Q 12-03-115JEE MainJEE Main 2023 (29 Jan, Shift 2)Medium
A null point is found at $200\ \text{cm}$ in potentiometer when cell in secondary circuit is shunted by $5\ \Omega$. When a resistance of $15\ \Omega$ is used for shunting, null point moves to $300\ \text{cm}$. The internal resistance of the cell is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 5
The terminal voltage $\dfrac{\varepsilon R}{R+r}$ is balanced, so it is proportional to the balancing length:
$$\frac{5/(5+r)}{15/(15+r)}=\frac{200}{300}\ \Rightarrow\ 3\times5(15+r)=2\times15(5+r)$$
$$225+15r=150+30r\ \Rightarrow\ r=5\ \Omega$$
Solution by Sreeraj P, M.Sc Physics