In the potentiometer, when the cell in the secondary circuit is shunted with 4Ω resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12Ω resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is
Q. In the potentiometer, when the cell in the secondary circuit is shunted with 4Ω resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12Ω resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ____ Ω.

(A) 4

(B) 12

(C) 6

(D) 3

Correct Answer: 4 Ω

Explanation

In a potentiometer, the balancing length is proportional to the potential difference across the external resistance connected to the cell.

Let the emf of the cell be E and internal resistance be r.

When the cell is shunted with resistance 4Ω, the potential difference is:

V₁ = E × 4 / (4 + r)

When the cell is shunted with resistance 12Ω, the potential difference is:

V₂ = E × 12 / (12 + r)

Since balancing length is proportional to potential difference:

V₁ / V₂ = l₁ / l₂ = 120 / 180 = 2 / 3

Substituting values:

(4 / (4 + r)) ÷ (12 / (12 + r)) = 2 / 3

(4 × (12 + r)) / (12 × (4 + r)) = 2 / 3

3(12 + r) = 2(4 + r)

36 + 3r = 8 + 2r

r = 4 Ω

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