P = 2Ω, Q = 4Ω
D
3 i2 114 (d)
30Ω
20Ω 4
E 2 2
I= = = = 0.888A
C B R+r/4 2+1/4 2.25
10 0
V V
114 (b)
5
Resistors are connected in series. So current
through each resistor will be same

P a g e | 317
 12-8 8-4 4-0 4 4 4 114 (d)
⇒i = = = ⇒ = =
R3 R2 R1 R3 R2 R1 9
Potential gradient is given by
So, R1:R2:R3∷1:1:1
V IR
114 (b) k= = (∵V=I R)
l l
6

( )
Let the resistance of the two heaters be denoted
I×ρl/A ρl
by R1 and R2. = ∵R=
l A
2
V Iρ
Then, R1 = =
P1 A
2
V -3
0.01×10 ×10 ×10
9 -2
R2 = ∴ k=
8
= 10 Vm
-1
P2 -2
10 ×10
-4

In parallel combination of resistances, 115 (d)

0
1 1 1 V 100
= + R = -G = -3 - 25 = 9975Ω
RP R1 R2 ig 10×10

PP P1 P2 115 (a)
2 = 2 + 2
V V V 1
Resistivity of combination
Pp = P1 + P2
= (1+2+3+…+n)
Given, P1 = 1000 W
n(n+1)
=
P2 = 1000 W 2

115 (b)
∴ Pp = 1000 + 1000
2
2 r 2
Hence, Pp = 2000 W Here P1 = i × , P2 = i ×3r,
3

( )
114 (a) 2 r 32
7 P3 = i +r = i r
2 2
l
By using R = ρ. ; here A = π(r2-r1)
2 2

[ ]
A 2 2r×2r 2
and P4 = i =ir
2r+2r
Outer radius r2 = 5cm
So it is obvious that P2 > P3 > P4 > P1
Inner radius r1 = 5 - 0.5 = 4.5 cm
115 (c)
r2 5 mm 3
r1 ' P 60
10 cm In series P = = = 20 watts
n 3

115 (b)
-8 5 4
So R = 1.7×10 × Resistance in the arms AC and BC are in series,
π{(5×10 ) -(4.5×10 ) }
-2 2 -2 2

-5
= 5.6×10 Ω

114 (a)
8
m ∝ q⇒m ∝ it

P a g e | 318
 A t
I
= Constant
N
3Ω 3Ω
3
V t1 t
∴ = 2
N1 N2
B 3Ω C

N2 9
⇒ = ×t = ×16 =14.4 min.
N1 1 10
∴ R’=3+3=6Ω
115 (b)
Now, R’ and 3Ω are in parallel, 7
Resistivity of the material of the rod
6×3
∴ Req = = 2Ω
6+3 RA
ρ=
l
Now, V=IR
3×10 ×π(0.3×10 )
-3 -2 2

3 =
⟹ I = = 1.5A 1
2
-9
= 27×10 π Ωm
115 (d)
5 Resistance of disc,
The residence of conductor
Resistivity of rod ×Thickness
ρl 1 R=
R= = 2 Area of cross-section
A πr
-3
-9 10
l = 27×10 π×
π×(1×10 )
-2 2
or R ∝ 2
r
-7
= 2.7×10 Ω
()
2 2
R1 l r 2 2 8
∴ = 1 × 22 = × =
R2 l 2 r1 1 1 1 115 (b)
8
Thermal potential between the ends of the roads The circuit can be shown as given below
are same. So, heat conducted per second
D C
Q 1
H= ∝
t R
O 15Ω
O'
H R 1
∴ 1 = 2 = = 1:8
H2 R1 8
A B
115 (b)
6 2
Vt The equivalent resistance between D and C.
We know, = Q = ms dθ
4.2 R
15×(15+15)
Let,N = initial number of turns RDC =
15+(15+15)
R = resistance of the coil 15×30
=
15+30
L ρ×N×2πr
⇒R = ρ =
A A 15×30
= = 10Ω
2 45
Vt
= Q = ms dθ
4.2×ρ×N×2πr Now, between A and B, the resistance of upper

P a g e | 319
 part ADCB,
∴ V = E- ( ) E
R+r
r
R1 = 15 + 10 + 15 = 40Ω
Putting the numerical values, we have
Between A and B, the resistance of middle part
AOB E = 2V, r = 0.1Ω, R = 3.9Ω

R2 = 15 + 15 = 30Ω V = 2- ( 2
3.9+0.1
×0.1)
Therefore, equivalent resistance between A and B
V = 1.95 Volt
1 1 1 1
' = + + 116 (a)
R R 1
R 2
R 3
1
(220)2
1 1 1 R1 = = 484 Ω
= + + 100
40 30 15
(220)2
3+4+8 And R2 = = 242 Ω
= 200
120
220
15 So, i = = 0.3 A
= 484+242
120
Hence, total power consumed will be
120
'
⟹ R = = 8Ω
15 2
P = i R = (0.3)2×(484+242) = 65.34 ≈ 65 W
115 (a) 116 (b)
9 2
m 2.0124 1×10
-3
i= = = 0.5 A n=
15
-19 = 6.25×10
-3
Zt 1.118×10 ×3600 1.6×10
⇒ Error = 0.54 - 0.5 = 0.04 A 116 (b)
3
116 (a) L
0 Resistance of the wire, R = ρ
A
When a cell of emf E is connected to a resistance
of 3.9Ω, then the emf E of the cell remains When the wire is enlongated to n- fold, its length
constant, while voltage V goes on decreasing on becomes L' = nL
taking more and more current from the cell.
As the volume of the wire remains constant

E ' ' ' AL A

∴ A L = AL⇒A = =
L' n
r = 0.1 Ω
New resistance,
'
' L (nL) 2 L 2
R =ρ =ρ =nρ =nR
R = 3.9 Ω A' (A/n) A
( )

116 (b)
4
∴ V = E - ir Let the resistances be R1, R2 and R3

Where, r is internal resistance. R1 1

∴ = ⟹R1 = k, R2 = 2k
R2 2
E
Also, current i =
R+r

P a g e | 320
 1 1 1 1 2
In parallel = + + In the first case I is 3r, in second case it is r, in
R R1 R2 R3 3
r
third case it isand in fourth case the net
1 1 1 1 3
= + +
1 k 2k R3 3r
resistance is .
2
1 1 1
= 1- =
R3 k 2k RIII < RII < RIV < RI

2k-2-1 2k-3 ∴ PIII < PII < PIV < PI

= =
2k 2k
116 (a)
2k 8
R3 =
2k-3 As the resistance of metal increases on increasing
the temperature, so resistance of metal conductor
If k=1, then R3 is found to be negative, which is in left arm will increase on heating. For meter
impossible. bridge
If k=2, then R1 = 2, R2 = 4 R3 = 4 R1 l
=
R2 100-l
R2 = R3, not satisfying the condition of the
question that all resistance are unequal. As R1 increases l also increases.

If k = 3, then R1 = 3, R2 = 6 Hence, balancing point shifts towards right.

R3 = 2Ω 116 (a)
9
∴ Largest resistance =6Ω Mass of substance liberated at cathode m = zit

116 (a) Where, z = electro chemical equivalent

5 -7 -1
1 1 1 tt = 3.3×10 kg .C
In parallel = + ⇒tp - 1 2
tp t1 t2 t1+t2
i = current flowing = 3 A,
5×10 50
= = = 3.33 min = 3 min. 20sec t = 2s
5+10 15
-7
116 (b) m = 3.3×10 ×3×2
6 7
Electric power consumed by kettle P = 220×4W = 19.8×10 kg

Heat required 117 (a)

0
H = 1000×1(100-20) = 1000×80 cal R1 (1+αt1) 10
= ⇒
= 4200×80 J R2 (1+αt2) R2
-3
(1+5×10 ×20)
H = -3 ⇒R2 ≈ 15Ω
P= ⇒H = P×t (1+5×10 ×120)
t
i1 R 30 15
∴ 220×4×t = 4200×80⇒t = 6.3 minutes Also = 2⇒ = ⇒i = 20 mA
i2 R1 i2 10 2
116 (a)
117 (a)
7
P=iR
2 1 5
1 kWh = 1000 W×3600 sec = 36×10 W-
Current is same, so P ∝ R sec (or J)

P a g e | 321
117 (a) 117 (c)
2 6
Here same current is passing throughout the If resistance of ammeter is r then
length of the wire, hence V ∝ R ∝ l
20 = (R+r)4⇒R + r = 5⇒R < 5Ω
V1 l 6 300
⇒ = 1⇒ = ⇒V2 = 1 V 117 (b)
V2 l2 V2 50
7
Current through each arm DAC and DBC = 1A
117 (d)
3 VD - VA = 2 and VD - VB - 3⇒VA - VB = +1V
Before connecting voltmeter potential difference
across 400Ω resistance is 117 (b)
10,000 8 2
V (250)2
V R= = 3 = 62.5Ω
P 10

A 400 B 800 118 (c)

0
2×2
Total resistances = 1 + 2 + + 4 = 8Ω
6V 2+2

118 (b)
400 1
Vi = ×6 = 2V Current in the given circuit
(400+800)
50
i= = 2A
After connecting voltmeter equivalent resistance (5+7+10+3)
between A and
Potential difference between A and B,
400×10,000 VA - VB = 2×12
B=
(400+10,000) = 384.6Ω
⇒VA - 0 = 24V⇒VA = 24 V
Hence, potential difference measured by
voltmeter 118 (b)
2
384.6 V
2
(15)2 225×(R+2)
Vf = ×6 = 1.95V P= ⇒150 = =
(384.6+800) Req [2R/(R+2)] 2R
Error in measurement 450
= Vi - Vf = 2 - 1.95 = 0.05V ⇒R = = 6Ω
75

117 (d) 118 (d)

4 3
V 100±0.5 Total external equivalent resistance Req = 4Ω
R= = = 10 ± 0.25Ω
i 10±0.2
E 10
117 (b) Current supply by cell i = = = 2A
Req+r (4+1)
5
P
i
P = VI, I =
V ∴ (VA-VB) = (R -R ) = 22 (2-4) = -2V
2 2 1
500 W
or I = = 5A 118 (a)
100 V
4
Now, 5 R = 100 r= ( ) ( )
l1-l2
l2
R=
25
100
2 = 0.5 Ω

or R = 20Ω.
118 (a)
P a g e | 322
5 Let Q is divided into two parts. If one part is q = 34×10 ×4.2 J
3

then other will (Q-q).

H = VIt = Vq
Let these parts are kept a (Q-q) 3
H 34×4.2×10
∴V = = = 1.49 V
distance r. q 5
10 /1.044

q 118 (c)
| r | 7
iR i.ρ neAvdρ
E= = = ⇒vd ∝ E [Straight line]
L A A

( )
The electrostatic force between them 2 EA
2
2
P=iR= R⇒P ∝ E [Symmetric parabola]
ρ
q(Q-q)
F = K. 2 ………(i)
r 2
Also P ∝ i (parabola)
(where K=constant –(1/4πε0) Hence all graphs a,b,d are correct and c is
incorrect
On differentiating Eq. (i) w.r.t. q, we get

[ ]
118 (b)
dF d Kq
= (Q-q) 8
dq dq r2 When we move in the direction of the current in a
uniform conductor, the potential difference
dF K d
= 2 [Qq-q ]
2
or decreases linearly. When we pass through the cell,
dq r dq
from it’s negative to it’s positive terminal, the
dF K potential increases by an amount equal to it’s
or = 2 [Q-2q] ……(ii)
dq r potential difference. This is less than it’s emf, as
there is some potential drop across it’s internal
But we know that, when force is maximum then resistance when the cell is driving current
dF 118 (d)
=0
dq 9
At cold junction, current flows from copper to
Then from Eq. (ii), we have nickel and from iron to copper, and at hot junction
K from nickel to iron, thus the contributions add
2 [Q-2q] = 0 or Q - 2q = 0
r 119 (d)
Q 0
Or Q = 2q or q = The equivalent circuit is shown as
2 D

Q Q 3
So, the other part = Q - = 3Ω
2 2 C E
A 3 B
Hence, the each part have the same charge.
3 3

Q F
2
3
118 (c)
6 -3 5
m 1×10 10 We can emit the resistance in the arm DF as
q= = -8 = C;
z 1.044×10 1.044 balance condition is satisfied.
Given, H = 43 k cal. Therefore, the 3Ω resistances in arm CD and DE

P a g e | 323
 are in series. Charge passed to liberate copper
'
∴ R = 3 + 3 = 6Ω = 9×20×60 = 10800 C

Similarly, for arms CF and FE, R’’=6Ω Charge passed to liberate silver
'
R and R'' are in parallel = 10×15×50 = 9000 C

1 1 1 2 1 As 9000 C charge liberates 10.8 g of Ag.

∴ ''' = + = =
R 6 6 6 3
So 10800 C charge liberates
R’’’=3Ω
10.8×10800
= = 10.8×1.2 g of Ag.
Now, R’’’ and 3Ω resistances are in parallel 9000

1 1 1 Using Faraday’s second law of electrolysis

∴ = +
R 3 3
mCu E 63.5/2
= Cu =
⟹R = 1.5Ω mAg EAg 108/1

Moreover, V across AB=3V and resistance in the 63.5×10.8×1.2

or mCu = = 3.81 g.
arm=3Ω 2×108

∴ Current through the arm will be 119 (d)

7
3V R1R2 3
= = 1A. Rp = 2 = and R1 = 3Ω then
3Ω R1+R 5

119 (a) 3×R2 3

2 = or 15R2 = 9 + 3R2
Both R and 2R are in parallel [V- constant] 3+R2 5

2
V P1 R H R 2 3
So using P = ⇒ = 2⇒ 1 = 2 = or 12R2 = 9 or R2 = Ω
R P2 R1 H2 R1 1 4

119 (c)
119 (a)
8
4 Let G be resistance of galvanometer and ig the
By the concept of balanced Wheatstone bridge,
the given circuit can be redrawn as follows current through it. Let V is maximum potential
difference, then from Ohm’s law
30
5 10 15 | V |

a b
A B
ig

10 20 30

60

G S

30×60 Voltmeter
⇒RAB = = 20 Ω
(30+60)

119 (a)
5
For applying Faraday’s second law of electrolysis
the same charge should be passed through copper
and silver voltmeters.
P a g e | 324
 V 120 (c)
ig =
G+R 4
In a meter bridge the ratio of two resistances is
V
⟹R = -G R l
ig
' = '
R l
Given G = 10Ω, ig = 0.01A
Where l and l' are balancing lengths.
V = 10 volt
ρl ρl
Resistance R = = 2
10 A πr
∴ R= - 10 = 990Ω
0.01 '
In material remains same ρ = ρ
Thus, on connecting a resistance R of 990 Ω in '
series with the galvanometer, the galvanometer Given, l = 2l
will become a voltmeter of range zero to 10V. r
'
r =
120 (c) 2
0 ρl
'
ρ2l 8ρl
'
e R ∴ R = ' = 2 =
Potential gradient x = .
()
2
(R+Rh+r) L A r πr
π
2
3 20
= × = 0.2 '
R = 8R
(20+10+0) 10

120 (b) Therefore, the new balancing point is expected to

1 be 8l.
Mass of O2 ions Chemical equivalent of O2
= 120 (b)
Mass of Ag ions Chemical equivalent of Ag
5
0.8 8 5 10 Ω
⇒ = ⇒m = 10.8 gm
m 108
A
120 (b) 10 10 Ω10
2
E = V + ir⇒V = -ri + E 10 Ω 5Ω
D

Comparing it with y = mx + c; Slope (m) = -r

and intercept = E
10Ω in series with 10Ω will gives
120 (c)
(10+10)=20Ω
3
R
I= and 10Ω in series with 10Ω will gives
R2+r
(10+10)=20Ω
(since finally no current flows through capacitor)

∴ Potential difference across 5Ω 20 5Ω

ER2 A D
R2,V = IR2 = 20
R2+r

∴ Charge on the capacitor

20Ω in parallel with 20Ω will gives
CER2
Q = CV =
R2+r ( 20×20
20+20
=
400
40
= 10Ω )
P a g e | 325
 121 (b)
5Ω 10 5Ω 0
As temperature increases resistance of filament
A D
also increase

121 (b)
Resistance in series between points A and D
1
ig S
=5+10+5 = ⇒i G = (i-ig)S
i G+S g
=20Ω
∴ igG = (0.03-ig)4r …(i)
120 (a)
6 and igG = (0.06-ig)r …(ii)
a = 1,b = 2,c = 3
from (i) and (ii)
ρ.L ρ.c
⇒Rmax = =
A a.b 0.12 - 4ig = 0.06 - ig⇒ig = 0.02A

121 (d)
3
Resistance of the bulb= ( 1.5× )
1.5
4.5
1
= 0.5 = Ω
2

1
1×
ρ.L'' ρ.a 2 1
Rmin = = Resistance of the circuit R = = Ω
A'' b.c 1 3
1+
2
ρ.c

() () E-V
2 2 2
Rmax a.b c c c c 3 9 Now , r= R
⇒ = = × ⇒ 2 = = =
Rmin ρ.a a a a a 1 1 V
b.c
8 E-15 1
= × or E = 13.5 volt
120 (a) 3 1.5 3
7
l R A A R 121 (b)
R = ρ ⇒ 1 = 2 [ρ,L constant]⇒ 1 = 2 = 2 4
A R2 A1 A2 R1
P 60 3
P = Vi⇒i = = = amp
Now, when a body dipped in water, loss of weight V 220 11

= VσLg = ALσLg 121 (d)

5
Let E and r be the emf and internal resistance of a
(Loss of weight)1 A1
So,
(Loss of weight)2 = A2 = 2;so A has more battery respectively

loss of weight

120 (c)
8
1 1 1 1 3 1
= + + = ⇒R = ohm
R 1 1 1 1 3
In the first case current flowing in the circuit
Now such three resistance are joined in series,
hence total E
I1 =
R1r
1 1 1
R= + + = 1ohm
3 3 3 Or E = I1(R1 + r)

P a g e | 326
 E1 R 1
Eeq Req
A B A B

 i
E3 R 2

E2 E2
In the second case current flowing in the circuit

E
I2 = E1R2+E3R1 2×4+2×4
R2+r Eeq = = = 2V and
R1+R2 4+4
Or E = I2(R2+r)
4 2+2
Req = = 2Ω. Current i = = 2A from A to
Equating equations (i) and (ii), we get 2 2
B through E2
I1(R1+r) = I2(R2+r)⇒I1R1 + I1r = I2R2 + I2r
121 (b)
I1R1 - I2R2 = (I2-I1)r⇒(I2-I1)r = I1R1 - I2R2 9
E 10
i= ⇒0.5 = ⇒10 = 0.5R + 1.5⇒R
I1R1-I2R2 R+r R+3
r= = 17Ω
I2-I1

121 (d) 122 (d)

6 0
The given circuit having parallel and series
combination of resistance 3Ω, we can calculate as

R1 = 3 + 3 = 6Ω
This is a series connection. Further, whatever
current enters A has to pass . I = 2 A. R2 = 3 + 3 = 6Ω

The total resistance = 6 + 9 + 5 = 20Ω. The 1 1 1

= +
effective potential across the resistances is R3 R1 R2
20Ω×2A = 40V. But ( + 12 - 4)V is opposing the
1 1
potential difference caross AB therefore the = + = 3Ω
6 6
potential difference applied across AB is
40V + 8V = 48V R4 = 3Ω
121 (a) 1 1 1
7 = +
R R3 R4
ABCD forms a balanced Wheatstone bridge. Hence
the resistances of arm BD will be ineffective. Now
1 1 2
we have resistance (2 + 2) Ω, (4 + 4)Ω and = + =
3 3 3
(2+2) Ω in parallel between A and B.
R = 1.5Ω
Potential difference across A and C =EMF of
battery= 8V 122 (b)
2
8 Remember mass of the metal deposited on
∴ Current i1 = = 1A
(4+4) cathode depends on the current through the
voltameter and not on the current supplied by the
121 (b)
battery. Hence by using
8
The given circuit can be redrawn
mParallel i
m = Zit, we can say = Parallel
mSeries iSeries

P a g e | 327
 5 122 (d)
⇒mParallel = ×1 = 2.5 gm.
2 9
Current in silver voltmeter.
Hence increase in mass = 2.5 - 1 = 1.5 g
m1 1
i1 = = = 0.5 A
3
Volta
i2
Volta z1t1 (11.2×10 )×(30×60)
-4

i1 2 2
3
m2
Current in copper voltmeter i2 =
10 10 z2t2

1.8
i2 = = 1.51 A
(6.6×10-4)×(30×60)
122 (d)
3 So, total current given by battery i1 + i2 = 2.01 A
V∝l
Energy supplied by battery= Eit
V l
⟹ =
E L 4
W = (12)×(2.01)×(30×60)J = 4.34×10 A
Where, l= balance point
123 (d)
L= length of potentiometer wire. 0
Tow bulbs of 100 W are connected in series, the
l total power
Or V = E
L
' 100×100 10000
P = =
30×E 30 100+100 200
or V = = E
100 100 '
P = 50 W
122 (c)
When two bulbs are connected in parallel the
4
R150 [1+α(150)] total power
=
R500 [1+α(500)] . Putting R150 = 133Ω and ''
P = 100 + 100 = 200 W
α = 0.0045/℃, we get R500 = 258Ω
123 (b)
1
122 (c) From the formula shunt resistance
5
As ammeter must be connected in series of 20Ω ig×g
resistor, and the voltmeter in parallel to 20Ω S=
i-ig
resistor, the correct arrangement is as shown in
figure(c). 2×12
⟹ S= = 8Ω (in parallel)
5-2
122 (b)
6 123 (d)
For maximum energy equivalent resistance of 2
combination should be minimum ρl ρ(4a) 2ρ
R = .So R1 = =
A ( )
2a (a) a
122 (b)
7 ρ(a) ρ ρ(2a) ρ
In the given case cell is in open circuit (i = 0) so R2 = = and R3 = =
(4a)(2a) 8a (4a)(a) 2a
voltage across the cell is equal to its e.m.f
∴ R1 > R3 > R2
122 (d)
8 2
123 (d)
R2 = n R = (2)2×5.5 = 22Ω 3

P a g e | 328
 i 2 1+0.00125×T
ig = ⇒ Required shunt ∴ =
10 1 1+0.00125×300
G 90
S= = = 10 Ω or 2.75=1+0.00125×T
(n-1) (10-1)

123 (c) 1.75

or T = = 1400K
4 0.00125
At room temperature, the free electrons in a
conductor move randomly with speed of the order 123 (a)
5 -1 7
of 10 ms . Since, the motion of the electrons is Power consumed by heater is 110 W, so by using
random there is no net charge flow in any V
2

direction. P =
R
123 (a) 110
i1 i 11
5 Heater
Let the resistance of P,Q and R be r. i2 R

The total resistance across the battery is 110 V 110 V

220 V
r 3
rtotal = r + = r.
2 2
2
V
Current through P, 110 = ⇒V = 110V. Also from figure
110
110
Power 12 8 i1 = = 1A
IP = = = . 110
rtotal 3 r
r
2 110
and i = = 10 A. So i2 = 10 - 1 = 9 A
11
Current through R,
Applying Ohms law for resistance R, V = iR
1 2
I = IP = .
2 r ⇒110 = 9×R⇒R = 12.22 Ω

Power dissipated in R is thus 123 (b)

8
PR = I r =
2
()
2
r
r =2 W
Rt = R0(1 + αt)

Initially, R0(1+30α) = 10Ω

123 (c)
6 Finally, R0(1+αt) = 11Ω
The resistance of a metallic wire at temperature
t°C is given by 11 1+αt
∴ =
10 1+30α
Rt = R0(1+α t) …(i)
⇒10 + (10×0.002×t) = 11 + 330×0.002
-1
Given, α = 0.00125K
1.66
⇒0.02t = 1 + 0.66 = 1.66⇒t = = 83℃
R300 = 1 Ω 0.02

From Eq. (i), we have 123 (b)

9
1 = Ro(1+0.00125×300) ..(ii) 2 H 80
H = i Rt⇒R = 2 = = 2Ω
it 4×10
and 2 = R0(1+0.00125×T) ..(iii)
124 (b)
0
An electric fuse of length l, radius r when used in
P a g e | 329
 series of the circuit can withstand only if the rate
– e– +
of heat produced due to current in it is equal to Ne+
the rate of heat lost due to radiation. If H is the
i
rate of lost per unit area of the fuse wire, then
2 2 2
H×2πrl = I R = I ρl/πr
(n+) (n )
2 ⇒i = ×e + - ×e
Iρ t t
or H = 2 3 ie, H is independent of l.
2π r
18 -19 18 -19
= 2.9×10 ×1.6×10 + 1.2×10 ×1.6×10
124 (c)
1 ⇒i = 0.66 A
E 1.5
Short circuit current iSC = ⇒3 = ⇒r = 0.5Ω
r r 124 (b)
7
124 (d) The resistance of a metallic wire at temperature
2 t°C is given by
2
H V H 1
= ⇒ ∝
t R t R Rt = R0(1+αt)

124 (a) Where α is coefficient of expansion.

3
dE Hence, resistance of wire increases on increasing
Thermo electric power P = = α + βθ
dθ V
the temperature. Also, from Ohm’s law, ratio of
i
Comparing it with y = mx + c, option (a) is is equal to R ie,
correct
V
124 (d) =R
i
4
In series, total resistance of 5 resistance= 5R V
Hence, on increasing the temperature the ratio
2 i
V
Power dissipated = =5 increases.
5R
2 124 (a)
V
or = = 25 8
R Let resistances are R1and R2,

( )
2
V Then S = R1 + R2
In parallel, total power = 5 = 5×25 = 125
R
W R1R2
And P =
R1+R2
124 (b)
5 n×R1R2
1 faraday (96500C) is the electricity which ∴ (R1+R2) = (From S=nP)
liberates that amount of substance which is equal R1+R2
to equivalent wt
⟹(R1+R2) = nR1R2
2

63.5
So liberated amount of Cu is
[ ]
2 2
2 R1+R2+2R1R2
⟹n =
R1R2
= 31.25 gm ≈ 32 gm

124 (b)
6
= [ R1 R2
+ +2
R2 R1 ]
(n+)(q+) (n )(q )
Net current i = i+ + i- = + - - We know,
t t

P a g e | 330
 Arithmetic Mean ≥ Geometric Mean 1 1 1 1
= ' + ' = '
RAB 2R 2R R
R1 R2
+
R2 R1 R1 R2 ' 2R
≥ × ⟹RAB = R = Ω
2 R2 R1 3

R1 R 125 (c)
⟹ + 2 ≥2 3
R2 R1 25 5l
Given, l1 = l + l = . Since volume or wire
100 4
So, n(minimum value) =2+2=4
remains unchanged on increasing length, hence
124 (c)
Al - A1×5l 4 or A1 = 4A/5
9
i 2
From vd = ⇒i ∝ vdA⇒i ∝ vdr Given, R = 10 = ρl/A, and
neA

125 (c) ρl1 ρ5l/4 25ρl

R1 = = =
0 A1 4A/5 16A
2 2
V V (200)2
P= ⇒R1 = 1 = = 1000Ω 25 250
R P1 40 or R1 = ×10 = = 15.6Ω
16 16
2
V (200)2
and R2 = 2 = = 400Ω 125 (c)
P2 100
4
Given, l1 = 1K, l2 = 3K,l3 = 5K
125 (d)
2 or m1 = 5m,m2 = 3m,m3 = 1m
Resistance R bisecting the circuit can be neglected
due to the symmetry of the circuit. We knows
Now, there are four triangles ρl
R=
A
Effective resistance of each triangle
l1 l2 l3
1 1 1 So R1:R2:R3 = : :
' = + A1 A2 A3
R R 2R
2 2 2
2+1 3 l l l
= = = 1 : 2 : 3
2R 2R V1 V2 V3
2 2 2
' 2 l1 l2 l3
∴R = R = : :
3 m1 m2 m3

Now the given circuit reduced to 1 9 25

= : : = 1 :15 :125
5 3 1
R' R' 125 (b)
5
V 3 2
R = -G = -3 - 20 = 10 - 20 = 80Ω
Ig 30×10
R' R'
125 (c)
6
Let ρ is the resistivity of the material
Therefore, effective resistance between A and B, Resistance for contact A - A
x ρ
RAA = ρ =
2x×4x 8x
P a g e | 331
 23
Similar for contacts B - B and C - C are respectively 6×10 ×9
n=
2x ρ 4ρ 64
RBB = ρ. = =
x×4x 2x 8x
4x 2ρ 16ρ 64
and RCC = ρ = = ∴ vd = 480× 23 -19
x×2x x 8x 6×10 ×9×1.6×10
It is clear maximum resistance will be for contact
480×64 -1
C-C ⟹vd = cms
6×9×1.6×10000
125 (a) 32 -1
7 ⟹vd = cms
At point A the slope of the graph will be negative. 900
Hence resistance is negative 32×10 -1
= mms
900
125 (d)
8 = 0.36mms
-1
Equivalent resistance of the circuit R = 9Ω
126 (d)
V 9
∴ Main current i = = = 1A 1
R 9 i i×4 1
vd = = 2 ie, vd ∝ 2
3Ω 2Ω 2Ω
nAe nπD e D

()
0.25 A 2 2
1A 0.5 A vd1 D 1 1
∴ = 22 = =
9V 8Ω 8Ω 4Ω vd2 D1 2 4
2Ω 2Ω 2Ω
126 (b)
2 -6 -6
W = qV = 6×10 ×9 = 54×10 J
After proper distribution, the current through 4Ω
126 (b)
resistance is 0.25 A
3
Resistance of bulb 1
125 (a)
9
Potential gradient = Change in voltage per unit
length

V2-V1
∴ 10 = ⇒V2 - V1 = 3 volt
30/100

126 (d)
0
Drift velocity is given by V
2
(220)2
R1 = =
P 100
I
vd =
nqA R1 = 484 Ω
Where I is current, n the number of electrons, A
Resistance of bulb 2
the area, q the charge. Given
(220 V)2 4840
I 480A -19 R2 = = = 806.6 Ω
= 2 , q = 1.6×10 C 60 W 6
A cm

Req = R1 + R2 = (220)2 ( 1
+
1
100 60 )
⇒ Current flowing I =
220 V
Req
=
220 100×60
(220)2 160 ( )
P a g e | 332
 I=
1 75
220 2
,I =( )
15
88
A
9 For a limited range of temperatures, the graph
between resistivity and temperature is a straight
2
line for a material like nichrome as shown in the
Power consumed by 100W bulb = I R1 figure

( )
2
15 (220)2 225
= × = = 14W
88 100 16

126 (b)
4
H = σitΔθ⇒ If i = 1A, Δθ = 1℃,t = 1sec then
H=σ

126 (b) 127 (c)

5 0
6×4 Total resistant= 2.5 + 0.5 = 3.0 kΩ = 3000 Ω
P.d. across the circuit = 1.2× = 2.88 volt
6+4
6
∴ Current, i = A;
2.88 3000
Current through 6 ohm resistance = = 0.48
6
A Reading of voltmeter = i×(2.5×1000)

126 (d) 6
= ×2500 = 5V
6 3000
Thermocouples are widely used type of
temperature sensor and are used as means to 127 (c)
convert thermal potential difference into electric 1
In the absence of electric field (or potential
potential difference, using different combinations
difference applied) across the conductor, the
of metals. If metals were arranged according to
average thermal velocity of electron is zero
their heating contrasts a series were formed
antimony, iron, zinc, silver, gold, lead, mercury, 127 (d)
copper, platinum and bismuth. The greater the 2
heating contrasts between metals, the greater the E M mpF
m = zq = It = It or I =
electromotive force (EMF). Antimony and F pF Mt
bismuth formed the best junction for emf.
' 2.60×1×96500
∴ I = = 0.968 A
126 (c) 108×(40×60)
7 % error in reading of ammeter
The resistance of ideal voltmeter is always

( ) ( )
infinite. '
I-I 0.90-0.986
= ' ×100 = ×100
126 (b) I 0.968
8
V
2
200×200 = -7%
R= ⇒R1 = = 400 Ω and
P 100
127 (c)
100×100 4
R2 = = 50 Ω. Maximum current rating When the key is in contact with 1, then energy
200
stored in
P
i=
V 1 2
the condenser = CE
2
100 200 i1 1
So i1 = and i2 = ⇒ =
200 100 i2 4 But when the key is thrown to contact 2, total heat

126 (b)

P a g e | 333
 1 2 q q
H = I (500+330) =
2
CE = 1 +1
2 q2 q2

H1 = I (500)
2
q

( )
⇒q2 =
q
H1 R1 1+ 1
= q2
H (R1+R2)
q

( )
500 1 ∴ q2 =
H1 =
-6
× ×5×10 ×(200)2 z
830 2 1+ 2
z1
-3
H1 = 60×10 J
127 (a)
127 (c) 8
Reading of galvanometer remains same whether
5 switch S is open or closed, hence no current will
Before adding, resistance is 5 ohms. After the
addition, the central one is a Wheatstone’s flow through the switch i.e. R and G will be in
network. series and same current will flow through them.
IR = IG
∴ Total resistance is
1 + (2 & 2 in parallel) + 1 = 3Ω 127 (d)
9
5 Density of copper,
∴ The ratio of resistances =
3 3 3 6 3
ρ = 9×10 kg/m = 9×10 g/m
127 (a) 23
Avogadro number, NA = 6.023×10
6 2
V
Resistance of bulb R = Mass of 1 mole of copper atom, M=63.5g
P

1 Thus, number of free electrons per volume is

R∝
P 23
NA 6.023×10 6 28 -3
n= ρ= ×9×10 = 8.5×10 m
Here PX = 40 W, and PY = 60 W M 63.5

∴ RX > RY 128 (c)

0
The temperature coefficient of the carbon is
So, potential drop across bulb X ie, of 40 W bulb
negative so the resistance of carbon decreases
will be greater and it will glow brighter.
with the increase of temperature.
127 (b)
128 (d)
7
We know m = zq 1
To convert a moving coil galvanometer into a
1 voltmeter, a high resistance is connected in series
⇒z ∝
q with it.

z2 q 128 (d)
∴ = 1
z1 q2 2
Due to high resistance of voltmeter, connected in
Total charge q = q1 + q2 series, the effective resistance of circuit will
increase and hence the current in circuit will
decrease. Due to which the ammeter and
voltmeter will not be damaged.

128 (c)

P a g e | 334
4 Pp = nP = 2×40 = 80 W dE
For neutral temperature (θn), =0
dθ
128 (a)
5 ⇒ a + 2b θn = 0
A is false because at neutral temperature thermo
emf is maximum. B is true a
⇒ θn = -
2b
128 (d)
6
From the curve it is clear that slopes at points
∴ θn = -
700
2 ( a
∵ =700℃
b )
A,B,C,D have following order A > B > C > D
= -350℃ < 0℃
And also resistance at any point equals to slope of
the V - i curve. But neutral temperature can never be negative
(less than zero), ie, θn ≮ 0℃.
So order of resistance at three points will be
RA > RB > RC > RD Hence, no neutral temperature is possible for this
thermocouple.
128 (c)
7 129 (a)
The equivalent circuit is given by 0
R1 - 1 2Ω,R2 = 1 4Ω;R3 = 1/6Ω

In parallel;
A
1 1 1 1
= + + = 2 + 4 + 6 = 12
4Ω Rp R1 R2 R¬3

1
D or Equivalent conductance, G = = 12S
Rp
2Ω 4Ω
129 (c)
1
Let S be the large and R be the smaller resistance.
4Ω and 2Ω resistances are in series on both sides.
From formula for meter bridge
∴ 4Ω + 2Ω = 6Ω

Then 6Ω and6Ω resistances are in parallel on

both sides
S= ( )
100-l
l
R=
100-20
20
R = 4R

1
R
1 1 2
= + = =
6 6 6
1
3
Again, S =( )( 100-l
100
R+15)

100-40
R=3Ω = (R+15) = 3 (R+15)
40 2
128 (c) 3
8 ∴ 4R = (R+15)
-8 2
V iR iρl iρ 1×1.7×10
E= = = = = -6
l l Al A 2×10 8R
- R = 15
-3 -1 3
= 8.5×10 Vm
5R
128 (d) ⟹ = 15
3
9 2
E = aθ + bθ (given)
R = 9Ω

129 (c)

P a g e | 335
2 Reading of voltmeter Tc+ 600°
320° =
2
E1r2+E2r1 18×1+12×2
= Eeq = = = 14V
r1+r2 1+2 640° = Tc + 600°

129 (a) Tc = 40℃

4
de 129 (d)
∵ Peltier coefficient π = T and t℃ = T - 273
dT 9 2
V
As resistance of a bulb R = ,
∴ e = a(T-273) + b(T-273)2 P

de 1 1 1
Differentiating w.r.t. T = a + 2b(T - 273) Hence R1:R2:R3 = : :
dT 100 60 60

de Now the combined potential difference across B1

π=T = T[a+2b(T-273)]⇒π
dT and B2is same as the potential difference across
= (t + 273)(a + 2bt) B3. Hence, W3 is more than W1 and W2, being in
129 (b) series, carry same current and R1 < R2, therefore
5 W1 < W2,
Positive ions get deposited on cathode
∴ W1 < W2 < W3
129 (a)
6 130 (a)
120
Current through the combination i = 0
(6+9) Given,
= 8A
-2
l = 5cm = 5×10 m,
So, power consumed by 6 Ω resistance
-5
ρ = 3.5×10 Ωm
P = (8) ×6 = 384 W
2

A = π(1×10 -0.25×10 )
-4 -4

129 (c)
-4
7 = 0.75π×10 m
As we know, for conductors, resistance ∝
-5
temperature. = 7.5π×10 m
-5 -2
From figure ρl 3.5×10 ×5×10
∴ R= = -5
R1 ∝ T1⇒tan θ ∝ T1⇒tan θ = kT1 …(i) A 7.5π×10
-3
and R2 ∝ T2⇒tan (90°-θ)∝T2⇒cot θ=kT2 …(ii) = 7.42×10 Ω

From equation (i) and (ii), 130 (d)

k(T2-T1) = (cot θ-tan θ) 1
At an instant approach the student will choose

( ) tan θ will be the right answer. But it is to be seen

2 2
cos θ sin θ (cos θ-sin θ)
(T2-T1) = -
sin θ cos θ
=
sin θcos θ here the curve makes the angle θ with the V-axis.
= 2cot 2θ So it makes an angle (90 - θ) with the i-axis

⇒(T2 - T1) ∝ cot 2θ So resistance = slope = tan (90-θ)=cot θ

129 (a) 130 (a)

8 2
Tc + Tf
Neutral temperature Tn =
2

P a g e | 336
 2
Sl Sl HAB R
R= = H ∝ R or = AB
A V HBC RBC

∴
∆R
R
∆l
= 2 = +2.0%
l =
(1/2r)2
(1/r)2 ( 1 1
as R∝ ∝ 2
A r )
130 (c) 1
3 = or HBC = 4HAB
4
R×R
Rs = R + R = 2R and RP = = R/2
R+R 130 (c)
2 8
H1 IR R 2R 4 P = 100 watt, V = 125 V
= 2 S = S = = .
H2 I RP RP R/2 1
P 100
Since P = VI, ∴ I = = ampere
130 (c) V 125
4
V
2
Mass of chlorine liberated = zIt
H= t
R
-6 100 -3
= 0.367×10 × ×60 = 0.0176×10 kg
Since supply voltage is same and equal amount of 125
heat is produced, therefore = 17.6mg

R1 R R t 131 (a)
= 2 or 1 = 1 …(i) 0
t1 t2 R2 t2 8 -19 5
2×10 ×2×1.6×10 ×10
J = nqv = n(ze)v =
R1 l (10-2)3
But R ∝ l⇒ = 1 …(ii) = 6.4A/m
2
R2 l2

l1 t 131 (d)
By (i) and (ii), = 1 …(iii) 1
l2 t2 P P
In series PS = ⇒10 = ⇒P = 30 W
n 3
2 l 3
Now l2 = l1⇒ 1 =
3 l2 2 In parallel PP = nP = 3×30 = 90 W

3 15 131 (b)
∴ By equation (iii), = ⇒t = 10 minutes
2 t2 2 3
Cells are joined in parallel when internal
130 (c) resistance is higher than external resistance.
6 [R << r]
The best conductor of electricity is one whose
resistance is least. As R = ρl/A. Therefore for the E
i=
given value of l and A,R ∝ ρ. Hence, silver is the r
R+
best conductor of heat and electricity n

130 (a) 131 (b)

7 5
Current flowing through both the bars is equal. Since the value of R continuously increases, both α
and β must be positive.
Now, the heat produced is given by
Actually the components of the given equation are
2
H = I Rt as follows

P a g e | 337
 Rt  t produced per sec.
2 -1
R0 I (R+r) = (0.2)2(21+4) = 1 J s

t
132 (d)
0
Potential gradient of a potentiometer
131 (b) -7
Iρ 0.2×4×10
6 K= = -7
In series : Potential difference ∝ R A 8×10

3 = 0.1V/m
When only S1 is closed V1 = E = 0.75E
4
132 (b)
6 1
When only S2 is closed V2 = E = 0.86E V
7 Let the current in the circuit = i =
R
And when both S1 and S2 are closed combined
Across the cell,
resistance of 6R and 3R is 2R
E = V + ir⇒r =
E-V
=
E-V
=
E-V
( )
R
∴ V3 = ()
2
3
E = 0.67E⇒V2 > V1 > V3
132 (b)
i V/R V

131 (a) 2
Charge Q = I t = 1.6 ×60 = 96 C
7
Applying Kirchhoff’s voltage law in the given loop +2
Let n be the number of Cu ions, then
1 4V i 8V 2
P Q Q 96 20
ne = Q⇒n = = -19 = 3×10
4V e 2×1.6×10
i
9
132 (b)
3
Maximum number of resistances
n-1 3-1
1 =2 =2 =4
-2i + 8 - 4 - 1×i - 9i = 0⇒i = A
3
132 (b)
1 4
Potential difference across PQ = ×9 = 3V
3 Shunt is a low resistance used in parallel with the
galvanometer to make it ammeter.
131 (b)
8 The circuit is shown in figure.
Number of free electrons per unit volume,
i i g
23 G
N 6.023×10 3
n= ρ= -3 ×9×10
M 63.5×10

i
∴ vd =
nAe (i- i g )
S
-3 Ammeter
1.5×63.5×10
= 3 23-7 -19
6.023×10 ×9×10 ×10 ×1.6×10
-3 -1
= 1.1×10 ms
Voltage across galvanometer = voltage across
131 (c) shunt
9
Chemical energy consumed per sec = heat energy
P a g e | 338
 igG m = zq
ie., igG = (i-ig)S or S =
i-ig
mzn z
= Zn
Given, G = 22.8Ω, i = 20A, ig = 1A mCu zCu

1×22.8 22.8 0.13 32.5

∴ S= = = 1.2Ω ∴ =
20-1 19 mCu 31.5

132 (d) mCu =0.126 g.

5
1 R1 P 60 3 132 (b)
PRated ∝ ⇒ = 2 = =
R R2 P1 40 2 8
It is the electric field that is set up which moves
132 (c) with the velocity of light in that medium
6
According to Faraday’s law of electrolysis
1 (b)
3
2 Given resistance of each part will be
9

P a g e | 339
Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

Assertion - Reasoning Type

This section contain(s) 0 questions numbered 1 to 0. Each question contains STATEMENT 1(Assertion) and
STATEMENT 2(Reason). Each question has the 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

a) Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1

b) Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1

c) Statement 1 is True, Statement 2 is False

d) Statement 1 is False, Statement 2 is True

Statement 1: Heat is generated continuously is an electric heater but its temperature becomes constant
after some time
Statement 2: At the stage when heat produced in heater is equal to the heat dissipated to its
surrounding the temperature of heater becomes constant
2

Statement 1: The drift velocity of electrons in a metallic wire will decrease, if the temperature of the
wire is increased
Statement 2: On increasing temperature, conductivity of metallic wire decreases

P a g e | 340
 Statement 1: A toaster produces more heat than a bulb, when connected in parallel.

Statement 2: Toaster has less resistance than a bulb

Statement 1: The resistance of super-conductor is zero

Statement 2: The super-conductors are used for the transmission of electric power

Statement 1: A person touching a high power line gets stuck with the line

Statement 2: The current carrying wires attract the man towards it

Statement 1: Duddell’s thermo galvanometer is suitable to measure direct current only.

-3
Statement 2: Thermopile can measure temperature differences of the order of 10 ℃.

Statement 1: A potentiometer is preferred over that of a voltmeter for measurement of emf of a cell.

Statement 2: Potentiometer is preferred as it does not draw any current from the cell.

Statement 1: The temperature coefficient of resistance is positive for metals and negative for p-type
semiconductor
Statement 2: The effective charge carries in metals are negatively charged whereas in p-type
semiconductor they are positively charged
9

Statement 1: The 200 W bulbs glow with more brightness than 100 W bulbs

Statement 2: A 100 W bulbs has more resistance than a 200 W bulbs

10

Statement 1: a

Statement 2: r

11

Statement 1: Rapidly changing temperatures can be measured by thermocouples.

Statement 2: The thermal capacity of the junction of a thermocouple is very small.

12

Statement 1: There is no current in the metals in the absence of electric field

Statement 2: Motion of free electrons is random

13
P a g e | 341
 Statement 1: Voltmeter measures current more accurately than ammeter

Statement 2: Relative error will be small if measured from voltameter

14
-1
Statement 1: The power dissipated in a conductor of resistivity ρ is proportional to ρ .
l
Statement 2: ρ
The expression for resistance of a conductor is given by R = .
A
15

Statement 1: In a simple battery circuit the point of lowest potential is positive terminal of the battery

Statement 2: The current flows towards the point of the higher potential as it flows in such a circuit
from the negative to the positive terminal
16

Statement 1: The possibility of an electric bulb fusing fusing is higher at the time of switching ON and
OFF
Statement 2: Inductive effects produce a surge at the time of switch-OFF switch-ON

17

Statement 1: Length is immaterial for fuse wire

Statement 2: Fuse wire consists low melting point

18

Statement 1: The equivalent resistance in series combination is larger than the largest resistance.

Statement 2: The equivalent resistance of the parallel combination is smaller than the smallest
resistance.
19

Statement 1: A heater coil is cut into two equal parts and only one part is used in the heater. Heat
generated will be doubled.
2
Statement 2: Vt
Heat generated is given by the expression Q = .
R
20

Statement 1: The temperature dependence of resistance is usually given as R = R0(1+α∆t). The

resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from
-3
27℃ to 227℃. This implies that α = 2.5×10 /℃
Statement 2: R = R0(1 + α∆t) is valid only when the change in the temperature ∆T is small and
∆R = (R-R0) ≪ R0
21

Statement 1: In all conductors, for studying the thermoelectric behaviour of the metals, lead is taken as
a reference metal
Statement 2: In lead, the Thomson effect is negative

22

P a g e | 342
 Statement 1: Two electric bulbs of 50 and 100 W are given. When connected in series 50 W bulb glows
more but when connected parallel 100 W bulb glows more
Statement 2: In series combination, power is directly proportional to the resistance of circuit. But in
parallel combination, power is inversely proportional to the resistance of the circuit
23

Statement 1: In a given thermo-couple, the temperature of cold junction is 15℃, while the neutral
temperature is 270℃. The value of the temperature of inversion is 720℃.
Statement 2: Neutral temperature is the arithmetic mean to temperature of cold junction and
temperature of inversion.
24

Statement 1: Though the same current flows through the line wires and the filament of the bulb but
heat produced in the filament is much higher than that in line wires
Statement 2: The filament of bulbs is made of a material of high resistance and high melting point

25

Statement 1: When a 40 W, 220 V lamp and 100 W, 220 V lamp are connected in series across 440 V
supply then 40 W lamp will fuse.
Statement 2: 40 W and 100 W lamps can tolerate 440 V.

26

Statement 1: The e.m.f. of the driver cell in the potentiometer experiment should be greater than the
e.m.f. of the cell to be determined
Statement 2: The fall of potential across the potentiometer wire should not be less than the e.m.f. of the
cell to be determined
27

Statement 1: A potentiometer of longer length is used for accurate measurement

Statement 2: The potential gradient for a potentiometer of longer length with a given source of e.m.f.
becomes small
28

Statement 1: a

Statement 2: b

29

Statement 1: A silver voltmeter and a zinc voltmeter are connected in series and current i is passed for
a time t liberated w kg of zinc, the weight of the silver deposited is 3.5 w.
Statement 2: m1 z w
= 1 = 1.
m2 z2 w2
30 Consider the following statements A and B are identify the correct answers given below :

Statement 1: Peltier coefficient is numerically equal to the potential difference across the junctions of
the thermocouple through which current is flowing.
Statement 2: According to Thomson, energy is neither absorbed nor evolved at the junction of a
thermocouple but is observed or evolved only along the lengths of both the conductors.
31

P a g e | 343
 Statement 1: Fuse wire must have high resistance and low melting point

Statement 2: Fuse is used for small current flow only

32

Statement 1: In the given circuit if lamp B or C fuses then light emitted by lamp A decreases

Statement 2: Voltage on A decreases

33

Statement 1: Electric appliances with metallic body have three connections, whereas an electric bulb
has a two pin connection
Statement 2: Three pin connections reduce heating of connecting wires

34

Statement 1: When temperature of cold junction of a thermocouple is lowered, the value of neutral
temperature of this thermocouple is raised
Statement 2: When the difference of temperature of two junctions is raised, more thermo e.m.f. is
produced
35

Statement 1: The electric bulbs glows immediately when switch is on

Statement 2: The drift velocity of electrons in a metallic wire is very high

36

Statement 1: Thermocouple acts as a heat engine

Statement 2: When two junctions of thermocouple are at different temperature, thermo e.m.f. is
produced
37

Statement 1: Bending a wire does not effect electrical resistance

Statement 2: Resistance of wire is proportional to resistivity of material

38

Statement 1: A certain charge is liberated by 0.8 g of oxygen, 10.8 g of can liberate same amount of
charge.
Statement 2: From Faraday’s first law of electrolysis, the amount of charge liberated is proportional to
the mass of substance taken.
39

Statement 1: In the following circuit emf is 2V and internal resistance of the cell is 1 Ω and R = 1Ω,
then reading of the voltmeter is 1V

P a g e | 344
 Statement 2: 2
V = E - ir where E = 2V, i = = 1A and R = 1 Ω
2
40

Statement 1: Leclanche cell is used when constant supply of electric current is not required

Statement 2: The e.m.f. of a Leclanche cell falls if it is used continuously

41

Statement 1: If the length of the conductor is doubled, the drift velocity will become half of the original
value (keeping potential difference unchanged)
Statement 2: At constant potential difference, drift velocity inversely proportional to the length of the
conductor
42

Statement 1: An electric bulb is first connected to a dc source and then to an ac source having then
same brightness in both the cases
Statement 2: The peak value of voltage for an A.C. source is 2 times the root mean square voltage

43

Statement 1: A laser beam of 0.2 W power can drill holes through a metal sheet, whereas 1000 W torch
-light cannot
Statement 2: The frequency of laser light is much higher than that of torch light

44

Statement 1: Two unequal resistance are connected in series with a cell, then potential drop across a
larger resistance is more
Statement 2: The current will be same in both unequal resistance

45

Statement 1: A

Statement 2: All the resistances are in parallel to each other

46

Statement 1: In meter bridge experiment, a high resistance is always connected in series with a
galvanometer
Statement 2: As resistance increases current through the circuit increases

47

Statement 1: a

Statement 2: The given resistor are in parallel to one another

48 A

P a g e | 345
 Statement 1: a

Statement 2: r

49

Statement 1: A car battery is of 12 V. eight dry cells of 1.5 V connected in series can give 12V. still such
cells are not used in starting a car.
Statement 2: It is easier to start a car engine on a warm day than on a rainy day.

50

Statement 1: A domestic electrical appliance, working on a three pin will continue working even if the
top pin is removed
Statement 2: The third pin is used only as a safety device

51

Statement 1: Two bulbs of same wattage, one having a carbon filament and other having a metallic
filament are connected in series. The metallic filament bulb will glow more brightly than
carbon filament bulb
Statement 2: Carbon is a semiconductor

52

Statement 1: An electric bulb becomes dim, when an electric heater in parallel circuit is switched on

Statement 2: Dimness decreases after sometime

53

Statement 1: In a Meter Bridge experiment, null point for an unknown resistance is measured. Now,
the unknown resistance is put inside an enclosure maintained at a higher temperature.
The null point can be obtained at the same point as before by decreasing the value of the
standard resistance
Statement 2: Resistance of a metal increases with increase in temperature

54

Statement 1: Neutral temperature of a thermocouple does not depend upon temperature of cold
junction
Statement 2: Its value is constant for the given metals of the couple

55

Statement 1: If three identical bulbs are connected in series as shown in figure, then on closing the
switch bulb C is short circuited and hence illumination of bulbs A and B is decreases

Statement 2: Voltage on A and B decreases

P a g e | 346
Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

: ANSWER KEY :

1) a 2) b 3) a 4) b

5) c 6) d 7) a 8) b

9) a 10) a 11) a 12) a

13) a 14) b 15) d 16) a

17) a 18) b 19) a 20) d

21) c 22) a 23) d 24) a

25) c 26) a 27) a 28) c

29) a 30) c 31) c 32) a

33) c 34) d 35) c 36) b

37) a 38) a 39) a 40) a

41) a 42) d 43) c 44) a

45) c 46) c 47) a 48) a

49) b 50) a 51) d 52) b

53) d 54) b 55) d

P a g e | 347
Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

: HINTS AND SOLUTIONS :

2 (b) exercise his nervous control to get himself free

from the high power line
On increasing temperature of wire the kinetic
energy of free electrons increase and so they 7 (a)
collide more rapidly with each other and hence
their drift velocity decreases. Also when When a voltmeter is connected across the two
temperature increases, resistivity increases and poles of a call, it draws a small current from the
resistivity is inversely proportional to cell. So, it measures terminal potential difference
conductivity of material between the two poles of the cell, which is always
less than the emf of the cell.
3 (a)
On the other hand, when a potentiometer is used
Heat produced per second in an electrical for the measurement of emf of a cell it does not
appliance is given by draw any current from the cell. Hence, it
2
accurately measures the emf of a cell. Thus, a
V 1 potentiometer is preferred over a voltmeter.
H= or H ∝ .
R R
8 (b)
As toaster has less resistance than that of bulb
hence, it is clear that more heat will be produced The temperature co-efficient of resistance for
in comparison with a bulb, when it connected in metal is positive and that for semiconductor is
parallel with the bulb. negative.

4 (b) In metals free electrons (negative charge) are

charge carries while in P-type semiconductors,
Here assertion and reason both are correct but holes (positive charge) are majority charge
the reason is not the correct explanation of carriers
assertion
9 (a)
5 (c)
2
V
Because there is no special attractive force that The resistance, R = ⇒R ∝ 1/P
P
keeps a person stuck with a high power line. The
actual reason is that a current of the order of 0.05 i.e., higher the wattage of a bulb, lesser is the
A or even less is enough to bring disorder in our resistance and so it will glow bright
nervous system. As a result of it, the affected
person may lose temporarily his ability to 10 (a)

P a g e | 348
 s 17 (a)

12 (a) The maximum current that can be tolerated by a

fuse wire in independent of its length. So, length is
It is clear that electrons move in all directions immaterial for fuse wire.
haphazardly in metals. When an electric field is
applied, each free electron acquire a drift velocity. A safely fuse is basically a piece of wire having a
There is a net flow of charge, which constitute suitable high resistivity and low melting point so
current. In the absence of electric field this is that it breaks if the current in it exceeds the safe
impossible and hence, there is no current limit

13 (a) 18 (b)

Voltmeter measures current indirectly in terms of Let R 1, R 2 be two resistances. Suppose R 2 >R
mass of ions deposited and electrochemical 1then if they are connected in series combination,

( )
m
equivalent of the substance I= . Since value of
Zt
the equivalent resistance R 1is,

m and Z are measured to 3rd decimal place and Rs = R1 = R2

5th decimal place respectively. The relative error
Hence, it is obvious that Rs will be larger than
in the measurement of current by voltmeter will
highest resistance R2
be very small as compared to that when measured
by ammeter directly Again when the resistances are connected in
14 (b) parallel the equivalent resistance is,

From the formula 1 1 1

= =
Rp R1 R2
2
V
P= …(i) R1R2
R Rp =
R1+R2
ρl
But R = …(ii)
A It is also clear that Rp will be smaller than the
smallest resistance R1.
Now, from Eqs. (i) and (ii) , we have
2 2 19 (a)
V VA
P= =
ρl ρl The heat generated in heater coil is
A
2
Vt
1 Q1 =
⇒P ∝ ⇒P ∝ ρ
-1
R
ρ
When half of the coil is used in heater then the
15 (d) resistance halved, and so the heat generated is
It is quite clear that in a battery circuit, the point 2
Vt Vt
2

of lowest potential is the negative terminal of the Q2 = =2 = 2Q1

R R
battery and the current flows from higher 2
potential to lower potential
Hence, the heat generated b half of coil will be
16 (a) doubled.
The possibility of an electric bulb fusing is higher 21 (c)
at the time of switching ON and switching OFF
because inductive effect produces a surge at the Thomson e.m.f. in lead is practically zero
time of switching ON and OFF
22 (a)
P a g e | 349
 Resistance of 50 W bulb is two times the tolerate 220 V.
resistance of 100 W bulb. When bulbs are
connected in series, 50 W bulb will glow more as 26 (a)
2
P - i R (current remains same in series). In parallel If either the e.m.f. of the driver cell or potential
2
the 100 W bulb will glow more as P = V /R difference across the whole potentiometer wire is
(potential difference remains same in parallel) lesser than the e.m.f. of the experimental cell, then
balance point will not obtained
23 (d)
27 (a)
Here, temperature of cold junction
1
θc = 15℃ Sensitivity ∝
Potential gradient
Neutral temperature θn = 270℃ ∝ (Length of wire)

If θi is temperature of inversion, then 28 (c)

From the relation for V - i graph

θi+θc
θn =
2 R = tan θ=R0(1+αT)
or θi = 20θn - θc = 2×270 - 15° Where, θ is angle made by V - i graph with i axis.
= 540×15 = 525℃ So, R1 = tan θ=R0(1+αT1)…(i)
24 (a) R2 = tan (90°-θ)=cot θ
As filament of bulb and line wire are in series,
= R0(1 + aT2)
hence current through both is same. Because
2
i Rt So,cot θ - tan θ=R0α(T2-T1)
H= and resistance of the filament of the
4.2
bulb is much higher than that of line wires, hence cos θ sin θ
or - = R0α(T2 - T1)
heat produced in the filament is much higher than sin θ cos θ
that of line wires 2 2
cos θ-sin θ 2cos 2θ
= = R0 α(T2 - T1)
25 (c) sin θcos θ sin 2θ
Hence, T2 - T1 ∝ cot 2θ
Resistance of 40 W lamp is given by
2
29 (a)
V (220)2
R= = = 1210Ω
P 40 According to Faraday’s second law of electrolysis,
the masses of silver and zinc deposited will be
Resistance of 100 W lamp is given by
proportional to the respective equivalent weights.
2
V (220)2 We know that equivalent of silver is 108 and that
R= = 484 Ω of zinc is nearly 32. Hence, silver deposited is
P 100

Current through series combination is given by z1 w

= 1
z2 w2
440
i= = 0.26 A
(1210+484) 32 w
⇒ =
108 w2
Hence, potential drop across 40 W lamp
108
= 1210×0.26 = 314.6 V ∴ w2 = w = 3.375w = 3.5w
32
Hence, 40 W bulb will fuse because lamp can 31 (c)
P a g e | 350
 Assertion is true but reason is false. Fuse wire l
Resistance wire R = ρ , where ρ is resistivity of
must have high resistance because in series A
current remains same, therefore according to material which does not depend on the geometry
2
i Rt of wire. Since when wire is bent resistivity, length
Joule’s law H = , heat produced is high if R is
4.2 and area of cross-section do not change, therefore
high. The melting point must be slow so that wire resistance of wire also remain same
may melt with increase in temperature. As the
current equal to maximum safe value flows 38 (a)
through the fuse wire, it heats up, melts and From the Faraday’s first law
breaks the circuit
m = zit = z q
32 (a)
Here, z stands for electrochemical equivalent of
When lamp B or C gets fused equivalent substance.
resistance of B and C increases. In series voltage
distributes in the ratio of resistance, so voltage m z w
Hence, 1 = 1 = 1
appearing across B increases or in other words m2 z2 w2
voltage across A decreases
0.8 8 0.8×108
∴ = ⇒m2 =
33 (c) m2 108 8

The metallic body of the electrical appliances is = 10.8 g

connected to the third pin which is connected to
the earth. This is actually a safety precaution and 39 (a)
avoids eventual electric shock. In this process, the
2
extra charge flowing through the body is passed Here, E = 2V, i = = 1A and r = 1Ω
2
to earth.
Therefore, V = E - ir = 2 - 1×1 = 1V
The three pin connections do not affect on
heading of connecting wires 40 (a)

34 (d) The e.m.f. of a Leclanche cell falls because of the

partial polarisation due to accumulation of
Here assertion and reason are not correct
hydrogen gas. In case Leclanche cell is used in
35 (c) experiment where current is drawn after short
breaks, then during each break hydrogen gas
In a conductor there are large number of free escapes and Mn2O3 converts into MnO2 by taking
electrons. When we close the circuit, the electric oxygen from the atmosphere. As a result, the cell
field is established instantly with the speed of regains its original e.m.f
electromagnetic wave which causes electrons to
drift at every portion of the circuit. Due to which 41 (a)
the current is set up in the entire circuit instantly.
Drift velocity of free electrons is given by
The current which is set up does not wait for the
electrons flow from one end of the conductor to eE
the another end. It is due to this reason, the Vd = τ
m
electric bulb glows immediately when switch is on
V
Here, E =
36 (b) l

Here reason is not the correct explanation of the eV

∴ vd = τ
assertion, which is correct ml

37 (a)
P a g e | 351
 or vd ∝ (
1 eV
l m
=τ is constant)
42 (d)
1 1 1
= +
Voltage of dc source is constant but in ac, peak RAB R+R R+R
value of voltage is 2 times the rms voltage. Hence
bulb will glow with more brightness when Or
connected to an ac source of the same voltage 1 1 1
= = ⇒RAB = R
RAB 2R R
43 (c)
This is balanced Wheatstone bridge hence,
A laser beam is a beam of light which is light resistance in branch 𝑀 𝑁 is not taken into
amplification by simulated emission of radiation consideration. Hence, the equivalent resistance
between points 𝐴 and 𝐵 is given by
The energy per unit area of the laser beam is very
high as compared to the torch light

44 (a) 46 (c)
If a series combination of two resistance R1 and The resistance of the galvanometer is fixed. In
R2( > R1) is connected across a cell then current meter bridge experiments, to protect the
in both resistances will be same. galvanometer from a high current, high resistance
is connected to the galvanometer in order to
protect it from damage

47 (a)

The equivalent circuit is represented as

∴ I =constant

V
or =constant
R

V1 R Now, the resistances 2R, 2R and R are connected

∴ = 1
V2 R2 in parallel combination. Hence, equivalent
resistance is given by
R1
As R1 < R2 or <1
R2 1 1 1 1 6 3 R
= + = = = ⇒RP =
RP 2R 2R 2R 2R R 3
V1
∴ < 1 or V1 < V2
V2 48 (a)

Hence, on larger resistance, the potential drop S

will be maximum
49 (b)
45 (c)
To start a car, a very high current is required. A
The equivalent circuit is represented as, car battery has very low internal resistance, so it
can provide the required high current. When eight
dry cells are joined in series, the internal
resistance of the combination becomes very high.
P a g e | 352
 Due to this high internal resistance, small current than that of the electric bulb, when a heater
will be drawn from it. connected in parallel to the bulb switched on, it
draws more current due to its lesser resistance,
Hence, such cells cannot be used to start a car. consequently, the current through the bulb
On a warm day, the internal resistance of car decreases and so it becomes dim.
battery decreases and so large current can be When the heat coil becomes sufficiency hot, its
drawn from the battery. But on a chilly day, the resistance becomes more and then it draws a little
reverse process occurs. lesser current. Consequently, the current through
50 (a) the electric bulbs recovers

The electrical appliances with metallic body like Hence, dimness of the bulb decreases
heater, press etc. have three pin connections. Two 53 (d)
pins are for supply line and third pin is for earth
connection for safety purposes Resistance of metallic wire increase with rise of
temperature
51 (d)
54 (b)
When two bulbs are connected in series then
resistance of the circuit increase therefore voltage Neutral temperature is the temperature of hot
in each will decrease, so the brightness and junction, at which the thermos e.m.f. produced in
temperature also when the temperature decrease the thermocouple becomes maximum. It is
the resistance of carbon filament slightly independent of cold junction and depends on the
increases and the resistance of metal filament will nature of materials of two metals used to form
decrease. Therefore, the bulb consisting of carbon thermocouple
filament will glow more than metallic filament
2
bulb from relation P = i R. Carbon is a 55 (d)
semiconductor. It is found in IVA group of When switch S is closed, bulb C is short circuited,
periodic table so voltage V distributes only in two parts i.e.
52 (b) voltage on bulb A and B increases as compared
previously. Hence illumination of bulb A and B
The electric power of a heater is more than that of increases
1
a bulb As P ∝ , the resistance of heater is less
R

P a g e | 353
P a g e | 354
Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

Matrix-Match Type

This section contain(s) 0 question(s). Each question contains Statements given in 2 columns which have to be
matched. Statements (A, B, C, D) in columns I have to be matched with Statements (p, q, r, s) in columns II.

1. Match the List I with the List II from the combination shown. In the left side (List I) there are four different
conditions and in the right side (List II), there are ratios of heat produced in each resistance for each
condition:
Column-I Column- II

(A) Two wires of same resistance are connected in (p) 1 :2

series and same current is passed through
them
(B) Two wires of resistance R and 2R ohm are (q) 4 :1
connected in series and same P.D. is applied
across them
(C) Two wires of same resistance are connected in (r) 1 :1
parallel and same current is flowing through
them
(D) Two wires of resistances in the ratio 1 :2 are (s) 2 :1
connected in parallel and same P.D. is applied
across them
CODES :

A B C D

a) B a c d

b) c d c d

c) b d a c

d) a b d c

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Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

: ANSWER KEY :

1) b

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Session: 2023-24 Total Questions: 1385

JEE/NEET PHYSICS

3.CURRENT ELECTRICITY

: HINTS AND SOLUTIONS :

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