Q 12-03-175JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
A cell, shunted by a $8\ \Omega$ resistance, is balanced across a potentiometer wire of length $3$ m. The balancing length is $2$ m when the cell is shunted by $4\ \Omega$ resistance. The value of internal resistance of the cell will be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 8
When the cell (emf $E$, internal resistance $r$) is shunted by $R$, the potentiometer balances its terminal voltage $V = \dfrac{ER}{R + r}$, and the balancing length is proportional to $V$.
$$\frac{3}{2} = \frac{\dfrac{8}{8 + r}}{\dfrac{4}{4 + r}} = \frac{2(4 + r)}{8 + r}$$
$$3(8 + r) = 4(4 + r) \Rightarrow 24 + 3r = 16 + 4r \Rightarrow r = 8\ \Omega$$
Solution by Sreeraj P, M.Sc Physics