Dr. V.G.

Girhepunje

VERTEX ACADEMY
72, New Diamond Nagar Kharbi Road, Nagpur-440024
Contact : 9923403431 / 7038855627
DPP on Current Electricity
Date : 02/12/2022 TEST ID: 40
Time : 20:00:00 PHYSICS
Marks : 2000
3.CURRENT ELECTRICITY ,9.CURRENT ELECTRICITY

Single Correct Answer Type the ratio of 1 ∶ 2 are connected in parallel to a

1. When 1 g hydrogen (ECE = 1.044× constant voltage source. The powers
10−8 kg C −1 ) forms water, 34 kcal heat is dissipated in them have the ratio
liberated. The minimum voltage required to a) 1 :2 b) 1 :1 c) 2 :1 d) 1 :4
decompose water is 7. The resistance of a conductor increases with
a) 0.75 V b) 3 V c) 1.5 V d) 4.5 V a) Increase in length
2. In an experiment to measure the internal b) Increase in temperature
resistance of a cell by potentiometer, it is c) Decrease in cross-sectional area
found that the balance point is at a length of d) All of these
2𝑚 when the cell is shunted by a 5Ω 8. In a conductor if 3000 coulomb of charge
resistance; and is at a length of 3𝑚 when the enters and 3000 coulomb of charge exits in
cell is shunted by a 10Ω resistance. The time 10 minutes, then the current is
internal resistance of the cell is, then a) 5 ampere b) 10 ampere
a) 1.5Ω b) 10Ω c) 15Ω d) 1Ω c) 2.5 ampere d) Zero
3. Two wires of equal diameters, of resistivities 9. The charge supplied by source varies with time
𝜌1 and 𝜌2 and lengths 𝑙1 and 𝑙2 , respectively, 𝑡 as 𝑄 = 𝑎𝑡 − 𝑏𝑡 2 . The total heat produced in
are joined in series. The equivalent resistivity resistor 2𝑅 is
of the combination is R

𝜌1 𝑙1 + 𝜌2 𝑙2 𝜌1 𝑙2 + 𝜌2 𝑙1
a) b) +
𝑙1 + 𝑙2 𝑙1 − 𝑙2 Source
-
R 2R
𝜌1 𝑙2 + 𝜌2 𝑙1 𝜌1 𝑙1 − 𝜌2 𝑙2
c) d)
𝑙1 + 𝑙2 𝑙1 − 𝑙2
𝑎3 𝑅 𝑎3 𝑅
4. In the circuit shown figure potential difference a) b)
6𝑏 27𝑏
between 𝑋 and 𝑌 will be 𝑎3 𝑅 d) None of these
c)
X Y 3𝑏
10. 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
120 V to
a) The conducting property of the conductor
a) Zero b) 20 V c) 60 V d) 120 V b) The resistance of the conductor is small
5. A battery has e.m.f. 4 𝑉 and internal resistance c) The electron number density of the
𝑟. When this battery is connected to an conductor is small
external resistance of 2 𝑜ℎ𝑚, a current of d) The electron number density of the
1 𝑎𝑚𝑝. flows in the circuit. How much current conductor is enormous
will flow if the terminals of the battery are 11. What length of the wire of specific resistance
connected directly 48 × 10−8 Ω 𝑚 is needed to make a resistance
a) 1 𝑎𝑚𝑝 b) 2 𝑎𝑚𝑝 c) 4 𝑎𝑚𝑝 d) Infinite of 4.2 Ω (diameter of wire = 0.4 𝑚𝑚)
6. Two electric bulbs whose resistances are in a) 4.1 𝑚 b) 3.1 𝑚 c) 2.1 𝑚 d) 1.1 𝑚

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12. For a thermocouple the neutral temperature is 10 5V

270℃ when its cold junction is at 20℃. What

will be the neutral temperature and the A B

temperature of inversion when the

2V
temperature of cold junction is increased to X

40℃
a) 5 Ω b) 10 Ω c) 15 Ω d) 20 Ω
a) 290℃, 580℃ b) 270℃, 580℃
20. The resistance of a 5 𝑐𝑚 long wire is 10 Ω. It is
c) 270℃, 500℃ d) 290℃, 540℃
uniformly stretched so that its length becomes
13. A railway compartment is lit up by thirteen
20 𝑐𝑚. The resistance of the wire is
lamps each taking 2.1 A at 15 V. The heat
a) 160 Ω b) 80 Ω c) 40 Ω d) 20 Ω
generated per second in each lamp will be
21. Equal potentials are applied on an iron and
a) 4.35 cal b) 5.73 cal c) 7.5 cal d) 2.5 cal
copper wire of same length. In order to have
14. If potential 𝑉 = 100 ± 0.5 𝑉𝑜𝑙𝑡 and current 𝐼 =
the same current flow in the two wires, the
10 ± 0.2 𝑎𝑚𝑝 are given to us, then what will be
ration 𝑟 (iron)/𝑟 (copper) of their radii must
the value of resistance
be (Given that specific resistance of iron =
a) 10 ± 0.7 𝑜ℎ𝑚 b) 5 ± 2 𝑜ℎ𝑚
1.0 × 10−7 𝑜ℎ𝑚 − 𝑚 and specific resistance of
c) 0.1 ± 0.2 𝑜ℎ𝑚 d) None of these
copper = 1.7 × 10−8 𝑜ℎ𝑚 − 𝑚)
15. Two identical cells weather connected in
a) About 1.2 b) About 2.4
parallel or in series gives the same current
c) About 3.6 d) About 4.8
when connected to an external resistance
22. The thermo emf of a thermo-couple is found to
1.5 Ω. Find the value of internal resistance of
depend on temperature 𝑇(in degree Celsius)
each cell.
𝑇2
a) 1 Ω b) 0.5 Ω c) Zero d) 1.5 Ω as 𝐸 = 4𝑇 − 200, where 𝑇℃ is the temperature
16. The potential difference across 8Ω resistance of the hot junction. The neutral and inversion
is 48V as shown in figure. The value of temperature of the thermocouple are (in
potential difference across points 𝐴 and 𝐵 will degree celsius)
be a) 100, 200 b) 200, 400 c) 300, 600 d) 400, 800
23. A 100 𝑜ℎ𝑚 galvanometer gives full scale
A 3Ω
deflection at 10 𝑚𝐴. How much shunt is
required to read 100 𝑚𝐴
20 Ω 30 Ω 60 Ω a) 11.11 𝑜ℎ𝑚 b) 9.9 𝑜ℎ𝑚
c) 1.1 𝑜ℎ𝑚 d) 4.4 𝑜ℎ𝑚
24. To convert a 800 𝑚𝑉 range 𝑚𝑖𝑙𝑙𝑖 𝑣𝑜𝑙𝑡𝑚𝑒𝑡𝑒𝑟 of
24 Ω 8Ω 48 V
resistance 40 Ω into a galvanometer of 100 𝑚𝐴
range, the resistance to be connected as shunt
B 1Ω is
a) 10 Ω b) 20 Ω c) 30 Ω d) 40 Ω
a) 62 V b) 80 V c) 128 V d) 160 V 25. Kirchoff’s second law for the analysis of circuit
17. A galvanometer has a resistance of 3663Ω. A is based on
shunt S is connected across it such that (1/34) a) Conversion of charge
of the total current passes through the b) Conversion of energy
galvanometer. Then the value of shunt is c) Conversion of both charge and energy
a) 3663Ω b) 111Ω c) 107.7Ω d) 3555.3Ω d) Conversion of momentum of electron
18. The temperature at which thermo emf is zero, 26. Dimensions of a block are 1 𝑐𝑚 × 1 𝑐𝑚 ×
is 100𝑐𝑚. If specific resistance of its material is
a) Temperature of inversion 3 × 10−7 𝑜ℎ𝑚 − 𝑚, then the resistance
b) Temperature of cold junction between the opposite rectangular faces is
c) Neutral temperature a) 3 × 10−9 𝑜ℎ𝑚 b) 3 × 10−7 𝑜ℎ𝑚
d) None of the above c) 3 × 10−5 𝑜ℎ𝑚 d) 3 × 10−3 𝑜ℎ𝑚
19. If 𝑉𝐴𝐵 = 4𝑉 in the given figure, then resistance 27. In a copper voltameter, mass deposited in 6
𝑋 will be minutes is 𝑚 gram. If the current-time graph
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 for the voltameter is as shown here, then the a) 4.5 V b) 1.5 V c) 2.67 V d) 13.5 V
E.C.E of the copper is 32. A uniform wire of resistance 9 Ω is cut into 3
equal parts. They are connected in the form of
2.0
equilateral triangle 𝐴𝐵𝐶. A cell of e. m. f. 2 𝑉
I ampere

1.0 and negligible internal resistance is connected

across 𝐵 and 𝐶. Potential difference across 𝐴𝐵
0 2 4 6 is
t (min)
a) 1 𝑉 b) 2 𝑉 c) 3 𝑉 d) 0.5 𝑉
a) 𝑚/5 b) 𝑚/300 33. Two copper wires have their masses in the
c) 5 𝑚 d) 𝑚/18000 ratio 2 : 3 and the lengths in the ratio 3 : 4 the
28. The kirchoff;s forst law (∑𝑖 = 0)and second ratio of their resistance is
law (∑𝑖𝑅 = ∑𝐸)where the symbols have their a) 4 : 9 b) 27 : 32 c) 16 : 9 d) 27 :128
usual meanings, are respectively based on 34. The length of the resistance wire is increased
a) Conservation of charge, conversion of by 10%. What is the corresponding change in
momentum the resistance of wire?
b) Conservation of energy, conservation of a) 10 % b) 25 % c) 21 % d) 9 %
charge 35. A bulb of 220 V and 300 W is connected across
c) Conservation of momentum, conservation of 110 V circuit, the percentage reduction in
charge power is
d) Conservation of charge, conservation of a) 100% b) 25% c) 70% d) 75%
energy 36. A cell supplies a current 𝑖1 through a resistance
29. Which of the following set up can be used to 𝑅1 and a current 𝐼2 through a resistance 𝑅2 .
verify the Ohm’s law? The internal resistance of a cell is
a) A b) (𝑖1 + 𝑖2 )
a) 𝑅2 − 𝑅1 b) 𝑅 𝑅
V A V 𝑖1 − 𝑖2 1 2
𝑖1 𝑅2 − 𝑖2 𝑅1 𝑖2 𝑅2 − 𝑖1 𝑅1
c) d)
𝑖1 − 𝑖2 𝑖1 − 𝑖2
c) A d) V
37. The resistors 𝑃, 𝑄 and 𝑅 in the circuit have
equal resistance. The battery, of negligible
internal resistance, supplies a total power of
V A 12 W. What is the power dissipated by heating
in resistor 𝑅?
P

30. A thermocouple is made from two metals,

Antimony and Bismuth. If one junction of the Q R

couple is kept hot and the other is kept cold,

then, an electric current will a) 2 W b) 4 W c) 3 W d) 6 W
a) Flow from Antimony to Bismuth at the hot 38. A heater draws a current of 2𝐴 when
junction connected to a 250 𝑉 source. The rate of
b) Flow from Bismuth to Antimony at the cold energy dissipation is
junction a) 500 𝑊 b) 1000 𝑊 c) 250 𝑊 d) 125 𝑊
c) Not flow through the thermocouple 39. Potentiometer measures the potential
d) Flow from Antimony to Bismuth at the cold difference more accurately than a voltmeter
junction because
31. A torch bulb rated at 4.5 W, 1.5 V is connected a) It has a wire of high resistance
as shown in figure. The emf of the cell needed b) It has a wire of low resistance
to make the bulb glow at full intensity if c) It does not draw current from external
circuit
d) It draws a heavy current from external
circuit
40. Which of the following statement is correct

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 a) Electric field is zero on the surface of current 𝛼1 respectively. These are joined in series. The
carrying wire effective temperature coefficient of resistance
b) Electric field is non-zero on the axis of is
hollow current carrying wire 𝛼 + 𝛼2
a) 1 b) √𝛼1 𝛼2
Surface integral of magnetic field for any 2
closed surface is equal to 𝜇0 times of total 𝛼1 𝑅1 + 𝛼2 𝑅2 √𝑅1 𝑅2 𝛼1 𝛼2
c) c) d)
algebraic sum of current which are crossing 𝑅1 + 𝑅2 √𝑅 2 + 𝑅 2 1 2
through the closed surface 47. Variation of current and voltage in a conductor
d) None has been shown in the diagram below. The
41. Who among the following scientists made the resistance of the conductor is
statement –“Chemical change can produce
6
electricity” V
5
a) Galvani b) Faraday 4
c) Coulomb d) Thomson 3
2
42. A fuse wire of circuit cross-section and having 1
diameter of 0.4 mm, allows 3 A of current to 1 2 3 4 5 6
i
pass through it. But if another fuse wire of
same material and circular cross-section and a) 4 𝑜ℎ𝑚 b) 2 𝑜ℎ𝑚 c) 3 𝑜ℎ𝑚 d) 1 𝑜ℎ𝑚
having diameter of 0.6 mm is taken, then the 48. The resistivity of a potentiometer wire is 40 ×
amount of current passed through the fuse is 10−8 𝑜ℎ𝑚 − 𝑚 and its area of cross-section is
a) 3 A 3
8 × 10−6 𝑚2 . If 0.2 𝑎𝑚𝑝 current is flowing
b) 3 × √ A through the wire, the potential gradient will be
2
3/2 3 a) 10−2 𝑣𝑜𝑙𝑡/𝑚 b) 10−1 𝑣𝑜𝑙𝑡/𝑚
c) 3 × ( 3) A d) 3 × (2) A
2 c) 3.2 × 10−2 𝑣𝑜𝑙𝑡/𝑚 d) 1 𝑣𝑜𝑙𝑡/𝑚
43. A galvanometer acting as a voltmeter should 49. An electric heater kept in vacuum is heated
have continuously by passing electric current. Its
a) Low resistance in series with its coil temperature
b) Low resistance in parallel with its coil a) Will go on rising with time
c) High resistance in series in series with its b) Will stop after sometime as it will loose heat
coil to the surroundings by conduction
d) High resistance in parallel with its coil c) Will rise for sometime and there after will
44. Silver and copper voltameter are connected in start falling
parallel with a battery of 𝑒.m.f. 12 𝑉. In d) Will become constant after sometime
30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠, 1𝑔 of silver and 1.8𝑔 of copper are because of loss of heat due to radiation
liberated. The power supplied by the battery is 50. Three resistors are connected to form the
(𝑍𝐶𝑢 = 6.6 × 10−4 𝑔/𝐶 and 𝑍𝐴𝑔 = 11.2 × sides of a triangle 𝐴𝐵𝐶, the resistance of the
10−4 𝑔/𝐶) sides 𝐴𝐵, 𝐵𝐶 and 𝐶𝐴 are 40 𝑜ℎ𝑚, 60 𝑜ℎ𝑚 and
a) 24.13 𝐽/𝑠𝑒𝑐 b) 2.413 𝐽/𝑠𝑒𝑐 100 𝑜ℎ𝑚 respectively. The effective resistance
c) 0.2413 𝐽/𝑠𝑒𝑐 d) 2413 𝐽/𝑠𝑒𝑐 between the points 𝐴 and 𝐵 in 𝑜ℎ𝑚 will be
45. In the figure, current through the 3Ω resistor is a) 32 b) 64 c) 50 d) 200
0.8 𝑎𝑚𝑝𝑒𝑟𝑒, then potential drop through 4 Ω 51. The resistance of the following circuit figure
resistor is between 𝐴 and 𝐵 is
3 E
2Ω 2Ω
4 C D
2Ω
6
2Ω 2Ω
2Ω 2Ω
+ – F

a) 9.6 𝑉 b) 2.6 𝑉 c) 4.8 𝑉 d) 1.2 𝑉 B

A
46. Two wires of resistance 𝑅1 and 𝑅2 have 2Ω
temperature coefficient of resistances 𝛼1 and a) (3/2) Ω b) 2 Ω c) 4 Ω d) 8 Ω
Page|4
52. In the figure given below, the current passing c) 𝛼 is positive and 𝛽 is negative
through 6Ω resistor is d) 𝛼 is negative and 𝛽 are positive
6 59. Consider the circuit shown in the figure. The
current 𝐼3 is equal to
1.2 A
28 54

4
6V
a) 0.40 𝑎𝑚𝑝𝑒𝑟𝑒 b) 0.48 𝑎𝑚𝑝𝑒𝑟𝑒
I3
c) 0.72 𝑎𝑚𝑝𝑒𝑟𝑒 d) 0.80 𝑎𝑚𝑝𝑒𝑟𝑒
53. The plates of a charged condenser is connected 8V 12 V
to a voltmeter. If the plates are moved apart, a) 5 𝑎𝑚𝑝 b) 3 𝑎𝑚𝑝
the reading of voltmeter will c) −3 𝑎𝑚𝑝 d) −5/6 𝑎𝑚𝑝
a) Increase 60. In the circuit shown in the figure the potential
b) Decrease 40 Ω X Y
difference between X and Y will be
c) Remain unchanged
d) Information is insufficient
54. All the edges of a block with parallel faces are
unequal. Its tangent edge is twice its shortest 20 Ω
||
edge. The ratio of the maximum to minimum 120 V 20 Ω

resistance between parallel faces is a) Zero b) 20 V c) 60 V d) 120 V

a) 8 b) 4 61. Six equal resistances each of 4Ω are connected
c) 2 d) None of these to form a figure. The resistance between two
55. −1
A charge of 2 × 10 C move at 30 revolutions corners 𝐴 and 𝐵 is
per second in a circle of diameter 80 cm. The
current linked with the circuit is 4Ω
A B
a) 0.02 A b) 20 A c) 0.60 A d) 60 A 4Ω
56. Current provided by a battery is maximum
4Ω
when O
4Ω
a) Internal resistance equal to external
resistance 4Ω
b) Internal resistance is greater than external D
4Ω
C

resistance
a) 4 Ω b) 4/3 Ω c) 12 Ω d) 2 Ω
c) Internal resistance is less than external
62. Two identical batteries each emf 𝐸 =2V and
resistance
internal resistance 𝑟 = 1Ω are available to
d) None of these
produce heat in an external resistance by
57. An ammeter of 5 𝑜ℎ𝑚 resistance can read
passing a current through it. The maximum
5 𝑚𝐴. If it is to be used to read 100 𝑣𝑜𝑙𝑡, how
Joulean power that can be developed across 𝑅
much resistance is to be connected in series
using these batteries is
a) 19.9995 Ω b) 199.995 Ω 8
c) 1999.95 Ω d) 19995 Ω a) 1.28 W b) 2.0 W c) 9 W d) 3.2 W
58. The resistance 𝑅𝑡 of a conductor varies with 63. When current is passed in antimony-bismuth
temperature 𝑡 as shown in the figure. If the couple, then
variation is represented by 𝑅𝑡 = 𝑅0 [1 + 𝛼𝑡 + a) The junction becomes hot when the current
𝛽𝑡 2 ], then is from bismuth to antimony
Rt b) The junction becomes hot when current
flows from antimony to bismuth
c) Both junctions becomes hot
d) Both junctions becomes cold
t 64. In the circuit shown in figure, the current
a) 𝛼 and 𝛽 are both negative drawn from the battery is 4𝐴. If 10 Ω resistor
b) 𝛼 and 𝛽 are both positive is replaced by 20 Ω resistor, then current

Page|5
 drawn from the circuit will be
2.0 V
1 3
j' j
10
10
--
--
--
4A 7 21 -G
--
-
G
--
--
+ – 1.5 V
--
--
-
a) 1 𝐴 b) 2 𝐴 c) 3 𝐴 d) 0 𝐴
65. Electromotive force is the force, which is able 10 Ω
to maintain a constant
a) 2.5 Ω b) 2.0 Ω c) 1.54 Ω d) 1.0 Ω
a) Current b) Resistance
71. Three resistance 𝑃, 𝑄, 𝑅 each of 2Ω and an
c) Power d) Potential difference
unknown resistance 𝑆 form the four arms of a
66. In a given network, each resistance has value
wheatstone bridge circuit. When a resistance
of 6Ω. The point X is connected to point A by a
of 6Ω is connected in parallel to 𝑆 the bridge
copper wire of negligible resistance and point
gets balanced. What is the value of 𝑆
Y is connected to point B by the same wire.
a) 2 Ω b) 3 Ω c) 6 Ω d) 1 Ω
The effective resistance between X and Y will
72. The figure here shows a portion of a circuit.
be
What are the magnitude and direction of the
current 𝑖 in the lower right-hand wire
6Ω 6Ω A 6Ω
X Y
B

a) 18Ω b) 6 Ω c) 3 Ω d) 2 Ω
67. In the circuit shown below, the cell has an
e.m.f. of 10 𝑉 and internal resistance of 1 𝑜ℎ𝑚.
The other resistances are shown in the figure.
The potential difference 𝑉𝐴 − 𝑉𝐵 is a) 7 𝐴
E=10V
r=1
b) 8 𝐴
1 c) 6 𝐴
4 A 2
d) 2 𝐴
C
2 B 4 73. A series combination of two resistors 1 Ω each
is connected to a 12 𝑉 battery of internal
a) 6 𝑉 b) 4 𝑉 c) 2 𝑉 d) −2 𝑉
resistance 0.4 Ω. The current flowing through
68. A milliammeter of range 10 𝑚𝐴 has a coil of
it will be
resistance 1 Ω. To use it as voltmeter of range
a) 3.5 𝐴 b) 5 𝐴 c) 6 𝐴 d) 10 𝐴
10 𝑣𝑜𝑙𝑡, the resistance that must be connected
74. A silver and a zinc voltmeter are connected in
in series with it, will be
series and a current 𝐼 is passed through them
a) 999 Ω b) 99 Ω
for a time 𝑡, liberating 𝑤 gram of zinc. The
c) 1000 Ω d) None of these
weight of silver deposited is nearly
69. A thermoelectric refrigerator works on
a) 1.7 𝑤 g b) 2.4 𝑤 g c) 3.5 𝑤 g d) 1.2 𝑤 g
a) Joule effect b) Seeback effect
75. The total power dissipated in Watts in the
c) Peltier effect d) Thermonic emission
circuit shown here is
70. 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 a) 16 b) 40 c) 54 d) 4
internal resistance of the cell is 76. The junction of Ni-Cu thermo couple are
maintained at 0℃ and 100℃. The seeback emf

Page|6
 developed in the temperature is copper wire become three times its value at
𝑎Ni −Cu = 16.3 × 10−6 V℃−1 0℃ (Temperature coefficient of resistance for
𝑏Ni −Cu = −0.021 × 10−6 V℃−1 copper = 4 × 10−3 𝑝𝑒𝑟 ℃)
a) 2.73 × 103 V b) 1.42 × 10−3 V a) 400℃ b) 450℃ c) 500℃ d) 550℃
c) 3.68 × 10 V−3 d) 2.23 × 103 V 82. Three resistors each of 2 𝑜ℎ𝑚 are connected
77. If θi is the inversion temperature, θn is the together in a triangular shape. The resistance
natural temperature, θc is the temperature of between any two vertices will be
the cold junction then a) 4/3𝑜ℎ𝑚 b) 3/4 𝑜ℎ𝑚 c) 3 𝑜ℎ𝑚 d) 6 𝑜ℎ𝑚
a) 𝜃𝑖 + θ𝑐 = θ𝑛 b) 𝜃𝑖 − θ𝑐 = 2θ𝑛 83. The equivalent resistance between the points
𝜃 + θ𝑐 A and B will be (each resistance is
c) 𝑖 = θ𝑛 d) 𝜃𝑐 − θi = 2θ𝑛 15 Ω)
2
78. When the key 𝐾 is pressed at time 𝑡 = 0, which 15Ω
of the following statements about the current 𝐼 D C

in the resistor 𝐴𝐵 of the given circuit is true 15Ω

15Ω
A B
15Ω
2V K 1000Ω 15Ω 15Ω
15Ω
1000Ω

1F C
A B
15Ω

a) 𝐼 = 2 𝑚𝐴 at all 𝑡 a) 30 Ω b) 8 Ω c) 10 Ω d) 40 Ω
b) 𝐼 oscillates between 1 𝑚𝐴 and 2𝑚𝐴 84. What is the current (𝑖) in the circuit as shown
c) 𝐼 = 1 𝑚𝐴 at all 𝑡 in figure
At 𝑡 = 0, 𝐼 = 2 𝑚𝐴 and with time it goes to i R2 = 2
d)
1 𝑚𝐴

R3 = 2
79. For what value of 𝑅 in the circuit as shown in 3V R1 = 2
figure, current passing through 4Ω resistance
will be zero. R4 = 2
B C D a) 2 𝐴 b) 1.2 𝐴 c) 1 𝐴 d) 0.5 𝐴
85. Current through wire 𝑋𝑌 of circuit shown is
1Ω X 2Ω
2Ω 4Ω R
1 2
6V
3Ω Y 4Ω
A F E
9V 3V
50V
a) 1 Ω b) 2 Ω c) 3 Ω d) 4 Ω a) 1 𝐴 b) 4 𝐴 c) 2 𝐴 d) 3 𝐴
80. A wire of resistor 𝑅 is bent into a circular ring 86. To get the maximum current from a parallel
of radius 𝑟. Equivalent resistance between two combination of 𝑛 identical cells each of internal
points 𝑋 and 𝑌 on its circumference, when resistance 𝑟 and external resistance 𝑅, when
angle 𝑋𝑂𝑌 is 𝛼, can be given by a) 𝑅 ≫ 𝑟 b) 𝑅 ≪ 𝑟
X c) 𝑅 = 𝑟 d) None of these
87. Metals have
W  O Z
a) Zero resistivity b) High resistivity
Y
c) Low resistivity d) Infinite resistivity
𝑅𝛼 𝑅 88. The 𝑉-𝑖 graphs A and B are drawn for two
a) 2 (2𝜋 − 𝛼) (2𝜋 − 𝛼) b) voltameters. Identify each graph
4𝜋 2𝜋
4𝜋
c) 𝑅(2𝜋 − 𝛼) d) (2𝜋 − 𝛼)
𝑅𝛼
81. At what temperature will the resistance of a

Page|7
 I I 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
95. If the ratio of the concentration of electron to
1.7 V V 7
V that of holes in a semiconductor is 5 and the
(A) (B)
7
𝐴 for water voltameter and 𝐵 for 𝐶𝑢 ratio of current is 4, then what is the ratio of
a)
voltameter their drift velocities
𝐴 for 𝐶𝑢 voltmeter and 𝐵 for water 4 5 4 5
b) a) b) c) d)
voltameter 5 4 7 8
c) Both 𝐴 and 𝐵 represents 𝐶𝑢 voltameter 96. Faraday’s 2nd law states that mass deposited on
d) None of these the electrode is directly proportional to
89. The resistance between the points 𝐴 and 𝐶 in a) Atomic mass
the figure below is b) Atomic mass × Velocity
𝑅Ω
c) Atomic mass/Valency
A B d) Valency
𝑅Ω 97. In the circuit shown the cells 𝐴 and 𝐵 have
𝑅Ω
negligible resistance. For 𝑉𝐴 = 12𝑉, 𝑅1 = 500Ω
𝑅Ω E 𝑅Ω and 𝑅 = 100Ω the galvanometer (G) shows no
deflection. The value of 𝑉𝐵 is
𝑅Ω 𝑅Ω
D C
𝑅Ω
4 2 8𝑅
a) 𝑅 Ω b) Ω c) 𝑅Ω d) a) 4𝑉 b) 2𝑉 c) 12𝑉 d) 6𝑉
3 3 3
90. According to Joule’s law, if the potential 98. Electric bulb 50 𝑊-100 𝑉 glowing at full power
difference across a conductor having a are to be used in parallel with battery
material of specific resistance remains 120 𝑉, 10 Ω. Maximum number of bulbs that
constant, then the heat produced in the can be connected so that they glow in full
conductor is directly proportional to power is
1 1 a) 2 b) 8 c) 4 d) 6
a) 𝜌 b) 𝜌2 c) d) 99. When 1 A current flows for 1 min through a
√𝜌 𝜌
silver voltmeter, it deposits 0.067 g of silver on
91. The emf is thermocouple changes sign at 600
the cathode, then how much charge will flow to
K. If the neutral temperature is 210℃, the
deposit 108 g of silver?
temperature of cold junction is
a) 180 K b) 117 K c) 93℃ d) 90℃ a) 10.6 × 104 Cg −1
eq b) 9.67 × 104 Cg−1eq
4 −1 4 −1
92. The resistance of an ideal voltmeter is c) 8.7 × 10 Cg eq d) 4.3 × 10 Cg eq `
a) Zero b) Very low 100. A 10 Ω electric heater operates on a 110V line.
c) Very large d) Infinite The rate at which heat is developed in watts is
93. In the circuit shown below, the power a) 1310 W b) 670 W c) 810 W d) 1210 W
developed in the 6Ω resistor is 6 watt. The 101. The current in a conductor varies with time 𝑡
power in watts developed in the 4Ω resistor is as 𝐼 = 2𝑡 + 3𝑡 2 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
102. Find equivalent resistance between 𝐴 and 𝐵

a) 16 b) 9 c) 6 d) 4
94. A heater of 220 V heats a volume of water in 5

Page|8
 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 𝑜ℎ𝑚c) 1.0 𝑜ℎ𝑚 d) 0.5 𝑜ℎ𝑚
108. 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
a) 𝑅
3𝑅 109. A voltmeter of resistance 1000Ω gives full
b) scale deflection when a current of 100 𝑚𝐴
4
𝑅 flows through it. The shunt resistance required
c)
2 across it to enable it to be used as an ammeter
d) 2𝑅 reading 1 𝐴 at full scale deflection is
103. The 𝑉 − 𝐼 graph for a wire of copper of length a) 10000Ω b) 9000Ω c) 222Ω d) 111Ω
𝐿 and cross-section ares 𝐴 is shown in 110. In Wheatstone’s bridge 𝑃 = 9 𝑜ℎ𝑚, 𝑄 =
adjoining figure. The slope of the graph will be 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
44
a) 24 𝑜ℎ𝑚 b) 𝑜ℎ𝑚 c) 26.4 𝑜ℎ𝑚d) 18.7 𝑜ℎ𝑚
9
a) Less if the experiment is repeated at a higher 111. In the circuit shown 𝑃 ≠ 𝑅, the reading of the
temperature galvanometer is same with switch 𝑆 open or
b) More if a wire of silver having the same closed. Then
dimension is used
c) Doubled if the length of the wire is doubled
d) Halved if length of the wire is halved
104. The electric bulbs have tungsten filaments of
same length. If one of then gives 60 𝑤𝑎𝑡𝑡 and
other 100 𝑤𝑎𝑡𝑡, then
a) 100 𝑤𝑎𝑡𝑡 bulb has thicker filament a) 𝐼𝑅 = 𝐼𝐺 b) 𝐼𝑃 = 𝐼𝐺 c) 𝐼𝑄 = 𝐼𝐺 d) 𝐼𝑄 = 𝐼𝑅
b) 60 𝑤𝑎𝑡𝑡 bulb has thicker filament 112. A battery of 𝑒𝑚𝑓 E produces currents 𝐼1 and 𝐼2
c) Both filaments are of same thickness when connected to external resistances 𝑅1 and
d) It is possible to get different wattage unless 𝑅2 respectively. The internal resistance of the
the lengths are different battery is
105. The resistance of the filament of an electric 𝐼1 𝑅2 − 𝐼2 𝑅1 𝐼1 𝑅2 + 𝐼2 𝑅1
bulb changes with temperature. If an electric a) b)
𝐼2 − 𝐼1 𝐼1 − 𝐼2
bulb rated 220 volt and 100 watt is connected 𝐼1 𝑅1 + 𝐼2 𝑅2 𝐼1 𝑅1 − 𝐼2 𝑅2
to (220 × .8) 𝑣𝑜𝑙𝑡 sources, then the actual c) d)
𝐼1 − 𝐼2 𝐼2 − 𝐼1
power would be 113. A moving coil galvanometer has 150 equal
a) 100 × 0.8 𝑤𝑎𝑡𝑡 divisions. Its current sensitivity is 10 divisions
b) 100 × (0.8)2 𝑤𝑎𝑡𝑡 per milliampere and voltage sensitivity is 2
c) Between 100 × 0.8 𝑤𝑎𝑡𝑡 and 100 𝑤𝑎𝑡𝑡 divisions per millivolt. In order that each
Between 100 × (0.8)2 𝑤𝑎𝑡𝑡 and 100 × division reads 1V, the resistance in Ohm’s
d)
0.8 𝑤𝑎𝑡𝑡 needed to be connected in series with the coil
106. The maximum power dissipated in an external will be
resistance R, when connected to a cell of emf E a) 103 b) 105 c) 99995 d) 9995
and internal resistance r, will be 114. In voltaic air cell if 5g zinc is consumed, how
𝐸2 𝐸2 𝐸2 𝐸2 many ampere hours shall we get?
a) b) c) d)
𝑟 2𝑟 3𝑟 4𝑟 a) 2.05 b) 8.2
107. A battery of 𝑒. m. f. 10 𝑉 and internal resistance c) 4.1 d) 5 × 5.38 × 10−3
Page|9
115. The total current supplied to the circuit by the 121. A galvanometer of resistance 100Ω is
battery as shown figure 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
122. In the network of resistors shown in the
a) 1A b) 6A c) 4A d) 2A adjoining figure, the equivalent resistance
116. A certain electrical conductor has a square between 𝐴 and 𝐵 is
cross-section, 2.0 mm on side, and is 12 m long. 3 3 3 3 3 3
The resistance between its ends is 0.072Ω. The A B

resistivity of its material is equal to

3 3 3
a) 2.4 × 10−6 Ωm b) 1.2 × 10−6 Ωm 3 3 3

c) 1.2 × 10−8 Ωm d) 2.4 × 10−8 Ωm a) 54 𝑜ℎ𝑚 b) 18 𝑜ℎ𝑚 c) 36 𝑜ℎ𝑚 d) 9 𝑜ℎ𝑚

117. Two resistances when connected in parallel 123. When a charged particle of charge 𝑒 revolves
across a cell of negligible internal resistance in circular orbit of radius 𝑟 with frequency 𝑛,
consume 4 times the power they would then orbital current will be
consume when connected in series. If one 𝑒𝑣 𝑒𝑣 𝑒𝑣 𝑒𝑣
a) 2 b) c) d)
resistance is 5 Ω , the other is 𝜋𝑟 4𝜋𝑟 2𝜋𝑟 4𝜋𝑟 2
a) 1 Ω b) 2.5 Ω c) 5 Ω d) 10 Ω 124. A lead acid accumulatory (storage battery) is
118. A dry cell of emf 1.5 V and internal resistance connected to a battery charge for over night
0.10Ω is connected across a resistor in series charging. Which of the following observations
with a very low resistance ammeter. When the will indicate that the battery was partly
circuit is switched on, the ammeter reading charged during the next morning
settles to a steady rate of 2A. Find (i) chemical a) The density of acid has decreased
energy consumption of the cell (ii) energy b) The density of acid has increased
dissipation inside the cell (iii) energy c) The acid has changed colour
dissipation inside the resistor (iv) power d) The acid level has dropped
output of source is 125. To verify Ohm’s law, a student is provided with
a) (i) 3 W (ii) 0.4 W (iii) 2.6 W (iv) 2.6 W a test resistor 𝑅𝑇 , a high resistance 𝑅1 , a small
b) (i) 0.4 W (ii) 3 W (iii) 2.6 W (iv) 2.6 W resistance 𝑅2 , two identical galvanometers
c) (i) 2.6 W (ii) 0.4 W (iii) 9 W (iv) 1 W 𝐺1 and 𝐺2 and a variable voltage source V. the
d) None of the above correct circuit to carry out the experiment is
119. In the given figure, when key 𝐾 is opened, the a) G1 b) G1

reading of the ammeter 𝐴 will be

R2 R1
10V
G2 G2
+ –
RT RT RT R2
5
E A D

4 A
B C
K V V

10
a) 50 𝐴 b) 2 𝐴 c) 0.5 𝐴 d) 𝐴 c) d) R2
9 G1 G1
120. A cell in secondary circuit gives null deflection
for 2.5𝑚 length of potentiometer having 10𝑚
G2 G2
length of wire. If the length of the
RT RT
potentiometer wire is increased by 1 𝑚
without changing the cell in the primary, the R2 R1

position of the null point now is

a) 3.5 𝑚 b) 3 𝑚 c) 2.75 𝑚 d) 2.0 𝑚
V V

P a g e | 10
126. If a 2 𝑘𝑊 boiler is used everyday for 1 hour, batteries wear out
then electrical energy consumed by boiler in a) The light intensity gets reduced with no
thirty days is change in its colour
a) 15 unit b) 60 unit c) 120 unit d) 240 unit b) Light colour changes first to yellow and then
127. The electric resistance of a certain wire of iron red with no change in intensity
is 𝑅. If its length and radius are both doubled, c) It stops working suddenly while giving white
then light
a) The resistance will be doubled and the d) Colour changes to red and also intensity gets
specific resistance will be halved reduced
b) The resistance will be halved and the 133. Following figure shows cross-sections through
specific resistance will remain unchanged three long conductors of the same length and
c) The resistance will be halved and the material, with square cross-section of edge
specific resistance will be doubled lengths as shown. Conductor 𝐵 will fit snugly
d) The resistance and the specific resistance, within conductor 𝐴, and conductor 𝐶 will fit
will both remain unchanged snugly within conductor 𝐵. Relationship
128. What is the equivalent resistance of the circuit between their end to end resistance is
4V, 1  2
+ – 2
2 A
a

4 A B C

V
a) 𝑅𝐴 = 𝑅𝐵 = 𝑅𝐶
a) 6 Ω b) 7 Ω c) 8 Ω d) 9 Ω b) 𝑅𝐴 > 𝑅𝐵 > 𝑅𝐶
129. When a current of 1 ampere is passed through c) 𝑅𝐴 < 𝑅𝐵 < 𝑅𝐶
a conductor whose ends are maintained at d) Information is not sufficient
temperature difference of 1℃, the amount of 134. The colour code for a resistor of resistance
heat evolved or absorbed is called 3.5𝑘Ω with 5% tolerance is
a) Peltier coefficient a) Orange, green, red and gold
b) Thomson coefficient b) Red, yellow, black and gold
c) Thermoelectric power c) Orange, green, orange and silver
d) Thermo 𝑒.m.f. d) Orange, green, red and silver
130. There is a current of 1.344 𝑎𝑚𝑝 in a copper 135. A wire of length 100 𝑐𝑚 is connected to a cell
wire whose area of cross-section normal to the of 𝑒𝑚𝑓 2 𝑉 and negligible internal resistance.
length of the wire is 1 𝑚𝑚2 . If the number of The resistance of the wire is 3 Ω. The
free electrons per 𝑐𝑚3 is 8.4 × 1022, then the additional resistance required to produce a
drift velocity would be potential drop of 1 𝑚𝑖𝑙𝑙𝑖 𝑣𝑜𝑙𝑡 per 𝑐𝑚 is
a) 1.0 𝑚𝑚/𝑠𝑒𝑐 b) 1.0 𝑚/𝑠𝑒𝑐 a) 60 Ω b) 47 Ω c) 57 Ω d) 35 Ω
c) 0.1 𝑚𝑚/𝑠𝑒𝑐 d) 0.01 𝑚𝑚/𝑠𝑒𝑐 136. The voltage of clouds is 4 × 106 V with respect
131. The equivalent resistance of the following to ground. In a light ning strike lasting 100 ms,
infinite network of resistance is a charge of 4 C is delivered to the ground. The
2 2 2
A power of lightning strike is
2 2 2
a) 160 MW b) 80 MW c) 20 MW d) 500 Kw
137. A coil takes 15 min to boil a certain amount of
2 2 2
B water; another coil takes 20 min for the same
process. Time taken to boil the same amount
a) Less than 4Ω
of water when both coils are connected in
b) 4Ω
series
c) More than 4Ω but less than 12Ω
a) 5 min b) 8.6 min c) 35 min d) 30 min
d) 12Ω
138. A 100 𝑉 voltmeter of internal resistance 20 𝑘Ω
132. Flash light equipped with a new set of
in series with a high resistance 𝑅 is connected
batteries, produces bright white light. As the
P a g e | 11
 to a 110 𝑉 line. The voltmeter reads 5 𝑉, the 146. The temperature at which thermal electric
value of 𝑅 is power of a thermo couple becomes zero is
a) 210 𝑘Ω b) 315 𝑘Ω c) 420 𝑘Ω d) 440 𝑘Ω called
139. The relation between Faraday’s constant 𝐹, a) Inversion temperature
electron charge 𝑒 and avogadro number 𝑁 is b) Neutral temperature
a) 𝐹 = 𝑁/𝑒 b) 𝐹 = 𝑁𝑒 c) 𝑁 = 𝐹 2 d) 𝐹 = 𝑁 2 𝑒 c) Junction temperature
140. A 2V battery, a 990 Ω resistor and a d) Null temperature
potentiometer of 2m length, all are connected 147. A potentiometer wire of length 10 m and
in series of the residence of potentiometer resistance 20 Ω is connected is series with a
wire is 10 Ω, then the potential gradient of the 15V battery and an external resistance 40 Ω. A
potentiometer wire is secondary cell of emf E in the secondary circuit
a) 0.05𝑉𝑚−1 b) 0.5𝑉𝑚−1 is balanced by 240 cm long the potentiometer
c) 0.01𝑉𝑚−1 d) 0.1 𝑉𝑚−1 wire. The emf E of the cell is
141. Three moving coil galvanometers A, B and C a) 2.4V b) 1.2V c) 2.0V d) 3V
are made of coils of three different material 148. A thin wire of resistance 4 Ω is bent to form a
having torsional constant 1.8 × 10−8 , 2.8 × circle. The resistance across any diameter is
10−8 and 3.8 × 10−8 respectively. If the three a) 4 Ω b) 2 Ω c) 1 Ω d) 8 Ω
galvanometers are identical in all other 149. The effective resistance between points 𝐴 and
respect, then in which of the above cases 𝐵 in figure
sensitivity is maximum?
4Ω
a) A b) C
c) B d) Same in each case 6Ω
142. Drift velocity 𝑣𝑑 varies with the intensity of 12 Ω
3Ω 5Ω
electric field as per the relation
A 24 Ω B
1
a) 𝑣𝑑 ∝ 𝐸 b) 𝑣𝑑 ∝
𝐸 a) 10 Ω b) 12 Ω c) 9.85 Ω d) 10.85 Ω
c) 𝑣𝑑 = constant d) 𝑣𝑑 ∝ 𝐸 2
150. Figure shows a network of eight resistors, each
143. For measurement of potential difference,
equal to 2 Ω, connected to a 3V battery of
potentiometer is preferred in comparison to
negligible internal resistance. The current 𝐼 in
voltmeter because
the circuit is
a) Potentiometer is more sensitive than
voltmeter 3V

b) The resistance of potentiometer is less than

I

voltmeter A B C D

c) Potentiometer is cheaper than voltmeter

d) Potentiometer does not take current from
the circuit
E F

144. If a wire of resistance 20Ω is covered with ice a) 0.25A b) 0.50A c) 0.75A d) 1.0A
and a voltage of 210 𝑉 is applied across the 151. The resistance of a wire of uniform diameter 𝑑
wire, then the rate of melting of ice is and length 𝐿 is 𝑅. The resistance of another
a) 0.85𝑔/𝑠 b) 1.92𝑔/𝑠 wire of the same material but diameter 2𝑑 and
c) 6.56𝑔/𝑠 d) All of these length 4 𝐿 will be
145. In a Wheatstone’s bridge, three resistances a) 2𝑅 b) 𝑅 c) 𝑅/2 d) 𝑅/4
𝑃, 𝑄 and 𝑅 connected in the three arms and the 152. The neutral temperature 𝑡𝑛 = 285°C is
fourth arm is formed by two resistances constant for a Cu-Fe thermocouple. When the
𝑆1 and 𝑆2 connected in parallel. The condition cold junction is at 0°C, the value of inversion
for the bridge to be balanced will be temperature is 𝑡𝑖 = 570°C but if the cold
𝑃 2𝑅 𝑃 𝑅(𝑆1 + 𝑆2 )
a) = b) = junction is at 10°C, the inversion temperature
𝑄 𝑆1 + 𝑆2 𝑄 𝑆1 𝑆2 (𝑡𝑖 ) will be
𝑃 𝑅(𝑆1 + 𝑆2 ) 𝑃 𝑅
c) = d) = a) 550°C b) 560°C c) 570°C d) 580°C
𝑄 2𝑆1 𝑆2 𝑄 𝑆1 + 𝑆2
P a g e | 12
153. What is the equivalent resistance between P Q

points 𝐴 and 𝐵 in the circuit if figure, if 𝑅 = S

3 Ω? R G

A B
V
R R
R
a) 𝐼𝑅 = 𝑅𝐺 b) 𝐼𝑃 = 𝐼𝐺 c) 𝐼𝑄 = 𝐼𝐺 d) 𝐼𝑄 = 𝐼𝑅
C D
R R 161. An ammeter gives full scale deflection when
current 1.0 𝐴 is passed in it. To convert it into
a) 8 Ω b) 9 Ω c) 12 Ω d) 15 Ω 10 𝐴 range ammeter, the ratio of its resistance
154. If the resistance of a conductor is 5Ω at 50℃ and the shunt resistance will be
and 7Ω at 100℃ then the mean temperature a) 1 : 9 b) 1 : 10 c) 1 : 11 d) 9 : 1
coefficient of resistance of the material is 162. The cell has an emf of 2V and the internal
a) 0.008/℃ b) 0. .006/℃ resistance of 3.9 Ω, the voltage across the cell
c) 0.004/℃ d) 0.001/℃ will be
155. A 100 W bulb 𝐵1 and two 60 W bulbs a) 1.95 V b) 1.5V c) 2V d) 1.8V
𝐵2 and 𝐵3 are connected to a 250 V source as 163. 𝐼 − 𝑉 characteristic of a copper wire of length
shown in figure. Now 𝑊1 , 𝑊2 and 𝑊3 are the 𝐿 and area of cross-section 𝐴 is shown in
output powers of the bulbs figure. The slope of the curve becomes
𝐵1 , 𝐵2 and 𝐵3 respectively, then

a) More if the experiment is performed at

a) 𝑊1 > 𝑊2 = 𝑊3 b) 𝑊1 > 𝑊2 > 𝑊3 higher temperature
c) 𝑊1 < 𝑊2 = 𝑊3 d) 𝑊1 < 𝑊2 < 𝑊3 b) More if a wire of steel of same dimension is
156. A cell of internal resistance 3 𝑜ℎ𝑚 and used
𝑒𝑚𝑓 10 𝑣𝑜𝑙𝑡 is connected to a uniform wire of c) More if the length of the wire increased
length 500 𝑐𝑚 and resistance 3 𝑜ℎ𝑚. The d) Less if the length of the wire increased
potential gradient in the wire is 164. The effective resistance across the points 𝐴
a) 30 𝑚𝑉/𝑐𝑚 b) 10 𝑚𝑉/𝑐𝑚 and 𝐼 is
c) 20 𝑚𝑉/𝑚 d) 4 𝑚𝑉/𝑐𝑚
157. 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
a) 2 Ω b) 1 Ω c) 0.5 Ω d) 5 Ω
158. A 4𝜇 F conductor is charged to 400 V and then
165. The resistance between the terminal points 𝐴
its plates are joined through a resistance of 1
and 𝐵 of the given infinitely long circuit will be
kΩ. The heat produced in the resistance is 1 1 1
a) 0.18 J b) 0.21 J c) 0.25 J d) 0.32 J A

159. A battery is made by connecting 6 cells each Upto

1 1
having capacity 5 Ah at 1.5V. The battery will infinity
have capacity equal to B
1 1 1
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 a) (√3 − 1) b) (1 − √3) c) (1 + √3) d) (2 + √3)
160. In the circuit shown as 𝑃 ≠ 𝑅 and the reading 166. The power dissipated across resistance 𝑅
of the galvanometer 𝐺 is same with switch which is connected across a battery of
open or closed. Then potential 𝑉 is 𝑃. If resistance is doubled, then

P a g e | 13
 the power becomes a) 9 b) 6 c) 3 d) 1.5
a) 1/2 b) 2 c) 1/4 d) 4 175. Three equal resistors are connected as shown
167. A bulb of 220 V and 300 W is connected across in figure. The maximum power consumed by
110 V circuit. The percentage reduction in each resistor is 18 W. Then maximum power
power is consumed by the combination is
a) 100% b) 25% c) 70% d) 75%
168. A 50 𝑜ℎ𝑚 galvanometer gets full scale
deflection when a current of 0.01 𝐴 passes
a) 18 W b) 27 W c) 36 W d) 54 W
through the coil. When it is converted to a 10 𝐴
176. The 𝑒𝑚𝑓 of a thermocouple, one junction of
ammeter, the shunt resistance is
which is kept at 0℃, is given by 𝑒 = 𝑎𝑡 + 𝑏𝑡 2 .
a) 0.01 Ω b) 0.05 Ω c) 2000 Ω d) 5000 Ω
The Peltier co-efficient will be
169. A 100 𝑊 bulb produces an electric field of
a) (𝑡 + 273)(𝑎 + 2𝑏𝑡) b) (𝑡 + 273)(𝑎 − 2𝑏𝑡)
2.9 𝑉/𝑚 at a point 3 𝑚 away. If the bulb is
c) (𝑡 − 273)(𝑎 − 2𝑏𝑡) d) (𝑡 − 273)(𝑎 + 2𝑏𝑡)
replaced by 400 𝑊 bulb without disturbing
177. Two cells having emf 4V, 2V and internal
other conditions, then the electric field
resistances 1 Ω, 1 Ω are connected as shown in
produced at the same point is
figure below. Current through 6 Ω resistance is
a) 2.9 𝑉/𝑚 b) 3.5 𝑉/𝑚 c) 5 𝑉/𝑚 d) 5.8 𝑉/𝑚
170. Kirchhoff’s first law 𝑖. 𝑒. ∑ 𝑖 = 0 at a junction is 4 V 1Ω 2 V1Ω
based on the law of conservation of
a) Charge b) Energy
c) Momentum d) Angular momentum 3 Ω40 Ω
171. At steady state, energy stored in capacitor is 2Ω
2 F
6Ω
A B
1 2 2
a) 𝐴 b) 𝐴 c) 1A d) 𝐴
2Ω 3 3 9
178. If the cold junction is held at 0℃, the same
thermo-emf 𝑉 of a thermocouple varies as 𝑉 =
2V 1
10 × 10−6 𝑡 − × 10−6 𝑡 2 , where 𝑡 is the
40
a) 4 × 10−6 J b) 2 J temperature of the hot junction in ℃. The
c) 4 J d) Zero neutral temperature and the maximum value
172. In a meter bridge experiment, the ratio of the of thermo-emf are respectively
left gap resistance to right gap resistance is a) 200℃; 2 mV b) 400℃; 2 mV
2:3, the balance point from left is c) 100℃; 1 mV d) 200℃; 1 mV
a) 60 cm b) 50 cm c) 40 cm d) 20 cm 179. Potentiometer wire of length 1m is connected
173. When two identical batteries of internal in series with 490Ωresistance and 2V battery.
resistance 1 Ω each are connected in series If 0.2m Vcm−1 is the potential gradient, then
across a resistor 𝑅, the rate of heat produced resistance of the potentiometer wire is
in 𝑅 is 𝐽1 . When the same batteries are a) 4.9Ω b) 7.9Ω c) 5.9Ω d) 6.9Ω
connected I parallel across 𝑅, the rate is 𝐽2 . If 180. A storage cell is charged by 5 𝑎𝑚𝑝 D.C. for
𝐽1 = 2.25 𝐽2 then the value of 𝑅 in Ω is 18 ℎ𝑜𝑢𝑟𝑠. Its strength after charging will be
a) 4 b) 6 c) 4.8 d) 5.16 a) 18 𝐴𝐻 b) 5 𝐴𝐻 c) 90 𝐴𝐻 d) 15 𝐴𝐻
174. A galvanometer, having a resistance of 181. The ammeter 𝐴 reads 2 𝐴 and the voltmeter 𝑉
50 Ω, gives a full scale deflection for a current reads 20 𝑉. The value of resistance 𝑅 is
of 0.05A. The length in meter of a resistance (Assuming finite resistance’s of ammeter and
wire of area of cross-section 2.97 × 10−2 cm2 voltmeter)
that can be used to convert the galvanometer R
A
into a ammeter which can read a maximum of
5A current is
(Specific resistance of the wire = 5 × V
10−7 Ωm) a) Exactly 10 𝑜ℎ𝑚
P a g e | 14
 b) Less than 10 𝑜ℎ𝑚 189. The resistance of a galvanometer is 90 𝑜ℎ𝑚. If
c) More than 10 𝑜ℎ𝑚 only 10 percent of the main current may flow
d) We cannot definitely say through the galvanometer, in which way and of
182. In the circuit shown in the figure, the current what value, a resistor is to be used
through a) 10 𝑜ℎ𝑚 in series b) 10 𝑜ℎ𝑚 in parallel
3Ω 2Ω 2Ω c) 810 𝑜ℎ𝑚 in series d) 810 𝑜ℎ𝑚 in parallel
190. If a high power heater is connected to electric
9V 8Ω 8Ω 4Ω mains, then the bulbs in the house become dim,
because there is a
2Ω 2Ω 2Ω a) Current drop b) Potential drop
c) No current drop d) No potential drop
a) The 3Ω resistor is 0.50 𝐴
191. In the given circuit the current 𝐼1 is
b) The 3Ω resistor is 0.25 𝐴
30 
c) The 4Ω resistor is 0.50 𝐴
d) The 4Ω resistor is 0.25 𝐴 I1
40 
183. If current in an electric bulb changes by 1%,
I3
then the power will change by I2
40 
40V

1
a) 1% b) 2% c) 4% d) % 80V
2
184. The internal resistance of a cell of e.m.f. 12𝑉 is a) 0.4 𝐴 b) −0.4 𝐴 c) 0.8 𝐴 d) −0.8 𝐴
5 × 10−2 Ω. It is connected across an unknown 192. Which of the following graphs shows the
resistance. Voltage across the cell, when a variation of thermoelectric power with
current of 60 𝐴 is drawn from it, is temperature difference between hot and cold
a) 15 𝑉 b) 12 𝑉 c) 9 𝑉 d) 6 𝑉 junction in thermocouples
185. 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 a) b)
potential gradient along the wire is
(0.1 mVcm−1 ), the resistance unplugged in the  

box is
a) 260 Ω b) 760 Ω c) 960 Ω d) 1060 Ω
186. The electromotive force of a primary cell is c) d)
2 𝑣𝑜𝑙𝑡. When it is short-circuited it gives a
current of 4 𝑎𝑚𝑝𝑒𝑟𝑒. Its internal resistance in  

𝑜ℎ𝑚 is 193. Two tangent galvanometer A and B are

a) 0.5 b) 5.0 c) 2.0 d) 8.0 identical except in their number of turns. They
187. The two bulbs, one of 60W and other 200W are are connected in series. On passing a current
connected in series to a 200 volt line, then through them, deflections of 60° and 30° are
a) The potential drop across two bulbs in the produced. The ratio of the number of units A
same and B is
b) The potential drop across the 60 W bulb is a) 1:3 b) 3:1 c) 1:2 d) 2:1
greater than the potential drop across the 194. A battery is charged at a potential of 15 𝑉 in 8
200 W bulb hours when the current flowing is 10 𝐴. The
c) The potential drop across the 200 W bulb is battery on discharge supplies a current of 5 𝐴
greater than the 60 W bulb for 15 hours. The mean terminal voltage
d) The potential drop across both the bulbs is during discharge is 14 𝑉. The "𝑊𝑎𝑡𝑡 − ℎ𝑜𝑢𝑟"
200 volt efficiency of battery is
188. The relation between Faraday constant (𝐹), a) 80% b) 90% c) 87.5% d) 82.5%
chemical equivalent (𝐸) and electrochemical 195. An electrical cable having a resistance of 0.2 Ω
equivalent (𝑍) is delivers 10 kW at 200 𝑉 DC to a factory. What
𝑍 𝐸 𝐸 is the efficiency of transmission
a) 𝐹 = 𝐸𝑍 b) 𝐹 = c) 𝐹 = d) 𝐹 = 2
𝐸 𝑍 𝑍
P a g e | 15
 a) 65% b) 75% c) 85% d) 95% 2Ω
196. Find the equivalent resistance across 𝐴𝐵
A
2 3Ω
2
2
2 𝐶 = 0.2 μF 4Ω
2
B
a) 1 Ω b) 2 Ω c) 3 Ω d) 4 Ω
6V 2.8Ω
197. 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 a) Zero b) 0.6A c) 0.9A d) 1.5A
resistance of the wire will be 203. 𝑂ℎ𝑚′𝑠 law is true
a) Equal b) One fourth a) For metallic conductors at low temperature
c) One eight 1 b) For metallic conductors at high temperature
d) 𝑡ℎ c) For electrolytes when current passes
16
198. When a battery connected across a resistor of through them
16 Ω, the voltage across the resistor is 12 V. d) For diode when current flows
When the same battery is connected across a 204. What will happen when a 40 𝑤𝑎𝑡𝑡, 220 𝑣𝑜𝑙𝑡
resistor of 10 Ω, voltage across it is 11V. The lamp and 100 𝑤𝑎𝑡𝑡, 220 𝑣𝑜𝑙𝑡 lamp are
internal resistance of the battery (in ohm) is connected in series across 40 𝑣𝑜𝑙𝑡 supply
10 20 25 30 a) 100 𝑤𝑎𝑡𝑡 lamp will fuse
a) b) c) d)
7 7 7 7 b) 40 𝑤𝑎𝑡𝑡 lamp will fuse
199. When the temperature difference between hot c) Both lamps will fuse
and cold junctions of a thermo-couple is 100 K d) Neither lamp will fuse
an emf of 1 V is generated. Assume the cold 205. Two similar accumulators each of emf E and
junction is heated by 20 K, the percentage internal resistance r are connected as shown in
change in thermo emf is the following figure. Then, the potential
a) 20% b) 30% c) 40% d) 25% difference between x and y is
200. The current in the 1Ω resistor shown in the
circuit is
X y
| |
4Ω

1Ω 6V
a) 2E b) E
4Ω
c) Zero d) None of these
206. Electric power is transmitted over long
2 distances through conducting wires at high
a) 𝐴 b) 3A c) 6A d) 2A
3 voltage because
201. In the arrangement shown in figure, the a) High voltage travels faster
current through 5Ω resistor is b) Power loss is large
2Ω c) Power loss is less
2Ω
12V 5Ω 12V d) Generator produce electrical energy at a
very high voltage
12 207. Given figure shows a rectangular block with
a) 2A b) Zero c) A d) 1A
7 dimensions 𝑥, 2𝑥 and 4𝑥. Electrical contacts
202. In the given figure the steady state current in can be made to the block between opposite
the circuit is 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 −

P a g e | 16
 𝐶 : Front and rear faces) 214. 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
R

R1
a) 𝐴 − 𝐴 R2
E
b) 𝐵 − 𝐵
E1
c) 𝐶 − 𝐶
d) Same for all three pairs
208. A milli voltmeter of 25 milli volt range is to be a) 𝐸𝑅1 /𝑅2 b) 𝐸𝑅2 /𝑅1
converted into an ammeter to 25 ampere c) 𝐸(𝑅1 + 𝑅2 )/𝑅2 d) 𝐸𝑅1 /(𝑅1 + 𝑅2 )
range. The value (in ohm) of necessary shunt 215. In the above question if potential difference is
will be applied, the drift velocity at temperature 𝑇 is
a) 0.001 b) 0.01 c) 1 d) 0.05 a) Inversely proportional to 𝑇
209. The resistance across 𝐴 and 𝐵in the figure b) Proportional to √𝑇
below will be c) Zero
d) Finite but independent of 𝑇
216. A battery of 𝑒.m.f. 3 volt and internal
A B
resistance 1.0 𝑜ℎ𝑚 is connected in series with
R
R R copper voltmeter. The current flowing in the
circuit is 1.5 amperes. The resistance of
a) 3𝑅 b) 𝑅
voltmeter will be
𝑅 d) None of these
c) a) Zero b) 1.0 𝑜ℎ𝑚 c) 1.5 𝑜ℎ𝑚 d) 2.0 𝑜ℎ𝑚
3
210. A thermocouple develops 40 𝜇𝑉/𝑘𝑒𝑙𝑣𝑖𝑛. If hot 217. The electrolyte used in Lechlanche cell is
and cold junctions are at 40℃ and 20℃ a) Copper sulphate solution
respectively, then then emf developed by a b) Ammonium chloride solution
thermopile using such 150 thermocouples in c) Dilute sulphuric acid
series shall be d) Zinc sulphate
a) 150𝑚𝑉 b) 80𝑚𝑉 c) 144𝑚𝑉 d) 120𝑚𝑉 218. The wiring of a house has resistance 6 Ω. A 100
211. The emf of the battery shown in figure, is W bulb is glowing as shown in figure. If a
geyser of 1000 W is switched on, the change in
2Ω 2Ω 1Ω
potential drop across the bulb is nearly

E 4Ω 2Ω 1Ω
1A
a) Nil b) 12 V c) 24 V d) 32 V
a) 12 V b) 13 V c) 16 V d) 18 V
219. In above question, if length is doubled, the drift
212. Voltmeters 𝑉1and 𝑉2 are connected in series
velocity
across a DC line. 𝑉1reads 80V and has a
a) Is doubled b) Is halved
resistance of 200ΩV −1 and V2 has a total
c) Remains same d) Becomes zero
resistance of 32k Ω. The line voltage is
220. Four resistances are connected in a circuit in
a) 240 V b) 220 V c) 160 V d) 120 V
the given figure. The electric current flowing
213. The electro chemical equivalent of metal is
through 4 𝑜ℎ𝑚 and 6 𝑜ℎ𝑚 resistance is
3.3 × 10−7 kgC −1 . The mass of the metal
respectively
liberated at the cathode 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

P a g e | 17
 4 6
I A C
4 6
B

V
20V
a) 𝐴 b) 𝐵
a) 2 𝑎𝑚𝑝 and 4 𝑎𝑚𝑝 b) 1 𝑎𝑚𝑝 and 2 𝑎𝑚𝑝
c) 𝐶 d) None of these
c) 1 𝑎𝑚𝑝 and 1 𝑎𝑚𝑝 d) 2 𝑎𝑚𝑝 and 2 𝑎𝑚𝑝
227. The current in the given circuit is
221. An aluminium (resistivity𝜌 = 2.2 × 10−8 Ω −
m) wire of a diameter 1.4 mm is used to make
a 4Ω esistor. The length of the wire is RA = 3 RB = 6
4.8V
a) 220 m b) 1000 m c) 280 m d) 1 m
222. 𝐵1 , 𝐵2 and 𝐵3 are the three identical bulbs RC = 6
connected to a battery of steady emf with key
a) 8.31 𝐴 b) 6.82 𝐴 c) 4.92 𝐴 d) 2 𝐴
𝐾 closed. What happens to the brightness of
228. In a meter bridge experiment, null point is
the bulbs, 𝐵1 and 𝐵2 when the key is opened?
obtained at 20cm from one end of the wire
B1
when resistance X is balanced against another
resistance Y. If X<Y, then where will be the
K B2 new position of the null point from the same
end, if one decides to, balance a resistance of
B3 4X against Y?
a) 50 cm b) 80 cm c) 40 cm d) 70 cm
229. In the circuit shown in the figure, the current
Brightness of the bulb 𝐵1 increases and that
a) flowing in 2 Ω resistance
of 𝐵2 decreases
b) Brightness of the bulbs 𝐵1 and 𝐵2 increases 10 2
Brightness of the bulb 𝐵1 decreases and 𝐵2
c) 1.4A
G
increases
d) Brightness of the bulbs 𝐵1 and 𝐵2 decreases 25 5
223. In a potentiometer arrangement, a cell of emf
1.5V gives a balance point at 27cm length of a) 1.4 𝐴 b) 1.2 𝐴 c) 0.4 𝐴 d) 1.0 𝐴
wire. If the cell is replaced by another cell and 230. In the figure shown, the total resistance
balance point shifts to 54cm, the emf of the between 𝐴 and 𝐵 is
second cell is 2 C 1 1 1 1 1
a) 3V b) 1.5V c) 0.75V d) 2.25V A
224. If 1 A current is passed through CuSO4 solution 8 8 4

for 10 s, the number of copper atoms deposited

at the cathode will be
B 2 D 1 1 1 1 1

a) 8 × 1019 b) 3.1 × 1019 a) 12 Ω b) 4 Ω c) 6 Ω d) 8 Ω

c) 6.2 × 10 19 d) 1.6 × 1020 231. A voltmeter having a resistance of 998 𝑜ℎ𝑚 is
225. The current in the arm 𝐶𝐷 of the circuit will be connected to a cell of e.m.f. 2 𝑣𝑜𝑙𝑡 and internal
B resistance 2 𝑜ℎ𝑚. The error in the
measurement of e.m.f. will be
i2
i1 a) 4 × 10−1 𝑣𝑜𝑙𝑡 b) 2 × 10−3 𝑣𝑜𝑙𝑡
O A −3
c) 4 × 10 𝑣𝑜𝑙𝑡 d) 2 × 10−1 𝑣𝑜𝑙𝑡
i3
232. When a current is passed in a conductor, 3℃
C
rise in temperature is observed. If the strength
D of current is increased by two times, then rise
a) 𝑖1 + 𝑖2 b) 𝑖2 + 𝑖3 in temperature will approximately be
c) 𝑖1 + 𝑖3 d) 𝑖1 − 𝑖2 + 𝑖3 a) 36℃ b) 27℃ c) 18℃ d) 9℃
226. Resistance as shown in figure is negative at 233. Two bulbs 𝑋 and 𝑌 having same voltage rating

P a g e | 18
 and of power 40 W and 60 W respectively are 5 10 15
P
connected in series across a potential
difference of 300 V, then A 10 10 B

Q
10 20 30

a) 20 Ω b) 30 Ω c) 90 Ω d) 110 Ω
240. Four identical resistors of 4 Ω each are joined
a) 𝑋 will glow brighter in circuit as shown in figure. The cell 𝐵 has emf
b) Resistance of 𝑌 will be greater than 𝑋 2V and its internal resistance is negligible. The
c) Heat produced in 𝑌 will be greater than 𝑋 ammeter reading is
d) Voltage drop in 𝑋 will be greater than 𝑌
4Ω
234. An electric heater of 1.08 Kw is immersed in
water. After the water has reached a B
temperature of 100℃, how much time will be + -
A
required to produce 100 g of steam? 4Ω
4Ω
a) 420 s b) 210 s c) 105 s d) 50 s
235. An electric kettle takes 4 𝐴 current at 220 𝑉.
4Ω
How much time will it take to boil 1 𝑘𝑔 of
water from room temperature 20℃? The 3 1 1
a) A b) 2A c) A d) A
temperature of boiling water is 100℃ 8 2 8
a) 0.63 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 b) 6.3 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 241. In producing chlorine through electrolysis
c) 12.6 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 d) 12.8 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 100 𝑤𝑎𝑡𝑡 power at 125 𝑉 is being consumed.
236. In the circuit shown in figure the heat How much chlorine per minute is liberated?
produced in the 5Ω resistor due to the current E.C.E. chlorine is 0.367 × 10−6 𝑘𝑔/𝑐𝑜𝑢𝑙𝑜𝑚𝑏
flowing through it is 100Js −1 .The heat a) 24.3 𝑚𝑔 b) 16.6 𝑚𝑔 c) 17.6 𝑚𝑔 d) 21.3 𝑚𝑔
generated in the 4Ω resistor is 242. Two resistance of 10 Ω and 20 Ω and an
inductor of inductance 5 𝐻 are connected to a
4Ω 6Ω battery of 2 𝑉 through a key 𝑘 as shown in the
i
5Ω
i figure. At time 𝑡 = 0, when the key 𝑘 is closed
the initial current through the battery is
a) 10 Js −1 b) 20 Js −1 c) 30 Js −1 d) 40 Js −1
237. A galvanometer can be converted into a
voltmeter by connecting
a) Low residence in parallel
b) Low residence in series
c) High residence in parallel 2 1
d) High residence in series a) 0.2 𝐴 𝐴 b) c) 𝐴 d) 0
15 15
238. The resistivity of a wire 243. Thirteen resistances each of resistance RΩ are
a) Increase with the length of the wire connected in the circuit as shown in the figure.
b) Decreases with the area of cross-section The effective resistance between points A and
c) Decreases with the length and increases B is
with the cross-section of wire R
A
R

d) None of the above statement is correct

239. In the arrangement of resistance shown below,
the effective resistance between points 𝐴 and
R
R R R

𝐵 is R

R
R R R

R R

P a g e | 19
 4𝑅 2 a) b)
a) Ω b) 2RΩ c) RΩ d) 𝑅Ω E E
3 3
244. A cylindrical conductor has uniform cross-
section. Resistivity of its material increases
t t
linearly from left end to right end. If a constant
current is flowing through it and at a section c) d) E
E
distance 𝑥 from left end, magnitude of electric
field intensity is 𝐸, which of the following
graphs is correct t
t
a) E b) E
251. An ammeter with internal resistance 90Ω
reads 1.85 𝐴 when connected in a circuit
containing a battery and two resistors 700Ω
O x O x and 410Ω in series. Actual current will be
a) 1.85 𝐴 b) Greater than 1.85 𝐴
c) E d) c) Less than 1.85 𝐴 d) None of these
E
252. 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
O x O x primary circuit. The null point corresponding
𝐿
245. Two identical cell send the same current in 2Ω to a cell of e.m.f. 1𝑉 is obtained at a distance .
2
resistance, whether connected in series or in If the resistance in the primary circuit is
parallel. The internal resistance of the cell
doubled then the position of new null point
should be will be
1
a) 1 Ω b) 2 Ω c) Ω d) 2.5 Ω a) 0.4 𝐿 b) 0.5 𝐿 c) 0.6 𝐿 d) 0.8 𝐿
2
253. If an increase in length of copper wire is 0.5%
246. When current flows through a conductor, then
due to stretching, the percentage increase in
the order of drift velocity of electrons will be
its resistance will be
a) 1010 𝑚/𝑠𝑒𝑐 b) 10−2 𝑐𝑚/𝑠𝑒𝑐
a) 0.1% b) 0.2% c) 1% d) 2%
c) 104 𝑐𝑚/𝑠𝑒𝑐 d) 10−1 𝑐𝑚/𝑠𝑒𝑐
254. If the potential difference across the internal
247. A bulb rated at (100𝑊 − 200𝑉) is used on a
resistance 𝑟1 is equal to the emf E of the
100𝑉 line. The current in the bulb is
1 1 battery, then
a) 𝑎𝑚𝑝 b) 4 𝑎𝑚𝑝 c) 𝑎𝑚𝑝 d) 2 𝑎𝑚𝑝 -
------- ----------
- -
------- ----------
r -
4 2 -
-
-
-
E r
1
-
-
-
-
- E
-
-
-
2
-
-
-
-
248. 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

i

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 R

249. On increasing the temperature of a conductor, 𝑟1

its resistance increases because the a) 𝑅 = 𝑟1 + 𝑟2 b) 𝑅 =
𝑟2
a) Relaxation time increases 𝑟2
c) 𝑅 = 𝑟1 − 𝑟2 d) 𝑅 =
b) Mass of electron increases 𝑟1
c) Electron density decreases 255. A wire has a resistance of 6Ω. It is cut into two
d) Relaxation time decreases parts and both half values are connected in
250. Two different metals are joined end to end. parallel. The new resistance is
One end is kept at constant temperature and a) 3Ω b) 6Ω c) 12Ω d) 1.5Ω
the other end is heated to a very high 256. What determines the emf between the two
temperature. The high depicting the thermo metals placed in an electrolyte?
𝑒.m.f. is a) Relative position of metals in the electro
chemical series
P a g e | 20
 b) Distance between them 𝑚 and relaxation time 𝜏) in a metal is
c) Strength of electrolyte proportional to
d) Nature of electrolyte 𝑒 𝑚 𝑒 𝑚
a) 𝜏 b) 𝜏 c) d)
257. A fuse wire with radius 1 mm blows at 1.5 A. 𝑚 𝑒 𝑚𝜏 𝑒𝜏
The radius of the fuse wire of the same 264. If 𝑅1 and 𝑅2 are respectively the filament
material to blow at 3 A will be resistances of a 200 𝑤𝑎𝑡𝑡 bulb and 100 𝑤𝑎𝑡𝑡
a) 31/4 mm b) 41/3 mm c) 31/2 mm d) 21/3 mm bulb designed to operate on the same voltage,
then
258. In the given figure, the equivalent resistance
a) 𝑅1 is two times 𝑅2 b) 𝑅2 is two times 𝑅1
between the points 𝐴 and 𝐵 is
R2 = 4  c) 𝑅2 is four times 𝑅1 d) 𝑅1 is four times 𝑅2
265. In the circuit shown below 𝐸1 = 4.0 𝑉, 𝑅1 =
R1 = 2 R4 = 2  2 Ω, 𝐸2 = 6.0 𝑉, 𝑅2 = 4 Ω and 𝑅3 = 2 Ω. The
A B current 𝐼1 is
R3 = 4  R1 = 2 Ω

E1 = 4 V
a) 8 Ω b) 6 Ω c) 4 Ω d) 2 Ω I1
R3 = 2 Ω
259. The amount of heat generated in 500Ω
resistance, when the key is thrown over from
I2
contact 1 to 2, as shown in figure is R2 = 4 Ω

5 F E2 = 6 V
a) 1.6 𝐴 b) 1.8 𝐴 c) 1.25 𝐴 d) 1.0 𝐴
330 Ω 266. In the following Wheatstone bridge 𝑃/𝑄 =
500 Ω 𝑅/𝑆. If key 𝐾 is closed, then the galvanometer
1
2 will show deflection
P Q

E = 200 V K

R S
a) 10℃ b) 7.5℃ c) 5.0℃ d) 2.5℃
260. A current 𝑖 passes through a wire of length 𝑙,
radius of cross-section 𝑟 and resistivity 𝜌. The a) In left side b) In right side
rate of heat generation is c) No deflection d) In either side
𝑖 2 𝑙𝜌 𝑙𝜌 2 2 267. The current in a simple series circuit is 5.0.A.
a) b) 𝑖 2 ( ) c) 𝑖 𝑙 𝜌/𝑟 d) 𝑖𝑙 𝜌/𝑟
𝜋𝑟 2 𝜋𝑟 2 when an additional resistance of 2.0 Ω is
261. The current flowing in a coil of resistance 90 Ω inserted, the current drops to 4.0 A. the
is to be reduced by 90%. What value of original resistance of the circuit in ohm was
resistance should be connected in parallel with a) 1.25 b) 8 c) 10 d) 20
it 268. In ballistic galvanometer, the frame in which
a) 9 Ω b) 90 Ω c) 1000 Ω d) 10 Ω the coil is wound is non-metallic to
262. In the circuit shown, the current through the a) Avoid the production of induced emf
4 Ω resistor is 1 amp when the points 𝑃 and 𝑀 b) Avoid the production of eddy currents
are connected to a d.c. voltage source. The c) Increase the production of eddy currents
potential difference between the points 𝑀 and d) Increase the production of induced emf
𝑁 is 269. 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
a) 0.5 𝑉 b) 3.2 𝑉 c) 1.5 𝑉 d) 1.0 𝑉 c) The drift velocity will decrease
d) The drift velocity will increase
263. The mobility of free electrons (charge 𝑒, mass
270. The temperature of cold junction of thermo-
P a g e | 21
 couple is 0℃. If the neutral temperature is the power dissipated in 𝑅𝐿 will decrease by a
270℃, then the inversion temperature is factor of 9
a) 540℃ b) 520℃ c) 640℃ d) 580℃ 278. Two electric bulbs marked 40 W, 220 V and 60
271. A current of 2.0 𝑎𝑚𝑝𝑒𝑟𝑒 passes through a cell W, 220 V when connected in series, across
of e.m.f. 1.5 𝑣𝑜𝑙𝑡 having internal resistance of same voltage supply of 220 V, the effective
0.15 𝑜ℎ𝑚. The potential difference measured, power is 𝑃1 and when connected in parallel the
in 𝑣𝑜𝑙𝑡, across both the ends of the cell will be 𝑃
effective power is 𝑃2 . Then 𝑃1 is
2
a) 1.35 b) 1.50 c) 1.00 d) 1.20
a) 0.5 b) 0.48 c) 0.24 d) 0.16
272. A potentiometer has uniform potential
279. When the length and area of cross-section both
gradient across it. Two cells connected in
are doubled, then its resistance
series (i) to support each other and (ii) to
a) Will become half
oppose each other are balanced over 6𝑚 and
b) Will be doubled
2𝑚 respectively on the potentiometer wire.
c) Will remain the same
The e.m.f.'s of the cells are in the ratio of
d) Will become four times
a) 1 : 2 b) 1 : 1 c) 3 : 1 d) 2 : 1
280. A wire of a certain material is stretched slowly
273. The current flowing in a copper voltmeter is
by ten percent. Its new resistance and specific
3.2 𝐴. The number of copper ions (𝐶𝑢2+ )
resistance become respectively
deposited at the cathode per minute is
a) Both remain the b) 1.1 times, 1.1 times
a) 0.5 × 1020 b) 1.5 × 1020
same
c) 3 × 1020 d) 6 × 1020
c) 1.2 times , 1.1 times d) 1.21 times, same
274. If two electric bulbs have 40 𝑊 and 60 𝑊
281. The dimensions of 1 𝜀𝑜 𝐸 2 (𝜀𝑜 :permittivity of
rating at 220 𝑉, then the ratio of their 2
resistances will be free space; 𝐸: electric field) is
a) 9 :4 b) 4 :3 c) 3 :8 d) 3 :2 a) [MLT] b) [ML2 T −2 ]
275. For a certain thermocouple, if the temperature c) [ML−1 T−2 ] d) [ML2 T −1 ]
of the cold junction is 0℃, the neutral 282. A voltmeter essentially consists of
temperature and inversion temperature are a) A high resistance, in series with a
285℃ and 570℃ respectively. If the cold galvanometer
junction is brought to 10℃, then the new b) A low resistance, in series with a
neutral and inversion temperatures are galvanometer
respectively c) A high resistance in parallel with a
a) 285℃ and 560℃ b) 285℃ and 570℃ galvanometer
c) 295℃ and 560℃ d) 275℃ and 560℃ d) A low resistance in parallel with a
276. Resistance in the two gaps of a 𝑚𝑒𝑡𝑒𝑟 bridge galvanometer
are 10 𝑜ℎ𝑚 and 30 𝑜ℎ𝑚 respectively. If the 283. There are two electric bulbs of 40 𝑊 and
resistances are interchanged the balance point 100 𝑊. Which one will be brighter when first
shifts by connected in series and then in parallel
a) 33.3 𝑐𝑚 b) 66.67 𝑐𝑚c) 25 𝑐𝑚 d) 50 𝑐𝑚 a) 40 𝑊 in series and 100 𝑊 in parallel
277. For the circuit shown in the figure b) 100 𝑊 in series and 40 𝑊 in parallel
40 𝑊 both in series and parallel will be
c)
uniform
100 𝑊 both in series and parallel will be
2k Ω d)
R1 uniform
I 24 V
284. An energy source will supply a constant
current into, the load, if its internal resistance
24 V 6k Ω R2 RL 1.5k Ω
is
a) Equal to the resistance of the load
a) The current 𝐼 through the battery is 7.5mA b) Very large as compared to the load
b) The potential difference across 𝑅𝐿 is 18 V resistance
c) Ratio of powers dissipated in 𝑅1 and 𝑅2 is 3 c) Zero
d) If 𝑅1 and 𝑅2 are interchanged, magnitude of d) Non-zero but less than the resistance of the
P a g e | 22
 load between 𝑎𝑏 will be
285. The figure shows a network of currents. The a

magnitude of current is shown here. The A B

current I will be c d

1A X Y
b
10 A I

6A a) From 𝑎 to 𝑏
b) From 𝑏 to 𝑎
2A
c) From 𝑏 to 𝑎 through 𝑐
a) 3A b) 9A c) 13A d) 19A d) From 𝑎 to 𝑏 through 𝑐
286. The length of a conductor is doubled and its 293. For obtaining chlorine by electrolysis a current
radius is halved, its specific resistance is of 100 KW and 125 V is used. (Electro chemical
a) Unchanged b) Halved equivalent of chlorine is 0.367 × 106 kgC −1 ).
c) Doubled d) Quadrupled The amount of chlorine obtained in one min
287. By increasing the temperature, the specific will be
resistance of a conductor and a semiconductor a) 1.7616 g b) 17.616 g
a) Increases for both b) Decreases for both c) 0.17161 kg d) 1.7616 kg
c) Increases, decreases d) Decreases, increases 294. Certain wire has resistance of 10Ω. If its is
288. A wire 100𝑐𝑚 long and 2.0 𝑚𝑚 diameter has a stretched by 1/10th of its length, then its
resistance of 0.7 𝑜ℎ𝑚, the electrical resistivity resistance is nearly
of the material is a) 9 Ω b) 10 Ω c) 11 Ω d) 12 Ω
a) 4.4 × 10−6 𝑜ℎ𝑚 × 𝑚 295. Two uniform wires 𝐴 and 𝐵 are of the same
b) 2.2 × 10−6 𝑜ℎ𝑚 × 𝑚 metal and have equal masses. The radius of
c) 1.1 × 10−6 𝑜ℎ𝑚 × 𝑚 wire 𝐴 is twice that of wire 𝐵. The total
d) 0.22 × 10−6 𝑜ℎ𝑚 × 𝑚 resistance of 𝐴 and 𝐵 when connected in
289. Constant current is flowing through a linear parallel is
conductor of non-uniform area of cross- a) 4 Ω when the resistance of wire 𝐴 is 4.25 Ω
section. The charge flowing per second b) 5 Ω when the resistance of wire 𝐴 is 4.25 Ω
through the area of conductor at any cross- c) 4 Ω when the resistance of wire 𝐵 is 4.25 Ω
section is d) 5 Ω when the resistance of wire 𝐵 is 4.25 Ω
a) Proportional to the area of cross- section 296. When a copper voltmeter is connected with a
b) Inversely proportional to the area of cross- battery of emf 12V, 2 g of copper is deposited
section in 30 min. If the same voltmeter is connected
c) Independent of the area of cross-section across 6 V battery, the mass of copper
d) Dependent on the length of conductor deposited in 45 min would be
290. Two identical heaters rated a) 1 g b) 1.5 g c) 2 g d) 2.5 g
220 𝑣𝑜𝑙𝑡, 1000 watt are placed in series with 297. A 50V battery is connected across a 10 Ω
each other across 220 𝑣𝑜𝑙𝑡 lines. If resistance resistor and a current of 4.5 A flows. The
does not change with temperature, then the internal resistance of the battery is
combined power is a) 10 Ω b) 0.5 Ω c) 1.1 Ω d) 5 Ω
a) 1000 𝑤𝑎𝑡𝑡 b) 2000 𝑤𝑎𝑡𝑡 298. What is the ratio of heat generated in 𝑅 and 2𝑅
c) 500 𝑤𝑎𝑡𝑡 d) 4000 𝑤𝑎𝑡𝑡
291. 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℃ a) 2 :1
292. In the Wheatstone’s bridge (shown in figure) b) 1 :2
𝑋 = 𝑌 and 𝐴 > 𝐵. The direction of the current c) 4 :1
P a g e | 23
 d) 1 :4
299. An electric kettle boils some water in 16 min. I E
Due to some defect, it becomes necessary to C
B
remove 10% turns of heating coil of the kettle. D
Now, how much time will it take to boil the A
V

same of water? a) 𝐴𝐵 b) 𝐵𝐶 c) 𝐶𝐷 d) 𝐷𝐸
a) 17.7 min b) 14.4 min c) 20.9 min d) 13.7 min 307. Water of volume 2 𝑙𝑖𝑡𝑟𝑒 in a container is
300. The temperature of the cold junction of a heated with a coil of 1 𝑘𝑊 at 27℃. The lid of
thermocouple is 0℃ and the temperature of the container is open and energy dissipates at
the hot junction is 𝑇℃. The emf is 𝐸 = 16𝑇 − rate of 160 𝐽/𝑠. In how much time temperature
0.04𝑇 2 𝜇V. The inversion temperature 𝑇𝑖 is will rise from 27℃ to 77℃ [Given specific heat
a) 200℃ b) 400℃ c) 100℃ d) 300℃ of water is 4.2 𝑘𝐽/𝑘𝑔]
301. In a 𝐴𝑔 voltameter 2.68 𝑔 of silver is deposited a) 8 𝑚𝑖𝑛 20 𝑠 b) 6 𝑚𝑖𝑛 2 s
in 10 𝑚𝑖𝑛. The heat developed in 20Ω resistor c) 7 𝑚𝑖𝑛 d) 14 𝑚𝑖𝑛
during the same period will be 308. Two conductors made of the same material are
20 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
a) 192 𝑘𝐽 b) 192 𝐽 c) 200 𝐽 d) 132 𝑘𝐽 309. Bulb 𝐵1 (100 W-250 V) and bulb 𝐵2 (100 W-
302. A moving coil galvanometer of resistance 200 V) are connected across 250 V. What is
100Ωis used as an ammeter using a resistance potential drop across 𝐵2 ?
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 a) 200 V b) 250 V c) 98 V d) 48 V
303. The potential difference across the terminals 310. Resistors 𝑃 and 𝑄 are connected in the gaps of
of a battery is 50𝑉 when 11𝐴 current is drawn the meter bridge. The balancing point is
1
and 60𝑉 when 1𝐴 current is drawn. The obtained 3 m from the zero end. If a 6 Ω
𝑒. 𝑚. 𝑓. and the internal resistance of the
resistance is connected in series with 𝑃 the
battery are 2
balance point shifts to m from the same end.
a) 62𝑉, 2Ω b) 63𝑉, 1Ω c) 61𝑉, 1Ω d) 64𝑉, 2Ω 3
304. 5 cells, each of emf 0.2𝑉 and internal 𝑃 and 𝑄 are
resistance 1Ω are connected to an external a) 4,2
circuit of resistance of 10Ω. Find the current b) 2,4
through external circuit c) Both (a) and (b)
1 1 1 1 d) Neither (a) nor (b)
a) 𝐴 b) 𝐴 c) 𝐴 d) 𝐴 311. In the given circuit diagram the current
2.5 10 15 2
305. If 2 A of current is passed through CuSO4 through the battery and the charge on the
solution for 32 s, then the number of copper capacitor respectively in steady state are
ions deposited at the cathode will be
a) 4 × 1020 b) 2 × 1020
c) 4 × 1019 d) 2 × 1019
306. From the graph between current 𝐼 and voltage
𝑉 shown below, identify the portion
corresponding to negative resistance

P a g e | 24
 a) 1𝐴 and 3 𝜇𝐶 b) 17 𝐴 and 0 𝜇𝐶 emf of first cell is 1.2V. the emf of second cell is
6 12 a) 2.7 V b) 2.1 V c) 3 V d) 4.2V
c) 7 𝐴 and 7 𝜇𝐶 d) 11𝐴 and 3𝜇𝐶
319. A block has dimensions 1 𝑐𝑚, 2 𝑐𝑚, 3 𝑐𝑚. Ratio
312. An aluminium (Al) rod with area of cross-
of the maximum resistance to minimum
section 4 × 10−6 m2 has a current of 5 A
resistance between any point of opposite faces
flowing through it. Find the drift velocity of
of this block is
electron in the rod. Density of Al= 2.7 ×
a) 9 : 1 b) 1 : 9 c) 18 : 1 d) 1 : 6
103 kgm−3 and atomic wt.=27u. Assume that
320. Figure shows a network of three resistance.
each Al atom provides one electron.
When some potential difference is applied
a) 8.6 × 10−4 ms−1 b) 1.3 × 10−4 ms−1
−2 −1 across the network, the thermal powers
c) 2.8 × 10 ms d) 3.8 × 10−3 ms−1
dissipated by 𝐴,𝐵 and 𝐶 in the ratio
313. A steady current 𝑖 is flowing through a 3R
conductor of uniform cross-section. Any A C
segment of the conductor has B R
a) Zero charge
6R
b) Only positive charge
c) Only negative charge a) 2 : 3 : 4 b) 2 : 4 : 3
d) Charge proportional to current 𝑖 c) 4 : 2 : 3 d) 3 : 2 : 4
314. A brass rectangular plate 12cm × 3cm is to be 321. A copper wire of length 𝐿 and radius 𝑟 is nickel
electroplated with copper. If we wish to coat it plated till its final radius become 𝑅 but length
with a layer of 0.02 mm thick both sides, how remains 𝐿. If the resistivity of nickel and
much time will it take with a constant current copper be ρn and ρc respectively, the
conductance of the nickelled wire is
of 5A? Given ECE of copper is 33 × 10−5 g C −1
𝜋𝑟 2 𝜋(𝑅 2 − 𝑟 2 )
and density of copper is 8.9 g cm−3 . a) b)
a) 388 s b) 776 s c) 400 s d) 800 s 𝐿. ρc 𝐿. ρn
2 2 2 𝐿ρc 𝐿. ρn
315. 𝑥 g of Ag is deposited by passing 4 A of 𝜋 𝑟 (𝑅 − 𝑟 )
c) [ + ] d) 2 +
current of for 1 h. How many gram of Ag will 𝐿 ρc ρn 𝜋𝑟 𝜋(𝑅 2 − 𝑟 2 )
be deposited by passing 6 A for 40 min? 322. In charging a battery of motor-car, the
a) 2𝑥 g b) 4𝑥 g c) 𝑥 g d) 5𝑥 g following effect of electric current is used
316. The resistance of a wire is 10Ω. Its length is a) Magnetic b) Heating
increased by 10% by stretching. The new c) Chemical d) Induction
resistance will now be 323. The 𝑉 − 𝑖 graph for a conductor at
a) 12Ω b) 1.2Ω c) 13Ω d) 11Ω temperatures 𝑇1 and 𝑇2 are as shown in the
317. In the circuit shown, the reading of ammeter figure. (𝑇2 − 𝑇1 ) is proportional to
when switch 𝑆 is open and when switch 𝑆 is V
T2

closed respectively are T1



i

a) cos 2𝜃 b) sin 𝜃 c) cot 2𝜃 d) tan 𝜃

324. The electron of hydrogen atom is considered
to be revolving round in circular orbit of
radius ℎ2 /𝑚𝑒 2 with velocity 𝑒 2 ⁄ℎ, where ℎ =
ℎ/2𝜋. The current 𝑖 is
4𝜋 2 𝑚𝑒 5 4𝜋 2 𝑚𝑒 2 4𝜋 2 𝑚2 𝑒 5 4𝜋 2 𝑚𝑒 5
a) 3 𝐴 and 4 𝐴 a) b) c) d)
ℎ2 ℎ3 ℎ3 ℎ3
b) 4 𝐴 and 5 𝐴 325. What is the equivalent resistance across the
c) 5 𝐴 and 6 𝐴 points 𝐴 and 𝐵 in the circuit given below?
d) 6 𝐴 and 7 𝐴
318. 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
P a g e | 25
 C
continuously from the higher potential end
to the lower potential end of the conductor
10 Ω c) The free electrons acquire a constant drift
B
velocity from the lower potential end to the
16 Ω
higher potential end of the conductor
10 Ω
d) The free electrons are set in motion from
12 Ω their position of rest
10 Ω 2.5 Ω A 331. For the circuit shown in the figure the
potential difference between A and B will be
D E (in volt)
1Ω
a) 8 Ω b) 12 Ω c) 16 Ω d) 32 Ω
326. Two wires have lengths, diameters and
specific resistances all in the ratio of 1 : 2. The
resistance of the first wire is 10Ω. Resistance 2V 2V

3Ω
of the second wire in ohm will be 1Ω

a) 5 b) 10 c) 20 d) Infinite
327. 𝑛 identical cells, each of emf 𝐸 and internal a) 2 b) 1.5 c) 1.0 d) Zero
resistance 𝑟, are connected in series a cell 𝐴 is 332. A lamp having tungsten filament consumes 50
joined with reverse polarity. The potential W. Assume the temperature coefficient of
difference across each cell, except 𝐴 is resistance for tungsten is 4.5 × 10−3 ℃−1 and
2𝑛𝐸 (𝑛 − 2)𝐸 temperature of the surrounding is 20℃. When
a) b) the lamp burns, the temperature of its filament
𝑛−2 𝑛
(𝑛 − 1)𝐸 2𝐸 becomes 2500℃ , then the power consumed at
c) d)
𝑛 𝑛 the moment switch is on, is
328. A strip of copper and another of germanium a) 608 W b) 710 W c) 215 W d) 580 W
are cooled from room temperature to 80 𝐾. 333. All bulbs in figure, are identical. Which bulb
The resistance of lights brightly?
a) Each of these increases
b) Each of these decreases
c) Copper strip increases and that of
germanium decreases
a) 1 b) 2 c) 3 d) 4
d) Copper strip decreases and that of
334. The temperature of cold, hot junction of a
germanium increases
thermocouple is 0℃ and 𝑇℃ respectively. The
329. Two batteries, one of emf 18 𝑣𝑜𝑙𝑡 and internal 1
resistance 2Ω and the other of emf 12 𝑣𝑜𝑙𝑡 and thermo-emf produced is 𝐸 = 𝐴𝑇 − 2 𝐵𝑇 2 . If
internal resistance 1Ω, are connected as 𝐴 = 16, 𝐵 = 0.080, the temperature of
shown. The voltmeter 𝑉 will record a reading inversion will be
of a) 100℃ b) 300℃ c) 400℃ d) 500℃
V 335. In a thermo-couple, one junction which is at
0℃ and the othe at 𝑡℃ the emf is given by 𝐸 =
18V 2
𝑎𝑡 2 − 𝑏𝑡 2 . The neutral temperature is given by
a) 𝑎/𝑏 b) 2 𝑎/3𝑏 c) 3𝑎/2𝑏 d) 𝑏/2𝑎
12V 1
336. If voltage across a bulb rated 220 Volt-100
a) 15 𝑣𝑜𝑙𝑡 b) 30 𝑣𝑜𝑙𝑡 c) 14 𝑣𝑜𝑙𝑡 d) 18 𝑣𝑜𝑙𝑡 Watt drops by 2.5% of its rated value, the
330. When a potential difference is applied across percentage of the rated value by which the
the ends of a linear metallic conductor power would decrease is
a) The free electrons are accelerated a) 20% b) 2.5% c) 5% d) 10%
−19
337. An electron (charge = 1.6 × 10 𝑐𝑜𝑢𝑙𝑜𝑚𝑏) is
continuously from the lower potential end to
the higher potential end of the conductor moving in a circle of radius 5.1 × 10−11 𝑚 at a
b) The free electrons are accelerated frequency of 6.8 × 1015 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠/𝑠𝑒𝑐. The

P a g e | 26
 equivalent current is approximately the two points at 30 𝑐𝑚 and 60 𝑐𝑚 point on a
a) 5.1 × 10−3 𝑎𝑚𝑝 b) 6.8 × 10−3 𝑎𝑚𝑝 meter scale fitted along the wire. The potential
c) 1.1 × 10−3 𝑎𝑚𝑝 d) 2.2 × 10−3 𝑎𝑚𝑝 difference between 𝐵 and 𝐶 will be
338. When the resistance of 9 Ω is connected at the a) 3 𝑣𝑜𝑙𝑡 b) 0.4 𝑣𝑜𝑙𝑡 c) 7 𝑣𝑜𝑙𝑡 d) 4 𝑣𝑜𝑙𝑡
ends of a battery, its potential difference 348. A wire of resistance 10 Ω is bent to form a
decreases from 40 𝑣𝑜𝑙𝑡 to 30 𝑣𝑜𝑙𝑡. The internal circle. 𝑃 and 𝑄 are points on the circumference
resistance of the battery is of the circle dividing it into a quadrant and are
a) 6 Ω b) 3 Ω c) 9 Ω d) 15 Ω connected to a battery of 3 𝑉 and internal
339. Two electric bulbs have ratings respectively of resistance 1 Ω as shown in the figure. The
25 W, 220 V and 100 W, 220 V. If the bulbs are currents in the two
connected in series with a supply of 440, which parts of the circle are
bulb will fuse?
P
a) 25 W bulb b) 100 W bulb
c) Both of these d) None of these
340. The drift velocity does not depend upon 3V
Q
1Ω
a) Cross-section of the wire
b) Length of the wire
6 18 5 15
c) Number of free electrons a) 23 𝐴 and 23 𝐴 b) 26 𝐴 and 26 𝐴
d) Magnitude of the current 4 12
c) 25 𝐴 and 25 𝐴
3 9
d) 25 𝐴 and 25 𝐴
341. A wire of resistance 𝑅 is elongated 𝑛 − fold to
349. Sensitivity of potentiometer can be increased
make a new uniform wire. The resistance of
by
new wire
a) Increasing the e.m.f. of the cell
a) 𝑛𝑅 b) 𝑛2 𝑅 c) 2𝑛𝑅 d) 2𝑛2 𝑅
b) Increasing the length of the potentiometer
342. How much energy in kilowatt hour is
wire
consumed in operating ten 50 watt bulbs for
c) Decreasing the length of the potentiometer
10 hours per day in a month (30 days)
wire
a) 1500 b) 5,000 c) 15 d) 150
d) None of the above
343. Two bulbs are working in parallel order. Bulb
350. The 𝑉 − 𝑖 graph for a conductor makes an
𝐴 is brighter than bulb 𝐵. If 𝑅𝐴 and 𝑅𝐵 are their
angle 𝜃 with 𝑉-axis. Here 𝑉 denotes the
resistance respectively then
voltage and 𝑖 denotes current. The resistance
a) 𝑅𝐴 > 𝑅𝐵 b) 𝑅𝐴 < 𝑅𝐵
of conductor is given by
c) 𝑅𝐴 = 𝑅𝐵 d) None of these
a) sin 𝜃 b) cos 𝜃 c) tan 𝜃 d) cot 𝜃
344. The arrangement as shown in figure is called
351. The resistance of the series combination of
as
two resistance is S. When they are joined in
parallel, the total resistance is P. If 𝑆 = 𝑛𝑃,
Total P.D. then the minimum possible value of 𝑛 is
a) 4 b) 3 c) 2 d) 1
Variable P.D. 352. A galvanometer having a coil resistance of 60 Ω
a) Potential divider b) Potential adder shows full scale deflection when a current of
c) Potential substracter d) Potential multiplier 1.0 amp passes through it. It can be converted
345. The current density (number of free electrons into an ammeter to read currents upto 5.0 amp
per m3 ) in metallic conductor is of the order of by
a) 1022 b) 1024 c) 1026 d) 1028 a) Putting in parallel a resistance of 240 Ω
b) Putting in series a resistance of 15 Ω
346. Resistance of rod is 1 Ω.It is bent in form of
square. What is resistance across adjoint c) Putting in series a resistance of 240 Ω
corners? d) Putting in parallel a resistance of 15 Ω
3 3 353. The Avogadro’s number is 6 × 1023 per gm
a) 1 Ω b) 3 Ω c) Ω d) Ω mole and electronic charge is 1.6 × 10−19 𝐶.
16 4
347. The potential gradient along the length of a The Faraday’s number is
uniform wire is 10 𝑣𝑜𝑙𝑡/𝑚𝑒𝑡𝑟𝑒. 𝐵 and 𝐶 are a) 6 × 1023 × 1.6 × 10−19

P a g e | 27
 6 × 1023 resistance of 20 Ω. The resistance to be
b)
1.6 × 10−19 connected in series to convert it into a
2 voltmeter of maximum reading 3V is
c)
6 × 10 × 1.6 × 10−19
23
a) 49 Ω b) 80 Ω c) 40 Ω d) 30 Ω
1.6 × 10−19 362. A heating coil is labelled 100 𝑊, 220 𝑉. The
d)
6 × 1023 coil is cut in half and the two pieces are joined
354. 2, 4 and 6 𝑆 are the conductance of three in parallel to the same source. The energy now
conductors. When they are joined in parallel, liberated per second is
their equivalent conductance will be a) 200 𝐽 b) 400 𝐽 c) 25 𝐽 d) 50 𝐽
a) 12 𝑆 b) (1/12)𝑆 363. The thermo-emf of a thermocouple is 25𝜇 V/℃
c) (12/11)𝑆 d) (11/12)𝑆 at room temperature. A galvanometer of 40 Ω
355. Three resistances each of 4Ω are connected in resistance, capable of detecting current as low
the form of an equilateral triangle. The
as 10−5 A, is connected with the thermocouple.
effective resistance between any two corners
The smallest temperature difference that can
is
be detected by this system is
a) (3/8) Ω b) (8/3) Ω c) 8 Ω d) 12 Ω
a) 16℃ b) 12℃ c) 8℃ d) 20℃
356. Above neutral temperature, thermo 𝑒.m.f. in a
364. For what value of 𝑅 the net resistance of the
thermocouple
circuit will be 18 𝑜ℎ𝑚𝑠
a) Decreases with rise in temperature R
b) Increases with rise in temperature
c) Remains constant 10 10

d) Changes sign 10

357. A cell of emf 𝐸 and internal resistance 𝑟
10 10 10
supplies currents for the same time 𝑡 through A B
external resistance 𝑅1 =100 Ω and 𝑅2 = 40 Ω a) 8 Ω b) 10 Ω c) 16 Ω d) 24 Ω
separately. If the heat developed in both the 365. A combination of two resistance of 2 W and
cases in the same, then the internal resistance 2/3 W connected in parallel is joined across a
of the cell is given by battery of emf of 3 V and of negligible internal
a) 28.6 Ω b) 70 Ω c) 63.3 Ω d) 140 Ω resistance. The energy given out per sec will be
358. If 2.2 𝑘𝑖𝑙𝑜𝑤𝑎𝑡𝑡 power is transmitted through a 1
a) 2 × 3 ×3 J
1 1
b) 2 × 3 × 3 × 3 J
10 𝑜ℎ𝑚 line at 22000 𝑣𝑜𝑙𝑡, the power loss in
c) 2 × 3 J d) 3 × 3 × 2 J
the form of heat will be
366. A galvanometer of resistance 25Ω giving full
a) 0.1 𝑤𝑎𝑡𝑡 b) 1 𝑤𝑎𝑡𝑡 c) 10 𝑤𝑎𝑡𝑡 d) 100 𝑤𝑎𝑡𝑡
scale deflection for a current of
359. A lead-acid battery of a car has an emf of 12 V.
10 𝑚𝑖𝑙𝑙𝑖𝑎𝑚𝑝𝑒𝑟𝑒, is to be changed into a
If the internal resistance of the battery is 0.5 Ω,
voltmeter of range 100 𝑉 by connecting a
the maximum current that can be drawn from
resistance of ′𝑅′ in series with galvanometer.
the battery will be
The value of resistance 𝑅 in Ω is
a) 30 A b) 20 A c) 6 A d) 24 A
a) 10000 b) 10025 c) 975 d) 9975
360. The potential difference across the 100Ω
367. A conductor with rectangular cross-section has
resistance in the following circuit is measured
dimensions (𝑎 × 2𝑎 × 4𝑎) as shown in figure.
by a voltmeter of 900 Ω resistance. The
Resistance across 𝐴𝐵 is 𝑅1 , across 𝐶𝐷 is 𝑅2 and
percentage error made in reading the potential
across 𝐸𝐹 is 𝑅3 .Then
difference is
900 
V
10 

100 
a) 𝑅1 = 𝑅2 = 𝑅3 b) 𝑅1 > 𝑅2 > 𝑅3
c) 𝑅2 > 𝑅3 > 𝑅1 d) 𝑅1 > 𝑅3 > 𝑅2
10
a) b) 0.1 c) 1.0 d) 10.0 368. There are 𝑛 similar conductors each of
9 resistance 𝑅. The resultant resistance comes
361. A milliammeter of range 0-30mA has internal

P a g e | 28
 out to be 𝑥 when connected in parallel. If they a) 0.005 𝑉/𝑐𝑚 b) 0.05 𝑉/𝑐𝑚
are connected in series, the resistance comes c) 0.02 𝑉/𝑐𝑚 d) 0.2 𝑉/𝑐𝑚
out to be 376. In the following star circuit diagram (figure),
a) 𝑥/𝑛2 b) 𝑛2 𝑥 c) 𝑥/𝑛 d) 𝑛𝑥 the equivalent resistance between the points 𝐴
369. When a resistance of 2𝑜ℎ𝑚 is connected across and 𝐻 will be
the terminals of a cell, the current is A i
0.5 𝑎𝑚𝑝𝑒𝑟𝑒. When the resistance is increased r r 72°
D r E
to 5 𝑜ℎ𝑚, the current is 0.25 𝑎𝑚𝑝𝑒𝑟𝑒. The B r C

internal resistance of the cell is r r

F J
a) 0.5 𝑜ℎ𝑚 b) 1.0 𝑜ℎ𝑚 c) 1.5 𝑜ℎ𝑚 d) 2.0 𝑜ℎ𝑚 r
H
r
370. Two electric bulbs rated 𝑃1 watt 𝑉 volts and 𝑃2 r r
G I
watt 𝑉 volts are connected in parallel and 𝑉
volts are applied to it. The total power will be a) 1.944 𝑟 b) 0.973 𝑟 c) 0.486 𝑟 d) 0.243 𝑟
377. An electric cable of copper has just one wire of
a) 𝑃1 + 𝑃2 𝑤𝑎𝑡𝑡 b) √𝑃1 𝑃2 𝑤𝑎𝑡𝑡
radius 9 mm. Its resistance is 5Ω. This single
𝑃1 𝑃2 𝑃1 + 𝑃2
c) 𝑤𝑎𝑡𝑡 d) 𝑤𝑎𝑡𝑡 copper wire of cable is replaced by 6 different
𝑃1 + 𝑃2 𝑃1 𝑃2
well insulated copper wires each of radius 3
371. Which arrangement of four identical resistance
mm. The total resistance of the cable will now
should be used to draw maximum energy from
be equal to
a cell of voltage 𝑉
a) 7.5 Ω b) 45 Ω c) 90 Ω d) 270 Ω
a) 378. Which of the following statements is correct
a) Liquids obey fully the 𝑜ℎ𝑚′𝑠 law
b) Liquids obey partially the 𝑜ℎ𝑚′𝑠 law
b) c) There is no relation between current and
p.d. for liquids
d) None of the above
379. On passing 96500 𝑐𝑜𝑢𝑙𝑜𝑚𝑏 of charge through
c) a solution 𝐶𝑢𝑆𝑂4 the amount of copper
liberated is
a) 64 𝑔𝑚 b) 32 𝑔𝑚 c) 32 𝑘𝑔 d) 64 𝑘𝑔
d) 380. The charge on the capacitor of capacitance 𝐶
shown in the figure below2Ω will be
372. The heat developed in an electric wire of E

resistance 𝑅 by a current 𝐼 for a time 𝑡 is

R2
𝐼 2 𝑅𝑡 𝐼2 𝑡 𝐼2 𝑅 𝑅𝑡
a) 𝑐𝑎𝑙 b) 𝑐𝑎𝑙 c) 𝑐𝑎𝑙 d) 𝑐𝑎𝑙 I
4.2 4.2𝑅 4.2 𝑡 4.2 𝐼 2 R1
373. The internal resistance of a primary cell is 4 Ω .
C
It generates a current of 0.2 A in an external
resistance of 21 Ω. The rate at which chemical 𝐶𝐸𝑅1 𝐶𝐸𝑅2 𝐶𝐸𝑅2
energy is consumed in providing the current is a) 𝐶𝐸 c) b) d)
𝑅1 + 𝑟 𝑅2 + 𝑟 𝑅1 + 𝑟
a) 0.42 J s −1 b) 0.84 J s −1 381. The reciprocal of resistance is
c) 1 J s −1 d) 5 J s −1 a) Conductance b) Resistivity
374. A galvanometer of resistance 25 Ω measures c) Voltage d) None of the above
10−3A. shunt required to increase range upto 382. If the electronic charge is 1.6 × 10−19 C, then
2A is the number of electrons passing through a
a) 12.5 Ω b) 0.125 Ω c) 0.125 Ω d) 1.25 Ω section of wire per second, when the wire
375. A 2 𝑣𝑜𝑙𝑡 battery, a 15 Ω resistor and a carries a current of 2 A is
potentiometer of 100 cm length, all are a) 1.25 × 1017 b) 1.6 × 1017
connected in series. If the resistance of c) 1.25 × 10 19 d) 1.6 × 1019
potentiometer wire is 5Ω, then the potential 383. The resistance will be least in a wire with
gradient of the potentiometer wire is dimension
P a g e | 29
 a) 𝐿/2,2𝐴 b) 2𝐿, 𝐴 d) Random error
c) 𝐿, 𝐴 d) None of these 388. When a 12Ω resistor is connected with a
384. Two bars of radius 𝑟 and 2𝑟 are kept in contact moving coil galvanometer then its deflection
as shown. An electric current 𝑖 is passed reduces from 50 divisions to 10 divisions. The
through the bars. Which one of the following is resistance of the galvanometer is
correct? a) 24 Ω b) 36 Ω c) 48 Ω d) 60 Ω
389. 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
Heat produced in bar 𝐵𝐶 is 4 times the heat arrangement is found from the expression
a) r
produced in bar 𝐴𝐵
b) Electric field in both halves is equal
Current density across 𝐴𝐵 is doubled that of
c) R
across 𝐵𝐶
energy dissipated in the load
Potential difference across 𝐴𝐵 is 4 times that .
d) energy dissipatd in the compete circuit
of across 𝐵𝐶
Which of the following gives the efficiency in
385. In the circuit given, the current relation to a
this case?
balanced Wheatstone’s bridge is 𝑟 𝑅 𝑟 𝑅
a) b) c) d)
P R
𝑅 𝑟 𝑅+𝑟 𝑅+𝑟
390. A resistor 𝑅1 dissipates power 𝑃 when
connected to a certain generator. If the resistor
S Q
𝑅2 is put in series with 𝑅1 , the power
dissipated by 𝑅1
𝑃 𝑅 𝑃 𝑆 a) Decreases
a) = b) =
𝑄 𝑆 𝑄 𝑅 b) Increases
𝑃 𝑄 𝑃 𝑆 c) Remains the same
c) = d) =
𝑆 𝑅 𝑅 𝑄 Any of the above depending upon the
d)
386. Figure below shows a thick copper rod 𝑋 and a relative values of 𝑅1 and 𝑅2
thin copper wire 𝑌 joined in series. They carry 391. In a Wheatstone’s network 𝑃 = 2Ω, 𝒬 =
a current which is sufficient to make 𝑌 much 2Ω, R = 2Ω and S = 3Ω. The resistance with
hotter than 𝑋 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
392. When a current passes through a wire whose
Which one of the following is correct? different parts are maintained at different
Number density of Mean time between temperatures, evolution or absorption of heat
Conduction collisions of the all along the length of wire is known as
electrons electrons a) Joule effect b) Seebeck effect
a) Same in 𝑋 and 𝑌 less in 𝑋 than in 𝑌 c) Peltier effect d) Thomson effect
b) Same in 𝑋 and 𝑌 same in 𝑋 and 𝑌 393. If a rod has resistance 4Ω and if rod is turned
c) Same in 𝑋 and 𝑌 more in 𝑋 than in 𝑌 as half circle, then the resistance along
d) more in 𝑋 and 𝑌 less in 𝑋 than in 𝑌 diameter is
387. In meter bridge or wheatstone bridge for a) 1.56Ω b) 2.44Ω c) 4Ω d) 2Ω
measurement of resistance, the known and the 394. If 𝑁 is the Avogadro’s number and 𝑒 is the
unknown resistance are interchanged. The electronic charge then the Faraday’s constant
error so removed is 𝐹 is equal to
a) End correction 1
a) 𝑁𝑒 b) 𝑁 2 𝑒 c) 𝑁𝑒 2 d)
b) Index error 𝑁𝑒
c) Due to temperature effect 395. A battery of emf E and internal resistance r is

P a g e | 30
 connected to an external resistance R the voltmeter is 𝑉1 when only 𝑆1 is closed, reading
condition for maximum power transfer is of voltmeter is 𝑉2 when only 𝑆2 is closed and
a) r<R b) r>R c) r=1/R d) R=R reading of voltmeter is 𝑉3 when both 𝑆1 and 𝑆2
396. Three bulbs of 40 𝑊, 60 𝑊, 100 𝑊 are are closed. Then
arranged in series with 220 𝑣𝑜𝑙𝑡 supply. Which 3R

bulb has minimum resistance R S1

a) 100 𝑊 b) 40 𝑊 6R

c) 60 𝑊 d) Equal in all bulbs S2

V
397. If two identical heaters each rated as (1000 W-
220 V) are connected in parallel to 220 V, then E
the total power consumed is a) 𝑉3 > 𝑉2 > 𝑉1 b) 𝑉2 > 𝑉1 > 𝑉3
a) 200 W b) 2500 W c) 250 W d) 2000 W c) 𝑉3 > 𝑉1 > 𝑉2 d) 𝑉1 > 𝑉2 > 𝑉3
398. Combination of two identical capacitors, a 403. Two cells of equal 𝑒. 𝑚. 𝑓. and of internal
resistor 𝑅 and a DC voltage source of voltage 6 resistance 𝑟1 and 𝑟2 (𝑟1 > 𝑟2 ) are connected in
V is used in an experiment on𝐶 − 𝑅 circuit. It is series. On connecting this combination to an
found that for a parallel combination of the external resistance 𝑅, it is observed that the
capacitor the time in which the voltage of the potential difference across the first cell
fully charged combination reduces to half its becomes zero. The value of 𝑅 will be
original voltage is 10s. For series combination 𝑟1 + 𝑟2 𝑟1 − 𝑟2
a) 𝑟1 + 𝑟2 b) 𝑟1 − 𝑟2 c) d)
the time needed for reducing the voltage of the 2 2
fully charged series combination by half is 404. When 1 g hydrogen (ECE=1.044 ×
a) 200s b) 10s c) 5s d) 2.5s 106−8 kg C −1 ) forms water, 34 kilo cal heat is
399. In an electroplating experiment, 𝑚 𝑔𝑚 of liberated. The minimum voltage required to
silver is deposited when 4 𝑎𝑚𝑝𝑒𝑟𝑒 of current decompose water is
flows for 2 𝑚𝑖𝑛𝑢𝑡𝑒. The amount (in 𝑔𝑚) of a) 0.75 V b) 1.5 V c) 3.0 V d) 4.5 V
silver deposited by 6 𝑎𝑚𝑝𝑒𝑟𝑒 of current for 405. If an electric current is passed through a nerve
40 𝑠𝑒𝑐𝑜𝑛𝑑 will be of a man, then man
a) 4 𝑚 b) 𝑚/2 c) 𝑚/4 d) 2 𝑚 a) Begins to laugh
400. A metal wire is subjected to a constant b) Begins to weep
potential difference. When the temperature of c) Is excited
the metal wire increases, the drift velocity of d) Becomes insensitive to pain
the electron in it 406. 𝐴𝐵 is a potentiometer wire of length 100 𝑐𝑚
a) increases, thermal velocity of the electron and its resistance is 10 𝑜ℎ𝑚. It is connected in
decreases series with a resistance 𝑅 = 40 𝑜ℎ𝑚 and a
b) Decreases, thermal velocity of the electron battery of e.m.f. 2 𝑉 and negligible internal
decreases resistance. If a source of unknown e.m.f. 𝐸 is
c) increases, thermal velocity of the electron balanced by 40 𝑐𝑚 length of the potentiometer
increases wire, the value of 𝐸 is
2V
d) Decreases, thermal velocity of the electron R

increases
401. Find the true statements 40 cm
B
a) Ohm’s law is applicable to all conductors of A

electricity
b) In an electrolyte solution, the electric E
current is mainly due to the movement of a) 0.8 𝑉 b) 1.6 𝑉 c) 0.08 𝑉 d) 0.16 𝑉
electrons 407. Two bulbs 40 W and 60 W and rated voltage
c) The resistance of an incandescent lamp is 240 V are connected in series across a
lesser when the lamp is switched on potential difference of 420 V. Which bulb will
d) Specific resistance of a wire depends upon work at above its rated voltages?
its dimension a) 40 W bulb b) 60 W bulb
402. In the circuit shown in the figure reading of
P a g e | 31
 c) Both will work d) None of these 10−4 per℃ ?
408. A galvanometer of resistance 36 Ω is changed a) 400 W b) 990 W c) 250 W d) 1500 W
into an ammeter by using a shunt of 4 Ω. The 418. The current flowing through a wire depends
fraction 𝑓0 of total current passing through the on time as 𝐼 = 3𝑡 2 + 2𝑡 + 5. The charge
galvanometer is flowing through the cross-section of the wire
1 1 1 1 in time from 𝑡 = 0 to 𝑡 = 2 sec. is
a) b) c) d)
40 4 140 10 a) 22 C b) 20 C c) 18 C d) 5 C
409. A resistor has a colour code of green, blue, 419. In a Wheatstone’s bridge all the four arms
brown and silver. What is its resistance? have equal resistance 𝑅. If the resistance of the
a) 5600Ω ± 10% b) 560Ω ± 5% galvanometer arm is also 𝑅, the equivalent
c) 560Ω ± 10% d) 56Ω ± 5% resistance of the combination as seen by the
410. One junction of thermocouple is at 0℃ and the battery is
other is at 𝑇℃. The thermo emf (in volts) is a) 𝑅/2 b) 𝑅 c) 2 𝑅 d) 𝑅/4
given by 420. In the circuit shown, if a conducting wire is
𝐸 = 20 × 10−6 𝑇 − 0.02 × 10−6 𝑇 2 connected between points 𝐴 and 𝐵, the current
The maximum value of 𝐸 is in this wire will
a) 5 mV b) 1 m V c) 10 m V d) Zero
411. 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 %
412. If a wire is stretched to make it 0.1% longer, its
resistance will a) Be zero
a) Increase by 0.2% b) Decrease by 0.2% b) Flow from B to A
c) Decrease 0.05% d) Increase by 0.05% c) Flow from A to B
413. 𝑊𝑎𝑡𝑡-ℎ𝑜𝑢𝑟 meter measures Flow in the direction which will be decided
a) Electric energy b) Current d)
by the value of 𝑉
c) Voltage d) Power 421. The thermo-emf of a thermocouple varies with
414. An electric heater boils 1 kg of water in a time the temperature θ of the hot junction as 𝐸 =
𝑡1 . Another heater boils the same amount of 𝑎θ + 𝑏θ2 in volts where the ratio 𝑎/𝑏 is 700℃.
water in a time 𝑡2 . When the two heaters are If the cold junction is kept at 0℃, then the
connected in parallel, the time required by neutral temperature is
them together to boil the same amount of a) 700℃.
water is b) 350℃.
𝑡1 + 𝑡2 𝑡1 𝑡2 c) 1400℃.
a) 𝑡1 + 𝑡2 b) 𝑡1 𝑡2 c) d)
2 𝑡1 + 𝑡2 d) No neutral temperature is possible for this
415. The accurate measurement of emf can be thermocouple
obtained using 422. Out of five resistances of resistance 𝑅 Ω each 3
a) Multimeter b) Voltmeter are connected in parallel and are joined to the
c) Voltameter d) Potentiometer rest 2 in series. Find the resultant resistance
416. In the circuit shown, the current though 8 𝑜ℎ𝑚 3 7 7 8
is same before and after connecting 𝐸. The a) ( ) 𝑅 Ω b) ( ) 𝑅 Ω c) ( ) 𝑅 Ω d) ( ) 𝑅 Ω
7 3 8 7
value of 𝐸 is 423. 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
424. The tolerance level of a resistor with the
a) 12 𝑉 b) 6 𝑉 c) 4 𝑉 d) 2 𝑉
colour code red, blue, orange, gold is
417. The power of heater is 750 W at 1000℃. What
a) ±5% b) ±10% c) ±20% d) ±40%
will be its power at 200℃ if 𝑎 = 4 ×
425. In the absence of applied potential, the electric
P a g e | 32
 current flowing through a metallic wire is zero potentiometer wire is 4 m. What is the
because potential difference across two points on the
a) The electrons remain stationary wire separated by 50cm?
The electrons are drifted in random a) 2.5 V b) 5.0 V c) 1.25 V d) 4.0 V
b) direction with a speed of the order of 10−2 432. Two bulbs 25 W, 220 V and 100 W, 220 V are
cm s −1 given. Which has higher resistance?
c) The electrons move in random direction a) 25 W bulb
with a speed of the order close to that of b) 100 W bulb
velocity of light c) Both bulbs will have equal resistance
d) Electrons and ions move in opposite d) Resistance of bulbs cannot be compared
direction 433. A moving coil galvanometer has a resistance of
426. Which of the plots shown in figure may 50Ω and gives full scale deflection for 10 𝑚𝐴.
represent the thermal energy produced in a How could it be converted into an ammeter
resistor in a given time as a function of the with a full scale deflection for 1𝐴
electric current? a) 50/99 Ω in series b) 50/99 Ω in parallel
c) 0.01 Ω in series d) 0.01 Ω in parallel
434. When a current I is passed through a wire of
constant resistance, it produces a potential
difference 𝑉 across its ends. The graph drawn
a) 𝑎 b) 𝑏 c) 𝑐 d) 𝑑 between log I and log V will be
427. 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 a) b)
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
428. A current of 0.01mA passes through the c) d)
potentiometer wire of a resistivity of
109 Ω-cm and area of cross-section 10−2 cm2.
The potential gradient is 435. An electric bulb rated 220 V, 100 W is
a) 109 Vm−1 b) 1011 Vm−1 connected in series with another bulb rated
c) 1010 Vm−1 d) 108 Vm−1 220 V, 60 W. If the voltage across the
429. You are given several identical resistances combination is 220 V, the power consumed by
each of value 𝑅 = 10Ω and each capable of the 100 W bulb will be about
carrying maximum current of 1 ampere. It is a) 25 W b) 14 W c) 60 W d) 100 W
required to make a suitable combination of 436. Two wires that are made up of two different
these resistances to produce a resistance of 5Ω materials whose specific resistance are in the
which can carry a current of 4 ampere. The ratio 2 : 3, length 3 : 4 and area 4 : 5. The ratio
minimum number of resistances of the type 𝑅 of their resistances is
that will be required for this job a) 6 : 5 b) 6 : 8 c) 5 : 8 d) 1 : 2
a) 4 b) 10 c) 8 d) 20 437. The equivalent resistance of 𝑛 resistors each of
430. Two wires of the same material and having same resistance when connected in series is R.
same uniform area of cross-section are If the same resistances are connected in
connected in series in an electrical circuit. The parallel, the equivalent resistance will be
masses of the wires are 𝑚 and 2𝑚. When a a) 𝑅/𝑛2 b) 𝑅/𝑛 c) 𝑛2 𝑅 d) 𝑛𝑅
current 𝐼flows in the circuit, the heats 438. The resistance of an incandescent lamp is
produced by them in a given time are in ratio a) Greater when switched off
a) 2 : 1 b) 1 : 2 c) 4 : 1 d) 1 : 4 b) Smaller when switched on
431. The potential gradient along the length of a c) Greater when switched on
uniform wire is 10Vm−1 .The length of the d) The same whether it is switched off or

P a g e | 33
 switched on with a power of 500 W in 220 V line are
439. The relation between voltage sensitivity (𝜎𝑉 ) connected in series in a 110 V line. The power
and current sensitivity (𝜎𝑖 ) of a moving coil generated by each bulb will be
galvanometer is (resistance of galvanometer is a) 31.25 W b) 40 W c) 60 W d) 3.125 W
G). 447. If 𝐸 = 𝑎𝑡 + 𝑏𝑡 2 , what is the neutral
𝜎𝑖 𝜎v 𝐺 𝐺 temperature
a) = 𝜎v b) = 𝜎i c) = 𝜎i d) = 𝜎V 𝑎 𝑎 𝑎 𝑎
𝐺 G 𝜎v 𝜎i
a) − b) + c) − d) +
440. Every atom makes one free electron in copper. 2𝑏 2𝑏 𝑏 𝑏
If 1.1 A Current is flowing in the wire of copper 448. Which of the following is not a correct
having 1 mm diameter, then the drift statement
velocity(approx.) will be (density of a) Resistivity of electrolytes decreases on
copper=9 × 103 kg m−3and atomic weight of increasing temperature
copper=63) b) Resistance of mercury falls on decreasing its
a) 0.1 mms−1 b) 0.2 mms−1 temperature
c) 0.3 mms −1 d) 0.2 mms−1 When joined in series a 40 𝑊 bulb glows
c)
441. If an ammeter is connected in parallel to a more than a 60 𝑊 bulb
circuit, it is likely to be damaged due to excess Resistance of 40 𝑊 bulb is less than the
d)
a) Current b) Voltage resistance of 60 𝑊 bulb
c) Resistance d) All of these 449. A resistance of 4Ω and a wire of length
442. A torch bulb rated as 4.5 𝑊, 1.5 𝑉 is connected 5 𝑚𝑒𝑡𝑟𝑒𝑠 and resistance 5Ω are joined in
as shown in the figure. The 𝑒. 𝑚. 𝑓. of the cell series and connected to a cell of e.m.f. 10 𝑉 and
needed to make the bulb glow at full intensity internal resistance 1Ω. A parallel combination
is of two identical cells is balanced across
4.5 W 300 𝑐𝑚 of the wire. The e.m.f. 𝐸 of each cell is
1.5 V 4 10V
1
1Ω
3m
E(r=2.67Ω) 5, 5m
E
G
a) 4.5 𝑉 b) 1.5 𝑉 c) 2.67 𝑉 d) 13.5 𝑉 E

443. The internal resistances of two cells shown are a) 1.5 𝑉 b) 3.0 𝑉 c) 0.67 V d) 1.33 𝑉
0.1 Ω and 0.3 Ω. If 𝑅 = 0.2Ω, the potential 450. In the circuit shown, the currents 𝑖1 and 𝑖2 are
difference across the cell
i 1 12Ω
2V, 0.1 2V, 0.3
2Ω
A B 4Ω
i2

0.2
12V, 1Ω
a) 𝐵 will be zero
a) 𝑖1 = 3A, 𝑖2 = 1A b) 𝑖1 = 1A, 𝑖2 = 3A
b) 𝐴 will be zero
c) 𝑖1 = 0.5A, 𝑖2 = 1.5 A d) 𝑖1 = 1.5 A, 𝑖2 = 0.5 A
c) 𝐴 and 𝐵 will be 2𝑉
451. The resistance of a bulb filament is 100 Ω at a
d) 𝐴 will be > 2𝑉 and 𝐵 will be < 2𝑉
temperature of 100°C. If its temperature
444. A source of a primary cell is 2V. what is the
coefficient of resistance be 0.005 per°C, its
short circuited it provides 4A current, then the
resistance will become 200 Ω at a temperature
internal resistance of cell will be
of
a) 8 Ω b) 2.0 Ω c) 4 Ω d) 0.5 Ω
a) 300°C b) 400°C c) 500°C d) 200°C
445. The material of wire of potentiometer is
452. What will be the equivalent resistance
a) Copper b) Steel
between the two points 𝐴 and 𝐷
c) Manganin d) Aluminium
446. Two electric bulbs, each designed to operate

P a g e | 34
 10 10 10 are joined in parallel. Maximum current is
A B
taken from this combination across as external
10 10 resistance of 3 Ω resistance. If the total number
of cells used are 24 and internal resistance of
C D
10 10 10 each cell is 0.5 Ω, then
a) 𝑚 = 8, 𝑛 = 3 b) 𝑚 = 6, 𝑛 = 4
a) 10 Ω b) 20 Ω c) 30 Ω d) 40 Ω
c) 𝑚 = 12, 𝑛 = 2 d) 𝑚 = 2, 𝑛 = 12
453. In the circuit shown in figure, the points 𝐹 is
459. An electric current passes through a circuit
grounded. Which of the following is wrong
containing two wires of the same material
statement?
connected in parallel. If the lengths of the wires
5Ω are in the ratio of 4/3 and radius of the wires
B C are in the ratio of 2/3, then the ratio of the
current passing through the wires will be
2Ω 3Ω a) 3 b) 1/3
A D c) 8/9 d) None of these
10 V 4 Ω 3V
460. If in a voltaic cell, 5 g of zinc is consumed, we
F E will get how many ampere hour (given that
ECE of zinc is 3.38 × 10−7 kgC −1)
a) 𝐷 is at 5V a) 2.05 b) 8.2
b) 𝐸 is at zero potential c) 4.1 d) 5 × 3.338 × 10−7
c) The current in the circuit will be 0.5 A 461. By using only two resistances coils-singly, in
The potential at 𝐸 is same whether or not 𝐹 series or in parallel one should be able to
d)
is rounded obtain resistance of 3,4,12 and 16 ohm. The
454. In the circuit shown, the cell is ideal, with separate resistance of the coil are
emf=10V. Each resistance is of 2Ω. The a) 3 and 4 b) 4 and 12
potential difference across the capacitor is c) 12 and 16 d) 16 and 13
B
R
G
C =3 F
H 462. The current in the primary circuit of a
potentiometer is 0.2A. the specific resistance
R R
R R and cross-section of the potentiometer wire
A
are 4 × 10−7 Ωm and 8 × 10−7 𝑚2 respectively.
F
D
10 V
Potential gradient will be equal to
a) 12 V b) 10 V c) 8 V d) zero a) 0.2 V/m b) 1V/m c) 0.3V/m d) 0.1V/m
455. Faraday’s laws of electrolysis are related to 463. In the given figure, battery 𝐸 is balanced on
a) The atomic number of positive ion 55 𝑐𝑚 length of potentiometer wire but when
b) The equivalent weight of electrolyte a resistance of 10 Ω is connected in parallel
c) The atomic number of negative ion with the battery then it balances on 50 𝑐𝑚
d) The velocity of positive ion length of the potentiometer wire then internal
456. In a Wheatstone bridge, 𝑃 = 90Ω, 𝑄 = resistance 𝑟 of the battery is
2V
110Ω, 𝑅 = 40Ω and 𝑆 = 60Ω and a cell of 4 V
emf. Then the potential difference between the 1m
diagonal along which a galvanometer is B
connected is A
E r
a) −0.2 V b) +0.2 V c) −1 V d) +1 V
457. If three bulbs 60𝑊, 100𝑊 and 200 𝑊 are a) 1 Ω b) 3 Ω c) 10 Ω d) 5 Ω
connected in parallel, then 464. 𝐴, 𝐵, 𝐶 and 𝐷 are four resistances of
a) 200 𝑊 bulb will glow more 2Ω, 2Ω, 2Ω and 3Ω respectively. They are used
b) 60 𝑊 bulb will glow more to form a Wheatstone bridge. The resistance 𝐷
c) 100 𝑊 bulb will glow more is short circuted with a resistances R in order
d) All the bulbs will glow equally to get the bridge balanced. The value of 𝑅 will
458. The 𝑛 rows each containing m cells in series be

P a g e | 35
 a) 4 Ω b) 6 Ω c) 8 Ω d) 3 Ω mark. The end-corrections are 1cm and 2cm
465. The resistance of a 10 m long wire is 10 Ω. Its respectively for the ends A and B. the
length is increased by 25% by stretching the determined value of x is
wire uniformly. The resistance of wire will
change to (approximately) X 10 Ω
a) 12.5 Ω b) 14.5 Ω c) 15.6 Ω d) 16.6 Ω
466. 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
A B
current flowing in 𝐴𝐹𝐶𝐸𝐵 will be
C
a) 10.2 Ω b) 10.6 Ω c) 10.8 Ω d) 11.1 Ω
R
473. A current of 1 𝑚𝐴 is flowing through a copper
R R
F wire. How many electrons will pass a given
R A
B point in one 𝑠𝑒𝑐𝑜𝑛𝑑
D
R
E [𝑒 = 1.6 × 10−19 𝐶𝑜𝑢𝑙𝑜𝑚𝑏]
3𝑉 𝑉 𝑉 2𝑉 a) 6.25 × 1019 b) 6.25 × 1015
a) b) c) d) c) 6.25 × 1031 d) 6.25 × 108
𝑅 𝑅 2𝑅 𝑅
467. A student has 10 resistors of resistance ′𝑟 ′ . 474. A current of 16 𝑎𝑚𝑝𝑒𝑟𝑒 flows through molten
The minimum resistance made by him from 𝑁𝑎𝐶𝑙 for 10 𝑚𝑖𝑛𝑢𝑡𝑒. The amount of metallic
given resistors is sodium that appears at the negative electrode
𝑟 𝑟 𝑟 would be
a) 10 𝑟 b) c) d)
10 100 5 a) 0.23 𝑔𝑚 b) 1.15 𝑔𝑚 c) 2.3 𝑔𝑚 d) 11.5 𝑔𝑚
468. The masses of the three wires of copper are in 475. Neutral temperature of a thermocouple is
the ratio 5 : 3 : 1 and their lengths are in the defined as the temperature at which
ratio 1 : 3 : 5 the ratio of their electrical a) The thermo 𝑒.m.f. changes sign
resistance is b) The thermo 𝑒.m.f. is maximum
a) 5 : 3 : 1 b) √125 ∶ 15 ∶ 1 c) The thermo 𝑒.m.f. is minimum
c) 1 : 15 : 125 d) 1 : 3 : 5 d) The thermo 𝑒.m.f. is zero
469. Two resistors 400 Ω and 800 Ω are connected 476. In the electric circuit shown each cell has an
in series with 6 V battery. The potential emf of 2V and internal resistance of 1Ω. The
difference measured by voltmeter of 10k Ω external resistance is 2Ω. The value of the
across 400 Ω resistor is current I is(in ampere)
a) 2 V b) 1.95 V c) 3.8 V d) 4 V 2Ω
470. Identify the incorrect statement regarding a
superconducting wire
a) Transport current flows through its surface
b) Transport current flows through the entire I I
area of cross-section of the wire | | |
2V ,1 Ω
c) It exhibits zero electrical resistivity and
expels applied magnetic field a) 2 b) 1.25 c) 0.4 d) 1.2
d) It is used to produce large magnetic field 477. 12 cells each having same emf are connected
471. The resistance of a cell does not depend on in series with some cells wrongly connected.
a) Current drawn from the cell The arrangement is connected in series with
b) Temperature of electrolyte an ammeter and two cells which are in series.
c) Concentration of electrolyte Current is 3 𝐴 when cells and battery aid each
d) The 𝑒.m.f. of the cell other and is 2 𝐴 when cells and battery oppose
472. A meter bridge is set-up as shown in figure, to each other. The number of cells wrongly
determine an unknown resistance X using a connected is
standard 10 Ω resistor. The galvanometer a) 4 b) 1 c) 3 d) 2
shows null point when tapping key is at 52cm 478. An infinite sequence of resistances is shown in

P a g e | 36
 the figure. The resultant resistance between 𝐴 resistance are 𝛼1 𝑎𝑛𝑑 𝛼2 . The respective
and 𝐵 will be, when 𝑅1 = 1 𝑜ℎ𝑚 and 𝑅2 = temperature coefficients of their series and
2 𝑜ℎ𝑚 parallel combinations are nearly
R1 R1 R1 R1 R1 𝛼1+ 𝛼2 𝛼1+ 𝛼2
A a) , 𝛼1 + 𝛼2 b) 𝛼1+ 𝛼2 ,
2 2
𝛼1 𝛼2 𝛼1+ 𝛼2 𝛼1+ 𝛼2
R2 R2 R2 R2 R2 c) 𝛼1 + 𝛼2 , d) ,
𝛼1+ 𝛼2 2 2
B 487. The relaxation time in conductors
a) Infinity b) 1 Ω c) 2 Ω d) 1.5 Ω a) Increases with the increase of temperature
479. A tap supplies water at 22℃, a man takes of 1 b) Decreases with the increase of temperature
L of water per min at 37℃ from the geyser. c) It does not depend on temperature
The power of geyser is d) All of sudden changes at 400 𝐾
a) 525 W b) 1050 W c) 1775 W d) 2100 W 488. The resistance of a wire of iron is 10 𝑜ℎ𝑚 and
480. In the circuit as shown in the figure, the heat temp. coefficient of resistance is 5 × 10−3 /℃.
produced by 6 𝑜ℎ𝑚 resistance due to current At 20℃ it carries 30 𝑚𝑖𝑙𝑙𝑖𝑎𝑚𝑝𝑒𝑟𝑒 of current.
flowing in it is 60 𝑐𝑎𝑙𝑜𝑟𝑖𝑒 per 𝑠𝑒𝑐𝑜𝑛𝑑. The heat Keeping constant potential difference between
generated across 3 𝑜ℎ𝑚 resistance per second its ends, the temperature of the wire is raised
will be to 120℃. The current in 𝑚𝑖𝑙𝑙𝑖𝑎𝑚𝑝𝑒𝑟𝑒 that
2 3 flows in the wire is
a) 20 b) 15 c) 10 d) 40
489. A 100 W bulb 𝐵1 and two 60 W bulb 𝐵2 and 𝐵3
6 4
are connected to a 250 V source as shown in
a) 30 𝑐𝑎𝑙𝑜𝑟𝑖𝑒 b) 60 𝑐𝑎𝑙𝑜𝑟𝑖𝑒 the figure. Now 𝑊1 , 𝑊2 and 𝑊3 are the out-put
c) 100 𝑐𝑎𝑙𝑜𝑟𝑖𝑒 d) 120 𝑐𝑎𝑙𝑜𝑟𝑖𝑒 powers of the bulbs 𝐵1 , 𝐵2 and 𝐵3 respectively.
481. 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. Then
The galvanometer resistance is a) 𝑊1 > 𝑊2 = 𝑊3 b) 𝑊1 > 𝑊2 > 𝑊3
a) 100 Ω b) 99 Ω c) 10 Ω d) 9.9 Ω c) 𝑊1 < 𝑊2 = 𝑊3 d) 𝑊1 < 𝑊2 < 𝑊3
482. Heat produced (cals) in a resistance 𝑅 when a 490. Three resistance 𝐴, 𝐵 and 𝐶 have values 3𝑅, 6𝑅
current 𝐼 amperes flows through it for 𝑡 and 𝑅 respectively. When some potential
seconds is given by the expression difference is applied across the network, the
𝐼 2 𝑅𝑡 𝐼𝑅 2 𝑡 4.2𝐼𝑅 𝐼𝑅𝑡 2 thermal powers dissipated by 𝐴, 𝐵 and 𝐶 are in
a) b) c) 2 d) the ratio
4.2 4.2 𝑡 4.2
483. The mass of a substance liberated when a 3R

charge ′𝑞′ flows through an electrolyte is A

R
proportional to C
6R
a) 𝑞 b) 1/𝑞 c) 𝑞 2 d) 1/𝑞 2
B
484. A Daniel cell is balanced on 125𝑐𝑚 length of a
a) 2 :3 :4 b) 2 :4 :3 c) 4 :2 :3 d) 3 :2 :4
potentiometer wire. Now the cell is short-
491. When a current 𝐼flows through a wire, the drift
circuited by a resistance 2 𝑜ℎ𝑚 and the
velocity of the electrons is 𝑣. When current 2𝐼
balance is obtained at 100 𝑐𝑚. The internal
flows through another wire of the same
resistance of the Daniel cell is
material having double the length and double
a) 0.5 𝑜ℎ𝑚 b) 1.5 𝑜ℎ𝑚 c) 1.25 𝑜ℎ𝑚d) 4/5 𝑜ℎ𝑚
the area of cross-section, the drift velocity of
485. With a potentiometer null point were obtained
the electrons will be
at 140 𝑐𝑚 and 180 𝑐𝑚 with cells of emf 1.1 𝑉 𝑣 𝑣 𝑣
and one unknown 𝑋 volt. Unknown emf is a) b) c) d) 𝑣
8 4 2
a) 1.1 𝑉 b) 1.8 𝑉 c) 2.4 𝑉 d) 1.41 𝑉 492. In the given figure when galvanometer shows
486. Two conductors have the same resistance at no deflection current flowing through 5Ω
0° C but their temperature coefficients of resistance will be

P a g e | 37
 B 496. By ammeter, which of the following can be
8Ω 2Ω measured
2.1 A a) Electric potential b) Potential difference
A
G
C c) Current d) Resistance
20 Ω
497. Which factor is immaterial for the wire used in
5Ω
electric fuse?
D a) Length b) Radius
c) Material d) Current
a) 0.5 A b) 0.6 A c) 1.5 A d) 2.0 A 2
498. A current of ( ) 𝐴 produces a deflection of 60°
493. A thermo couple develops 200𝜇 𝑉 between 0℃ 3
and 100℃. If it develops 64 𝜇 𝑉 and 76 𝜇 𝑉 in a tangent galvanometer. The reduction
respectively between (0℃ − 32℃) and factor is
(32℃ − 70℃) then what will be the thermo 2 2 2
a) ( ) 𝐴 b) 2A c) ( ) 𝐴 d) ( ) 𝐴
𝑒𝑚𝑓 it develops between 70℃ and 100℃ 3 3 √3
a) 65 𝜇 𝑉 b) 60 𝜇 𝑉 c) 55 𝜇 𝑉 d) 50 𝜇 𝑉 499. In Seebeck series 𝑆𝑏 appears before 𝐵𝑖. In a
494. A wire of diameter 0.02 𝑚𝑒𝑡𝑟𝑒 contains 1028 𝑆𝑏 − 𝐵𝑖 thermocouple current flows from
free electrons per cubic metre. For an electrical a) 𝑆𝑏 to 𝐵𝑖 at the hot junction
current of 100 𝐴, the drift velocity of the free b) 𝑆𝑏 to 𝐵𝑖 at the cold junction
electrons in the wire is nearly c) 𝐵𝑖 to 𝑆𝑏 at the cold junction
a) 1 × 10−19 𝑚/𝑠 b) 5 × 10−10 𝑚/𝑠 d) None of the above
c) 2 × 10−4 𝑚/𝑠 d) 8 × 103 𝑚/𝑠 500. Two wires of the same dimensions but
495. How many calories of heat will be produced resistivities 𝜌1 and 𝜌2 are connected in series.
approximately in a 210 W electric bulb in 5 The equivalent resistivity of the combination is
min? 𝜌1 + 𝜌2
a) b) 𝜌1 + 𝜌2
a) 80000 cal b) 63000 cal 2
c) 1050 cal d) 15000 cal c) 2(𝜌1 + 𝜌2 ) d) √𝜌1 𝜌2

P a g e | 38
 Dr. V.G.Girhepunje

VERTEX ACADEMY
72, New Diamond Nagar Kharbi Road, Nagpur-440024
Contact : 9923403431 / 7038855627
DPP on Current Electricity
Date : 02/12/2022 TEST ID: 40
Time : 20:00:00 PHYSICS
Marks : 2000
3.CURRENT ELECTRICITY ,9.CURRENT ELECTRICITY

: ANSWER KEY :
1) c 2) b 3) a 4) d 153) a 154) a 155) d 156) b
5) b 6) c 7) d 8) a 157) c 158) d 159) c 160) a
9) b 10) d 11) d 12) c 161) d 162) a 163) a 164) b
13) c 14) d 15) d 16) d 165) c 166) a 167) d 168) b
17) b 18) a 19) d 20) a 169) d 170) a 171) a 172) c
21) b 22) d 23) a 24) a 173) d 174) c 175) b 176) a
25) b 26) b 27) b 28) d 177) a 178) d 179) a 180) c
29) a 30) d 31) d 32) a 181) c 182) d 183) b 184) c
33) b 34) c 35) d 36) d 185) b 186) a 187) b 188) c
37) a 38) a 39) c 40) c 189) b 190) b 191) b 192) a
41) a 42) c 43) c 44) a 193) b 194) c 195) d 196) a
45) c 46) c 47) d 48) a 197) d 198) b 199) a 200) d
49) d 50) a 51) a 52) b 201) a 202) d 203) a 204) d
53) a 54) b 55) c 56) a 205) c 206) c 207) c 208) a
57) d 58) b 59) d 60) d 209) c 210) d 211) b 212) a
61) b 62) b 63) b 64) d 213) a 214) c 215) a 216) b
65) d 66) d 67) d 68) a 217) b 218) c 219) b 220) d
69) c 70) c 71) b 72) b 221) c 222) c 223) a 224) b
73) b 74) c 75) c 76) b 225) b 226) a 227) d 228) a
77) c 78) d 79) b 80) a 229) d 230) d 231) c 232) b
81) c 82) a 83) b 84) a 233) a 234) b 235) b 236) b
85) c 86) b 87) c 88) a 237) d 238) d 239) a 240) a
89) c 90) d 91) c 92) d 241) c 242) c 243) d 244) b
93) b 94) b 95) b 96) c 245) b 246) b 247) a 248) c
97) b 98) c 99) b 100) d 249) d 250) d 251) b 252) c
101) b 102) c 103) c 104) a 253) c 254) c 255) d 256) a
105) d 106) d 107) d 108) d 257) b 258) b 259) c 260) a
109) d 110) c 111) a 112) d 261) d 262) b 263) a 264) b
113) d 114) c 115) c 116) d 265) b 266) d 267) b 268) b
117) c 118) a 119) b 120) c 269) c 270) a 271) d 272) d
121) d 122) d 123) c 124) b 273) d 274) d 275) a 276) d
125) c 126) b 127) b 128) c 277) d 278) c 279) c 280) d
129) b 130) c 131) c 132) d 281) c 282) a 283) a 284) c
133) a 134) a 135) c 136) a 285) c 286) a 287) c 288) b
137) c 138) c 139) b 140) c 289) c 290) c 291) b 292) b
141) a 142) a 143) d 144) c 293) b 294) d 295) a 296) b
145) b 146) b 147) b 148) c 297) c 298) a 299) b 300) b
149) c 150) d 151) b 152) b 301) a 302) a 303) c 304) c

P a g e | 39
305) b 306) c 307) a 308) d
309) c 310) b 311) d 312) b
313) a 314) b 315) c 316) a
317) b 318) a 319) a 320) c
321) c 322) c 323) c 324) d
325) a 326) b 327) d 328) d
329) c 330) c 331) d 332) a
333) a 334) c 335) b 336) c
337) c 338) b 339) a 340) b
341) b 342) d 343) b 344) a
345) d 346) c 347) a 348) a
349) b 350) d 351) a 352) d
353) a 354) a 355) b 356) a
357) c 358) a 359) d 360) c
361) b 362) b 363) a 364) c
365) d 366) d 367) d 368) b
369) b 370) a 371) b 372) a
373) c 374) a 375) a 376) b
377) a 378) b 379) b 380) c
381) a 382) c 383) a 384) a
385) d 386) c 387) a 388) c
389) d 390) a 391) d 392) d
393) c 394) a 395) d 396) a
397) d 398) d 399) b 400) a
401) c 402) b 403) b 404) b
405) c 406) d 407) b 408) d
409) c 410) a 411) b 412) a
413) a 414) d 415) d 416) c
417) b 418) a 419) b 420) b
421) d 422) b 423) c 424) a
425) c 426) d 427) d 428) d
429) c 430) b 431) b 432) a
433) b 434) a 435) b 436) c
437) a 438) c 439) a 440) a
441) a 442) d 443) a 444) d
445) c 446) a 447) a 448) d
449) b 450) c 451) b 452) c
453) b 454) c 455) b 456) a
457) a 458) c 459) b 460) c
461) b 462) d 463) a 464) b
465) c 466) b 467) b 468) c
469) b 470) b 471) d 472) b
473) b 474) c 475) b 476) d
477) b 478) c 479) b 480) d
481) b 482) a 483) a 484) a
485) d 486) d 487) b 488) a
489) d 490) c 491) d 492) b
493) b 494) c 495) d 496) c
497) a 498) c 499) b 500) a

P a g e | 40
 Dr. V.G.Girhepunje

VERTEX ACADEMY
72, New Diamond Nagar Kharbi Road, Nagpur-440024
Contact : 9923403431 / 7038855627
DPP on Current Electricity
Date : 02/12/2022 TEST ID: 40
Time : 20:00:00 PHYSICS
Marks : 2000
3.CURRENT ELECTRICITY ,9.CURRENT ELECTRICITY

: HINTS AND SOLUTIONS :

Single Correct Answer Type A R B
C
(I - I1)

1 (c) R 2R

1×10−3 105
I1
𝑚 I

𝑞= = = C; F
1.044×10−8
D
𝑧 1.044 E

Given, 𝐻 = 43 k cal. Where, 𝑡 = 𝑡0 , 𝐼 = 0

= 34 × 103 × 4.2 J In loop 𝐵𝐶𝐷𝐸𝐵 𝐼1 (2𝑅) − (𝐼 − 𝐼1 )𝑅 = 0
𝐻 = 𝑉𝐼𝑡 = 𝑉𝑞 or 3𝐼1 = 𝐼
𝐻 34×4.2×103 𝐼 𝑎 − 2𝑏𝑡
∴𝑉= = = 1.49 V ∴ 𝐼1 = =
𝑞 105 /1.044
3 3
2 (b) 𝑡0
𝑙 −𝑙 𝑙 −2
𝑟 = ( 1𝑙 2 ) × 𝑅 ′ = ( 12 ) × 5 …(i) ∴ 𝐻 = ∫ 𝐼12 (2𝑅)𝑑𝑡
2
𝑙 −3 0
and 𝑟 = ( 13 ) × 10 …(ii) 𝑡0
2𝑅
on solving (i) and (ii), 𝑟 = 10 Ω = ∫ (𝑎 − 2𝑏𝑡)2 𝑑𝑡
3 (a) 9
0
𝜌1 𝑙1 𝜌2 𝑙2 𝑡0
𝑅1 = and 𝑅2 = . In series 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 2𝑅
𝐴 𝐴
𝜌𝑒𝑞. (𝑙1 + 𝑙2 ) 𝜌1 𝑙1 𝜌2 𝑙2 𝜌1 𝑙1 + 𝜌2 𝑙2 = [ ∫ (𝑎2 − 4𝑏 2 𝑡 2 − 4𝑑𝑡)𝑑𝑡 ]
= + ⇒ 𝜌𝑒𝑞 = 9
0
𝐴 𝐴 𝐴 𝑙1 + 𝑙2 𝑡𝑜
2𝑅 4𝑏2 𝑡 2 4𝑏𝑎𝑡 2
4 (d) = [{𝑎2 𝑡 + − } ]
9 3 2 0
As circuit is open, therefore no current flows 2𝑅 4𝑏2 𝑡03
through circuit. Hence potential difference across = 9
[𝑎2 𝑡0 + 3
− 2𝑏𝑎𝑡02 ]
𝑎
𝑋 and 𝑌=EMF of battery =120V but 𝑡0 = [from Eqn.(i)]
2𝑏
2𝑅 2 𝑎 4𝑏2 𝑎 3 𝑎2
5 (b) 𝐻= 9
[𝑎 × 2𝑏
+ 3 8𝑏3
− 2𝑎𝑏 4𝑏2 ]
𝐸 4 2𝑅 𝑎3 𝑎3 𝑎3 𝑎3 𝑅
𝑖= ⇒1= = 𝑟 = 2Ω = [ + − ]=
𝑅+𝑟 2+𝑟 9 2𝑏 6𝑏 2𝑏 27𝑏
Short circuit, is when terminals of battery are 10 (d)
connected directly, then current which flows is In a conductor, the electron number density 𝑖𝑒,
𝐸 4 number of electrons per unit volume of a
𝑖𝑆𝐶 = = = 2𝐴
𝑟 2 conductor is very large (≈ 1028 m−3 ), so large
6 (c) current in a conductor is obtained irrespective of
𝑃 𝑅 2
We know that 𝑃1 = 𝑅2 = 1 their small drift speed.
2 1
8 (a) 11 (d)
𝑞 3000𝐶 𝑅𝜋𝑟 2 4.2 × 3.14 × (0.2 × 10−3 )2
Current, 𝐼 = = 10×60𝑠 = 5𝐴 𝑙= = = 1.1𝑚
𝑡 𝜌 48 × 10−8
9 (b) 12 (c)
𝑄 = 𝑎𝑡 − 𝑏𝑡 2 No charge in neutral temperature but
𝑑𝑄
𝐼= = 𝑎 − 2𝑏𝑡 … (i) temperature of inversion is
𝑑𝑡 𝑡𝑖 = 2𝑡𝑛 − 𝑡𝑐 ⇒ 𝑡𝑖 = 2 × 270 − 40 = 500℃

P a g e | 41
13 (c) across 𝑅2 (= 66Ω) is 48 V, hence
From Joule’s law, for a current carrying conductor
𝑅 48 × 20
at a definite temperature the rate of production of 𝑉𝐴𝐵 = 48 × = = 160V
heat is given by 𝑅2 6

17 (b)
𝐻 = 𝑉𝑖𝑡
Shunt
Where 𝑖 is current, 𝑉 the potential difference and 𝐼𝑔 𝐺
𝑆=
𝑡 the time. 𝐼 − 𝐼𝑔
𝐼𝑔 1
Here, 𝐼 = 34
Given, 𝑉 = 1.5 volt, 𝑖 = 2.1 A, 𝑡 = 1s
1 𝑆
∴ =
∴ 𝐻 = 15 × 2.1 × 1 = 31.5 J 34 𝑆 + 𝐺
𝐺 3663
Also, 1 cal = 4.2 J ∴ 𝑆= = = 111Ω
33 33
19 (d)
31.5
∴𝐻= = 7.5 cal 5𝑋 + 2 × 10
4.2 𝑉𝐴𝐵 = 4 = ⇒𝑋
𝑋 + 10
14 (d) 𝐸2 𝑟1 + 𝐸1 𝑟2
= 20Ω, [𝑣 = ]
𝑉 100 ± 0.5 𝑟1 + 𝑟2
𝑅= = = 10 ± 0.25Ω 20 (a)
𝑖 10 ± 0.2
15 (d) 𝑅1 𝑙1 2 10 5 2 1
=( ) ⇒ =( ) = = 𝑅2 = 160Ω
Let 𝑛 cells be in series and 𝑚 in parallel, then 𝑅2 𝑙2 𝑅2 20 16
𝑛𝐸 𝐸 21 (b)
=
𝑅 + 𝑛𝑟 𝑅 + 𝑟
𝑚 𝑟iron 𝜌iron 1 × 10−7
𝑟 =√ =√ = 2.4
⟹ 𝑛 [𝑅 + ] = 𝑅 + 𝑛𝑟 𝑟Copper 𝜌copper 1.7 × 10−8
𝑚
⟹ 𝑛𝑅𝑚 + 𝑛𝑟 = 𝑅𝑚 + 𝑚𝑛𝑟 22 (d)
⟹ 6 + 2𝑟 = 3 + 4𝑟 𝑑𝐸 2𝑇 𝑇
=4− 4− ;
⟹ 2𝑟 = 3 𝑑𝑇 200 100
⟹ 𝑟 = 1.5Ω At neutral point, 𝑇 = 𝑇𝑛 .
𝑑𝑇
16 (d) = 0 = 4 − 𝑇𝑛 /100
𝑑𝑇
Effective value of resistance of parallel or 𝑇𝑛 = 400℃, 𝑇𝑖 = 2𝑇𝑛 − 𝑇0
combination of 20Ω, 30 Ω, 60Ω is 𝑅1 , where = 2 × 400 − 0 = 800℃
1 1 1 1 3+2+1 6 1 23 (a)
= + + = = = 𝑆
𝑅1 20 30 60 60 60 10 𝑖𝑔 = 𝑖 ⇒ 10 × 10−3
𝐺+𝑆
𝑅1 = 10Ω 𝑆
= × 100 × 10−3
100 + 𝑆
Similarly effective value of parallel combination of 1000
90 𝑆 = 1000 ⇒ 𝑆 = = 11.11 Ω
24Ω and 8Ω resistance is given by 90
24 (a)
24 × 8 𝑖 𝐺 𝑖. 𝐺 𝐺 100 × 10−3 × 40
𝑅2 = = 6Ω =1+ ⇒ =1+ ⇒
24 + 8 𝑖𝑔 𝑆 𝑉𝑔 𝑆 800 × 10−3
40
48V =1+
𝑆
10 Ω 6Ω 1Ω ⇒ 𝑆 = 10Ω
3Ω
A B 25 (b)
Kirchhoff’s second law is ∑𝑉 = 0
Hence, the circuit may be redrawn as shown in It states that the algebric sum of the potential
the adjacent figure, where total resistance across differences in any loop including those associated
𝐴 and 𝐵, 𝑅 = 3 + 10 + 6 + 1 = 20Ω. As potential emf’s and those of resistive elements, must equal
zero.
P a g e | 42
 This law represents ‘conservation of energy’. 1.5 1
Resistance of the bulb= ( 1.5 × 4.5) = 0.5 = 2 Ω
26 (b) 1
1× 1
Length 𝑙 = 1 𝑐𝑚 = 10−2 𝑚 Resistance of the circuit 𝑅 = 2
1 =
3
Ω
1+
2
𝐸−𝑉
Now , 𝑟= 𝑅
𝑉
8 𝐸−15 1
= 1.5 × 3 or 𝐸 = 13.5 volt
1 cm

3
100 cm
32 (a)
1 cm The circuit can be drawn as follows
Area of cross-section 𝐴 = 1 𝑐𝑚 × 100 𝑐𝑚 A

= 100 𝑐𝑚2 = 10−2 𝑚2

3 3
10−2
Resistance 𝑅 = 3 × 10−7 × 10−2 = 3 × 10−7 Ω
i1
27 (b) B C
i2 3
𝑚 = 𝑍𝑖 𝑡 and 𝑖 𝑡 =Area of given curve i
= Area of triangle + Area of rectangle
1 2V
⇒ 𝑖 𝑡 = × (2 × 60) × 1 + (6 − 2) × 60 × 1
2 3×(3+3)
Equivalent resistance 𝑅 = 3+(3+3) = 2Ω
= 300
𝑚 𝑚 2 3
Current 𝑖 = 2 = 1𝐴. So, 𝑖1 = 1 × (3+6) = 3 𝐴
1
∴𝑍= =
𝑖 𝑡 300 1
28 (d) Potential difference between 𝐴 and 𝐵 = 3 × 3 =
Kirchhoff’s Ist law or KCL states that the algebraic 1𝑣𝑜𝑙𝑡
sum of current meeting at any junction is equal to 33 (b)
zero. In other words we can say that “the sum of The resistance of one wire
all the currents directed towards a junction in a 𝑙1
𝑅1 = 𝜌
circuit is equal to the sum of all the currents 𝐴1
directed away from that junction.” Thus, no and the resistance of second wire
charge has been accumulated at any junction i.e., 𝑙2
𝑅2 = 𝜌
charge is conserved, and hence, we can say that 𝐴2
KCL (∑𝑖 = 0) is based on conservation of charge. Ratio of their resistance
Kirchhoff’s IInd law or KVL states that algebraic 𝑅1 𝑙1 𝐴2
= ×
sum of changes in potential around any closed 𝑅2 𝐴1 𝑙2
resistor loop must be zero. In other words ∵Mass=density×volume
“around any closed loop, voltage drops are equal ∵Mass=density ×area ×length
to voltage rises”. No energy is gained or lost in 𝑅1 𝑙1 2 𝐴 𝑙
Or 𝑅2
= (𝑙2
) × 𝜌 𝐴2 𝑙2
circulating a charge around a loop, thus, we can 𝜌 1 1

𝑅1 𝑙 2 𝑚
say that KVL is based on conservation of energy. Or = (𝑙1 ) × 𝑚2
𝑅2 2 1
29 (a) 𝑅1 9 3 𝑅1 27
Or 𝑅 = 16 × 2 or 𝑅 = 32
Ammeter is always connected in series and 2 2

voltmeter in parallel. 𝑅1 : 𝑅2 = 27: 32

30 (d) 34 (c)
When a circuit is made up on any two metals in If volume and density remains same, then
thermoelectric series, the current flows across the resistance of wire
cold junction from the later occurring metal in the 𝑅 ∝ 𝑙2
series to the one occurring earlier. In Where 𝑙is the length of the wire
thermoelectric series Bismuth comes earlier than 10 2 11 2
∴ 𝑅 ′ = (1 + 𝑙) = ( ) 𝑅
Antimony. So, at cold junction current flows from 100 10
Antimony to Bismuth and at hot junction it flows Hence,
from bismuth to Antimony. 𝑅′ − 𝑅 21
= = 21%
31 (d) 𝑅 100
Therefore, change in resistance of wire =21%

P a g e | 43
35 (d) convert a galvanometer into voltmeter, a high
𝑉2 220 ×220 resistance must be connected in series with it so
Power 𝑃 = 𝑅
⇒ 300 = 𝑅
that it draws negligible current from the circuit.
22 × 22
𝑅= 44 (a)
3
110 ×110 ×3 The current taken by the silver voltameter
Again 𝑃 = 22 ×22
𝑛 1
𝑃 = 75 𝐼1 = = −4
= 0.496 𝐴
𝑍𝑡 11.2 × 10 × 30 × 60
75 × 100 and by copper voltameter
𝑃% = = 25%
300 1.8
The percentage reduction in power 𝐼2 = −4
= 1.515 𝐴
6.6 × 10 × 30 × 60
𝑃 = 100 − 25 = 75% The current 𝐼 = (𝐼1 + 𝐼2 ) = 2.011 𝐴
36 (d) Power 𝑃 = 𝐼𝑉 = 2.011 × 12 = 24.132 𝐽/𝑠𝑒𝑐
𝐸 = 𝑖1 (𝑅1 + 𝑟) = 𝑖2 (𝑅2 + 𝑟) 45 (c)
𝑖2 𝑅2 −𝑖1 𝑅1
Current through 6Ω resistance in parallel with 3Ω
On solving, 𝑟 = resistance = 0.4 𝐴
(𝑖1 −𝑖2 )
So total current = 0.8 + 0.4 = 1.2 𝐴
37 (a) Potential drop across 4Ω = 1.2 × 4 = 4.8𝑉
Let the resistance of 𝑃, 𝑄 and 𝑅 be 𝑟. 46 (c)
The total resistance across the battery is 𝑅𝑡1 = 𝑅1 (1 + 𝛼1 𝑡) and 𝑅𝑡2 = 𝑅2 (1 + 𝛼2 𝑡)
𝑟 3
𝑟total = 𝑟 + = 𝑟. Also 𝑅𝑒𝑞 = 𝑅𝑡1 + 𝑅𝑡2 ⇒ 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + (𝑅1 𝛼1 +
2 2
𝑅2 𝛼2 )𝑡
Current through 𝑃,
𝑅1 𝛼1 + 𝑅2 𝛼2
⇒ 𝑅𝑒𝑞 = (𝑅1 + 𝑅2 ) [1 + ( ) . 𝑡]
Power 12 8 𝑅1 + 𝑅2
𝐼𝑃 = √ = √3 = √ . 𝑅1 𝛼1 +𝑅2 𝛼2
𝑟total 𝑟 𝑟 So 𝛼𝑒𝑓𝑓 =
2 𝑅1 +𝑅2
Current through 𝑅, 47 (d)
1
1 2
𝐼 = 2 𝐼𝑃 = √𝑟. Slope of 𝑉 − 𝑖 curve = resistance. Hence 𝑅 = 1 =
1Ω
Power dissipated in 𝑅 is thus
2
48 (a)
𝑃𝑅 = 𝐼 2 𝑟 = ( 𝑟) 𝑟 =2 W V 𝑖𝑅 𝑖𝜌𝐿 𝑖𝜌
Potential gradient = 𝐿
= 𝐿
= 𝐴𝐿
= 𝐴
38 (a)
0.2 × 40 × 10−8
𝑃 = 𝑉𝑖 = 250 × 2 = 500 𝑊 = = 10−2 𝑉/𝑚
8 × 10−6
39 (c) 49 (d)
When we measure the emf of a cell by the After some time, thermal equilibrium will reach
potentiometer then no current flows in the circuit 50 (a)
in zero-deflection condition 𝑖𝑒, cell is in open 160 × 40
circuit. Thus, in this condition the actual value of a 𝑅= = 32Ω
160 + 40
cell is found. In this way, potentiometer is Series
C
equivalent to an ideal voltmeter of infinite
100+60 = 160
resistance. 60
100
42 (c)
For a fuse 𝐼 2 ∝ 𝑟 3
A B
𝐼12 𝑟13 40
∴ 2= 2 51 (a)
𝐼2 𝑟2
32 0.02 3 Since resistance connected in arms 𝐶𝐸, 𝐸𝐷, 𝐶𝐹
= ( ) and 𝐹𝐷 will form a balanced Wheatstone bridge,
𝐼22 0.03
therefore, the resistance of arm 𝐸𝐹 becomes
3 3/2
𝐼2 = 3 × ( 2) A ineffective. Now resistance of arm 𝐶𝐸𝐷 or 𝐶𝐹𝐷 =
43 (c) 2 + 2 = 4Ω. Effective resistance of these two
4×4
Voltmeter has high resistance and is always parallel arm= 4+4 = 2Ω
connected in parallel with the circuit. So, to

P a g e | 44
 Now resistance of arm 𝐴𝐶𝐷𝐵 = 2 + 2 + 2 = 6Ω, is Rt  t2
t
in parallel with resistance arm 𝐴𝐵 = 2Ω. Thus,
R0
effective resistance between 𝐴 and 𝐵

6×2 3 t
= = Ω
6+2 2 59 (d)
Suppose current though different paths of the
52 (b)
6×4 circuit is allows :
P.d. across the circuit = 1.2 × 6+4 = 2.88 𝑣𝑜𝑙𝑡 28 54
2.88
Current through 6 𝑜ℎ𝑚 resistance = 6
= 0.48 𝐴
6V
53 (a) 1
i3
2

The capacitance (C) of a capacity is defined as the

ratio of the charge (q) given to the rise in the 8V 12 V
potential (V) of the conductor. When the plates After applying 𝐾𝑉𝐿 for loop (1) and loop (2)
are moved apart, the charge remains constant, 1
We get 28𝑖1 = −6 − 8 ⇒ 𝑖1 = − 2 𝐴
hence
1
𝑞 = 𝐶𝑉 = constant … … … . . (i) and 54i2 = −6 − 12 ⇒ 𝑖2 = − 𝐴
3
Capacitance (C) of a parallel plate capacitor is 5
hence 𝑖3 = 𝑖1 + 𝑖2 = − 6 𝐴
given by
60 (d)
𝜀0 𝐴
𝐶= … … … . . (ii) As circuit is open, therefore no current flows
𝑑
Where A is area of plates and d the separation through circuit.
between them. Hence, potential difference across X and Y
From Eqs. (i) and (ii), we have =emf of battery =120V.
𝜀0 𝐴 61 (b)
𝑞= . 𝑉 = constant Equivalent circuit of this combination of
𝑑
When plates are moved apart, 𝑑 increases, hence resistances is as shown in figure. The effective
value of C decreases and in order that charge (q) resistance of arm
remains constant V increases.
4×4
54 (b) 𝐸𝐺 = = 2Ω
4+4
Let the length of various edges in increase order
be 𝑙, 𝑥, 2𝑙 respectively
4Ω
ρ2𝑙 2ρ ρ𝑙 ρ A B
𝑅max = = ; 𝑅min = = 4Ω 4Ω
𝑥𝑙 𝑥 2𝑙𝑥 2𝑥
𝑅max E G
∴ =4 4Ω 4Ω
𝑅min
4Ω
55 (c) C
D
𝑖 = 𝑞/𝑇 = 𝑞𝑣

= 2 × 10−2 × 30 = 0.6 A Total resistances between 𝐴 and 𝐵 will be

57 (d) 1 1 1 1 3 4
𝑉 100 = + + = or 𝑅 = Ω
By using 𝑅 = −𝐺 ⇒𝑅 = − 5 = 19,995Ω 𝑅 4 4 4 4 3
𝑖𝑔 5×10−3
58 (b) 62 (b)
Since the value of 𝑅 continuously increases, both For maximum power 𝑅ext = 𝑅int
𝛼 and 𝛽 must be positive. When batteries are connected in parallel.
Actually the components of the given equation are 1
𝑅int = = 0.5Ω
as follows 2
𝐸 2
Current 𝐼 = 𝑅 +𝑅 = 0.5+0.5 = 2Ω
ext int

P a g e | 45
 Maximum power is given by 12
𝑖= = 5𝐴
𝑃max = 𝑖 2 𝑅ext (1 + 1) + 0.4
⇒ 𝑃max = (2)2 × 0.5 =2 W 74 (c)
If we assume batteries to be connected in series, 𝑚Ag /𝑚Zn = 𝐸Ag /𝐸Zn = 108/31
then or 𝑚Ag = 𝑚Zn × 108/31= 𝑤 × 108/31
𝑅int = 1 + 1 = 2Ω 3.48 w = 3.5 w
2𝐸 2×2
Current 𝐼 = = =1A 75 (c)
𝑅ext +𝑅int 2+2
𝑉 2 (18)2
So, maximum power is now given by 𝑃= = = 54𝑊
𝑃max = 𝑖 2 𝑅ext = 1 × 2 = 2 W 𝑅 6
76 (b)
In either case 𝑃max = 2 W
𝐸Ni −Cu = 𝑎𝑡 + 𝑏𝑡 2
63 (b)
Based on Peltier effect = (16.3 × 10−6 )(100) + (−0.021 × 10−6 )
64 (d) × (100)2
Given circuit is a balanced Wheatstone bridge = 1.42 × 10−3 V
circuit. So there will be no change in equivalent 78 (d)
resistance. Hence no further current will be At time 𝑡 = 0 𝑖. 𝑒. when capacitor is charging,
drawn current
2
66 (d) 𝑖= = 2𝑚𝐴
Three resistances are in parallel, 1000
When capacitor is full charged, no current will
1 1 1 1 3
∴ ′= + + = pass through it, hence current through the circuit
𝑅 𝑅 𝑅 𝑅 𝑅
2
The equivalent resistance 𝑖= = 1𝑚𝐴
𝑅 6 2000
𝑅’ = Ω = = 2Ω 79 (b)
3 3
67 (d) Since no current is to flow in the 4Ω resistance,
Total external equivalent resistance 𝑅𝑒𝑞 = 4Ω hence resistance 4Ω becomes ineffective in
𝐸 10 current.
Current supply by cell 𝑖 = 𝑅 = (4+1) = 2𝐴
𝑒𝑞 +𝑟
𝑖 2 B C D
∴ (𝑉𝐴 − 𝑉𝐵 ) = (𝑅2 − 𝑅1 ) = (2 − 4) = −2𝑉
2 2
68 (a) 4Ω
𝑉 10 2Ω R
𝑅 = −𝐺 = − 1 = 999Ω
𝑖𝑔 10 × 10−3 6V
70 (c) A F E
9V 3V
𝑙1 75
𝑟 = 𝑅 ( − 1) = 10 ( − 1)
𝑙2 65
= 10 × 0.0154 = 1.54Ω Current through resistances 2 Ω is
71 (b)
𝑃 𝑅 2 2 9−6 3
By using 𝑄 = 𝑆
⇒2= 6𝑆 ⇒ 𝑆 = 3Ω 𝑖= = A
(6+𝑆) 2 2
72 (b) 3
In circuit 𝐴𝐵𝐶𝐷𝐸𝐹𝐴, 9 − 3 = (2 + 𝑅) × 2
By using Kirchhoff’s junction law as shown below
or 12 = 6 + 3𝑅 or 𝑅 = 2Ω

80 (a)
𝑅 𝑅𝛼 𝑙
Here 𝑅𝑋𝑊𝑌 = 2𝜋𝑟 × (𝑟𝛼) = 2𝜋
[∴ 𝛼 = 𝑟]
𝑅 𝑅
and 𝑅𝑋𝑍𝑌 = 2𝜋𝑟 × 𝑟(2𝜋 − 𝛼) = 2𝜋 (2𝜋 − 𝛼)

We get 𝑖 = 8𝐴
73 (b)
P a g e | 46
 𝑅𝛼 𝑅 3
𝑅𝑋𝑊𝑌 𝑅𝑋𝑍𝑌 2𝜋
× 2𝜋 (2𝜋 − 𝛼) Equivalent resistance of the circuit 𝑅 = 2 Ω
𝑅𝑒𝑞 = = 𝑅𝛼 𝑅(2𝜋−𝛼) 𝑉 3
𝑅𝑋𝑊𝑌 + 𝑅𝑋𝑍𝑌 + 2𝜋 ∴ Current through the circuit 𝑖 = 𝑅 = 3/2 = 2𝐴
2𝜋
𝑅𝛼 85 (c)
= (2𝜋 − 𝛼)
4𝜋 2 1Ω X 2Ω
81 (c) I1
1 Ixy 2
By using 𝑅𝑡 = 𝑅0 (1 + 𝛼𝑡)
3 × 𝑅0 = 𝑅0 (1 + 4 × 10−3 𝑡) ⇒ 𝑡 = 500℃ I2 3Ω Y 4Ω
82 (a)
Equivalent resistance of the combination A B
(2 + 2) × 2 8 4 −𝑖1 + 0 × 𝑖𝑥𝑦 + 3𝑖2 = 0 𝑖. 𝑒. 𝑖1 = 3𝑖2 …(i)
= = = Ω
2+2+2 6 3 Also −2(𝑖1 − 𝑖𝑥𝑦 ) + 4(𝑖2 + 𝑖𝑥𝑦 ) = 0
𝑖. 𝑒. 2𝑖1 − 4𝑖2 = 6𝑖𝑥𝑦 …(ii)
2 2 Also 𝑉𝐴𝐵 − 1 × 𝑖1 − 2(𝑖1 − 𝑖𝑥𝑦 ) = 0 ⇒ 50 = 𝑖1 +
2(𝑖1 − 𝑖𝑥𝑦 )
P
2
Q
= 3𝑖1 − 2𝑖𝑥𝑦 …(iii)
83 (b) Solving (i), (ii) and (iii), 𝑖𝑥𝑦 = 2𝐴
The circuit can be shown as given below 86 (b)
D 15Ω C Cells are joined in parallel when internal
resistance is higher than external resistance. [𝑅 <
15Ω
< 𝑟]
15Ω
15Ω O 15Ω 𝐸
𝑂′ 𝑖= 𝑟
15Ω 𝑅+𝑛
15Ω
A B
87 (c)
15Ω 𝑅𝐴
Resistivity 𝜌 =
𝑙
The equivalent resistance between D and C. ⟹ 𝜌∝𝑅
15 × (15 + 15) ∵ Metals have low resistance, therefore they have
𝑅𝐷𝐶 =
15 + (15 + 15) low resistivity.
15 × 30 88 (a)
=
15 + 30 𝐶𝑢 voltameter with soluble electrodes obeys
15 × 30 ohm’s law. In water voltameter, in the beginning
= = 10Ω
45 when 𝑉 is small (< 1.7 𝑣𝑜𝑙𝑡), very little current
Now, between A and B, the resistance of upper flows, the voltameter does not obey ohm’s law. As
part ADCB, soon as 𝑉 exceeds 1.7 𝑣𝑜𝑙𝑡 (back e.m.f.) the
𝑅1 = 15 + 10 + 15 = 40Ω current increases steadily according to ohm’s law
Between A and B, the resistance of middle part 89 (c)
AOB If a cell is connected between points 𝐴 and 𝐶, no
𝑅2 = 15 + 15 = 30Ω current will flow in arms 𝐵𝐸 and 𝐸𝐷. Therefore,
Therefore, equivalent resistance between A and B the resistance of arms 𝐵𝐸 and 𝐸𝐷 an be removed.
1 1 1 1
= + + Now resistance between points 𝐴 and 𝐶 will be
𝑅 ′ 𝑅1 𝑅2 𝑅3 the resistance of three parallel arms, each of
1 1 1
= + + resistance= 𝑅 + 𝑅 = 2𝑅
40 30 15
3+4+8 ∴ Total resistance 𝑅𝑝 will be
=
120
15 1 1 1 1 3 2𝑅
= = + + = or 𝑅 =
120 𝑅𝑃 2𝑅 2𝑅 2𝑅 2𝑅 𝑃
3
′
120
⟹ 𝑅 = = 8Ω
15 90 (d)
84 (a) 𝑉2 𝜌𝑙 𝑉2 𝐴𝑉 2 𝐴𝑉 2
𝑃 = 𝑅 but 𝑅 = 𝐴 ⇒ 𝑃 = 𝜌𝑙 / 𝐴 = 𝜌𝑙 . Since 𝑙 is

P a g e | 47
 1 When each bulb is glowing at full power,
constant as per given condition so 𝑃 ∝ 𝜌
50 1
91 (c) Current from each bulb = 𝑖 ′ = 100 = 2 𝐴
θ0 = 2θ𝑛 − θ𝑖 = 2 × 210 − (600 − 273) 1

= 420 − 327 = 93℃ 2

92 (d)
The resistance of an ideal voltmeter is considered n
i
as infinite 100 V
93 (b)
𝑉 2 𝑃1 𝑅2 6 4 2 120 V, 10 
𝑃= ⇒ = ⇒ = = ⇒ 𝑃2 = 9 𝑊 𝑛
𝑅 𝑃2 𝑅1 𝑃2 6 3 So main current 𝑖 = 2 𝐴
94 (b) 𝑛
𝑉2 Also 𝐸 = 𝑉 + 𝑖𝑟 ⇒ 120 = 100 + (2 ) × 10 ⇒ 𝑛 =
Heat produced by the heater 𝐻 = ×𝑡
𝑅 4
For 220V heater heat produced 99 (b)
(220)2 𝑚
𝐻1 = ×5 Charge → 𝑞 = 𝑧
𝑅
For 110V heater heat produced Where, 𝑧 = electro chemical equivalent
(110)2 𝑚 0.067
∴ 𝑧= = = 0.001117
𝐻2 = ×𝐿 𝐼𝑡 1 × 60
𝑅 108
Now, 𝐻1 = 𝐻2 𝑞= = 9.67 × 104 C/g eq
110 × 110 220 × 220 × 5 0.001117
𝑡= 100 (d)
𝑅 𝑅
The rate at which heat is developed
𝑡 = 20 min
𝑉 2 (110)2
95 (b) 𝐻= = = 1210 W
(𝑣𝑑 )𝑒 𝑖𝑒 𝑛ℎ 7 5 5 𝑅 10
𝑖 = 𝑛𝑒𝐴𝑣𝑑 ⇒ = × = × = 101 (b)
(𝑣𝑑 )ℎ 𝑖ℎ 𝑛𝑒 4 7 4 3 3
𝑡=3
96 (c)
𝑑𝑄 = 𝐼𝑑𝑡 ⇒ 𝑄 = ∫ 𝐼𝑑𝑡 = [2 ∫ 𝑡𝑑𝑡 + 3 ∫ 𝑡 2 𝑑𝑡]
𝑚 = 𝑧𝑞, 𝑧 = atomic mass / valence 𝑡=2
2 2
97 (b)
= [𝑡 2 ]32 + [𝑡 3 ]32 = (9 − 4) + (27 − 8) = 5 + 19
= 24𝐶

Since deflection in galvanometer is zero so

current will flow as shown in the above diagram
𝑉𝐴 12 12
Current 𝐼 = 𝑅 = 500+100 = 600
1 +𝑅
So 𝑉𝐵 = 𝐼𝑅
12
= × 100 = 2𝑉
600
98 (c)
102 (c)
Given circuit can be redrawn as follows

P a g e | 48
 𝑅
⇒ 𝑅𝑒𝑞 =
2
103 (c) 𝐸2
∴ 𝑃max =
On doubling the length of wire its resistance is 4𝑅
doubled and slope of 𝑉 − 𝐼 graph is doubled 𝐸2
or 𝑃max =
4𝑟
104 (a) 107 (d)
1 1 For maximum power 𝑟 = 𝑅
𝑃𝑅𝑎𝑡𝑒𝑑 ∝ and 𝑅 ∝
𝑅 [Thickness of filament]2 108 (d)
So 𝑃𝑅𝑎𝑡𝑒𝑑 ∝ (Thickness of filament)2 In case of stretching of wire 𝑅 ∝ 𝑙 2
105 (d) ⇒ If length becomes 3 times so Resistance
(220)2 (220 × 0.8)2 becomes 9 times 𝑖. 𝑒. 𝑅 ′ = 9 × 20 = 180Ω
𝑃1 = and 𝑃2 =
𝑅1 𝑅2 109 (d)
𝑃2 (220 × 0.8)2 𝑅1 𝑃2 2
𝑅1 𝑖
By using 𝑖 = 1 + 𝑆
𝐺
= × ⇒ = (0.8) ×
𝑃1 (220)2 𝑅2 𝑃1 𝑅2 𝑔
𝑖 1000 1000
Here, 𝑅2 < 𝑅1 (because voltage decreases from ⇒ =1+ ⇒𝑆= = 111Ω
100 × 10 −3 𝑆 9
220 𝑉 → 220 × 0.8 𝑉
110 (c)
It means heat produced → decreases) 𝑃 𝑅
𝑅 = 𝑆 ′ [For balancing bridge]
So 𝑅1 > 1 ⇒ 𝑃2 > (0.8)2 𝑃1 ⇒ 𝑃2 > (0.8)2 × 𝑄
2 B
100 𝑊 P = 9 Q = 11
𝑃 (220×0.8)𝑖2
Also 𝑃2 = 220𝑖1
, Since 𝑖2 < 𝑖1 [we expect] A C
1
𝑃
So 𝑃2 < 0.8 ⇒ 𝑃2 (100 × 0.8) 6
1 R = 4
r
Hence the actual power would be between 100 × D
(0.8)2 𝑊 and S'

(100 × 0.8) 𝑊 4 × 11 44
⇒ 𝑆′ = =
106 (d) 9 9
𝑅 1 1 1
𝑖= ⇒ ′= +
𝑟+𝑅 𝑆 𝑟 6
9 1 1
𝑃 = 𝑖2𝑅 ⇒ − =
44 6 𝑟
𝐸2𝑅 132
⟹ 𝑃= ⇒𝑟= = 26.4 Ω
(𝑟 + 𝑅)2 5
Power is maximum when r=R 111 (a)

P a g e | 49
 Reading of galvanometer remains same whether 114 (c)
switch 𝑆 is open or closed, hence no current will 𝑚 = 𝑧𝑞
flow through the switch 𝑖. 𝑒. 𝑅 and 𝐺 will be in 𝑚 5 × 10−3
⇒𝑞= =
series and same current will flow through them. 𝑧 3.387 × 10−7
𝐼𝑅 = 𝐼𝐺 𝑞 = 1.476 × 104C
112 (d) Since a current of 1 A for 1 h gives a charge of
Let 𝐸 and 𝑟 be the emf and internal resistance of a 3600 C,
battery respectively ∴ 1 A h = 3600 C
1.476 × 104
⇒𝑞= Ah = 4.1 Ah
3600
115 (c)
The given circuit may be redrawn as shown in
In the first case current flowing in the circuit adjacent figure. Resistance of parallel
𝐸 combination of 2Ω and 6Ω,
𝐼1 =
𝑅1 𝑟 - +
Or 𝐸 = 𝐼1 (𝑅1 + 𝑟) 6V 2Ω
A
1.5 Ω B
6Ω

3Ω
In the second case current flowing in the circuit
𝐸 2×6
𝐼2 = 𝑅1 = = 1.5Ω
𝑅2 + 𝑟 2+6
Or 𝐸 = 𝐼2 (𝑅2 + 𝑟)
Equating equations (i) and (ii), we get Now resistance of 𝐴𝐵𝐶 arm= 1.5 + 1.5 = 3Ω
𝐼1 (𝑅1 + 𝑟) = 𝐼2 (𝑅2 + 𝑟) ⇒ 𝐼1 𝑅1 + 𝐼1 𝑟 = 𝐼2 𝑅2 + 𝐼2 𝑟 3×3
𝐼1 𝑅1 − 𝐼2 𝑅2 = (𝐼2 − 𝐼1 )𝑟 ⇒ (𝐼2 − 𝐼1 )𝑟 and total network resistance 𝑅 = 3+3 = 1.5Ω
= 𝐼1 𝑅1 − 𝐼2 𝑅2 6𝑉
𝐼1 𝑅1 − 𝐼2 𝑅2 ∴ Total current supplied by the battery 𝑖 = 1.5Ω =
𝑟=
𝐼2 − 𝐼1 4A
113 (d)
Full scale deflection current 116 (d)
150 The resistance R of a particular conductor is
= mA = 15mA related to the resistivity 𝜌 of its material by
10
Ig R G 𝜌𝑙
𝑅=
𝐴
| | Or
V
𝑅𝐴
𝜌 = resistivity =
Full scale deflection voltage 𝑙
150 Given, R=0.072Ω
= mV = 75mV 𝐴 = 2𝑚𝑚 × 2𝑚𝑚 = 4 × 10−6 𝑚2 , 𝑙=12
2
Galvanometer resistance 0.072 × 4 × 10−6
∴ 𝜌=
75mV 12
G= = 5Ω = 2.4 × 10 −8
Ωm
15mA
Required full scale deflection voltage. 117 (c)
𝑉 = 1 × 150 = 150 volt 𝑉2 𝑉5 𝑉2 𝑅+5 4
+ =4×( ) or =
Let resistance to be connected in series is R. 𝑅 5 𝑅+5 5𝑅 𝑅+5
⟹ 𝑉 = 𝐼𝑔 (𝑅 + 𝐺) On solving, we get 𝑅 = 5Ω.
∴ 150 = 15 × 10−3 (𝑅 + 5) 118 (a)
⟹ 104 = 𝑅 + 5 (i) Rate of chemical energy consumption
⟹ 𝑅 = 10000 − 5 = 9995 = 1.5 × 2 =3 W

P a g e | 50
 (ii) Rate of energy dissipation inside the cell 122 (d)
= 2 × 2 × 0.1 = 0.4 W The network can be redrawn as follows
(iii) Rate of energy dissipation inside the resistor 3 3 3
A B
= (3 − 0.4) W = 2.6 W
⇒ 𝑅𝑒𝑞 = 9Ω
(iv) Power output of source = (3 − 0.4) W = 2.6
123 (c)
W
If a charged particle of charge q resolves in a
∗ 𝐸𝐼 represents rate of chemical energy
circular orbit of radius r with frequency v, then
consumption of the cell.
the orbital current is given by
∗ 𝐼 2 𝑟 represents the rate of energy dissipation 𝐼 = 𝑞𝑣
𝜔 ω
inside the cell. 𝑜𝑟 𝐼 = 𝑞 (∵= 2𝜋𝑣 ⟹ v = )
2𝜋 2π
ev
∗ (𝐸𝐼 − 𝐼 2 𝑟) represents the power output of the I= (∵ 𝑣 = 𝑟𝜔)
2πr
source of emf.
124 (b)
119 (b) During charging of lead-acid accumulator, the
The circuit will be as shown specific gravity of H2 SO4 increases.
10V 125 (c)
We will require a voltmeter, an ammeter, a test
resistor and a variable battery to verify Ohm’s
5
A law.
10 Voltmeter which is made by connecting a high
𝑖= = 2𝐴 resistance with a galvanometer is connected in
5
120 (c) parallel with the test resistor.
𝐿∝𝑙 Further, an ammeter which is formed by
𝐿1 𝑙1 connecting a low resistance in parallel with
= galvanometer is required to measure the current
𝐿2 𝑙2
10 2.5 through test resistor.
= The correct option is (c).
11 𝑙2
10𝑙2 = 2.5 × 11 126 (b)
2.5 × 11 Energy = 𝑃 × 𝑡 = 2 × 1 × 30 = 60 𝑘𝑊ℎ = 60 unit
𝑙2 = = 2.75𝑀 127 (b)
10
121 (d) 𝑙 𝑅2 𝑙2 𝑟12 2 1 2 1
𝑅∝ 2⇒ = × 2 = ( )×( ) =
𝑟 𝑅1 𝑙1 𝑟2 1 2 2
𝑅1
⇒ 𝑅2 = 2
, specific resistance doesn’t depend
upon length, and radius
128 (c)
Here, 𝑉 = 10 𝑉, 𝑅 = 10𝑘Ω, 𝐺 = 100 Ω = 0.1 𝑘Ω
The voltmeter is assumed to have infinite
𝑉 = 𝐼𝑔 (𝐺 + 𝑅)
resistance. Hence (1 + 2 + 1) + 4 = 8Ω
𝑉 10𝑉 10𝑉
𝐼𝑔 = = = = 0.99𝑚𝐴 129 (b)
𝐺 + 𝑅 (10 + 0.1)𝑘Ω 10.1𝑘Ω 𝐻 = 𝜎𝑖𝑡Δ𝜃 ⇒ If 𝑖 = 1𝐴, Δ𝜃 = 1℃, 𝑡 = 1𝑠𝑒𝑐
≈ 1 𝑚𝐴 then 𝐻 = 𝜎
130 (c)
𝑖 1.344
𝑣𝑑 = = −2
𝑛𝐴𝑒 10 × 1.6 × 10−19 × 8.4 × 1022
1.344
= = 0.01𝑐𝑚/𝑠 = 0.1𝑚𝑚/𝑠
10 × 1.6 × 8.4
Here, 𝐼 = 1𝐴, (𝐼 − 𝐼𝑔 )𝑆 = 𝐼𝑔 𝐺 131 (c)
𝐼𝑔 𝐺 1 × 10−3 × 100 Let the resultant resistance be 𝑅. If we add one
𝑆= =
𝐼 − 𝐼𝑔 1 − 1 × 10−3 more branch, then the resultant resistance would
100 × 10−3 100 1 be the same because this is an infinite sequence
≈ = Ω= Ω = 0.1Ω
1 1000 10
P a g e | 51
 R1 = 1 X 𝑉 2𝑡
A 𝐻= cal
𝑅𝐽
R2 = 2 R
Where 𝑉 is potential difference, 𝑡 the time and 𝑅
B the resistance.
Y
𝑅𝑅2
∴ + 𝑅1 = 𝑅 ⇒ 2𝑅 + 𝑅 + 2 = 𝑅 2 + 2𝑅 𝑅 = 𝑅1 + 𝑅2
𝑅 + 𝑅2
⇒ 𝑅 2 − 𝑅 − 2 = 0 ⇒ 𝑅 = −1 or 𝑅 = 2𝑜ℎ𝑚 𝑡 𝐻𝐽
Let 𝑅 = 𝑉 2 = constant = 𝑘
2×𝑅
𝑅 = 2+2+ ⇒ 2𝑅 + 𝑅 2 = 8 + 4𝑅 + 2𝑅
2+𝑅 ⇒ 𝑅1 = 𝑘𝑡1 and 𝑅2 = 𝑘𝑡2
4 ± √ 16 + 32
⇒ 𝑅 2 − 4𝑅 − 8 = 0 ⇒ 𝑅 = ∴ 𝑘𝑡 = 𝑘𝑡1 + 𝑘𝑡2
2
= 2 ± 2√3
Or 𝑡 = 15 + 20
𝑅 cannot be negative, hence 𝑅 = 2 + 2√3 = 5.46Ω
132 (d) 𝑡 = 35 min
As batteries wear out, temperature of filament of
flash light attains lesser value, therefore intensity 138 (c)
20×103
of radiation reduces. Also dominating wavelength R
V
(𝜆𝑚 ) in spectrum, which is the red colour,
5V
increases i
133 (a) 110V
All the conductors have equal lengths. Area of 110
Here 𝑖 = 20×103 +𝑅
2 2
cross-section of 𝐴 is {(√3𝑎) − (√2𝑎) } = 𝑎2 110
∵ 𝑉 = 𝑖𝑅 ⇒ 5 = ( ) × 20 × 103
Similarly area of cross-section of 𝐵 = Area of 20 × 103 + 𝑅
cross-section of 𝐶 = 𝑎2 105
⇒ 105 + 5𝑅 = 22 × 105 ⇒ 𝑅 = 21 ×
𝑙 5
Hence according to formula 𝑅 = 𝜌 𝐴 ; resistances
= 420 𝐾Ω
of all the conductors are equal 𝑖. 𝑒. 𝑅𝐴 = 𝑅𝐵 = 𝑅𝐶 139 (b)
134 (a) Faraday constant = 1 mole electron charge = 𝑁𝑒
The first two bands indicate the first two = 6.02 × 1023 × 1.6 × 10−19 = 96500
significant figures of the resistance in ohm. The 140 (c)
third band indicates the decimal multiplier and Potential gradient
the last band stands for the tolerance in percent 𝑒 𝑅
about the indicated value 𝑥= .
(𝑅 + 𝑅ℎ + 𝑟) 𝐿
135 (c) 2 10
𝑒 𝑅 ⟹ 𝑥= ×
Potential gradient 𝑥 = (𝑅+𝑅 +𝑟) . 𝐿 (990 + 10) 2
ℎ
10−3 2 3 = 0.01 Vm−1
⇒ −2 = × ⇒ 𝑅ℎ = 57Ω 141 (a)
10 (3 + 𝑅ℎ + 0) 1
Current sensitivity of a galvanometer is defined as
136 (a)
the deflection produced in the galvanometer
Work done in delivering 𝑞 coulomb of charge
when a unit current flows through it.
from clouds to ground.
If θ is the deflection in the galvanometer when
𝑊 = 𝑉𝑞
current 𝐼 is passed through it, then
= 4 × 106 × 4 = 16 × 106 J
𝜃 𝑛𝐵𝐴
The power of lighting strike is Current Sensitivity 𝐼𝑠 = =
𝐼 𝑘
𝑊 16 × 106 6 where 𝑘 be restoring torque per unit twist, 𝑛 be
𝑃= = = 160 × 10 W
𝑡 0.1 number of turns in the coil, B is strength of
= 160 MW
magnetic field in which coil is suspended, A be
137 (c)
area of coil.
From Joule’s law,
Since, restoring per unit twist (torsional constant)
is minimum for galvanometer A, hence more
P a g e | 52
 sensitive. 1 1 1 2
⟹ = + = =1
142 (a) 𝑅 2 2 2
𝑒 𝑉 𝑒 𝐸𝑙 ⟹ R = 1Ω
𝑣𝑑 = × 𝜏 or 𝑣𝑑 = . 𝜏 [∴ 𝑉 = 𝐸𝑙]
𝑚 𝑙 𝑚 𝑙 149 (c)
∴ 𝑣𝑑 ∝ 𝐸
Here resistances 4 Ω, 6Ω, 12Ω and 24 Ω are in
143 (d)
parallel. Their effective resistances, 𝑅𝑃 will be
Potentiometer works on null deflection method.
In balance condition no current flows in 1 1 1 1 1
= + + +
secondary circuit. 𝑅𝑃 4 6 12 24
144 (c)
6 + 4 + 2 + 1 13 24
𝑉2 (210)2 = = or 𝑅𝑃 =
𝐻= ×𝑡 = × 1 = 𝑚𝐿 24 24 13
𝑅 20
(210)2 Total resistance between 𝐴 and 𝐵
∴ = 𝑚 × 80 × 4.2 ⇒ 𝑚 = 6.56 𝑔/𝑠
20
145 (b) 24 128
=3+ +5= = 9.85 Ω
Here 𝑆 consist of 𝑆1 and 𝑆2 arranged in parallel, 13 13
hence
150 (d)
𝑆1 𝑆2 The resistance AB, BC and CD in series. The total
𝑆= resistance is
𝑆1 + 𝑆2
𝑅1 = 2 + 2 + 2 = 6Ω
𝑃 𝑅 𝑅(𝑆1 +𝑆2 )
So, the balance condition will be = = The resistance AE, EF and FD in series. The total
𝑄 𝑆 𝑆1 𝑆2
resistance is
146 (b) 𝑅2 = 2 + 2 + 2 = 6Ω
𝑑𝐸
Thermo-electric power 𝑃 = 𝑑𝜃 ; at 𝑡𝑛 , 𝐸 → The resistance BE and CF are in effective
∵ 𝑅1 and 𝑅2 are in parallel
maximum
∴ The total resistance
So 𝑃 → zero
6×6
147 (b) 𝑅= = 3Ω
6+6
Total resistance
The current in the circuit
Or 𝑅 = 20 + 40 𝑉 3
𝑅 = 60𝛺 𝐼 = = = 1.0𝐴
𝑅 3
Given G=15V 151 (b)
𝑉 15 𝑙 𝑙 𝑅1 𝑙1 𝑑2 2 𝐿 2𝑑 2
Current 𝐼 = = 𝑅∝ ∝ 2⇒ = ×( ) = ( ) =1
𝑅 60 𝐴 𝑑 𝑅2 𝑙2 𝑑1 4𝐿 𝑑
𝐼 = 0.25𝐴
⇒ 𝑅2 = 𝑅1 = 𝑅
𝑉
Potential gradient = 152 (b)
𝑙 𝑡𝑖 + 𝑡𝑐
20 × 0.25 Neutral temperature, 𝑡𝑛 =
= = 0.5Vm−1 2
10 𝑡𝑖 + 10°
PD across 240 cm ⇒ 285° =
2
𝐸 = 0.5 × 2.4 570° = 𝑡𝑖 + 10°
𝐸 = 1.2𝑉 or 𝑡𝑖 = 560°
148 (c) 153 (a)
Given that the resistance of the total wire is 4Ω. Effective resistance of three resistances between
C
𝐶 and
A B 𝑅 × 2𝑅 2
𝐷= = 𝑅
𝑅 + 2𝑅 3
D
Total resistance between 𝐴 and 𝐵
Here, ACB(2Ω) and ADB (2Ω) are in parallel. 2 8 8
So, the resistance across any diameter is = 𝑅 + 𝑅 + 𝑅 = 𝑅 = × 3 = 8Ω
3 3 3

P a g e | 53
154 (a) circuit is balanced. Hence, 𝐼𝑝 = 𝐼𝑄 and 𝐼𝑅 = 𝐼𝐺
Using 𝑅𝑇2 = 𝑅𝑇1 [1 + 𝛼(𝑇2 − 𝑇1 )]
⇒ 𝑅100 = 𝑅50 [1 + 𝛼(100 − 50)] However, as 𝑃 ≠ 𝑅, hence 𝐼𝑃 ≠ 𝐼𝑅
(7 − 5)
⇒ 7 = 5[1 + (𝛼 × 50)] ⇒ 𝛼 = = 0.008/℃ 161 (d)
250 𝑖𝑔 𝐺 𝐺 𝑖 = 𝑖𝑔 10 − 1 9
155 (d) 𝑆= ⇒ = = =
𝑉2
(𝑖 − 𝑖𝑔 ) 𝑆 𝑖𝑔 1 1
As resistance of a bulb 𝑅 = 𝑃 , 162 (a)
1 1 1
When a cell of emf E is connected to a resistance
Hence 𝑅1 : 𝑅2 : 𝑅3 = 100 : 60 : 60 of 3.9Ω, then the emf E of the cell remains
constant, while voltage V goes on decreasing on
Now the combined potential difference across taking more and more current from the cell.
𝐵1 and 𝐵2 is same as the potential difference
across 𝐵3 . Hence, 𝑊3 is more than 𝑊1 and 𝑊2 , E
being in series, carry same current and 𝑅1 < 𝑅2 ,
𝑟 = 0.1 Ω
therefore 𝑊1 < 𝑊2 ,

∴ 𝑊1 < 𝑊2 < 𝑊3 𝑅 = 3.9 Ω

( )
156 (b)
Potential gradient =
𝑒.𝑅 10×3
= (3+3)×5 ∴ 𝑉 = 𝐸 − 𝑖𝑟
(𝑅+𝑟).𝐿
Where, r is internal resistance.
= 1𝑉/𝑚 = 10 𝑚𝑉/𝑐𝑚 𝐸
Also, current 𝑖 = 𝑅+𝑟
157 (c)
𝐸
Resistance of the wire is given by ∴ 𝑉 = 𝐸−( )𝑟
𝑙 𝑙 2 𝜌𝑙 2 𝑅+𝑟
𝑅=𝜌 =𝜌 = (∵ 𝐴𝑙 = 𝑉) Putting the numerical values, we have
𝐴 𝐴𝑙 𝑉
𝐸 = 2𝑉, 𝑟 = 0.1𝛺, 𝑅 = 3.9𝛺
So,
2
𝑅 ∝ 𝑙2 (if density remains same) 𝑉 =2−( ) × 0.1
3.9 + 0.1
𝑅 ′ (2𝑙)2
𝑜𝑟 = =4 𝑉 = 1.95 Volt
𝑅 (𝑙)2
163 (a)
𝑅 ′ = 4𝑅
Slope of graph
Hence, change in resistance 𝐼 1
=4R-R=3R = =
𝑉 𝑅
change in resistance 3R If experiment is performed at higher temperature
Therefore, = = 3: 1
original resistance 1R then resistance increase and hence slope
158 (d) decrease, choice (a) is wrong.
1
The energy stored in the capacitor = 2 𝐶𝑉 2 ; Similarly in choice (b) and (c) resistance increase.
This energy will be converted into heat in the But for choice (d) resistance R increases, so slope
resistor. decreases
1 164 (b)
𝐻 = × 4 × 10−6 × 400 × 400 The given circuit is
2
= 32 × 10−2
= 0.32 J
159 (c)
When cells are connected in series: emf increase
but current capacity remains unchanged. The emf 1 1 1 1 1 4
= + + ⇒ = ⇒ Req. = 1Ω
of 6 cells in series = 6 × 1.5 = 9 V and current Req. 4 2 4 Req. 4
capacity = 5A h 165 (c)
Let equivalent resistance between 𝐴 and 𝐵 be 𝑅,
160 (a) then equivalent resistance between 𝐶 and 𝐷 will
As galvanometer deflection remains unaffected also be 𝑅
with switch 𝑆 open or closed, hence the bridge
P a g e | 54
 A
1
C 2.9𝑉𝑚−1 (Given)]
′
or 𝐸𝑟𝑚𝑠 = 2.9 × 2𝑉𝑚−1 = 5.8𝑉𝑚−1
R 1 R 170 (a)
Kirchhoff’s first law is based on the law of
1
B D conservation of charge
𝑅 171 (a)
𝑅′ = +2 = 𝑅 𝑉𝐴 − 𝑉𝐵 = emf of the cell = 2 V
𝑅+1
1 1
⇒ 𝑅 2 − 2𝑅 − 2 = 0 ∴ 𝑈 = 2 𝐶𝑉 2 = 2 × 2 × 10−6 × (2)2 = 4 × 10−6 J
2 ± √4 + 8 172 (c)
∴𝑅= = √3 + 1 𝑃 2
2 Here, 𝑄 = 3
166 (a)
The rate of dissipation of electric energy is called we know
𝑃 𝑙
electric power =
𝑊 = 𝑉𝑖𝑡 𝑄 100 − 𝑙
2 𝑙
The electric power dissipated will be is given by ⟹ =
𝑊 𝑉𝑖𝑡 3 100 − 𝑙
𝑃= = = 𝑉𝑖 ⟹ 𝑙 = 40 cm
𝑡 𝑡
𝑉2 173 (d)
= …(i) 2𝐸
𝑅 In series 𝑖 =
2+𝑅
When resistance is doubled, then let electric
power is 𝑃′ . 2𝐸 2
𝑉2 ∴ 𝐽1 = 𝑖 2 𝑅 = ( ) .𝑅
∴ 𝑃′ = 2𝑅
…(ii) 2+𝑅
From Eqs, (i) and (ii), we get
In parallel
1
𝑃′ = 𝑃
2 𝐸
1 𝑖=
So, power becomes 2 of initial value. 0.5 + 𝑅
167 (d) 2
𝐸
Resistance of bulb, 𝐽2 = 𝑖 2 𝑅 = ( ) .𝑅
0.5 + 𝑅
𝑉 2 220 × 220 484
𝑅= = = Ω
𝑃 300 3 𝐽1 4(0.5 + 𝑅)2
(110)2 110×110 = 2.25 =
New power, 𝑃′ = = = 75 watt 𝐽2 (2 + 𝑅)2
𝑅 484/3
reduction Solving we get, 𝑅 = 4 Ω
300−75
Of power = 300
×
100 = 75% 174 (c)
168 (b) Resistance of galvanometer
𝑆 0.01 𝑆 50 𝐺 = 50𝛺
𝑖𝑔 = 𝑖 ⇒ = ⇒𝑆= = 0.05Ω Full scale current 𝑖𝑔 = 0.05
𝐺+𝑆 10 50 + 𝑆 999
169 (d) 𝐴 = 2.97 × 10−2 cm2
The light from bulb spread out uniformly in all = 2.97 × 10−2 × 10−4 m2
directions. = 2.97 × 10−6 m2
For a 100 W bulb, intensity at a distance of 3 m is 𝑖 = 5𝐴
Power 100 𝜌 = 5 × 10−7 Ωm
𝐼= = 2
Area 4𝜋(3) Required resistance to convert the galvanometer
2 2 𝐼
As 𝐼 = 𝜀0 𝑐𝐸𝑟𝑚𝑠 ⇒ 𝐸𝑟𝑚𝑠 = …(i) into ammeter.
𝜀0 𝑐
𝑖𝑔 𝐺 0.05 × 50 2.5
For a 400 W bulb, intensity at the same point is 𝑅= = =
400 𝐼′ 𝑖 − 𝑖𝑔 5 − 0.05 4.95
𝐼′ = ⇒ 𝐸 ′2
𝑟𝑚𝑠 = 𝑙 50
4𝜋(3)2 𝜀0 𝑐 𝜌 =
𝐸𝑟𝑚𝑠 𝐼 ′
′2
400 4𝜋(3)2 𝐴 99
′2 = = × 50 𝐴 50 2.97 × 10−6 50 29.7
𝐸𝑟𝑚𝑠 𝐼 4𝜋(3)2 100 𝑙= × = × = ×
′2 2 99 𝜌 99 5 × 10−7 99 5
𝐸𝑟𝑚𝑠 = 𝐸𝑟𝑚𝑠 × 4 = (2.9)2 × 4 [∵ 𝐸𝑟𝑚𝑠 =
= 10 × 0.3 = 3m
P a g e | 55
176 (a) 𝑉1 𝑅1
𝑑𝑒
∴ =
∵ Peltier coefficient 𝜋 = 𝑇 𝑑𝑇 and 𝑡℃ = 𝑇 − 273 𝑉2 𝑅2
0.02 𝑅
∴ 𝑒 = 𝑎(𝑇 − 273) + 𝑏(𝑇 − 273)2 ⟹ =
𝑑𝑒
2 490 + 𝑅
Differentiating 𝑤. 𝑟. 𝑡. 𝑇 𝑑𝑇 = 𝑎 + 2𝑏(𝑇 − 273) ⟹ 𝑅 = 4.9Ω
𝑑𝑒 180 (c)
𝜋=𝑇 = 𝑇[𝑎 + 2𝑏(𝑇 − 273)] ⇒ 𝜋
𝑑𝑇 Strength = 5 × 18 = 90𝐴𝐻
= (𝑡 + 273)(𝑎 + 2𝑏𝑡) 181 (c)
177 (a) 2𝑅 > 20 ⇒ 𝑅 > 10Ω
The emf of the circuit is 182 (d)
𝐸 = 𝐸1 + 𝐸2 Equivalent resistance of the circuit 𝑅 = 9Ω
= 4𝑉 + 2𝑉 = 6𝑉 ∴ Main current 𝑖 =
𝑉 9
= = 1𝐴
𝑅 9
In the given circuit, 3Ω and 6Ω are connected in 3Ω 2Ω 2Ω
parallel, hence equivalent resistance is
0.25 A
1 1 1 3 1 1A 0.5 A
′
= + = =
𝑅 3 6 6 2 9V 8Ω 8Ω 4Ω
⟹ 𝑅 ′ = 2Ω
2Ω 2Ω 2Ω
Total resistance of circuit is
𝑅 = 1𝛺 + 1𝛺 + 2𝛺 = 6𝛺 After proper distribution, the current through 4Ω
From Ohm’s law 𝑉 = 𝑖𝑅 resistance is 0.25 𝐴
𝑉 6 183 (b)
⟹ 𝑖 = = = 1𝐴
𝑅 6 ∆𝑃 2∆𝑖
The 3Ω and 6Ω resistors are in parallel, hence 𝑃 = 𝑖2𝑅 ⇒ = [𝑅 → Constant]
𝑃 𝑖
𝑖1 𝑅1 = 𝑖2 𝑅2 = 𝑉 ⇒ % change in power = 2 × % change in current
∴ 𝑖1 × 3 = 𝑖2 × 6 = 2 × 1 = 2%
⟹ 𝑖1 = 2𝑖2 𝑎𝑛𝑑 𝑖1 + 𝑖2 = 1 184 (c)
2𝑖2 + 𝑖2 = 1 𝑉 = 𝐸 − 𝑖𝑟 = 12 − 60 × 5 × 10−2 = 9𝑉
1 185 (b)
3𝑖2 = 1 ⟹ 𝑖2 = 𝐴
3 Potential gradient along wire
178 (d) potential difference along wire
1 =
Given, 𝑉 = 10 × 10−6 𝑡 − 40 × 10−6 𝑡 2 length of wire
At neutral temperature 𝐼 × 40
∴ 0.1 × 10−3 = Vcm−1
𝑑𝑉 1000
=0 Current in wire.
𝑑𝑡
1 1 𝐸
∴ 10 × 10−6 − × 10−6 𝑡𝑛 = 0 𝐼= 𝐴 or 𝐼 =
20 400 𝑅 + 𝑅′
or 𝑡𝑛 = 200℃ 2 𝐼
∴ =
Also at neutral temperature, thermo-emf is 40 + 𝑅 400
maximum. Or 𝑅 = 800 − 40 = 760𝛺
Thus, 186 (a)
1 𝐸 = 𝑉 + 𝑖𝑟
𝑉𝑚𝑎𝑥 = 10 × 10−6 (200) − × 10−6 (200)2 𝐸 2
40 After short-circuiting, 𝑉 = 0 ⇒ 𝑟 = 𝑖 = 4 = 0.5Ω
= 2 × 10−3 − 1 × 10−3 = 1 mv 187 (b)
179 (a) In series, the current 𝐼, is same in two bulbs.
When a constant current is passed through a wire 𝑉2
Resistance𝑅 = 𝑃
of uniform cross-section, the potential difference
across any portion of the wire is directly And potential drop (𝑉) = 𝐼𝑅
proportional to the length of that portion. ∴ Potential difference across 60 W bulb is greater
Potential gradient=0.2 mVcm −1 than the potential difference across 200 W bulb.
∴ Potential difference across potentiometer wire 189 (b)
𝑖 𝐺 90
=0.02 × 1= 0.02V ∵ 𝑖𝑔 = 10% of 𝑖 = 10 ⇒ 𝑆 = (𝑛−1) = (10−1) = 10Ω
Total resistance = 𝑟 + 𝑅 = 490 + 𝑅
P a g e | 56
191 (b) 𝑃 2 10 × 1000 2
The circuit can be simplified as follows ∴𝑥=( ) 𝑅=( ) × 0.2
𝑉 200
B C = 500W = 0.5 kW
30
i1
i3 i3 Efficiency of transmission
A D power delivered by line 10 kW
40 40V = =
i2 power supplied to line 10 kW + 𝑥 kW
F E 10 10
40 80V = = = 0.95 = 95%
10 + 0.5 10.5
Applying 𝐾𝐶𝐿 at junction 𝐴 196 (a)
𝑖3 = 𝑖1 + 𝑖2 …(i) A A
Applying Kirchhoff’s voltage law for the loop 2
2
𝐴𝐵𝐶𝐷𝐴 2  2 2

−30𝑖1 − 40𝑖3 + 40 = 0 2 2

⇒ −30𝑖1 − 40(𝑖1 + 𝑖2 ) + 40 = 0 B B

⇒ 7𝑖1 + 4𝑖2 = 4 …(ii) 2×2

Applying Kirchhoff’s voltage law for the loop 𝑅𝐴𝐵 = = 1Ω
2+2
𝐴𝐷𝐸𝐹𝐴 197 (d)
−40𝑖2 − 40𝑖3 + 80 + 40 = 0 Let the resistance of the wire be 𝑅, then we know
⇒ −40𝑖2 − 40(𝑖1 + 𝑖2 ) = −120 that resistance is proportional to the length of the
⇒ 𝑖1 + 2𝑖2 = 3 …(iii) wire. So each of the four wires will have 𝑅/4
On solving equation (ii) and (iii) 𝑖1 = −0.4𝐴 resistance and they are connected in parallel. So
192 (a) the effective resistance will be
𝑑𝐸
Thermo electric power 𝑃 = 𝑑𝜃 = 𝛼 + 𝛽𝜃 1 4 𝑅
= ( ) 4 ⇒ 𝑅1 =
𝑅1 𝑅 16
Comparing it with 𝑦 = 𝑚𝑥 + 𝑐, option (a) is
198 (b)
correct
Here, V<E
193 (b)
∴ 𝐸 = 𝑉 + 𝐼𝑟
As the current and the other factors are same for
For first case
both the galvanometers
12
𝑁 ∝ tan 𝜃 𝐸 = 12 + 𝑟 … . (𝑖)
16
𝑁1 tan 60° √3 For second case
= =
𝑁2 tan 30° 1/√3 11
N1 𝐸 = 11 + 𝑟 … … (𝑖𝑖)
10
⟹ =3
N2 From Eqs. (i) and (ii),
194 (c) 12 11
12 + 𝑟 = 11 + 𝑟
The power of the battery, when charged, is given 16 10
by 20
⟹𝑟= Ω
𝑃1 = 𝑉1 𝐼1 7
199 (a)
The electrical energy dissipated is given by 𝐸1 =
Assume thermo emf 𝐸 is directly proportional to
𝑃1 𝑡1
temperature difference 𝑇.
𝑖. 𝑒. , 𝐸1 = 𝑉1 𝐼1 𝑡1 = 15 × 10 × 8 = 1200 𝑊ℎ
𝑖𝑒, 𝐸 = 𝑎𝑇
Similarly, the electrical energy dissipated during
1
the discharge a battery is given by, 𝑎= = 10−2 VK −1
100
𝐸2 = 𝑉2 𝐼2 𝑡2 = 14 × 5 × 15 = 1050 𝑊ℎ When cold junction is heated by 20 K, the
Hence, watt-hour efficiency of the battery is given temperature
by difference 𝑇 between junction becomes 80K, then
𝐸2
𝜂= × 100 = 0.875 × 100 = 87.5% thermo
𝐸1
emf
195 (d)
𝐸 = 𝑎𝑇 = 10−2 × 80 = 0.8 V
Let 𝑥 watt be the power loss in transmission line
Percentage change of emf
in the form of heat

P a g e | 57
 𝐸 − 𝐸′
= × 100 6/5Ω
𝐸
1 − 0.8
= × 100 = 20%
1
200 (d)
In the given circuit 4Ω resistors are connected in
parallel, this combination is connected in series
6V 2.8Ω
with 1Ω resistance.
6
2Ω 1Ω Hence, 𝑅’’ = 5 Ω + 2.8Ω
= 1.2Ω + 2.8Ω = 4.0Ω
From Ohm’s law,
𝑉 6
6V Current 𝑖 = = = 1.5𝐴
𝑅 4.0
1 1 1 2 1 203 (a)
∴ ′
= + = =
𝑅 4 4 4 2 Because with rise in temperature resistance of
⟹ 𝑅 ′ = 2Ω conductor increases, so graph between 𝑉 and 𝑖
Also, R’’=2 Ω +1Ω =3Ω becomes non linear
From Ohm’s law, 𝑉 = 𝑖𝑅 204 (d)
𝑉 6 𝑃 40 2
∴ 𝑖 = = = 2𝐴 Bulb (I):Rated current 𝐼1 = 𝑉 = 220 = 11 𝑎𝑚𝑝
𝑅 3
𝑉2 (220)2
201 (a) Resistance 𝑅1 = = = 1210Ω
𝑃 40
The circuit may be redrawn as shown in the 100 5
Bulb (II):Rated current 𝐼2 = 220 = 11 𝑎𝑚𝑝
adjacent figure
(220)2
Resistance 𝑅2 = 100
= 484 Ω
2×2
Here 𝐸eq = 12𝑉, 𝑟eq = 2+2 = 1Ω When both are connected in series across 40 𝑉
supply
𝐸eq 12 12
𝑖= = = = 2Ω R1 R2
𝑅 + 𝑟eq 5 + 1 6

5Ω + +
2Ω
12 V 2Ω 12 V 40 V
Total current through supply
40 40 40
𝐼= = = = 0.03𝐴
𝑃1 + 𝑃2 1210 + 484 1254
202 (d)
This current is less than the rated current of each
In the given circuit the resistors of 2Ω and 3Ω are
bulb. So neither bulb will fuse
connected in parallel hence, equivalent resistance
Short Trick : Since 𝑉𝐴𝑝𝑝𝑙𝑖𝑒𝑑 < 𝑉𝑅𝑎𝑡𝑒𝑑′ neither bulb
is
1 1 1 5 will fuse
= + = 205 (c)
𝑅′ 2 3 6
6 From Kirchhoff’s second law
∴ 𝑅′ = Ω
5 𝑉 = ∑𝑖𝑟 (for closed mesh)
Also in steady state, the circuit is shown as. Where V is potential difference, 𝑖 the current and
Resistor’s of 𝑟 the resistance.
6
Ω and 2.8Ω are connected in series. ∴ 𝐸 + 𝐸 = 𝐼𝑟 + 𝐼𝑟 = 2𝐼𝑟
5
𝐸
or 𝐼 = … . (𝑖)
𝑟
𝑉𝑥 − 𝑉𝑦 = 𝐸 − 𝐼𝑟
Putting the value of I from Eq (i), we get
𝐸
𝑉𝑥 − 𝑉𝑦 = 𝐸 − × 𝑉 = 0
𝑟
206 (c)

P a g e | 58
 𝑃2 𝑅 1 𝑖1
Power loss in transmission 𝑃𝐿 = 𝑉2
⇒ 𝑃𝐿 ∝ 𝑉 2 2𝑖 + 2 × − 4(𝑖 − 𝑖1 ) = 0
2
207 (c)
Let 𝜌 is the resistivity of the material 7𝑖1 = 4𝑖
Resistance for contact 𝐴 − 𝐴 7 7 7
𝑥 𝜌 or 𝑖 = 𝑖1 = × 2 = A
𝑅𝐴𝐴 = 𝜌 = 4 4 2
2𝑥 × 4𝑥 8𝑥
Similar for contacts 𝐵 − 𝐵 and 𝐶 − 𝐶 are Total resistance of the circuit between 𝐴 and 𝐻 is
respectively
2𝑥 𝜌 4𝜌 4 × 3 26
𝑅𝐵𝐵 = 𝜌. = = =2+ =
𝑥 × 4𝑥 2𝑥 8𝑥 4+3 7
4𝑥 2𝜌 16𝜌
and 𝑅𝐶𝐶 = 𝜌 𝑥×2𝑥 = 𝑥 = 8𝑥 7 26
EMF of cell is 𝐸 = × = 13V
2 7
It is clear maximum resistance will be for contact
𝐶−𝐶 212 (a)
208 (a) Resistance of 𝑉1 = 80 × 200
=16000Ω
Resistance of 𝑉2 =32000Ω
The current flowing in the circuit is given by
80 1
𝑖= = 𝐴
𝐺𝑆 𝑉𝐺 25 × 10−3 16000 200
= = Total resistance of the circuit
𝐺+𝑆 𝐼 25
𝐺𝑆 =16000+32000=48000Ω
= 0.001Ω
𝐺+𝑆 Line voltage= 𝑖𝑅
Here S << G So 48000
= = 240𝑉
𝑆 = 0.001 Ω 200
209 (c) 213 (a)
Resistance are in parallel Mass of substance liberated at cathode 𝑚 = 𝑧𝑖𝑡
𝑅 Where, 𝑧 = electro chemical equivalent
∴ 𝑅𝑒𝑞 =
3 = 3.3 × 10−7 kg . C −1
210 (d) 𝑖 = current flowing = 3 A,
The temperature difference is 20℃ = 20 𝐾. So 𝑡 = 2s
𝜇𝑉
that thermo 𝑒mf developed 𝐸 = αθ = 40 × 𝑚 = 3.3 × 10−7 × 3 × 2
𝐾
20𝐾 = 800 𝜇𝑉 = 19.8 × 107 kg
Hence total 𝑒mf = 150 × 800 = 12 × 104 𝜇𝑉 = 214 (c)
120 𝑚𝑉 Current through resistance 𝑅 will be zero if
211 (b) 𝐸 𝐸1 𝐸(𝑅1 + 𝑅2 )
The distribution of current is as shown in figure. = or 𝐸1 =
𝑅2 𝑅1 + 𝑅2 𝑅2
As per question,
215 (a)
2Ω 2Ω 1Ω When temperature is raised, the ions/atoms of
A B C D
the conductor start vibrating with increased
i i - i1 i1 i1/2
i1/2 amplitude of vibration and greater frequency. Due
E 4Ω 2Ω 1Ω to which the electrons moving towards the
positive end of conductor will suffermore rapid
1A
collisions and hence time of relaxation
H i G i1 F i 1/2 E1 (𝜏)decreaseds. As 𝑣𝑑 ∝ 𝜏, thus drift velocity
decrease. Therefore 𝑣𝑑 ∝ 1/𝑇
𝑖1
= 1 or 𝑖1 = 2A 216 (b)
2 3 = 1.5(1 + 𝑟) ⇒ 𝑟 = 1Ω
In a closed circuit 𝐴𝐶𝐹𝐺 217 (b)

P a g e | 59
 Just for your knowledge remember, voltaic cell Given, 𝐼 = 1 A, 𝑡 = 10 s, 𝑞 = 𝐼𝑡, 𝑞 = 10 C
uses dil. 𝐻2 𝑆𝑂4 ; Dry cell uses 𝑁𝐻4 𝐶𝑙 + 𝑍𝑛𝐶𝑙2
paste; Daniel cell uses dil. 𝐻2 𝑆𝑂4 ; Lead Charge of Cu2+ = 2𝑒 = 2 × 1.6 × 10−19 C
Accumulator uses dil. 𝐻2 𝑆𝑂4 and 𝑁𝑖-𝐹𝑒 cell or The number of copper atoms deposited at the
Alkaline Accumulator uses 𝐾𝑂𝐻 solution cathode
218 (c)
2202 10
𝑅bulb = = 484Ω; = = 3.1 × 1019
100 2 × 1.6 × 10−19
2202
𝑅geyser = = 48.4Ω 225 (b)
1000
When only bulb is on, According to Kirchhoff’s law 𝑖𝐶𝐷 = 𝑖2 + 𝑖3
220×484 226 (a)
𝑉bulb = 484+6 = 217.4 volt
At point 𝐴 the slope of the graph will be negative.
When geyser is also switched on, effective Hence resistance is negative
resistance of bulb and geyser 227 (d)
484 × 48.4
= = 44 Ω 6Ω and 6Ω are in series, so effective resistance is
484 + 48.4 12Ω which is in parallel with 3Ω, so
220×44
𝑉bulb = (44+) = 193.6 V 1 1 1 15
= + =
Hence, the potential drop = 217.4 × 193.6 𝑅 3 12 36
= 23.8 V≈ 24 V. 36
⇒𝑅=
219 (b) 15
𝑉 4.8 × 15
𝑣𝑑 ∝ 1/𝑙. Therefore, drift velocity is halued ∴𝐼= = = 2𝐴
𝑅 36
220 (d) 228 (a)
4×4 6×6 Meter bridge is an arrangement which works on
Equivalent resistance = + = 5𝑜ℎ𝑚 So the
4+4 6+6 Wheatstone’s principle, so the balancing
20
current in the circuit = = 4 𝑎𝑚𝑝𝑒𝑟𝑒 Hence the condition is
5
current flowing through each resistance = 𝑅 𝑙1
=
2 𝑎𝑚𝑝𝑒𝑟𝑒 𝑆 𝑙2
221 (c) Where 𝑙2 = 100 − 𝑙1
Length of the wire is Ist case 𝑅 = 𝑋, 𝑆 = 𝑌, 𝑙1 = 20 cm, 𝑙2 = 100 − 20 =
𝑅𝐴 80cm
𝑙= 𝑋 20
𝜌
−3 2
∴ = … (i)
4 × 𝜋 × (0.7 × 10 ) 𝑌 80
∴𝑙= = 280m IInd Case Let the position null point is obtained at
2.2 × 10−8
222 (c) a distance 𝑙 from same end.
When key 𝐾 is opened, bulb 𝐵2 will not draw any ∴ 𝑅 = 4𝑋, 𝑆 = 𝑌, 𝑙1 = 𝑙, 𝑙2 = 100 − 𝑙
current from the source, so that terminal voltage So, from Eq. (i)
of source increases. Hence, power consumed by 4𝑋 𝑙
=
bulb increases, so light of the bulb becomes more. 𝑌 100 − 𝑙
𝑋 𝑙
The brightness of bulb 𝐵1 decreases. ⟹ = … (ii)
223 (a) 𝑌 4(100 − 𝑙)
Therefore, form Eqs. (i) and (ii)
Given, 𝐸1 = 1.5𝑉, 𝑙1 = 27 𝑐𝑚,
𝑙 20
𝑙2 = 54 𝑐𝑚, 𝐸2 =? =
𝐸1 𝑙1 4(100 − 𝑙) 80
= 𝑙 1
𝐸2 𝑙2 ⟹ =
𝐸1 𝑙2 4(100 − 𝑙) 4
or 𝐸2 = ⟹ 𝑙 = 100 − 𝑙
𝑙1
1.5 × 54 ⟹ 2𝑙 = 100
or 𝐸2 = Hence, 𝑙=50cm
27
𝐸2 = 3𝑉 229 (d)
224 (b) Since it’s a balanced Wheatstone bridge, the

P a g e | 60
 circuit can be redrawn as Heat produced by heater per second = 1.08 ×
103 J
Heat taken by water to form steam 𝑚𝐿
= 100 × 540 cal
= 100 × 540 × 4.2 J
12𝐼 = 30(1.4 𝐼) ∴ 1.08 × 103 × 𝑡 = 100 × 540 × 4.2
100×540×4.2
12𝐼 = 42 − 30𝐼 or 𝑡 = 1.08×103 = 210 s
∴𝐼 =1𝐴 235 (b)
230 (d) Electric power consumed by kettle 𝑃 = 220 × 4𝑊
The last two resistances are out of circuit. Now 8Ω Heat required
is in parallel with (1 + 1 + 4 + 1 + 1)Ω 𝐻 = 1000 × 1(100 − 20) = 1000 × 80 𝑐𝑎𝑙
8
∴ 𝑅 = 8Ω||8Ω = = 4Ω ⇒ 𝑅𝐴𝐵 = 4 + 2 + 2 = 8Ω = 4200 × 80 𝐽
2 𝐻
231 (c) 𝑃 = ⇒𝐻 =𝑃×𝑡
𝑡
Error in measurement = Actual value – Measured ∴ 220 × 4 × 𝑡 = 4200 × 80 ⇒ 𝑡 = 6.3 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
value 236 (b)
Actual value = 2𝑉 As 5Ω resister is joined in parallel to series
998 
V combination of 4Ω and 6Ω (𝑖𝑒, total resistance
10Ω), V =constant.
i
𝑖1 𝑅2 10
and = = =2
𝑖2 𝑅1 5
+ – 2
2V 𝑖1
2 or 𝑖2 = 2
𝑖=
998 + 2
1 Now heat produced per second in5Ω resistor
= 𝐴
500
𝐻1 = 𝑖12 𝑅1 = 𝑖12 × 5 = 100Js −1 …(i)
Since 𝐸 = 𝑉 + 𝑖𝑟
1 998 and for 4Ω resistor
⇒ 𝑉 = 𝐸 − 𝑖𝑟 = 2 − ×2= 𝑉
500 500
998 2
∴ Measured value = 𝑉 𝑖
𝐻2 = 𝑖22 𝑅2 = ( 21 ) × 4 = 𝑖12 …(ii)
500
998
⇒ Error = 2 − 500 = 4 × 10−3 𝑣𝑜𝑙𝑡
232 (b) Simplifying Eqs. (i) and (ii), we get
𝐻 = 𝐼 2 𝑅𝑡 = 𝑚𝑐∆θ 𝐻2 1 1
∴ ∆θ ∝ 𝐼 2 = or 𝐻2 = × 100 = 20Js −1
100 5 5
∆θ2 𝐼 2 ∆θ2 2𝐼+𝐼 2
Hence ∆θ = ( 𝐼2 ) or 3
=( 𝐼
) 237 (d)
2 1
or ∆θ2 = 9 × 3 = 27℃. To convert a moving coil galvanometer into a
233 (a) voltmeter, a high resistance is connected in series
𝑉2 with it.
Resistance of bulb 𝑅 = 𝑃
238 (d)
1 Resistivity is the property of the material. It does
𝑅∝ not depend upon size and shape
𝑃
239 (a)
Here 𝑃𝑋 = 40 W, and 𝑃𝑌 = 60 W By the concept of balanced Wheatstone bridge,
∴ 𝑅𝑋 > 𝑅𝑌 the given circuit can be redrawn as follows

So, potential drop across bulb 𝑋 𝑖𝑒, of 40 W bulb

will be greater and it will glow brighter.

234 (b)

P a g e | 61
 30
5 10 15
R' R'

A B

10 20 30 R' R'

60

30 × 60 Therefore, effective resistance between A and B,

⇒ 𝑅𝐴𝐵 = = 20 Ω 1 1 1 1
(30 + 60)
= ′+ ′= ′
240 (a) 𝑅𝐴𝐵 2𝑅 2𝑅 𝑅
Here three resistance of 4Ω each are connected in 2𝑅
⟹ 𝑅𝐴𝐵 = 𝑅 ′ = Ω
4
parallel so that their combined resistance = 3 Ω. It 3
244 (b)
is in series with ammeter, battery and last 4Ω Let resistivity at a distance ′𝑥 ′ from left end be
resistance. 𝜌 = (𝜌0 + 𝑎𝑥). Then electric field intensity at a
𝑖𝜌
4 16 distance ′𝑥 ′ from left and will be equal to 𝐸 = 𝐴
=
∴ Net resistances 𝑅 = +4= Ω
3 3 𝑖(𝜌0 +𝑎𝑥)
𝐴
∴ Current in main circuit =ammeter reading Where 𝑖 is the current flowing through the
conductor. It means 𝐸 ∝ 𝜌 or 𝐸 varies linearly
𝐸 2V 3
𝑖= = 16 = A with distance ′𝑥 ′ . But at 𝑥 = 0, 𝐸 has non-zero
𝑅 Ω 8
3 value. Hence (𝑏) is correct
245 (b)
241 (c) 2𝐸
In series, 𝑖1 =
𝑃 = 100 watt, 𝑉 = 125 𝑉 2+2𝑟
𝐸 2𝐸
𝑃 100
Since 𝑃 = 𝑉𝐼, ∴ 𝐼 = 𝑉 = 125 𝑎𝑚𝑝𝑒𝑟𝑒 In parallel, 𝑖2 = 𝑟 = 4+𝑟
2+
2
Mass of chlorine liberated = 𝑧𝐼𝑡 2𝐸 2𝐸
Since 𝑖1 = 𝑖2 ⇒ 4+𝑟 = 2+2𝑟 ⇒ 𝑟 = 2Ω
100
= 0.367 × 10−6 × × 60 = 0.0176 × 10−3 𝑘𝑔 246 (b)
125
= 17.6𝑚𝑔 Order of drift velocity = 10−4 𝑚/𝑠𝑒𝑐 =
242 (c) 10−2 𝑐𝑚/𝑠𝑒𝑐
Initially the inductance will oppose the current 247 (a)
which tries to flow through the inductance. But 𝑉2 (200)2 4 × 104
𝑃= ⇒ 100 = ⇒𝑅= = 400 Ω
10Ω and 20Ω can conduct. The current will be 𝑅 𝑅 102
𝑉 100 1
2𝑉 1 Now, 𝑖 = 𝑅 = 400 = 4 𝑎𝑚𝑝
= 𝐴
30Ω 15 248 (c)
243 (d) 𝐸1 =? , 𝑙1 = 60𝑐𝑚; 𝐸2 = 3𝑉, 𝑙2 = 45𝑐𝑚
Resistance R bisecting the circuit can be neglected In balance condition
due to the symmetry of the circuit. 𝐸1 𝑙1 𝐸1 60
Now, there are four triangles = ⇒ = ⇒ 𝐸1 = 4 𝑣𝑜𝑙𝑡
𝐸2 𝑙2 3 45
Effective resistance of each triangle 249 (d)
1 1 1 𝑚 𝑙
= + Resistance of a conductor, 𝑅 =
𝑅 ′ 𝑅 2𝑅 𝑛𝑒 2 𝜏 𝐴
2+1 3 Where the symbols have their usual meaning
= = As the temperature increases, the relaxation time
2𝑅 2𝑅
2 𝜏 decreases
∴ 𝑅′ = 𝑅
3 250 (d)
Now the given circuit reduced to 1
𝐸 = 𝛼𝑡 + 𝛽𝑡 2 , graph between 𝐸 and 𝑡 will be a
2
parabola, such that first emf increases and then
decreases
251 (b)
In general, ammeter always reads less than the

P a g e | 62
 actual value because of its resistance two metals placed in an electrolyte.
252 (c) 257 (b)
𝐸 2.5 2.5×10 25 The temperature of the wire increase to such a
𝐼= = and 𝑉 = 𝐼. 𝑅 = =
𝑅𝑇 10+𝑅 10+𝑅 10+𝑅
value at which the heat produced per second
𝑉 25
𝑥= = equals heat lost per second due to radiation.
𝐿 (10 + 𝑅)𝐿
𝐸 = 𝑥. 𝑙1 𝜌𝑙
25 𝐿 𝑖 2 ( 2 ) = 𝐻 × 2𝜋𝑟𝑙
⇒1= × ⇒ 25 = 20 + 2𝑅 𝜋𝑟
(10 + 𝑅)𝐿 2
5 Where 𝐻 is heat lost per second per unit area due
⇒ 2𝑅 = 5 ⇒ 𝑅 = to radiation.
2
∴ Now the resistance is doubled
5 Hence, 𝑖 2 ∝ 𝑟 3
𝑅1 = × 2 = 5Ω
2 𝑖12 𝑟3
25 25 5 So, 𝑖22
= 𝑟13
∴𝑥= = = 2
(10 + 5)𝐿 15. 𝐿 3𝐿
2
𝐸 = 𝑥. 𝑙2 𝑖 3
Or 𝑟2 = 𝑟1 ( 2 )
5 3𝐿 𝑖1
⇒1= . 𝑙2 ⇒ 𝑙2 = = 0.6𝐿
3𝐿 2
253 (c) [Here ∶ 𝑟1 = 1 mm, 𝑖1 = 1.5 A, 𝑖2 = 3 A]
Approximate change in resistance=2 × % change
3 2/3
in length by stretching 𝑟2 = 1 × ( ) = 41/3 mm
1.5
254 (c)
258 (b)
The total emf = 𝐸 + 𝐸 = 2𝐸 𝑅2 𝑅3 4×4
Total resistance =𝑅 + 𝑟1 + 𝑟2 𝑅𝐴𝐵 = 𝑅1 + + 𝑅4 = 2 + + 2 = 6Ω
𝑅2 + 𝑅3 4+4
∴ Current flowing through the circuit 259 (c)
2𝐸
𝑖= When the key is in contact with 1, then energy
𝑅 + 𝑟1 + 𝑟2
stored in
According to question 𝐸 = 𝑖𝑟1 1
E the condenser = 2 𝐶𝐸 2
⟹ i= But when the key is thrown to contact 2, total heat
r1
𝐸 2𝐸 1
∴ = 𝐻 = 𝐼 2 (500 + 330) = 𝐶𝐸 2
𝑟1 𝑅 + 𝑟1 + 𝑟2 2
𝑅 = 𝑟1 − 𝑟2 𝐻1 = 𝐼 2 (500)
𝐻1 𝑅1
255 (d) =
Resistance of original wire is 𝐻 (𝑅1 + 𝑅2 )
𝑙 500 1
𝑅=𝜌 𝐻1 = × × 5 × 10−6 × (200)2
𝐴 830 2
𝜌 ,being the specific resistance of wire. When the 𝐻1 = 60 × 10−3 J
wire is cut in two equal halves then resistance 260 (a)
becomes 𝐻 𝑖 2 𝜌𝑙
𝐻 = 𝑖 2 𝑅𝑡 ⇒ = 𝑖 2 𝑅 =
𝜌𝑙/2 𝑅 𝑡 𝜋 𝑟2
𝑅′ = = 261 (d)
𝐴 2
𝑖 𝐺 90
Thus, the net resistance of parallel combination of 𝑖𝑔 = 10 ⇒ Required shunt 𝑆 = (𝑛−1) = (10−1) =
two halves is given by 10 Ω
𝑅′ × 𝑅′ 262 (b)
𝑅net = ′
𝑅 + 𝑅′ Given current through 4Ω resistance
𝑅′ 𝑅 6
= = = = 1.5Ω Is 1𝐴, so P.D. across upper Branch
2 2×2 4
𝑖. 𝑒. P.D. between 𝑃 and 𝑀 is 4𝑉
256 (a)
Hence P.D. between 𝑀 & 𝑁 is
The relative position of metals in the electro
1
chemical series determines the emf between the × 4 = 3.2𝑉
1 + 0.25
P a g e | 63
 bamboo.
269 (c)
𝑖 = 𝑛𝐴𝑒𝑣𝑑
𝑖 1
or 𝑣𝑑 = 𝑛𝐴𝑒 𝑖𝑒, 𝑣𝑑 ∝ 𝐴

As 𝐴 increases 𝑣𝑑 decreases, because 𝑖 remains

264 (b)
1 𝑃1 𝑅2 200 𝑅2 constant
𝑃∝ ⇒ = ⇒ = ⇒ 𝑅2 = 2𝑅1
𝑅 𝑃2 𝑅1 100 𝑅1 270 (a)
265 (b) 𝑇𝑖 + 𝑇𝑐
Applying Kirchhoff’s law for the loops (1) and (2) 𝑇𝑛 =
2
as shown in figure 𝑇𝑖 = 2𝑇𝑛 − 𝑇𝑐 = 540℃
R1 = 2Ω i1 271 (d)
E1 = 4V
𝑉 = 𝐸 − 𝑖𝑟 = 1.5 − 2 × 0.15 = 1.20 𝑉𝑜𝑙𝑡
i1
1 (i1 – i2) 272 (d)
𝐸1 𝑙1 + 𝑙2 (6 + 2) 2
= = =
i2
R3 = 2Ω
𝐸2 𝑙1 − 𝑙2 (6 − 2) 1
2 i2
R2 = 4Ω 273 (d)
E2 = 6V Charge supplied per minute = 3.2 × 60 = 192 𝐶
Charge 2𝑒 liberates one 𝐶𝑢+2 ion
For loop (1)
∴ No of 𝐶𝑢+2 ion liberate by 192 𝐶
−2𝑖1 − 2(𝑖1 − 𝑖2 ) + 4 = 0 192 192
⇒ 2𝑖1 − 𝑖2 = 2 …(i) = = = 6 × 1020
2𝑒 2 × 1.6 × 10−19
For loop (2) 274 (d)
−2(𝑖1 − 𝑖2 ) + 4𝑖2 − 6 = 0 1 𝑅1 𝑃2 60 3
⇒ −𝑖1 + 3𝑖2 = 3 …(ii) 𝑃𝑅𝑎𝑡𝑒𝑑 ∝ ⇒ = = =
𝑅 𝑅2 𝑃1 40 2
On solving equation (i) and (ii), 𝑖1 = 1.8𝐴 275 (a)
266 (d) Neutral temperature remains same.
Pressing the key does not disturb current in all
resistances as the bridge is balanced. Therefore, Inversion temperature, 𝑇𝑖 = 2𝑇𝑛 − 𝑇𝑐
deflection in the galvanometer in whatever
= 2 × 285 − 10 = 560℃
direction it was, will stay
267 (b) 276 (d)
Let the original resistance is R Ω. 100 − 𝑙
∴ 𝑉 = 𝐼𝑅 𝑆=( ).𝑅
𝑙
𝑉 = 5 × 𝑅 = 5𝑅 …(i) 100−𝑙
Initially, 30 = ( 𝑙 ) × 10 ⇒ 𝑙 = 25𝑐𝑚
When 2 Ω resistance is inserted, then total 100−𝑙
resistance Finally, 10 = ( 𝑙
)× 30 ⇒ 𝑙 = 75𝑐𝑚
=(R+2) Ω So, shift = 50𝑐𝑚
∴ 𝑉 = 𝐼’ (𝑅 + 2) = 4(𝑅 + 2) …(ii) 277 (d)
From Eqs. (i) and (ii), we get 6 × 1.5
𝑅total = 2 + = 3.2kΩ
5𝑅 = 4(𝑅 + 2) 6 + 1.5
∴ R=8 Ω 24V
(a)𝐼 = = 7.5mA = 𝐼𝑅1
268 (b) 3.2𝑘Ω
𝑅𝐿
The moving coil galvanometer have their coil 𝐼𝑅2 = ( )𝐼
𝑅𝐿 + 𝑅2
wound on a metallic (copper or aluminium)
1.5
frame, so as to make the motion dead beat due to 𝐼= × 7.5 = 1.5mA
7.5
the production of eddy currents. In the ballistic
𝐼𝑅𝐿 = 6mA
galvanometer, on the other hand, the damping is
(𝑏)𝑉𝑅𝐿 = (𝐼𝑅𝐿 )(𝑅𝐿 ) = 9𝑉
to be reduced to the minimum and hence the
frame is of a non-conducting material 𝑒𝑔, paper or
P a g e | 64
 𝑃𝑅1 (𝐼𝑅21 )𝑅1 (7.5)2 (2) 25 1
(𝑐) = = = ∴ [ εo 𝐸 2 ] = [M −1 L−3 T 4 A2 ] × [MLT −3 A−1 ]2
𝑃𝑅2 (𝐼𝑅2 )𝑅2 (1.5)2 (6) 3 2
2
(d) When 𝑅1 and 𝑅2 are interchanged, then = [ML−1 T −2 ]
𝑅2 𝑅𝐿 2 × 1.5 6
= = 𝑘Ω 282 (a)
𝑅2 + 𝑅𝐿 3.5 7
Now potential difference across 𝑅𝐿 will be A voltmeter is an instrument used to measure the
6 potential difference between two points in an
7 electrical circuit directly in volts. Voltmeter is
𝑉𝐿 = 24 [ 6 ] 3𝑉
6+7 essentially a galvanometer which is connected in
Earlier it was 9V parallel across two points in the circuit between
𝑉2 which the potential difference is to be measured.
Since, 𝑃 = 𝑅
or 𝑃 ∝ 𝑉2
The potential difference read by the voltmeter is
In new situation potential difference has been
slightly less than the actual value to be measured.
decreased three times. Therefore, power
Hence, the resistance of the voltmeter should be
dissipated will decrease by a factor of 9.
as high as possible so, that on connecting it in a
278 (c)
circuit across two points the potential difference
(Rated voltage)2
Resistance of a bulb = (Rated power)
may not fall appreciably.
(220)2 (220)2
𝑅𝐵1 = Ω and 𝑅𝐵2 = Ω | V |
40 60
When the bulbs are connected in series, a b

𝑅𝑆 = 𝑅𝐵1 + 𝑅𝐵2
i
(220)2 (220)2
g
1 1
= + = (220)2 [ + ]
40 60 40 60
60 + 40 100 (220)2
= (220)2 [ ] = (220)2 ( )= R
60 × 40 2400 24 G
𝑉𝑆2 24
∴ 𝑃1 = = (220)2 × = 24𝑊
𝑅𝑠 (220)2 Voltmeter

When the bulbs are connected in parallel

1 1 1 1 40 60 Hence, a voltmeter is made by connecting a high
= + ⇒ = + resistance in series with a pivoted type moving
𝑅𝑝 𝑅𝐵1 𝑅𝐵2 𝑅𝑃 (220)2 (220)2
coil galvanometer G. The value of R depends upon
1 100 (220)2
= or 𝑅𝑝 = the range of the required voltmeter.
𝑅𝑝 (200)2 100
283 (a)
2
𝑉𝑆 100 𝑅1 𝑃2 100 5 5
∴ 𝑃2 = = (220)2 × = 100 𝑊 = = = . Resistance of 40 𝑊 bulb is
𝑅𝑝 (200)2 𝑅2 𝑃1 40 2 2

𝑃1 24 𝑊 times than 100 𝑊. In series, 𝑃 = 𝑖 2 𝑅 and in

∴ = = 0.24 𝑉2
𝑃2 100 𝑊 parallel, 𝑃 = . So 40 𝑊 in series and 100 𝑊 in
𝑅
279 (c) parallel will glow brighter
𝑙 2𝑙 𝑙 284 (c)
𝑅1 ∝ ⇒ 𝑅2 ∝ 𝑖. 𝑒. 𝑅2 ∝
𝐴 2𝐴 𝐴 𝐸
∴ 𝑅1 = 𝑅2 𝐼=
𝑅+𝑟
280 (d) 𝐸
𝑅 𝑙 2 𝐼 = = constant
In Stretching of wire 𝑅 ∝ 𝑙 2 ⇒ 𝑅1 = (𝑙1 ) 𝑅
2 2 where, R=external resistance
𝑅1 100 2
If 𝑙1 = 100, then 𝑙2 = 110 ⇒ = ( ) r= internal resistance =0
𝑅2 110
285 (c)
⇒ 𝑅2 = 1.21 𝑅1
Regarding Kirchhoff’s junction rule, the circuit can
Resistivity doesn’t change with stretching
be redrawn as
281 (c)
∴ [εo ] = [M −1 L−3 T4 A2 ] and [E] = [MLT −3 A−1 ]

P a g e | 65
 Now length 𝑙1 = 𝑙 + 𝑙/10 = 11 𝑙/10

1A ∴ New area 𝐴1 = 𝐴𝑙/𝑙1

B
∴ New resistance,

𝑅1 = ρ𝑙1 ⁄𝐴1 = ρ(11/10)/(10/11)𝐴

10 A I
C 121 ρ𝑙 121
A = = × 10 = 12.1 Ω
6A 100 𝐴 100

295 (a)
D
𝑅𝐴 𝑟𝐵 4 𝑅𝐴 1 4 1
2A =( ) ⇒ =( ) = ⇒ 𝑅𝐵 = 16𝑅𝐴
𝑅𝐵 𝑟𝐴 𝑅𝐵 2 16
When 𝑅𝐴 and 𝑅𝐵 are connected in parallel then
equivalent resistance
Current in arm, 𝐴𝐵 = 10 − 6 = 4A 𝑅𝐴 𝑅𝐵 16
Current in arm, 𝐷𝐶 = 6 + 2 = 8A 𝑅𝑒𝑞 = = 𝑅
(𝑅𝐴 + 𝑅𝐵 ) 17 𝐴
Current in arm, 𝐵𝐶 = 4 + 1 = 5A If 𝑅𝐴 = 4.25Ω then 𝑅𝑒𝑞 = 4Ω 𝑖. 𝑒. option (a) is
Hence, 𝐼 = 5 + 8 = 13A correct
286 (a) 296 (b)
Specific resistance is independent of dimensions 𝑉
As 𝑚 = 𝑧 𝑙 𝑡 = 𝑧 ( 𝑅) 𝑡 𝑖𝑒, 𝑚 ∝ 𝑉𝑡
of conductor but depends on nature of conductor.
𝑚2 𝑉𝑡
∴ 𝑚1
= 𝑉2 𝑡2
288 (b) 1 1
𝑉𝑡 6×45×2
𝜌𝐿 𝜌×1 or 𝑚2 = 𝑉2 𝑡2 × 𝑚1 = 12×30
=1.5g.
𝑅= ⇒ 0.7 = 22 1 2
𝐴 (1 × 10−3 )2 297 (c)
7
𝜌 = 2.2 × 10−6 𝑜ℎ𝑚 − 𝑚 𝐸 − 𝑉 = 𝑖𝑟 or 𝑟 = (𝐸 − 𝑉)𝑖
289 (c)
and 𝑉 = 𝑖𝑅 = 4.5 × 10 = 45V
Current through a conductor is constant at even
cross-section of the conductor (50 − 45) 5
∴ 𝑟= =
4.5 4.5
290 (c)
𝑃
Power of the combination 𝑃𝑠 = 𝑛 =
1000
= 500W = 1.10 Ω
2
291 (b) 298 (a)
𝑖 2 𝑅𝑡 = 𝐶𝜃 = 3𝐶; 𝐶 = Thermal capacity Both 𝑅 and 2𝑅 are in parallel [𝑉 − constant]
when 𝑖1 = 2𝑖 ⇒ 𝐶𝜃1 = 4𝑖 2 𝑅𝑡 = 4 × 3𝐶 ⇒ 𝜃1 = 𝑉2 𝑃1 𝑅2 𝐻1 𝑅2 2
So using 𝑃 = ⇒ = ⇒ = =
12℃ 𝑅 𝑃2 𝑅1 𝐻2 𝑅1 1

292 (b) 299 (b)

𝑉 2𝑡
In the part 𝑐 𝑏 𝑑, We know, 4.2 𝑅 = 𝑄 = 𝑚𝑠 𝑑θ
𝑉𝑐 + 𝑉𝑑 Let,𝑁 = initial number of turns
𝑉𝑐 − 𝑉𝑏 = 𝑉𝑏 − 𝑉𝑑 ⇒ 𝑉𝑏 =
2 𝑅 = resistance of the coil
In the part 𝑐 𝑎 𝑑 𝐿 𝜌 × 𝑁 × 2𝜋𝑟
𝑉𝑐 + 𝑉𝑑 ⇒𝑅=𝜌 =
𝑉𝑐 − 𝑉𝑎 > 𝑉𝑎 − 𝑉𝑑 ⇒ > 𝑉𝑎 ⇒ 𝑉𝑏 > 𝑉𝑎 𝐴 𝐴
2 𝑉 2𝑡
293 (b) = 𝑄 = 𝑚𝑠 𝑑θ
𝑃 4.2 × 𝜌 × 𝑁 × 2𝜋𝑟
𝑚 = 𝑧 𝐼 𝑡 = 𝑧( )𝑡 𝑡
𝑉 = Constant
𝑁
−6 100×1000
= 0.367 × 10 ( 125 ) × 60 kg 𝑡1 𝑡2
∴ =
𝑁1 𝑁2
= 0.017616 kg = 17.616 g. 𝑁2 9
294 (d) ⇒ = × 𝑡1 = × 16 =14.4 min.
𝑁1 10
𝑅 = ρ𝑙 ⁄𝐴 = 10 300 (b)
𝐸 = 16𝑇 − 0.04𝑇 2
P a g e | 66
 At temperature of inversion, 𝐸 = 0 305 (b)
∴ 16𝑇𝑖 − 0.04𝑇𝑖2 = 0 𝑞 = 𝑖𝑡 = 𝑛 × 2𝑒
16
⇒𝑇= = 400℃ 𝑖𝑡 2 × 32
0.04 𝑛= = = 2 × 1020
301 (a) 2𝑒 2 × 1.6 × 10−19
𝑚 2.68 2.68 965 306 (c)
𝐼= = 108 = × ≈4𝐴
𝑍𝑡 × 10 × 60 108 6 For portion 𝐶𝐷 slope of the curve is negative 𝑖. 𝑒.
96500
Energy = 𝐼 2 𝑅𝑡 = 42 × 20 × 600 = 192 𝑘𝐽 resistance is negative
302 (a) 307 (a)
𝑉𝑎𝑏 = 𝑖𝑔 . 𝐺 = (𝑖 − 𝑖𝑔 )𝑆 Heat gained by water = Heat supplied by
S container heat lost
⇒ 𝑚𝑆∆𝜃 = 1000𝑡 − 160𝑡
2 × 4.2 × 1000 × 50
⇒𝑡= = 8 𝑚𝑖𝑛 20 𝑠𝑒𝑐
840
( i - ig ) 308 (d)
𝑙 1 𝐴 𝑑2
𝑅 = 𝜌 𝐴 and 𝑃 ∝ 𝑅 ⇒ 𝑃 ∝ 𝑙
⇒𝑃∝ 𝑙
⇒ 𝑃𝐴 = 2𝑃𝐵
g
i
G 309 (c)
i a ig b 𝑉2
As for an electric appliance 𝑅 = 𝑃
.
𝐺 For first bulb, its resistance
∴ 𝑖 = (1 + ) 𝑖𝑔 𝑉2 250×250
𝑆 𝑅2 = = = 625 Ω
𝑃1 100
Substituting the values we get,
𝑖 = 100.1mA For second bulb, its resistance
303 (c) 𝑉22 200 × 200
𝑅2 = =
Given 𝑉1 = 50 𝑣𝑜𝑙𝑡, 𝑖1 = 11𝐴; 𝑉2 = 60 𝑣𝑜𝑙𝑡, 𝑖2 = 𝑃2 100
1𝐴 = 400 Ω
If 𝑒. 𝑚. 𝑓. and internal resistance of battery are 𝐸 Now, in series potential divides in proportion to
and 𝑟 respectively then 𝑃. 𝐷. across terminals of resistance.
𝑅2
battery, So, 𝑉2 = (𝑅 𝑉
1 + 𝑅2 )
𝑉 = 𝐸 − 𝑖𝑟 Where 𝑉 is supply voltage.
We have 50 = 𝐸 − 11𝑟 …(i) ∴ Potential drop across bulb 𝐵2 .
And 60 = 𝐸 − 1𝑟 …(ii) 400
From (i) and (ii), 𝑉2 = × 250
(625 + 400)
𝐸 = 61𝑉 and 𝑟 = 1Ω = 97.56 V
304 (c) = 98 V
310 (b)
𝑃 1/3 1/3
= =
𝑄 1 − 1/3 2/3
⟹ P: Q = 1: 2
𝑃 = 𝑘, 𝑄 = 2𝑘
Here, emf of each cell, 𝜀 = 0.2𝑉 2
𝑃+6
Internal resistance of each cell, 𝑟 = 1Ω = 3 2
𝑄 1−3
External resistance, 𝑅 = 10Ω
The total emf of 5 cells = 5𝜀 = 5(0.2)𝑉 = 1𝑉 𝑃+6 2
⟹ =
Total internal resistance of 5 cells = 5𝑟 = 𝑄 1
5(1)Ω = 5Ω 𝑘+6 2
=
Total resistance of the circuit = 𝑅 + 5𝑟 = 10 + 2𝑘 1
5 = 15Ω ⟹ 𝑘 + 6 = 4𝑘
The current in the external circuit, ⟹𝑘=2
5𝜀 1𝑉 1 ∴ 𝑃 = 2Ω, Q = 4Ω
𝐼= = = 𝐴 311 (d)
𝑅 + 5𝑟 15Ω 15
P a g e | 67
 In steady state capacitor is fully charged and no 20
When switch is closed 𝑖 = 3+(2||2) = 5𝐴
current flows through it
318 (a)
Given that, the resultant voltage across the
1.2V, 0.2Ω
battery terminal=1.5V

∴ No current passes through 4Ω 𝐸, 0.4Ω

1 1 1 1
= + +
𝑅𝑒𝑓𝑓 1 2 3 Let I be the current in the circuit then total
6 + 3 + 2 11 resistance =0.6Ω
= =
6 6 Hence, 𝑉 = 𝐼𝑅
6 ⟹ 1.5 = I × 0.6
⇒ 𝑅𝑒𝑓𝑓 = Ω
11 1.5 5
6×11 ⟹I= ⟹I= A
Current = 6 = 11𝐴 0.6 2
𝑄 = 𝐶𝑉 = 0.5 × 10−6 × 6 = 3.0 × 10−6 𝐶 = 3𝜇𝐶 Now, applying Kirchhoff’s second law in the
313 (a) circuit
As steady current is flowing through the 0.4I + 0.2I + 1.2 - E=0
5
conductor, hence the number of electrons 0.6 × + 1.2 = 𝐸
2
entering from one end and outgoing from the
⟹ 𝐸 = 2.7𝑉
other end of any segment is equal. Hence charge
319 (a)
will be zero
𝑎 = 1, 𝑏 = 2, 𝑐 = 3
314 (b) 𝜌. 𝐿 𝜌. 𝑐
Mass of copper deposited, ⇒ 𝑅max = =
𝐴 𝑎. 𝑏
𝑚 = volume × density
= (area × thickness) × density
= [2 × (12 × 3) × 0.002] × 8.9 g
𝑚 [2(12×3)×0.002×8.91]
𝑡= = = 776s.
𝑧𝐼 33×10−5 ×5
315 (c) 𝜌. 𝐿′′ 𝜌. 𝑎
𝑅min = =
According to Faraday’s first law of electrolysis, 𝐴′′ 𝑏. 𝑐
𝜌.𝑐
mass deposited 𝑅max 𝑎.𝑏 𝑐 𝑐 𝑐2 𝑐 2 3 2 9
𝑚 = 𝑍𝐼𝑡 ⇒ = = × ⇒ =( ) =( ) =
𝑅min 𝜌.𝑎 𝑎 𝑎 𝑎2 𝑎 1 1
𝑏.𝑐
Where 𝑍 = electrochemical equivalent of
320 (c)
substance
Let current flow from 𝑏 to 𝑎 as shown
𝑚1 𝐼1 𝑡1
∴ = × 2
I
𝑚2 𝐼2 𝑡2 3R 3

𝑥𝑔 4A 60 min
Or 𝑚2
= 6A × 40 min or 𝑚2 = 𝑥g a
I

R I b

316 (a) 6R 1
3 I
Since 𝑅 ∝ 𝑙 2 ⇒ If length is increased by 10%,
2 2
resistance increases by almost 20% 2
Ratio of thermal power is ( 𝐼) 3𝑅 ∶ ( 𝐼) 6𝑅 ∶
1
3 3
Hence new resistance 𝑅 ′ = 10 + 20% of 10
20 𝐼2 𝑅
= 10 + × 10 = 12 Ω 4 2
100 or : : 1 or 4 : 2 : 3.
3 3
317 (b) 321 (c)
When switch 𝑆 is open total current trough Resistance of copper part of wire 𝑅𝑐 =
ρc .𝐿 ρ .𝐿
= 𝜋𝑟c 2
𝐴𝑐
ammeter
20 and
𝑖= = 4𝐴
(3 + 2)

P a g e | 68
 ρc .𝐿 Now, 𝑅4 and 16 Ω are in parallel
Resistance of nickel portion of wire 𝑅𝑛 = 𝐴𝑛
=
ρc .𝐿
1 1 1
𝜋(𝑅2 −𝑟 2 ) ∴ = +
𝑅 16 16
As these two resistances are in parallel, hence
⇒ 𝑅 = 3Ω
conductance
326 (b)
1 1 1 𝜋𝑟 2
of the nickelled wire 𝐶 = = + = + Give that,
𝑅 𝑅𝑐 𝑅𝑛 ρc .𝐿
𝜋(𝑅2 −𝑟 2 ) 𝑙1 𝑑1 𝜌1 1
= = =
ρn .𝐿 𝑙2 𝑑2 𝜌2 2
and 𝑅1 = 10Ω
𝜋 𝑟2 𝑅2 − 𝑟2
= [ + ] We know that, the resistance of the wire
𝐿 ρc ρn
𝜌𝑙 𝜌𝑙 4𝜌𝑙 𝑑 2
𝑅= = = 2 [∵ 𝐴 = 𝜋 ( ) ]
323 (c) 𝐴 𝑑 2 𝜋𝑑 2
𝜋 (2)
As we know, for conductors, resistance ∝
So, the resistance of first wire is
temperature.
4𝜌1 𝑙1
From figure 𝑅1 ∝ 𝑇1 ⇒ tan 𝜃 ∝ 𝑇1 ⇒ tan 𝜃 = 𝑘𝑇1 𝑅1 = . . (𝑖)
𝜋𝑑12
…(i)
and the resistance of the second wire is
and 𝑅2 ∝ 𝑇2 ⇒ tan(90° − 𝜃) ∝ 𝑇2 ⇒ cot 𝜃 = 𝑘𝑇2
4𝜌2 𝑙2
…(ii) 𝑅2 = . . (𝑖𝑖)
𝜋𝑑22
From equation (i) and (ii), 𝑘(𝑇2 − 𝑇1 ) =
On dividing Eq. (ii) by Eq (i)
(cot 𝜃 − tan 𝜃)
𝑅2 𝜌2 𝑙2 𝑑12
cos 𝜃 sin 𝜃 (cos2 𝜃 − sin2 𝜃) = × ×
(𝑇2 − 𝑇1 ) = ( − )= 𝑅1 𝜌1 𝑙1 𝑑22
sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
= 2 cot 2𝜃 𝑅2 2 2 1 2
⟹ = × ×( )
⇒ (𝑇2 − 𝑇1 ) ∝ cot 2𝜃 10 1 1 2
324 (d) 𝑅2
⟹ = 1 ⟹ R 2 = 10Ω
𝑒 𝑒 𝑒𝑣 10
𝑖= = = 327 (d)
𝑡 2𝜋𝑟/𝑣 2𝜋𝑟
When one call is wrongly connected in series, the
𝑒2 emf of cells decrease by 2 𝐸, but internal
Here, 𝑣 = and 𝑟 = ℎ2 /𝑚𝑒 2
ℎ
resistance of cells remains the same for all the
𝑒(𝑒 2 ⁄ℎ) 𝑒 3 × 𝑚𝑒 2 𝑚𝑒 5 cells.
∴𝑖= = =
2𝜋(ℎ2 /𝑚𝑒 2 ) 2𝜋ℎ3 2𝜋ℎ3 (𝑛−2)𝐸
Current in the circuit is 𝑖 = ×𝑟
𝑛𝑟
4𝜋 2 𝑚𝑒 5
𝑖= Potential difference across each cell is
ℎ3

325 (a) (𝑛 − 2)𝐸 2𝐸

𝑉 = 𝐸 − 𝐼𝑟 = 𝐸 − ×𝑟 =
1 1 1 5 1 𝑛𝑟 𝑛
= + = = ⇒ 𝑅1 = 2Ω
𝑅1 10 2.5 10 2
328 (d)
Now 2Ω and 10 Ω are in series For conductors, resistance ∝ Temperature and for
1
semiconductor, resistance ∝ Temperature
𝑅2 = 10 + 2 = 12Ω
329 (c)
𝑅2 and 12Ω are in parallel Reading of voltmeter
𝐸1 𝑟2 + 𝐸2 𝑟1 18 × 1 + 12 × 2
1 1 1 = 𝐸𝑒𝑞 = = = 14𝑉
= + ⇒ 𝑅3 = 6Ω 𝑟1 + 𝑟2 1+2
𝑅3 12 12 331 (d)
Now 𝑅3 and 6Ω are in series The equivalent circuit is given by

3Ω
𝑅4 = 10 + 6 = 16Ω 2V

P a g e | 69
 338 (b)
The internal resistance of battery is given by
3Ω 2V 𝐸 40 9 × 10
A B 𝑟 = ( − 1) 𝑅 = ( − 1) × 9 = = 3Ω
𝑉 30 30
Since, 𝑉𝐴 = 𝑉𝐵 339 (a)
220 ×220
Potential difference is zero. Resistance of 25 W bulb = 25
= 1936 Ω
332 (a) It’s safe current
220
= 1936 = 0.11 A
Let 𝑅0 = resistance of filament at room 220 × 220
temperature Resistance of 100 W bulb = = 484 Ω
100
220
𝑅𝑡 = resistance of filament at 2500℃ It’s safe current = 484 = 0.48 A
Similarly powers, 𝑃0 and 𝑃𝑡 . when connected in series to 440 V supply, then
Here, voltage remains the same. the current
𝑉2 440
𝑃0 = 𝑖=
𝑅0 (1936 + 484)
𝑉2 𝑉2
or 𝑅0 = , 𝑅𝑡 = = 0.18 A
𝑃0 𝑃𝑡
Thus, current is greater for 25 W bulb, so it will
also 𝑅𝑡 = 𝑅0 [1 + 𝛼(2500 − 20)]
fuse.
and 𝑃0 = 𝑃𝑡 [1 + 𝛼(2500 − 20)]
341 (b)
= 50[1 + 4.5 × 10−3 (2500 − 20)] 𝐿
= 608W Resistance of the wire, 𝑅 = 𝜌 𝐴
333 (a) When the wire is enlongated to 𝑛 −fold, its length
Maximum current flows through bulb (1) becomes L′ = nL
Therefore, it will lights brightly. As the volume of the wire remains constant
334 (c) 𝐴𝐿 𝐴
∴ 𝐴′ 𝐿′ = 𝐴𝐿 ⇒ 𝐴′ = =
Let temperature of cold junction be 0℃ and that 𝐿′ 𝑛
𝐿′ (𝑛𝐿) 𝐿
of hot junction be 𝑇℃. The relation for thermo- New resistance, 𝑅 ′ = 𝜌 =𝜌 = 𝑛2 𝜌 = 𝑛2 𝑅
𝐴′ (𝐴/𝑛) 𝐴
emf is given by 342 (d)
1 𝑤𝑎𝑡𝑡×ℎ𝑜𝑢𝑟
𝐸 = 𝐴𝑇 − 𝐵𝑇 2 Energy consumed in 𝑘𝑊ℎ =
2 1000
10×50×10
Given, 𝐴 = 16, 𝐵 = 0.08 ⇒ For 30 days, 𝑃 = × 30 = 150𝑘𝑊ℎ
1000
1
∴ 𝐸 = 16𝑇 − × 0.08 × 𝑇 2 343 (b)
2 1
Since, at temperature of inversion emf is zero, we In parallel 𝑃𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 ∝ Brightness ∝ 𝑅
have 𝑃𝐴 > 𝑃𝐵 [Given] ∴ 𝑅𝐴 < 𝑅𝐵
0 = 16𝑇 − 0.04 𝑇 2 345 (d)
16 The current density of electrons in a metallic
⇒𝑇= = 400℃
0.04 conductor = 1022 cm−3 = 1028 m−3
335 (b)
𝑑𝑇 𝑑 346 (c)
= (𝑎𝑡 2 − 𝑏𝑡 3 ) = 2𝑎𝑡 − 3𝑏𝑡 2 When rod is bent in the form of square, then each
𝑑𝑡 𝑑𝑡
𝑑𝐸 side has resistance of
When 𝑡 = 𝑡𝑛 (𝑖𝑒, neutral temperature), = 0
𝑑𝑡 1
2𝑎 Ω. As shown 𝑅1 , 𝑅2 and 𝑅3 are connected in
∴ 0 = 2𝑎𝑡𝑛 − 3𝑏𝑡𝑛2 or 𝑡𝑛 = 3𝑏
. 4
series, so their equivalent resistance
336 (c)
1
Resistance of bulb is constant 𝑅2 = Ω
4
𝑣 2 ∆𝑃 2∆𝑉 ∆𝑅 1
𝑅1 = Ω
𝑃= ⇒ = + 4 1
𝑅 𝑃 𝑉 𝑅 𝑅3 = Ω
∆𝑃 1
𝑅 = 1Ω
4
= 2 × 2.5 + 0 = 5% 4
𝑃
337 (c)
𝑖 = 𝑒𝑣 = 1.6 × 10−19 × 6.8 × 1015
= 1.1 × 10−3 𝑎𝑚𝑝
𝑅 ′ = 𝑅1 + 𝑅2 + 𝑅3
P a g e | 70
 1 1 1 3 And 𝑃 = 𝑅 1+𝑅2
𝑅 𝑅
= + + = Ω 1 2
4 4 4 4
𝑛 × 𝑅1 𝑅2
Now, R’ and 𝑅4 are connected in parallel, so ∴ (𝑅1 + 𝑅2 ) = (From 𝑆 = 𝑛𝑃)
equivalent resistance of the circuit is 𝑅1 + 𝑅2
𝑅 ′ × 𝑅4 ⟹ (R1 + R 2 )2 = 𝑛R1 R 2
𝑅= ′ 𝑅12 + 𝑅22 + 2𝑅1 𝑅2
𝑅 + 𝑅4 ⟹𝑛=[ ]
3 1 𝑅1 𝑅2
(4) (4)
= 𝑅1 𝑅2
3 1 =[ + + 2]
(4) + (4) 𝑅2 𝑅1
(16)
3 We know,
3
= = Ω Arithmetic Mean ≥ Geometric Mean
1 16 𝑅1 𝑅
347 (a) 𝑅2
+ 𝑅2 𝑅1 𝑅2
1
≥√ ×
Potential gradient = Change in voltage per unit 2 𝑅2 𝑅1
length 𝑅1 𝑅2
𝑉2 − 𝑉1 ⟹ + ≥2
∴ 10 = ⇒ 𝑉2 − 𝑉1 = 3 𝑣𝑜𝑙𝑡 𝑅2 𝑅1
30/100
So, n(minimum value) =2+2=4
348 (a)
352 (d)
In the following figure 𝑖 𝐺 5 60
Resistance of part 𝑃𝑁𝑄; =1+ ⇒ =1+ ⇒ 𝑆 = 15Ω
𝑖𝑔 𝑆 1 𝑆
M
i2
353 (a)
i P
𝐹 = 𝑁𝑒 = 6 × 1023 × 1.6 × 10−19
i1 354 (a)
N
3V, 1Ω
Q
𝑅1 − 1⁄2Ω , 𝑅2 = 1⁄4Ω ; 𝑅3 = 1/6Ω
1 1 1 1
In parallel; 𝑅𝑝
= 𝑅 + 𝑅 + 𝑅3 = 2 + 4 + 6 = 12
10 1 2
𝑅1 = 4
= 2.5Ω and
1
Resistance of part 𝑃𝑀𝑄; or Equivalent conductance, 𝐺 = = 12𝑆
𝑅𝑝
3
𝑅2 = × 10 = 7.5Ω
4 355 (b)
𝑅1 𝑅2 2.5 × 7.5 15
𝑅𝑒𝑞 = = = Ω The equivalent resistance between two corners of
𝑅1 + 𝑅2 (2.5 + 7.5) 8 equilateral triangle having resistance 𝑅 in each
3 24
Main Current 𝑖 = 15 = 23 𝐴 arm = 2𝑅/3 = 2 × 4/3 = 8/3Ω
+1
8
𝑅2 24 7.5 18
So, 𝑖1 = 𝑖 × (𝑅 +𝑅 ) = 23 × (2.5+7.5) 357 (c) = 23 𝐴
1 2
24 18 6 Current drawn from the cell in resistance 𝑅1 will
and 𝑖2 = 𝑖 − 𝑖1 = 23 − 23 = 23 𝐴
be 𝐼 = 𝐸/(𝑅1 + 𝑟)
349 (b) Therefore, heat produced in 𝑅1 𝑖𝑒,
The sensitivity of potentiometer can be increased 𝐸 2 𝑅1 𝑡
by decreasing the potential gradient 𝑖. 𝑒. by 𝐻 1 =
(𝑅1 + 𝑟)2
increasing the length of potentiometer wire 𝐸 2 𝑅2 𝑡
1 Heat produced in 𝑅2 𝑖𝑒, 𝐻2 = (𝑅 +𝑟)
[Sensitivity ∝ 𝑃.𝐺. ∝ Length] 2
2

As per question 𝐻1 = 𝐻2
350 (d)
𝐸 2 𝑅1 𝑡 𝐸 2 𝑅2 𝑡
At an instant approach the student will choose or (𝑅 +𝑟) 2 = (𝑅2 +𝑟)2
1
tan 𝜃 will be the right answer. But it is to be seen On solving we get;
here the curve makes the angle 𝜃 with the 𝑉-axis. 𝑟 = √𝑅1 𝑅2
So it makes an angle (90 − 𝜃) with the 𝑖-axis
= √100 × 40=63.25Ω
So resistance = slope = tan(90 − 𝜃) = cot 𝜃
358 (a)
351 (a)
For power transmission power loss in line 𝑃𝐿 =
Let resistances are 𝑅1 and 𝑅2 ,
𝑖2𝑅
Then 𝑆 = 𝑅1 + 𝑅2
If power of electricity is 𝑃 and it is transmitted at
P a g e | 71
 𝑃
voltage 𝑉, then 𝑃 = 𝑉𝑖 ⇒ 𝑖 = 𝑉 = 𝑖𝑅 = 10−5 × 40 = 4 × 10−4 V
∴ 4 × 10−4 = 25θ × 10−6
𝑃 2 𝑃2 𝑅 2.2 × 103 × 2.2 × 103 × 10
𝑃𝐿 = ( ) 𝑅 = 2 = 4
𝑉 𝑉 22000 × 22000 ∴θ= × 102 = 16℃
25
= 0.1𝑊
364 (c)
360 (c)
Before connecting the voltmeter, potential
difference across 100Ω resistance
10 100

Vi

V
100 10
𝑉1 = ×𝑉 = 𝑉
(100 + 10) 11
𝑅×16
Finally after connecting voltmeter across 100Ω ∴ + 10 = 18, on solving we get, 𝑅 = 16Ω
𝑅+16
equivalent resistance 365 (d)
100 × 900 Total power spend across two resistors connected
= 90Ω
(100 + 900) 𝑉2 𝑉2
in parallel to battery = 𝑅 + 𝑅
Final potential difference 1 2
900 3 × 3 3 × 3 36
= + = = 18
2 2/3 2
10 100 =3×3×2J
366 (d)
Vf
𝑉 100
𝑅 = −𝐺 = − 25 = 9975Ω
V 𝑖𝑔 10 × 10−3
367 (d)
90 9 ρ𝑙 ρ(4𝑎) 2ρ
𝑉𝑓 = ×𝑉 = 𝑉 𝑅 = . 𝑆𝑜 𝑅1 = =
(90 + 10) 10 𝐴 (2𝑎)(𝑎) 𝑎
𝑉𝑖 −𝑉𝑓
% error = 𝑉𝑖
× 100
10 9
ρ(𝑎) ρ ρ(2𝑎) ρ
𝑉 − 10 𝑉 𝑅2 = = and 𝑅3 = =
11 (4𝑎)(2𝑎) 8𝑎 (4𝑎)(𝑎) 2𝑎
= 10 × 100 = 1.0
𝑉
11 ∴ 𝑅1 > 𝑅3 > 𝑅2
361 (b)
𝑉 3 368 (b)
𝑅= −𝐺 = − 20 = 102 − 20 = 80Ω
𝐼𝑔 30 × 10−3 In parallel, 𝑥 =
𝑅
𝑅 = 𝑛𝑥
𝑛
362 (b)
In series, 𝑅 + 𝑅 + 𝑅 … 𝑛 times = 𝑛𝑅 = 𝑛(𝑛𝑥) =
When the heating coil is cut into two equal parts
𝑛2 𝑥
and these parts are joined in parallel, the
369 (b)
resistance of coil is reduced to one fourth, so
Let the e.m.f. of cell be 𝐸 and internal resistance
power consumed will become 4 times
be 𝑟
𝑖. 𝑒. 400 𝐽𝑠 −1 𝐸 𝐸
Then 0.5 = (𝑟+2) and 0.25 = (𝑟+5)
363 (a)
𝜇V 5+𝑟
Thermo-emf of thermocouple = 25 . On dividing, 2 = 2+𝑟 ⇒ 𝑟 = 1Ω
℃
Let θ be the smallest temperature difference. 370 (a)
Therefore, after connecting the thermocouple If resistances of bulbs are 𝑅1 and 𝑅2 respectively
with the galvanometer, thermo-emf then in parallel
μV 1 1 1 1 1 1
𝐸 = (25 ) × θ(℃) = + ⇒ 2 = 𝑉2 + 𝑉2
℃ 𝑅𝑃 𝑅1 𝑅2 ( ) ( ) ( )
𝑉
𝑃1 𝑃2
= 25θ × 10−6 V 𝑃𝑝

Potential drop developed across the galvanometer ⇒ 𝑃𝑃 = 𝑃1 + 𝑃2

P a g e | 72
371 (b) 378 (b)
For maximum energy equivalent resistance of In VI graph, we will not get a straight line in case
combination should be minimum of liquids
373 (c) 379 (b)
Chemical energy consumed per sec = heat energy 1 faraday (96500𝐶) is the electricity which
produced per sec. liberates that amount of substance which is equal
2 (𝑅 2 −1
𝐼 + 𝑟) = (0.2) (21 + 4) = 1 J s to equivalent wt
375 (a) 63.5
So liberated amount of 𝐶𝑢 is 2
𝑒 𝑅
Potential gradient = (𝑅+𝑅 +𝑟) . 𝐿 = 31.25 𝑔𝑚 ≈ 32 𝑔𝑚
ℎ
2 5 𝑉 𝑉 380 (c)
= × = 0.5 = 0.005 𝑅
(15 + 5 + 0) 1 𝑚 𝑐𝑚 𝐼=
376 (b) 𝑅2 + 𝑟
Resistance of 𝐶𝐷 𝑎𝑟𝑚 = 2𝑟 cos 72° = 0.62𝑟 (since finally no current flows through capacitor)
Resistance of CBFC branch ∴ Potential difference across
A 𝐸𝑅2
i 𝑅2 , 𝑉 = 𝐼𝑅2 =
r r 72° 𝑅2 + 𝑟
B r C D r E ∴ Charge on the capacitor
r 𝐶𝐸𝑅2
r
𝒬 = 𝐶𝑉 =
F H J 𝑅2 + 𝑟
r
r
r 381 (a)
r
G I The reciprocal of resistance is called conductance
1 1 1 1 2.62 382 (c)
= + = ( )
𝑅 2𝑟 0.62𝑟 𝑟 2 × 0.62 𝑖 = 𝑞/𝑡 = 𝑛𝑒/𝑡
1 2.62 1.24𝑟
= ∴𝑅= 2×1
𝑅 1.24𝑟 2.62 or 𝑛 = 𝑖𝑡⁄𝑒 = = 1.25 × 1019
1.24𝑟 1.6×10−19
′
Equivalent 𝑅 = 2𝑅 + 𝑟 = 2 × 2.62 + 𝑟
2.48 383 (a)
= 𝑟( + 1) = 1.946𝑟 𝑅 = ρ𝑙 ⁄𝐴 or 𝑅 ∝ 𝑙/𝐴. Thus, resistance is least in a
2.62
Because the star circuit is symmetrical about the wire of length 𝐿/2 and area of cross-section 2𝐴
line 𝐴𝐻
384 (a)
∴ Equivalent resistance between 𝐴 and 𝐻
1 1 1 Current flowing through both the bars is equal.
= ′+ ′ Now, the heat produced is given by
𝑅𝑒𝑞 𝑅 𝑅
′
𝑅 1.946 𝐻 = 𝐼 2 𝑅𝑡
⇒ 𝑅𝑒𝑞 = = 𝑟 = 0.973𝑟
2 2
377 (a) 𝐻𝐴𝐵 𝑅
or 𝐻 ∝ 𝑅 or 𝐻𝐵𝑐
= 𝑅𝐴𝐵
For one wire cable, 𝐵𝐶

(1⁄2𝑟)2 1 1
Resistance, 𝑅 ′ = ρ𝑙/π(9 × 10−3 )2 = 5Ω = (∵ 𝑅 ∝ ∝ )
(1⁄𝑟)2 𝐴 𝑟2
For other wire of cable,
1
=
Resistance, 𝑅 ′ = ρ𝑙/π(3 × 10−3 )2 4

= 92 × 5/32 = 45Ω or 𝐻𝐵𝐶 = 4𝐻𝐴𝐵

When six wires each of resistance 𝑅’ are 385 (d)

connected in parallel, their effective resistance For balanced Wheatstone bridge
𝑃 𝑆
will be =
𝑅 𝑄
𝑅′ 45 386 (c)
𝑅𝑝 = = = 7.5Ω
6 6 The number density (𝑛) of conduction electrons
in the copper is a characteristic of the copper and
P a g e | 73
 is about 1029 at room temperature for both the In balanced position
copper rod 𝑋 and the thin copper wire 𝑌.
Both 𝑋 and 𝑌 carry the same current 𝐼 since they P
Q
are joined in series
From 𝐼 = 𝑛𝐴𝑣𝑞
Where 𝑞 is the electron charge of 1.6 × 10−19 C,
S
𝑣 is the drift velocity in the conductor and 𝐴 is the
R
cross-sectional area of the conductor.
We may conclude that rod 𝑋 has a lower drift r

velocity of electrons compared to wire 𝑌 since rod

𝑋 has a larger cross-sectional area. This is so 𝑃 𝑅
because the electrons in 𝑋 collide more often with =
𝑄 𝑆′
one another and with the copper ions when 2 2
drifting towards the positive end. Thus, the mean =
2 𝑆′
time between collisions of the electrons is more in ∴ 𝑆 ′ = 2Ω
𝑋 than 𝑌. Now,
388 (c) 1 1 1
𝑖𝑆 50 × 12 = +
𝑆′ 𝑆 𝑟
𝑖𝑔 = ⇒ 10 = ⇒ 12 + 𝐺 = 60 ⇒ 𝐺 1 1 1 1 1 3−2
𝑆+𝐺 12 + 𝐺 or = ′ − = − =
= 48Ω 𝑟 𝑆 𝑆 2 3 6
389 (d) 1 1
=
Assuming current 𝐼 flows through the circuit 𝑟 6
r
𝑟 = 6Ω
395 (d)
We know that the current in the circuit
𝐸
𝐼=
R 𝑅+𝑟
2
Energy dissipated in load = 𝐼 𝑅 and power delivered to the resistance R is
Energy dissipated in the compete circuit = 𝐼 2 (𝑟 + 𝐸2𝑅
𝑃 = 𝐼2 𝑅 =
𝑅) (𝑅 + 𝑟)2
𝑑𝑃
𝐼2 𝑅 𝑅
∴ The efficiency = 𝐼2 (𝑅+𝑟) = 𝑅+𝑟 It is maximum when 𝑑𝑅 = 0
𝑑𝑃 (𝑟 + 𝑅)2 − 2𝑅(𝑟 + 𝑅)
Relation between resistance, mass and cross- = 𝐸2 [ ]=0
𝑑𝑅 (𝑟 + 𝑅)4
sectional area
𝑙×𝛼 𝑉 Or (𝑟 + 𝑅)2 = 2𝑅(𝑟 + 𝑅)
𝑅=𝑝 or 𝑅 = 𝜌 2 Or 𝑅 = 𝑟
𝛼×𝛼 𝛼
𝑚 𝜌 𝑚 396 (a)
𝑅=𝜌 2=[ ] 2
𝑑𝛼 𝑑 𝛼 2
𝑉𝑅𝑎𝑡𝑒𝑑 1
𝑚 𝑃𝑅𝑎𝑡𝑒𝑑 = ⇒𝑅∝ [V − constant]
𝐼∝ 2 𝑅 𝑃𝑅𝑎𝑡𝑒𝑑
𝛼
390 (a) So bulb of high power will have less resistance
Equivalent resistance in the second case = 𝑅1 + 397 (d)
𝑅2 = 𝑅 When a single heater (resistance 𝑅1 = 𝑅) is
1 connected to 220 V, then it will consume a power
Now, we know that 𝑃 ∝ 𝑅
𝑃1 =1000 W. If two such identical heaters are
Since in the second case the resistance (𝑅1 + 𝑅2 ) connected in parallel (total resistance 𝑅2 = 𝑅1 /2
is higher than that in the first case (𝑅1 ) = 𝑅/2) to some source, then it will consume
Therefore power dissipation in the second case power 𝑃2 .
will be decreased 𝑃2 𝑅1
391 (d) =
𝑃1 𝑅2
Let a resistance r ohm be shunted with resistance ⇒ 𝑃2 = 2𝑃1
S, so that the bridge is balanced. 𝑃2 = 2000 W
If S’ is the resultant resistance of S and r, then 398 (d)
P a g e | 74
 If 𝐶𝑒 be the effective capacitance, then V
C
r1 r2

C E E

R
R
2𝐸
C C 𝑉 = 𝐸 − 𝑖𝑟 = 𝐸 − 𝑟1 ( )
𝑟1 + 𝑟2 + 𝑅
But 𝑉 = 0
2𝐸𝑟1
⇒𝐸− =0
R 𝑟1 + 𝑟2 + 𝑅
⇒ 𝑟1 + 𝑟2 + 𝑅 = 2𝑟1
1
𝑉𝐶 = 𝑉0 ⇒ 𝑅 = 𝑟1 − 𝑟2
2
𝑞 𝑞0 404 (b)
= Here , 𝑚 = 1 g = p−3 kg;
𝐶𝑒 2𝐶𝑒
−
𝑡 𝑞0 𝑧 = 1.044 × 10−8 kg C −1
⟹ 𝑞0 (1 − 𝑒 𝑅𝐶𝑒 ) =
2 𝐻 = 34 k cal. = 34 × 1000 × 4.2 J
⟹ 𝑡 = 𝑅𝐶𝑒 𝐼𝑛2 𝑚 10−3 105
𝑞 = 𝑧 = 1.044×10−8 = 1.044 C
For parallel grouping 𝐻 34×1000×4.2
2𝐶 and 𝑉 = 𝑞 = (105 /1.044) = 1.5 V.
𝐶𝑒 =
2 405 (c)
∴ 𝑡2 = 2𝑅𝐶 𝐼𝑛2 Human body, though has a large resistance of the
For series grouping order, of 𝐾Ω(say 10𝑘Ω), is very sensitive to
𝐶
𝐶𝑒 = minute currents even as low as a few 𝑚𝐴.
2 Electrons, excites and disorders the nervous
𝑅𝐶
𝑡1 = 𝐼𝑛 2 system of the body and hence one fails to control
2
𝑡2 1 the activity of the body
= 𝑡2 = 2.5𝑠 406 (d)
𝑡1 4
399 (b) 𝑒 𝑅
𝐸= ×𝑙
By Faraday’s law, 𝑚 ∝ 𝑖𝑡 (𝑅 + 𝑅ℎ + 𝑟) 𝐿
𝑚1 𝑖1 𝑡1 𝑚 4 × 120 𝑚 2 10
∴ = ⇒ = ⇒ 𝑚2 = = × × 0.4
𝑚2 𝑖2 𝑡2 𝑚2 6 × 40 2 (10 + 40 + 0) 1
400 (a) = 0.16𝑉
With rise in temperature the thermal velocity of 407 (b)
240 ×240
the electron increases. Relaxation time and hence Resistance of 40 W bulb = 40
drift velocity will decrease. = 1440 Ω
401 (c) It’s safe current
240
= 1440 = 0.167 A
The temperature coefficient of the carbon is 240 ×240
negative so the resistance of carbon decreases Resistance of 60 W bulb = 60
with the increase of temperature. = 960 Ω
240
402 (b) It’s safe current = = 0.25 A
960
In series : Potential difference ∝ 𝑅 When connected in series to 420 V supply, then
3
When only 𝑆1 is closed 𝑉1 = 4 𝐸 = 0.75𝐸 the current
6
When only 𝑆2 is closed 𝑉2 = 7 𝐸 = 0.86𝐸 420 420
𝐼= =
1440 + 960 2400
And when both 𝑆1 and 𝑆2 are closed combined = 0.175 A
resistance of 6𝑅 and 3𝑅 is 2𝑅 Thus, current is greater for 40 W bulb, so it will
2
∴ 𝑉3 = ( ) 𝐸 = 0.67𝐸 ⇒ 𝑉2 > 𝑉1 > 𝑉3 fuse.
3
408 (d)
403 (b)
Let the voltage across any one cell is 𝑉, then
P a g e | 75
 𝑖𝑔 𝑆 4 4 1 be such that the null point is nearly in the middle
= = = =
𝑖 𝐺 + 𝑆 36 + 4 40 10 of the wire. In this position all resistance P, Q, R
409 (c) and S become nearly equal. The emf of cell
𝑅 = 56 × 10 ± 10% = 560 + 10% depends upon the size and area of electrodes.
410 (a) 416 (c)
For 𝐸 to be maximum Before connecting 𝐸, the circuit diagram is as
𝑑𝐸 shown in the figure
= 20 × 10−6 − 0.02 × 10−6 × 2𝑇 = 0
𝑑𝑇
⇒ 𝑇𝑛 = 500℃
∴ 𝐸max = 20 × 10−6 (500) − 0.02 ×
10−6 (500)2 =5 mV
411 (b)
Resistance of combination 𝑅𝑒 = 4𝑅 The equivalent resistance of the given circuit is
∆𝑅𝑒 ∆𝑅 𝑅𝑒𝑞 = 6 Ω + 8 Ω + 10 Ω = 24 Ω
=
𝑅𝑒 𝑅 12𝑉 1
Current in the circuit, 𝐼 = 24Ω = 2 𝐴
5 × 100
= = 5% Before connecting 𝐸, the current through 8 Ω is
100
412 (a) 1
𝐼= 𝐴
𝑆𝑙 𝑆𝑙 2 2
𝑅= = After connecting 𝐸, the current through 8 Ω is also
𝐴 𝑉
∆𝑅 ∆𝑙 1
∴ = 2 = +2.0% 𝐼= 𝐴
𝑅 𝑙 2
1
413 (a) ∴ 𝐸 = 𝐴 × 8Ω = 4𝑉
2
𝑊𝑎𝑡𝑡-ℎ𝑜𝑢𝑟 𝑚𝑒𝑡𝑒𝑟 measures electric energy
417 (b)
414 (d)
𝑅1000 = 𝑉 2 /750 and 𝑅200 = 𝑉 2 /𝑃;
Let 𝑅1 and 𝑅2 be the resistances of the coils
Now, 𝑅1000 = 𝑅200 (1 + 𝛼 × 800)
𝑉2 𝑉2
𝑉 2 𝑡1 𝑉 2 𝑡2 So, = (1 + 4 × 10−4 × 800)
𝐻= and 𝐻 = 750 𝑃
𝑅1 𝑅2 or 𝑃 = 750(1 + 0.32) = 990 W
𝑡1 𝑡2 𝑅2 𝑡2 418 (a)
⇒ = , 𝑖𝑒, = … (i) 𝑑𝑞
𝑅1 𝑅2 𝑅1 𝑡1 𝐼= = 3𝑡 2 + 2𝑡 + 5
𝑑𝑡
Now in parallel ∴ 𝑑𝑞 = (3𝑡 2 + 2𝑡 + 5)𝑑𝑡
𝑡=2
𝑅1 𝑅2 ∴ 𝑞 = ∫ (3𝑡 2 + 2𝑡 + 5)𝑑𝑡
𝑅′ = = 𝑅1
𝑅1 + 𝑅2 𝑡=0
3
3𝑡 2𝑡 2 2 2
𝑉 2𝑡 = + + 5𝑡 | = 𝑡 3 + 𝑡 2 + 5𝑡| = 22 𝐶
∴𝐻= ′ … (ii) 3 2 0 0
𝑅 420 (b)
𝑉 2𝑡 𝑉 2 𝑡1
Now, 𝑅′
= 𝑅1

𝑡 × (𝑅1 + 𝑅2 ) 𝑡1
⇒ = … (iii)
𝑅1 𝑅2 𝑅1

On using Eqs. (i) and (iii), we get

𝑉
Current though arm 𝐶𝐴𝐷, 𝐼 = 𝑎𝑚𝑝
𝑡1 𝑡2 8
𝑡= Potential difference between 𝐶 and 𝐴 = 𝑉𝐶 − 𝑉𝐴
𝑡1 + 𝑡2
𝑉 𝑉
= × 4 = 𝑣𝑜𝑙𝑡
415 (d) 8 2
𝑉
For greater sensitivity of meter bridge the Current through 𝐶𝐵𝐷, 𝐼 ′′ = 4 𝑎𝑚𝑝
resistance (R) taken in the resistance box should Potential difference between 𝐶 and 𝐵 = 𝑉𝐶 − 𝑉𝐵

P a g e | 76
 𝑉 𝑉 𝑙
Given 𝑙1 = 2 and
1 𝑟1 2 𝑟 1 𝑅
= 1 or 𝑟2 = 2 ⇒ 𝑅1
1
=8
= × 1 = 𝑣𝑜𝑙𝑡 𝑟2
4 4 2 1 2
𝐻1 𝑉 2 /𝑅1 𝑅2 8
Potential between 𝐴 and 𝐵 = 𝑉𝐴 − 𝑉𝐵 ∴ Ratio of heats = 2 = =
𝐻2 𝑉 /𝑅2 𝑅1 1
𝑉 𝑉 𝑉
∴ 𝑉𝐴 − 𝑉𝐵 = 𝑉𝐶 − 𝑉𝐵 − (𝑉𝐶 − 𝑉𝐴 ) = − = − 428 (d)
4 2 4
Potential gradient is given by
⇒ 𝑉𝐴 − 𝑉𝐵 < 0 or, 𝑉𝐴 < 𝑉𝐵
𝑉 𝐼𝑅
as 𝑉𝐴 < 𝑉𝐵 , so direction of current will be 𝐵 to 𝐴 𝑘= = (∵ 𝑉 = 𝐼 𝑅)
𝑙 𝑙
421 (d) 𝐼 × 𝜌𝑙/𝐴 𝜌𝑙
𝐸 = 𝑎θ + 𝑏θ2 (given) = (∵ 𝑅 = )
𝑙 𝐴
𝑑𝐸
For neutral temperature (θn ), = 0 𝐼𝜌
𝑑θ =
⇒ 𝑎 + 2𝑏 θ𝑛 = 0 𝐴
𝑎 0.01 × 10−3 × 109 × 10−2
⇒ θ𝑛 = − ∴ 𝑘= = 108 Vm−1
2𝑏 10−2 × 10−4
700 𝑎 429 (c)
∴ θ𝑛 = − (∵ = 700℃) Suppose 𝑛 resistors are used for the required job.
2 𝑏
= −350℃ < 0℃ Suppose equivalent resistance of the combination
But neutral temperature can never be negative is 𝑅′ and according to energy conservation it’s
(less than zero), 𝑖𝑒, θ𝑛 ≮ 0℃. current rating is 𝑖 ′
Hence, no neutral temperature is possible for this Energy consumed by the combination = 𝑛 ×
thermocouple. (Energy consumed by each resistance)
422 (b) 2
′2 ′ 2
𝑖′ 𝑅′
1 1 1 1 3 ⇒ 𝑖 𝑅 = 𝑛 × 𝑖 𝑅 ⇒ 𝑛 = ( ) × ( )
= + + = 𝑖 𝑅
𝑅𝑃 𝑅 𝑅 𝑅 𝑅
4 2 5
=( ) ×( )=8
1 10
430 (b)
𝑅1 𝑚1 𝑎22
= ×
𝑅 𝑅2 𝑎12 𝑚2
⇒ 𝑅𝑃 = Ω
3 𝑚1 𝑎2 2 1 1
⇒ 𝑅𝑆 = 𝑅 + 𝑅 = 2 𝑅Ω = ( ) ( ) = ( ) (1)2 =
𝑚2 𝑎1 2 2
𝑅 7𝑅 2
𝑄1 𝐼 𝑅1 𝑡 1
⇒ 𝑅net = 𝑅𝑃 + 𝑅𝑆 = 2𝑅 + = Ω = =
3 3 𝑄2 𝐼 2 𝑅2 𝑡 2
423 (c)
431 (b)
′
𝑅 1
𝑅 = = = 0.1Ω Here, potential gradient, 𝐾 = 10Vm−1;
𝑛 10
424 (a) Potential difference across length 𝑙 = 𝐾𝑙
The tolerance level of resistance is mostly 1%,
2%, 5% and 10%. In old days 20% was also = 10 × 0.5 = 5 V
common, but these are now rare. Now a days 5%
tolerance in treands. 432 (a)
425 (c) Power in electric bulb
At room temperature, the free electrons in a 𝑉2
conductor move randomly with speed of the 𝑃=
𝑅
order of 105 ms−1. Since, the motion of the
electrons is random there is no net charge flow in So, resistance of electric bulb
any direction.
𝑉2
426 (d) 𝑅=
Graph (d) represents the thermal energy 𝑃
produced in a resistor. Given, 𝑃1 = 25 W, 𝑃2 = 100 W,
427 (d)
𝑙 𝑙 𝑅 𝑙 𝐴 𝑙 𝑟 2 𝑉1 = 𝑉2 = 220 volt
𝑅1 = 𝜌 𝐴1 and 𝑅2 = 𝜌 𝐴2 ⇒ 𝑅1 = 𝑙1 . 𝐴2 = 𝑙1 (𝑟2 )
1 2 2 2 1 2 1

P a g e | 77
 Therefore, for same potential difference 𝑉 question); so 𝑟 = 𝑅/𝑛. When these resistances are
connected in parallel, the effective resistance
1
𝑅∝
𝑃 𝑅/𝑛
𝑅𝑝 = 𝑟/𝑛 = = 𝑅/𝑛2
𝑛
Thus, we observe that for minimum power,
resistance will be maximum and 𝑣𝑖𝑐𝑒 − 𝑣𝑒𝑟𝑠𝑎. 438 (c)
1
𝑅 ∝ 𝜏 ;where 𝜏 = Relaxation time
Hence, resistance of 25 W bulb is maximum and
100 W bulb is minimum. When lamp is switched on, temperature of
filament increases, hence 𝜏 decreases so 𝑅
433 (b) increases
𝑖𝑔 ×𝐺 10×10−3 ×50 50 439 (a)
𝑆= 𝑖−𝑖𝑔
= 1−10−3 ×10
= 99 Ω in parallel
𝜃 𝜃 σi
434 (a) 𝜎𝑖 = = . 𝐺 = 𝜎𝑉 𝐺 ⟹ = σv
𝑖 𝑖𝐺 G
According to ohm’s law 𝑉 = 𝑖𝑅 440 (a)
⇒ log 𝑒 𝑉 = log 𝑒 𝑖 𝑣𝑑 = 𝑖 ⁄𝑛 𝐴𝑒 ; where 𝑛 = 𝑁ρ /𝑀
+ log 𝑒 𝑅
⇒ log 𝑒 𝑖 = log 𝑒 𝑉 − log 𝑒 𝑅 = 6.023 × 1026 × 9 × 103 /63 = 0.860 × 1029
The graph between log 𝑒 𝐼 and log 𝑒 𝑉 will be a
straight line which cuts log 𝑒 𝑉 axis and it’s = 8.6 × 1028
gradient will be positive 22
and 𝐴 = 𝜋𝐷 2⁄4 = × (10−3 )2⁄4m2 ;
435 (b) 7

Resistance of bulb 1 11
= × 10−6 m2
14
1.1
𝑣𝑑 = 11
8.6 × 1028 × 14
× 10−6 × 1.6 × 10−19

1 100 × 10−4
= = = 1.0 × 10−4 m/s
9.6 × 10+3 96
𝑉 2 (220)2
𝑅1 = =
𝑃 100 = 0.1 mms−1
𝑅1 = 484 Ω
Resistance of bulb 2 441 (a)
(220 𝑉)2 4840 When ammeter is connected in parallel to the
𝑅2 = = = 806.6 Ω circuit, net resistance of the circuit decreases.
60 𝑊 6
1 1 Hence more current is drawn from the battery,
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 = (220)2 ( + )
100 60 which damages the ammeter
220 𝑉 220 100×60
⇒ Current flowing 𝐼 = 𝑅 = (220)2 ( 160 ) 442 (d)
𝑒𝑞 𝑃 4.5
1 75 15 Current in the bulb = 𝑉 = 1.5 = 3𝐴
𝐼= ( ),𝐼 = 𝐴 1.5
220 2 88 Current in 1Ω resistance = = 1.5𝐴
1
Power consumed by 100𝑊 bulb = 𝐼 2 𝑅1
Hence total current from the cell 𝑖 = 3 + 1.5 =
15 2 (220)2 225 4.5𝐴
=( ) × = = 14𝑊
88 100 16 By using 𝐸 = 𝑉 + 𝑖𝑟 ⇒ 𝐸 = 1.5 + 4.5 × (2.67) =
436 (c) 13.5𝑉
𝑙
Resistance = 𝜌 𝐴 443 (a)
𝑅1 𝜌1 𝑙1 𝐴2 2 3 5 5 Applying Kirchhoff’s law
∴ = × × = × × = 20
𝑅2 𝜌2 𝑙2 𝐴1 3 4 4 8
(2 + 2) = (0.1 + 0.3 + 0.2)𝑖 ⇒ 𝑖 = 𝐴
437 (a) 3
Effective resistance of 𝑛 resistance each of the Hence potential difference across 𝐴
20 4
resistance 𝑟 in series 𝑅𝑠 = 𝑟 × 𝑛 = 𝑅(as per = 2 − 0.1 × 3
= 3 𝑉 [less than 2𝑉]

P a g e | 78
 20 = 𝑅0 (1.005θ) …(ii)
Potential difference across 𝐵 = 2 − 0.3 × 3
=0
444 (d) 1+0.005θ
𝑉 Dividing Eq. (ii) by Eq.(i), we get 2 = 1.5
From the relation, current 𝑖 =
𝑟
2 3 = 1 + 0.005θ
or 4 =
𝑟
1 2
or 𝑟 = Ω ⇒ θ= = 400°C
2 0.005
445 (c)
452 (c)
Manganin or constantan are used for making the
Resistances at 𝐶 and 𝐵 are not in the circuit. Use
potentiometer wire
laws of resistances in series and parallel excluding
446 (a)
the two resistance
Resistance of each bulb 𝑅 = 𝑉 2/𝑃.
453 (b)
When connected in series total resistance of bulbs
Effective emf of circuit = 10 − 3 = 7 V
= 2𝑅
Current in each bulb , 𝐼 = 𝑉 ′ /2𝑅; Total resistance of circuit= 2 + 5 + 3 + 4 = 14Ω
Power generated by each bulb = 𝐼 2 𝑅
𝑉′
2
𝑉′ 𝑉′2 Current, 𝑖 = 7/14 = 0.5 A
=( ) ×𝑅 = = 2
2𝑅 4𝑅 (𝑉 /𝑃) Potential difference between 𝐴 and 𝐷 = 0.5 ×
(110)2 ×500
= 2 =31.25 W
10 = 5V
4×(220)
447 (a) Potential at 𝐷 = 10 − 5 = 5𝑉
𝐸 = 𝛼𝑡 + 𝑏𝑡 2 . At temperature of inversion 𝐸 is
minimum Hence, 𝐸 cannot be at zero potential, as there is
𝑖. 𝑒. , 𝐸 = 0 potential drop at 𝐸
𝑎
∴ 𝛼𝑡𝑖 + 𝑏𝑡𝑖2 = 0, 𝑖. 𝑒. , 𝑡1 = − 454 (c)
𝑏
448 (d) A fully charged capacitor draws no current.
The resistance of 40 𝑊 bulb will be more and Therefore, no current flows in arm 𝐺𝐻𝐹. So the 𝑅
60 𝑊 bulb will be less of arm 𝐻𝐹 is ineffective. The total resistance of
449 (b) the resistors in circuit
𝑉 𝑖𝑅 𝑟 𝑅
𝐸 = 𝑥𝑙 = = × 𝑙 ⇒ 𝐸 = × ×𝑙 (𝑅+𝑅)×𝑅
𝑙 𝐿 (𝑅 + 𝑅ℎ + 𝑟) 𝐿 is 𝑅 ′ = (𝑅+𝑅)×𝑅 + 𝑅
10 5
⇒𝐸= × × 3 = 3𝑉 (2 + 2) × 2 10
(5 + 4 + 1) 5
= +2= Ω
450 (c) (2 + 2) + 2 3
12 × 4
𝑅= + 2 = 5Ω 𝐸 10
Total current, 𝑖 = 𝑅′ = (10/3) = 3A
12 + 4
E 12
I= = = 2𝐴
𝑅+𝑟 6 In parallel circuit, the current divides in the
𝐼1 + 𝐼1 = 2𝐴 inverse ratio of resistance, so current in arm
1 𝐴𝐵𝐺𝐷 = 1A and current in arm 𝐴𝐷 = 2A
𝐼∝
𝑅
∴ 𝐼1 = 0.5𝐴, 𝐼2 = 1.5𝐴 Potential difference between 𝐺 and 𝐷
451 (b)
Let resistance for bulb filament at o°C be R 0 and = 𝑉𝐺 − 𝑉𝐷 = 1 × 2 = 2V
at a temperature θ°C its value be 200 Ω. Then, Potential difference between 𝐷 and 𝐹
100 = 𝑅0 (1 + α × 100) = 𝑅0 (1 + 0.005 × 100)
= 𝑉𝐷 − 𝑉𝐹 = 3 × 2 = 6V
= 𝑅0 (1.5) …(i)
∴ 𝑉𝐺 − 𝑉𝐹 = (𝑉𝐺 − 𝑉𝐷 ) + (𝑉𝐷 − 𝑉𝐹 )
and 200 = 𝑅0 (1 + α × θ) = 𝑅0 (1 + 0.005 × θ)
=2+6=8V

P a g e | 79
 Potential difference across capacitor = 𝑉𝐺 − 𝑉𝐹 = Given that,
8𝑉 𝑙1 4 𝑟1 2
= and =
𝑙2 3 𝑟2 3
456 (a) Here, 𝑙1 and 𝑙2 are the length of the wires while
Current through resistance 𝑃 and 𝑄, 𝑟1 and 𝑟2 are the radii of the wires.
4 1 Now, we know that
𝑖1 = = A 𝑉 = 𝐼𝑅 ⟹ 𝐼𝑅 = constant
90 + 110 50
⟹ 𝐼1 𝑅1 = 𝐼2 𝑅2
1 𝐼1 𝑅2
𝑉𝐴 − 𝑉𝐵 = 𝑃𝑖1 = 90 × = 1.8 V or = … . (𝑖)
50 𝐼2 𝑅1
But we know that the resistance of the wire is
Current through resistance 𝑅 and 𝑆,
𝜌𝑙
𝑅=
4 1 𝐴
𝑖2 = = A Hence, from Eq (i)
40 + 60 25
𝐼1 𝜌𝑙2 /𝐴2
1 =
𝑉𝐴 − 𝑉𝐷 = 𝑅𝑖2 = 40 × = 1.6 V 𝐼2 𝜌𝑙1 /𝐴1
25 Here,
𝑉𝐵 − 𝑉𝐷 = (𝑉𝐴 − 𝑉𝐷 ) − (𝑉𝐴 − 𝑉𝐵 ) 𝜌𝑙1 𝜌𝑙2
𝑅1 = , 𝑅2 = 𝑎𝑛𝑑 𝜌1 = 𝜌2 = 𝜌
𝐴1 𝐴2
= 1.6 − 1.8 = −0.2V Because both wires are of same material.
𝐼1 𝑙2 𝐴1
457 (a) ∴ =
𝐼2 𝑙1 𝐴2
In parallel 𝑃𝐶𝑜𝑛𝑠𝑢𝑚𝑒𝑑 ∝ 𝑃𝑅𝑎𝑡𝑒𝑑 𝐼1 𝑙2 𝜋𝑟12
458 (c) ⟹ =
𝐼2 𝑙1 𝜋𝑟22
In mixed grouping the current in the external
𝐼1 𝑙2 𝑟12
circuit will be maximum when the internal ⟹ =
𝐼2 𝑙2 𝑟22
resistance of the battery is equal to the external 𝑙 3 𝑟 2
resistance, Here, 𝑙2 = 4 𝑎𝑛𝑑 𝑟1 = 3
1 2
𝑚𝑟
𝑅= 𝐼1 3 2 2
𝑛 ⟹ = ×( )
𝐼2 4 3
m 𝐼1 1
Or 𝐼2
=3
460 (c)
𝑚 5 × 10−3
n 𝐼𝑡 = =
𝑧 3.387 × 10−7
5×104
= Ah = 4.1 Ah
3.387×60×60
461 (b)
3Ω
If we take 𝑅1 = 4Ω, 𝑅2 = 12Ω, then in series
resistance
Given, R=3Ω, 𝑟 = 0.5Ω 𝑅 = 𝑅1 + 𝑅2 = 4 + 12 = 16Ω
𝑚 In parallel, resistance
∴ 3 = × 0.5
𝑛 4 × 12
𝑚 𝑅= = 3Ω
⟹ =6 4 + 12
𝑛 So, R1 4Ω and R 2 = 12Ω
⟹ 𝑚 = 6𝑛
462 (d)
Total number of cells= 𝑚 × 𝑛 = 24 …(i)
Potential gradient of a potentiometer
From Eqs. (i) and (ii), we get …(ii)
𝐼𝜌 0.2 × 4 × 10−7
6𝑛 × 𝑛 = 24 𝐾= =
𝐴 8 × 10−7
⟹ 6n2 = 24
= 0.1V/m
⟹ 𝑛2 = 4 463 (a)
⟹ 𝑛 = 2, 𝑚 = 12
459 (b)
P a g e | 80
 𝑙1 − 𝑙2 55 − 50 parallel, their effective resistance 𝑅𝑝 will be
𝑟=( ) × 𝑅′ ⇒ 𝑟 = ( ) × 10 = 1Ω
𝑙2 50
464 (b) 400 × 10,000 5000
𝑅𝑃 = = Ω
The bridge will be balanced when the shunted 400 + 10,000 13
3×𝑆
resistance of value 2Ω 𝑖𝑒, 2 = . On solving 𝑆 = Total resistance of circuit
3+𝑆
6Ω
5000 15400
= + 800 = Ω
465 (c) 13 13
25 5𝑙
Given, 𝑙1 = 𝑙 + 𝑙 = . Since volume or wire 6 39
100 4 Current in the circuit, 𝑖 = 15400/13 = 7700 𝐴
remains unchanged on increasing length, hence
Potential difference across voltmeter=𝑖𝑅𝑃 =
𝐴𝑙 − 𝐴1 × 5𝑙 ⁄4 or 𝐴1 = 4𝐴/5 39 5000
7700
× 13
Given, 𝑅 = 10 = ρ𝑙/𝐴, and
= 1.95 A
ρ𝑙1 ρ5𝑙/4 25ρ𝑙
𝑅1 = = =
𝐴1 4𝐴/5 16𝐴 470 (b)
Supercurrent always flows on the surface of the
25 250
or 𝑅1 = × 10 = = 15.6Ω superconductor.
16 16 471 (d)
466 (b) The resistance of the cell is independent of 𝑒.m.f
The given circuit can be redrawn as follows 472 (b)
A
R
Using the concept of balanced wheat stone bridge,
F C we have
R R 𝑃 𝑅
R =
𝑄 𝑆
D 𝑥 10
R =
B 52 + 1 48 + 2
Equivalent resistance between 𝐴 and 𝐵 is 𝑅 and 10 × 53
𝑥=
current 𝑖 = 𝑅
𝑉 50
= 10.6Ω
467 (b) 473 (b)
To obtain minimum resistance, all resistors must 1 × 10−3
be connected in parallel. 𝑛= = 6.25 × 1015
1.6 × 10−19
Hence equivalent resistance of combination of 474 (c)
𝑟
combination = 10 Amount of metallic sodium appearing,
468 (c) 𝐴
𝑚 = 𝑍𝑖𝑡 = ( ) 𝑖𝑡
Given, 𝑙1 = 1K, 𝑙2 = 3K, 𝑙3 = 5K 𝑉𝐹
or 𝑚1 = 5m, 𝑚2 = 3m, 𝑚3 = 1m 23
=( ) × 16 × 10 × 60
We knows 1 × 96500
ρ𝑙 = 2.3 𝑔𝑚
𝑅= 476 (d)
𝐴
𝑙 𝑙 𝑙 net emf
So 𝑅1 : 𝑅2 : 𝑅3 = 𝐴1 ∶ 𝐴2 ∶ 𝐴3 Current =
1 2 3 net resistance
𝑙12 𝑙22 𝑙32 2+2+2 6
or 𝐼 = 1+1+1+2 = 5 = 1.2𝐴
= ∶ ∶
𝑉1 𝑉2 𝑉3 477 (b)
𝑙12 𝑙22 𝑙32 Let 𝑛 be the number of wrongly connected cells
= ∶ ∶
𝑚1 𝑚2 𝑚3 Number of cells helping one another = (12 − 𝑛)
1 9 25 Total e.m.f. of such cells = (12 − 𝑛)𝐸
= ∶ ∶ = 1 ∶ 15 ∶ 125
5 3 1 Total e.m.f. of cells opposing = 𝑛𝐸
469 (b)
Resultant e.m.f. of battery = (12 − 𝑛)𝐸 − 𝑛𝐸 =
Here, the resistance of 400 Ω and 10000 Ω are in
(12 − 2𝑛)𝐸
P a g e | 81
 Total resistance of cells = 12𝑟 100 × 1
1=
(∵ resistance remains same irrespective of (1 + 𝐺)
connections of cells) ⟹ 𝐺 = 99Ω
With additional cells 482 (a)
(a) Total e.m.f. of cells when additional cells help 𝐼 2 𝑅𝑡 𝐼 2 𝑅𝑡
𝐻= = cal
battery = (12 − 2𝑛)𝐸 + 2𝐸 𝐽 4.2
Total resistance = 12𝑟 + 2𝑟 = 14𝑟 484 (a)
∴
(12−2𝑛)𝐸+2𝐸
=3 …(i) 𝑙1 − 𝑙2 25
14𝑟 𝑟=( )𝑅 = ( ) 2 = 0.5 Ω
𝑙2 100
(b) Similarly when additional cells oppose the
485 (d)
battery 𝑉
(12−2𝑛)𝐸−2𝐸 𝐸 = 𝑙 ; 𝐸 is constant (volt. gradient)
=2 …(ii)
14𝑟 𝑉1 𝑉2 1.1 𝑉 180 × 1.1
Solving (i) and (ii), 𝑛 = 1 ⇒ = ⇒ = ⇒𝑉=
𝑙1 𝑙2 140 180 140
478 (c)
= 1.41 𝑉
Let the resultant resistance be 𝑅. If we add one
486 (d)
more branch, then the resultant resistance would
Let 𝑅0 be the initial resistance of both conductors.
be the same because this is an infinite sequence
∴ At temperature θ their resistances will be,
R1 = 1 X
A 𝑅1 = 𝑅0 (1 + 𝛼1 θ)
and 𝑅2 = 𝑅0 (1 + 𝛼2 θ)
R2 = 2 R
For series combination, 𝑅𝑠 = 𝑅1 + 𝑅2
B
𝑅𝑠 (1 + 𝛼𝑠 θ) = 𝑅0 (1 + 𝛼1 θ) + 𝑅0 (1 + 𝛼2 θ)
Y
Where, 𝑅𝑠0 = 𝑅0 + 𝑅0 = 2𝑅0
𝑅𝑅2
∴ + 𝑅1 = 𝑅 ⇒ 2𝑅 + 𝑅 + 2 = 𝑅 2 + 2𝑅 ∴ 2𝑅0 (1 + 𝛼2 θ) = 2𝑅0 + 𝑅0 θ(𝛼1 + 𝛼2 )
𝑅 + 𝑅2 𝛼 +𝛼
or 𝛼𝑠 = 1 2 2
⇒ 𝑅 2 − 𝑅 − 2 = 0 ⇒ 𝑅 = −1 or 𝑅 = 2𝑜ℎ𝑚
479 (b) For parallel combination,
𝑅1 𝑅2
Mass of water = volume × density 𝑅𝑝 =
= 1000 × 1 = 1000 g. 𝑅1 + 𝑅2
𝑅0 (1 + 𝛼1 θ)𝑅0 (1 + 𝛼2 θ)
Heat taken by water = 𝑚𝑐 ∆𝜃 𝑅𝑝0 (1 + 𝛼𝑝 θ) =
= 1000 × 1(37 − 22) cal 𝑅0 (1 + 𝛼1 θ) + 𝑅0 (1 + 𝛼2 θ)
𝑅 𝑅 𝑅0
= 1000 × 15 × 4.2 J Where, 𝑅𝑝0 = 𝑅 0+𝑅0 = 2
0 0
energy spent
Power of geyser= time
𝑅0 𝑅02 (1 + 𝛼1 θ + α2 θ + α1 α2 θ2 )
∴ (1 + 𝛼𝑝 θ) =
=
1000×15×4.2
1050 W. 2 𝑅0 (2 + 𝛼1 θ + α2 θ)
60
as 𝛼1 and 𝛼2 are small quantities.
480 (d)
∴ 𝛼1 𝛼2 is negligible.
2 3
𝛼 +𝛼 𝛼1 +𝛼2 𝛼 +𝛼
i1 or 𝛼𝑝 = 2+(𝛼1 +𝛼2 )θ = 2
[1 − ( 1 2 2 ) θ]
1 2

i
as (𝛼1 + 𝛼2 )2 is negligible
𝛼1 + 𝛼2
∴ 𝛼𝑝 =
i2 6 4 2
487 (b)
Resistance of upper branch 𝑅1 = 2 + 3 = 5Ω
Because as temperature increases, the resistivity
Resistance of lower branch 𝑅2 = 4 + 6 = 10Ω
𝑖 𝑅 10 increases and hence the relaxation time decreases
Hence 𝑖1 = 𝑅2 = 5
=2 1
2 1 for conductors (𝜏 ∝ 𝜌)
Heat generated across 3 Ω (H1 ) 𝑖12 × 3 4
= = =2 488 (a)
Heat generated across 6 Ω (H2 ) 𝑖22 × 6 2
𝑅1 (1 + 𝛼𝑡1 ) 10 (1 + 5 × 10−3 × 20)
∴ Heat generated across 3 Ω = 120 𝑐𝑎𝑙/𝑠𝑒𝑐 = ⇒ =
𝑅2 (1 + 𝛼𝑡2 ) 𝑅2 (1 + 5 × 10−3 × 120)
481 (b)
⇒ 𝑅2 ≈ 15Ω
Shunt is connected to the galvanometer 𝑖 𝑅 30 15
𝑖𝑆 Also 𝑖1 = 𝑅2 ⇒ 𝑖2
= 10 ⇒ 𝑖2 = 20 𝑚𝐴
𝑖𝑔 = 2 1
𝑆+𝐺 489 (d)

P a g e | 82
 Voltage across 𝐵3 is greatest hence 𝐵3 will show 495 (d)
maximum brightness. In series combination of Heat developed by 210 W electric bulb in 5 min is
bulbs, the bulb of lesser wattage will glow more given by
bright. Hence 𝑊2 > 𝑊1 .
𝑊 210 × 5 × 60
So , 𝑊1 < 𝑊2 < 𝑊3 . 𝐻= = = 15000 cal
490 (c) 𝐽 4.2
2𝑖 2 4
Thermal power in 𝐴 = 𝑃𝐴 = ( 3 ) 3𝑅 = 3 𝑖 2 𝑅 496 (c)
𝑖 2 2
Ammeter is used to measure the current through
Thermal power in 𝐵 = 𝑃𝐵 = (3) 6𝑅 = 3 𝑖 2 𝑅 the circuit
Thermal power in 497 (a)
𝐶 = 𝑃𝐶 = 𝑖 2 𝑅 Electric fuse is a type of over current protection
3R device. They are engineered to contribute a
A negligible amount of extra resistance to the
R
circuits they protect. This is largely accomplished
i C
6R by making the fuse wire as short as possible.
B Fuses are primarily rated as current amperes. A
⇒ 𝑃𝐴 : 𝑃𝐵 : 𝑃𝐶 fuse wire of certain material and gauge will blow
4 2 at a certain current no matter how long it is.
= ∶ ∶1=4∶2∶3
3 3 Since, length is not a factor in current rating the
491 (d) shorter it can be made the less resistance it will
𝐼 = 𝑛𝑒𝐴𝑣𝑑 have end to end.
1
or 𝑣𝑑 = 498 (c)
𝑛𝑒𝐴
𝐼 𝐼 2/√3 2
or 𝑣𝑑 ∝ 𝐾= = = 𝐴
𝐴 tan 𝜃 tan 60° 3
𝑣 ′ 𝑑 𝐼 ′ /𝐴′ 2𝐼/2𝐴 500 (a)
∴ = = =1
𝑣𝑑 𝐼/𝐴 𝐼/𝐴 The effective resistance of combined wire
or 𝑣 ′ 𝑑 = 𝑣𝑑 = 𝑣 𝑙+𝑙 𝜌1 𝑙 𝜌2 𝑙
492 (b) 𝜌( )= +
𝐴 𝐴 𝐴
Let 𝑖 be the current through arm 𝐴𝐷𝐶. Then (∵total length L=𝑙 + 𝑙)
current through arm 𝐴𝐵𝐶 = (2.1 − 𝑖). As there is Or
no deflection in the galvanometer, hence 𝜌1 + 𝜌2
𝜌=
2
(20 + 5)𝑖 = (8 + 2)(2.1 − 𝑖)

or 25𝑖 = 21 − 10𝑖 or 35𝑖 = 21

or 𝑖 = 21/35 = 3/5 = 0.6 A

493 (b)
By using 𝑒0100 = 𝑒032 + 𝑒32
70 100
+ 𝑒70
100 100
⇒ 200 = 64 + 76 + 𝑒70 ⇒ 𝑒70 = 60 𝜇𝑉
494 (c)
𝑖 100
By using 𝑣𝑑 = 𝑛𝑒𝐴 = 𝜋
1028 ×1.6×10−19 × ×(0.02)2
4
= 2 × 10−4 𝑚/𝑠𝑒𝑐

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