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MHT CET 2025 (23 Apr, Shift 2) · Potentiometer

Potentiometer question from MHT CET 2025 (23 Apr, Shift 2)

Q29MHT CET 2025 · 23 APR · SHIFT 2Medium

Potentiometer

The length of a potentiometer wire is ' $L$ '. A cell of e.m.f. ' $E$ ' is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{\mathrm{L}}{3}$ , then using the same cell null point is obtained at

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Answer: (b)

$k = \dfrac{V}{L}$ , so $E = \dfrac{V}{L}\cdot\dfrac{L}{4} = \dfrac{V}{4}$ .
New length $\dfrac{4L}{3}$ : $\dfrac{3V}{4L}\,x = \dfrac{V}{4}$
$x = \dfrac{L}{3}$
So the answer is (b) $\dfrac{L}{3}$ .

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