Physics - Current Electricity Question with Solution | TestHub

Let $V$ be the potential difference across the potentiometer wire.

Initially, the potential gradient is $k_1=\frac{V}{L}$.

The balancing length for the cell of e.m.f. $E$ is $l_1=\frac{L}{5}$.

Thus, $E=k_1{l}_1=\frac{V}{L}\times \frac{L}{5}=\frac{V}{5}$.

When the length of the wire is increased by $\frac{L}{2}$, the new length becomes $L^{\prime }=L+\frac{L}{2}=\frac{3L}{2}$.

Assuming the potential difference $V$ across the wire remains constant, the new potential gradient is $k_2=\frac{V}{L^{\prime }}=\frac{2V}{3L}$.

The new balancing length is $x$.

Thus, $E=k_{2}x=\frac{2V}{3L}x$.

Equating the two expressions for $E$:

$\frac{V}{5}=\frac{2V}{3L}x$

$x=\frac{3L}{10}$

Answer: $\frac{3\mathrm{L}}{10}$