Q 12-03-058JEE MainJEE Main 2026 (28 Jan, Shift 1)Medium
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 ______ $\Omega$
Answer: (A) 4
With shunt $R$, the terminal voltage balanced is $V = \dfrac{ER}{R + r}$, and $V \propto$ balancing length.
$$\frac{\frac{4}{4+r}}{\frac{12}{12+r}} = \frac{120}{180} = \frac23$$
$$\frac{4(12+r)}{12(4+r)} = \frac23 \Rightarrow 3(12 + r) = 6(4+r)\Rightarrow 36 + 3r = 24 + 6r$$
$$r = 4\ \Omega$$
Solution by Sreeraj P, M.Sc Physics