Quizrr Chapter-Wise Test for JEE Main - 2025
Q.1
Find the equivalent resistance between points A and B in the circuit shown in Fig.
(A) 1.4 ohm
(B) 1.8 ohm
(C) 1.6 ohm
(D) 2.4 ohm
Q.2
A conductor of resistance 3Ω is stretched uniformly till its length is doubled. The wire is now bent in the form
of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is:
(A) 92
(B) 83
(C) 2
(D) 1
Q.3
Six equal resistances are connected between points A, B and C as shown Fig. If R1 , R2 and R3 are the net
resistances between A and B , between B and C and between A and C respectively, then R1 : R2 : R3
will be equal to
(A) 6:3:2
(B) 1:2:3
(C) 5:4:3
(D) 4:3:2
Q.4
The total current supplied to the circuit by the battery is
(A) 4A
(B) 2A
(C) 1A
(D) 6A
Q.5
Variation of current passing through a conductor as the voltage applied across its ends is varied as shown in the
adjoining diagram. If the resistance (R) is determined at the points A, B, C and D , we will find that
(A) RC = RD
(B) RB > RA
(C) RC > RB
(D) RA > RB
Q.6
Twelve resistors each of resistance 16Ω are connected in the circuit as shown. The net resistance between AB
is
(A) 1Ω
(B) 2Ω
(C) 3Ω
(D) 4Ω
Q.7
n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum
to the minimum resistance?
(A) n
(B) 1/n2
(C) n2
(D) 1/n
Q.8
There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between
neighbouring joint connections is R0 . The net resistance of the whole grid between the points A and B as
shown is
(A) R0
(B) R20
(C) R30
(D) R40
Q.9
The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then
(A) the resistance and the specific resistance, will both remain unchanged
(B) the resistance will be doubled and the specific resistance will be halved
(C) the resistance will be halved and the specific resistance will remain unchanged
(D) the resistance will be halved and the specific resistance will be doubled
Q.10
Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm2 . Each is 1 meter long.
One rod is of copper with a resistivity of 1.7 × 10−6 ohm-centimeter, the other is of iron with a resistivity of
10−5 ohm-centimeter.
How much voltage (in volt) is required to produce a current of 1 ampere in the rods?
(A) 0.111
(B) 0.113
(C) 0.115
(D) 0.117
Q.11
Two wires of the same metal have same length, but their cross-sections are in the ratio 3 : 1. They are joined in
series. The resistance of thicker wire is 10Ω. The total resistance of the combination will be
(A) 10Ω
(B) 20Ω
(C) 40Ω
(D) 100Ω
Q.12
In the network shown, each resistance is equal to R. The equivalent resistance between adjacent corners A and
D is
(A) R
(B) 23 R
(C) 37 R
8
(D) 15 R
Q.13
The resistance of the four arms P, Q, R and S in a Wheatstone's bridge are 10ohm, 30ohm, 30ohm
and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the
galvanometer resistance is 50 ohm, the current drawn from the cell will be
(A) 0.2 A
(B) 0.1 A
(C) 2.0 A
(D) 1.0 A
Q.14
The value of the resistance of a carbon resistor has bands of different colours as mentioned in the figure is
______ G𝛺 ± 5%
(A) 2.0
(B) 2.1
(C) 2.2
(D) 2.3
Q.15
A moving coil galvanometer, having a resistance G. produces full scales deflection when a current Ig flows
through it. It can be converted into (i) an ammeter of range 0 to Io ( 𝐼𝑜 > 𝐼𝑔 ) by connecting a shunt
resistance 𝑅𝐴 to it and (ii) into a voltmeter of range 0 to 𝑉𝑉 = 𝐺𝐼0 by connecting a series resistance Rv to it.
Then,choose the correct option
(A) 2
I 0 −I g RA Ig
2
RA RV = G (
Ig
) and RV
= (
I 0 −I g
)
(B) RA RV = G
2
and πan
RA
=
Ig
Rv (I 0 −I g )
(C) 2 RA Ig
2
RA Rv = G and = ( )
Rv I 0 −I g
(D) 2
Ig RA I 0 −I g
2
RA RV = G (
I 0 −I g
) and RV
= (
Ig
)
Q.16
Two resistances are connected in the two gaps of a meter bridge. The balance point is 20 cm from the zero
end. When a resistance 15 Ω is connected in series with the smaller of two resistance, the null point get
shifted to 40 cm. What is the smaller of the two resistance values ?
(A) 8 Ω
(B) 9 Ω
(C) 10 Ω
(D) 12 Ω
Q.17
A potentiometer wire has length 4 m and resistance 6Ω. The resistance that must be connected in series with
the wire and a battery of emf 4 V so as to get a potential gradient 5 mV per cm on the wire is
(A) 6Ω
(B) 12Ω
(C) 18Ω
(D) 24Ω
Q.18
In a uniform magnetic field, a flexible wire is bent to form a circle with its plane perpendicular to the field.
Above figure represents the change in the radius of the coil 𝑟 with time 𝑡. Now, calculate the voltage induced in
the coil:
(A)
(B)
(C)
(D)
Q.19
The resistance of the series combination of two resistors is S. When they are joined in parallel, the equivalent
resistance becomes P. If S = nP, then the minimum possible value of n is
(A) 3
(B) 4
(C) 5
(D) 6
Q.20
Two cylindrical wires A and B have the same resistance. The ratio of their specific resistances and diameters
are 1: 2 each, then what is the ratio of the length of B to the length of A?
(A) 1
(B) 2
(C) 3
(D) 4
Answers & Solutions
Q.1 Answer:
1.4 ohm
Solution:
The network of resistances shown in Fig. 12.31 is not a balanced Wheatstone's bridge. In such a
case, to find the resistance between A and B , we connect a battery of voltage V across A and B and
find the current I drawn from the battery in terms of V and find the equivalent resistance from the
relation V = IR. Refer to Fig.
The currents in various branches are shown in Fig. 12.32. Applying loops rule to loops
AFBGHA, ACDF A and DEBF D we have
V − 1 (I − I1 ) − 2 (I − I1 + I2 ) = 0 (i)
2I1 + I2 − (I − I1 ) = 0 (ii)
and (I1 − I2 ) + 2 (I − I1 + I2 ) − I2 = 0 (iii)
Simplifying these equations we get
3I − 3I1 + 2I2 = V (iv)
I − 3I1 − I2 = 0 (v)
and 2I − 3I1 + 4I2 = 0 (vi)
Eliminating I1 from (v) and (vi), we have I2 = − I5 . Using this value of I1 in (v) we get I1 = 25 I .
Using these values of I1 and I2 in (iv) we get
7I = 5 V ⇒ VI = 75 = 1.4 ⇒ RAB = 1.4Ω
Q.2 Answer:
8
3
Solution:
The resistance of a conductor of length l, crosssectional area A and made of a material of resistivity
ρ is given by
ρl2 2
R = ρl A =
ρ
Al = ( V ) l
where V = Al is the volume of the conductor. Since ρ is a constant and volume V cannot change if
the conductor is stretched, it follows that R is proportional to l2 . Thus if l is doubled, R becomes
four times. Hence the new resistance is 3 × 4 = 12Ω. So, each side of the equilateral triangle has a
resistance of 4Ω as shown in Fig. 12.127. Therefore, the effective resistance between the ends of any
side of the triangle (such as side AB ) is equal to the resistance of a parallel combination of R1 =
4Ω and R2 = 4 + 4 = 8Ω, which is given by
1 ×R 2 4×8
Re = R
R 1 +R 2
= 4+12
= 83 Ω
Hence the correct choice is (b).
Q.3 Answer:
5:4:3
Solution:
Let the value of each resistance be r . The network can be redrawn as shown in the Fig.
(i) Net resistance R1 between points A and B
The series combination of resistances 3r and 2r ,
r r 5r
which has an equivalent resistance r1 = 3 + 2
= 6 , is in parallel with resistance r . Hence
r×r1 r× 5r 5r
R1 = = 6
=
r+r1 r+ 5r 11
6
(ii) Net resistance R2 between points B and C The series combination of resistances r and 3r ,
4r
which has an equivalent resistance r2 =r+ r
3
= 3
, is in parallel with resistance 2r . Hence
r r 4r
2 ×r2 2× 3 4r
R2 = = =
r
+r2 r
+ 4r 11
2 2 3
(iii) Net resistance R3 between points A and C
The series combination of resistances r and 2r ,
r 3r
which has an equivalent resistance r3
=r+ 2
= 2 ,
r
is in parallel with resistance 3 . Hence
r
×r3 r
× 3r 3r
R3 = 3
= 3 2
=
r
+r3 r 3r 11
+
3 3 2
5r 4r 3r
Hence R1 : R2 : R3 =
11
: 11
: 11
= 5 : 4 : 3.
Thus the correct choice is (c).
Q.4 Answer:
4A
Solution:
6
hence Req = 3/2; ∴I= 3/2
=4A
Q.5 Answer:
RA > RB
Solution:
From the curve it is clear that slopes at points A, B, C , D have following order A >B>C>
D. And also resistance at any point equals to slope of the V − i curve. So order of resistance at
three points will be RA > RB > RC > RD
Q.6 Answer:
4Ω
Solution:
3R R
3 ×3 R2
Rnet between AB = = = 4Ω
3R R 4R
3 +3
Q.7 Answer:
n2
Solution:
= nR
In series, Rs
1
= R1 + R1 + … n term
In parallel, R
p
∴ Rs /Rp = n2 /1 = n2
Q.8 Answer:
R0
3
Solution:
By principle of symmetry and superposition, 2 × 6I × R0 = IReq. ⇒ Req. =
R0
3
(Current 6I in AB is due to division in current entering
at A and current 6I is due to current returning from infinity of grid).
Q.9 Answer:
the resistance will be halved and the specific resistance will remain unchanged
Solution:
R = ρℓA1 , now ℓ2 = 2ℓ1
1
2 2
A2 = π (r2 ) = π (2r1 ) = 4πr12 = 4A1
(2ℓ1 )
∴ R2 = ρ4A 1
= ρℓ1
2A 1
= R2
∴ Resistance is halved, but specific resistance remains the same.
Q.10 Answer:
0.117
Solution:
Copper rod and iron rod are joined in series. ∴ R = RCu + RFe = (ρ1 + ρ2 ) Aℓ
(∵ R = ρ Aℓ )
From ohm's law V = RI = (1.7 × 10−6 × 10−2 + 10−5 × 10−2 ) ÷
0.01 × 10−4 volt
= 0.117 volt (∵ I = 1 A)
Q.11 Answer:
40Ω
Solution:
Length of each wire = ℓ; Area of thick wire (A1 ) = 3 A; Area of thin wire (A2 ) = A and
2
resistance of thick wire (R1 ) = 10Ω. Resistance (R) = ρ Aℓ ∝ A 1
(if ℓ is constant) ∴ R
R2
1
=
A2
A1
= 3AA
= 13
or, R2 = 3R1 = 3 × 10 = 30Ω
The equivalent resistance of these two resistors in series = R1 + R2 = 30 + 10 = 40Ω
Q.12 Answer:
8
15
R
Solution:
The equivalent circuit is as shown in figure. The resistance of arm AOD(= R + R) is in
parallel to the resistance R of arm AD . Their effective resistance
2R×R
R1 =
2R×R
= 23 R
The resistance of arms AB, BC and CD is
R2 = R + 23 R + R = 83 R
The resistance R1 and R2 are in parallel. The effective resistance between A and D is R3
=
2
R 1 ×R 2 R× 83 R 8
= 3
= R
2
R 1 +R 2 R+ 83 R 15
3
Q.13 Answer:
0.2 A
Solution:
Given :
V = 7∇ and r = 5Ω
Req = 40×120
40+120 Ω
V 7
I = R = 5+ 40×120 = 0.2 A
40+120
Q.14 Answer:
2.1
Solution:
From color codes for resistors.
𝑅 = 21 × 108 𝛺 ± 5 % = 2 . 1 × 109 𝛺 ± 5%
∴ 𝑅 = 2 . 1𝐺𝛺 ± 5%
Q.15 Answer:
𝑅𝐴 𝐼𝑔 2
𝑅𝐴 𝑅𝑣 = 𝐺2 and =
𝑅𝑣 𝐼0 - 𝐼𝑔
Solution:
( 𝐼0 - 𝐼𝑔 ) 𝑅4 = 𝐼𝑔 G and V0 = 𝐼0 G + R𝑣
𝐼0 G + RA
I0 = RA
V0 = 𝐼0 G + RV
G + RA G
V0 = I0 G ( given ) ⇒ 𝐼0 G + RV = I𝑔
RA
𝑅𝐴 𝐺 + 𝑅𝑉 𝑅𝐴 = 𝐺2 + 𝑅𝐴 𝐺
𝑅𝑉 𝑅𝐴 = 𝐺2
𝐼𝑔 𝐺 𝐼0 - 𝐼𝑔 𝐺
𝑅𝐴 = ; 𝑅𝑉 =
𝐼0 - 𝐼𝑔 𝐼𝑔
𝑅𝐴 𝐼2𝑔
=
𝑅𝑉 𝐼0 - 𝐼𝑔 2
Q.16 Answer:
9 Ω
Solution:
𝑅1 20 1
𝑅2
= 80
= 4 , It is clear that 𝑅1 is smaller resistance
∴ 𝑅2 = 4𝑅1
When 15 Ω is connected in series with 𝑅1 , then equivalent resistance is 𝑅1 + 15 , So
𝑅1 + 15 40 2
∴ 𝑅2
= 60
= 3
𝑅1 + 15 2
∴ 4𝑅1
= 3
∴ 𝑅1 = 9 Ω
Q.17 Answer:
6Ω
Solution:
𝐿 𝐴𝐵 = 4 m 𝑅𝐴𝐵 = 6Ω 𝐸=4 V
-3
𝐸 𝑅 5 × 10 4 6
𝑥= ⇒ -2 = ⇒ 𝑅 = 6Ω
𝑅 + 𝑅𝑠 𝐿 10 𝑅 + 64
Q.18 Answer:
Solution:
The cross sectional area of a wire is given by πr2 where 𝑟 is the radius of the wire.
𝑑𝜙 𝑑𝐴 𝑑 dr
The emf induced in the wire is given by𝜀 = 𝑑𝑡
= 𝐵 𝑑𝑡 = 𝐵 𝑑𝑡 πr2 = 2πBr dt
dr
i.e., ε ∝ r
dt
dr
At t = 0 - 1s, dt
= 0 . so ε = 0
The induced emf will be zero
dr
At t = 1 - 2 s, dt
= constant . so ε ∝ r
The graph will raise linearly.
dr
At t > 2s, dt
= 0 . so ε = 0
The curve will fall steeply to zero.
Q.19 Answer:
4
Solution:
In series combination, S = R1 + R2
R1 R2
In parallel combination, P = R1 + R2
∵ S = nP
R1 R2
∴ R1 + R2 = n R + R2
1
∴ R1 + R2 2 = nR1 R2
For minimum value, R1 = R2 = R
R+R2 = n R×R ⇒ 4R2 = nR2 ⇒n = 4
Q.20 Answer:
2
Solution:
𝜌𝑙 𝜌𝑙 × 4
Resistance of a wire 𝑅 = 𝜋𝑟2
=
𝜋𝐷2
∵ 𝑅A = 𝑅B
4𝜌A 𝐿A 4𝜌B 𝐿B
∴ 2 = 2
𝜋𝐷A 𝜋𝐷B
𝐿B 𝜌A 𝐷 B 2
or 𝐿A
= 𝜌B 𝐷 A
𝜌 2𝐷A 2 4 2
= 2𝜌A 𝐷
= 2
= 1
A A
