Solution

PHYSICS

JEE main - Physics

1. (a)

Explanation: When we move in the direction of the current in a uniform conductor, the potential difference decreases linearly.

R
When we pass through the cell, from its negative to its positive terminal, the potential increase by an amount equal to its
potential difference. This is less than its emf, as there is some potential drop across its internal resistance when the cell is

EE
driving current.
2.
(b) 570 Ω
Explanation:
AR
Measured value of R = 5% less than actual value of R.

E
Actual values of R = 30 Ω
So, measured value of R is R' = 30 - (5% of 30) = 30 - 5
30
⇒ R' = 28.5 Ω ...(i)
UT 100
×
C

Now, let us assume that internal resistance of voltmeter Rv. Replacing voltmeter with its internal resistance, we get following
circuit.
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N
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IN

It is clear that the measured value, R' should be equal to parallel combination of R and Rv . Mathematically,
RRV
′
R = = 28.5Ω
R+RV

30RV
Given, R = 30Ω ⇒ 30+R
= 28.5
V

⇒ 30RV = (28.5 × 30) + 28.5 RV

⇒ 1.5RV = 28.5 × 30
⇒ RV = 28.5×30

1.5
= 19 × 30 or RV = 570 Ω

3.
(c) 8Ω
Explanation: 8Ω
4. (a) 4.5 Ω
100 90 R+r 10 0.5 10
Explanation: R+r
=
R
⇒
R
=
9
⇒ 1 +
R
=
9
0.5 1
⇒
R
=
9
∴ R = 4.5 Ω

5.
(b) AB
Explanation: According to Ohm's law, the resistance is the voltage applied across resistor per unit current passing through it.
V
R=
I

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 Hence R = 1

I/V

⇒ R= 1

slope

Hence resistance is negative when the slope of the curve is negative.

Therefore resistance is negative in region AB.
6. (a) zero V
Explanation: zero V
7.
(b) None of these
Explanation: Reading will remain zero, whatever may be the value of ammeter resistance.
8.
4
65πaR
(d) 2592

Explanation: Given that,

J = ar2

R
2

i=∫J × 2πrdr
1

EE
R

2
2
∫ ar × 2πrdr
R

2
AR
= 2πa ∫ r dr
3

E
R

4 4
πa R R
=

=
2
[(

4
2
) − (
UT 3

=
) ]

4
C
πaR 65 65πaR
×
2 81 × 16 2592

9.
(b) 0.4Ω -m
IT
N

Explanation: Since, it is an n-type semiconductor and concentration of the holes has been ignored. So, its conductivity is given
as σ = n e eμe where ne is the number density of electron, e is the charge on electron and μ is its mobility.
e
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Substituting the given values, we get σ = 1019 × 1.6 × 10-19 × 1.6 = 2.56
ST

As, resistivity, ρ = = or ρ = 0.39 ≃ 0.4Ω -m 1

σ
1

2.56

10.
AL

(b) 20 A
IN

Explanation: Net Power, P

= 15 × 45 + 15 × 100 + 15 × 10 + 2 × 1000
= 15 × 155 + 2000 W
Power, P = VI ⇒ I = P

V
15×155+2000
∴ Imain = 220
= 19.66 A ≈ 20A

11. (a) r = R
Explanation: When the cells are connected in series, current l1 is given by:
l1 =
R+nr
nE
...(i)
When the cells connected in parallel, current I2 is given by:
I2 =
E
r
=
nE

nR+r
...(ii)
R+
n

As l1 = l2
So, R+nr
nE
=
nE

nR+r

∴ R + nr = nR + r
or (n - 1)r = (n - 1)R or r = R
12.
(d) 3 volt
Explanation: Suppose, E is the emf of the cell. Let R be the resistance of voltmeter. After shunting, let the resistance be equal

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 to x. Let r be the internal resistance of the cell and let the voltmeter reading be decreased n times.
As reading of voltmeter is decreased n = (3) times
Hence, Ex

r+ x
=
1

n
[
ER

r+ R
] ..(i)
Further, reading of ammeter is increased n(= 3) times,
Hence, =
E

r+ x
...(ii) r+ R
nE

Solving equations (i) and (ii). We get ; Ex

r+ x
=
E

n + 1
Ex 12
∴ =
r+ x 3 + 1

E1 r2 + E2 r1 r1 r2
13. (a) r1 + r2
and r1 + r2

Explanation: Let the equivalent resistance and emf of the parallel combination
be r and E
r1 r2
Then, = 1

r
+ or r =
1

r1
1

r2 r1 + r2

E1 E2
However, maximum currents drawn from the batteries are ( r1
) and ( r2
) . Hence, total current I is given by:
E1 E2 E1 r2 + E2 r1
I= r1
+
r2
=
r1 r2

R
To obtain the same current I from the equivalent source its emf E is given by:
E1 r2 + E2 r1 r1 r2 E1 r2 + E2 r1
E = Ir = ( )( )

EE
r1 r2 r1 + r2 r1 + r2

14. (a) more with 39 bulbs then with 40 bulbs

Explanation: The voltage terminals same for both the cases So, illuminance, H ∝( 1

R
.
)

Hence the combination of 39 bulbs have low resistance as compared to 40 bulb, so 39 bulb combination will glow more.
AR
15.

E
(d) 0.13 A, from Q to P

Explanation:
UT
C

Applying Kirchhoff's loop law in loops 1 and 2 in the directions shown in figure we have
IT
N

6 - 3(i1 + i2) - i2 = 0 ...(i)

9 - 2i1 + i2 - 3i1 = 0 ...(ii)
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ST

Solving Eqs. (i) and (ii) we get, i2 = 0.13 A

Hence, the current in 1Ω resister is 0.13 A from Q to P
16.
AL

(b) 0.36 A
IN

Explanation: Wheatstone network is balanced as per resistances of P, 2, R and S given. No current flow through galvanometer.
P and R are in series in upper branch.
Resistance = (P + R) = 10 + 15 = 25 Ω
Q and S are in series in lower branch
∴ Resistance = Q + S = 20 + 30 = 50 Ω

The upper and the lower branches are in parallel.

∴ Resistance =
25 × 50 25 × 50 50
= = Ω
25 + 50 75 3
V 6
∴ Current I = R
=
50/3
= 0.36 A

17.
(d) 3 × 1019
Explanation: 3 × 1019
18.
(b) 90 kW
Explanation: Given that the capacitance is C = 40 μF and the potential is V = 3000 V. Therefore the energy
E= 1

2
CV2 = 1

2
× 40 × 10
−6
× (3000)
2

⇒ E = 20 × 9
⇒ E = 180 J

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 If this energy is transferred in a time t = 2 ms. Then the power delivered is,
P= = E

t
= 90 kW180

−3
2×10

19.
(c) 100
Explanation: G reads 0 ⇒ p = d across X = 2V
∴ pd across 500 Ω resistance = 10 V
10
∴ i =
500

For X = V

i
=
2 × 500

10
= 100

20.
(c) 9/32
Explanation:
Given circuit is redrawn and can be simplified as

ER
RE
E
CA
UT
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Voltage
N

So, current drawn through cell is i = Net resistance of the circuit

V 15 9
= = = A
′′ 32
R (160/3)
eq
LE
ST

21. (a)
AL
IN

Explanation: In series, current through both the conductors will be same and increases with the increase in voltage; hence, the
graph between I and V will have increasing slope upto 25 volts. After that the resistance of the first conductor increases in such
a way that current through it remains constant at one ampere.
22.
–
(c) 1 + √3
Explanation: The equivalent circuit may be shown as

Where R may be treated as the total resistance

Hence 1 + + 1 = R
1×R

1+R
–
This gives R = 1 ± √3Ω
–
Since (1 − √3)Ω gives -ve value,
–
therefore R = (1 + √3)Ω

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23.
E1
(b) R1
ε1
Explanation: i = R1
as R2 is also shorted.

24.
(b) Depends upon the relative values of external and internal resistance.
Explanation: Let each cell of emf 'E' and internal resistance 'r'. There are n such cells. The external resistance be 'R'.
If cells are connected in series :
nE
current through 'R' = nr+R

If cells are connected in parallel :

nE
current through R = nr+R

Do draw a maximum current :

if nr + R > r + nR cells must be connected in parallel.
if r + nR > nr + R cells must be connected in series.
So, it depends on the relative values of internal and external resistances.

R
25. (a) 10 amp
Explanation: The value of resistance R in the circuit = 20 + I

EE
2

Now, I = 250

I
20 +
2

or I(20 + I

2
) = 250
or I2 + 40I - 500 = 0
AR
Solving, we get; I = 10 ampere.

E
26. 2
Explanation: UT
Flow of the current in the circuit is shown in the following figure:
C
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Writing KVL equations for loops I, II and III, we get

-9 + 3l + 6l1 = 0
4l2 + 2l3 + l2 - 6l1 = 0
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2l4 + 4 - 2l3 = 0
IN

Using KCL equation, we get

l = l1 + l2
l = l3 + l4
Solving above equations, we get: l2 = 0.5A, so potential difference across 4Ω resistor is:
V = 4l2 = 4 × 0.5 = 2 volt
27. 1.0
Explanation:

Effective emf,
E2 - E1 = 8 - 4 = 4V

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 Req = 0.5Ω + 1Ω + 4.5Ω +
6×3

3+6
Ω =8Ω
4
Current, I = 8
= 0.5 A

Current through 3 Ω resistance,

6 2 1
∴ I1 = (
3+6
) × 0.5 = 3
× 0.5 =
3
A

⇒ I1 =
x

3
=
1

3
∴ x=1
28. 11
Explanation:
Net emf, E = E1 - E2 = 12 volt (∵ E1 > E2)

ER
Total resistance of circuit R = 3 + 1 = 4Ω
I = =
E

R
= 3 Amp
12

∴ PD across R1 = V1 = IR1 = 3 × 3 = 9V
But from battery of emf, E1 = 20 volt, a potential difference of 9 V will fall across resistance R1
∴ PD across P and Q = 20 - 9 = 11 volt
29. 3840.0
RE
E
Explanation:
CA
Rate of energy dissipated
= 192J

R = 12Ω
1s
= i2 R ⇒ 192 = 42 × R UT
Energy = i2Rt = (8)2 × 12 × 5 = 3840
30. 9
IT
N

Explanation:
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The net external resistance of the circuit across AD is (5R/9). Using concept of maximum power transfer to load, the power
dissipated by resistors of hexagon is maximum when its resistance is equal to internal resistance
∴ = 5 or R = 9Ω
5R

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