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JEE Main 2026 (28 Jan, Shift 1) · Potentiometer

Potentiometer question from JEE Main 2026 (28 Jan, Shift 1)

Q19JEE Main 2026 · 28 JAN · SHIFT 1Medium

Potentiometer

In the potentiometer, when the cell in the secondary circuit is shunted with $4\ \Omega$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with $12\ \Omega$ resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is $\underline{\qquad}\ \Omega$

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Answer: (a)

$l = l_0\dfrac{R}{R+r}$ : $120 = l_0\dfrac{4}{4+r}$ , $180 = l_0\dfrac{12}{12+r}$ .
Dividing: $1.5 = \dfrac{3(4+r)}{12+r} \Rightarrow 18+1.5r = 12+3r$
$r = 4\ \Omega$
So the answer is (a) 4.

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