To determine the internal resistance of a cell by using a potentiometer, the null point is obtained at 1 m when cell is shunted by 3 Ω resistance and at 1.5 m length when cell is shunted by 6 Ω resistance. The internal resistance of the cell is (a) 2 Ω (b) 4 Ω (c) 6 Ω (d) 8 Ω

Answer: (c). Shunted cell: l = l₀R/R+r. 1 = l₀3/3+r and 1.5 = l₀6/6+r Dividing: 1.5 = (2(3+r))/(6+r) ⇒ 9+1.5r = 6+2r r = 6 Ω So the answer is (c) 6 Ω.

To determine the internal resistance of a cell by using a… | MHT CET 2026 (12 May, Shift 2)

Q16MHT CET 2026 · 12 MAY · SHIFT 2Medium

Potentiometer

To determine the internal resistance of a cell by using a potentiometer, the null point is obtained at $1$ m when cell is shunted by $3\ \Omega$ resistance and at $1.5$ m length when cell is shunted by $6\ \Omega$ resistance. The internal resistance of the cell is

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Answer: (c)

Shunted cell: $l = l_0\dfrac{R}{R+r}$ .
$1 = l_0\dfrac{3}{3+r}$ and $1.5 = l_0\dfrac{6}{6+r}$
Dividing: $1.5 = \dfrac{2(3+r)}{6+r} \Rightarrow 9+1.5r = 6+2r$
$r = 6\ \Omega$
So the answer is (c) $6\ \Omega$ .

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Potentiometer question from MHT CET 2026 (12 May, Shift 2)

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