The length of a potentiometer wire is L. A ce | vedclass

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(C) Let the potential difference across the potentiometer wire be $V$. The potential gradient $k_1$ for the original wire of length $L$ is $k_1 = \fra

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$\frac{3L}{10}$

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(C) Let the potential difference across the potentiometer wire be $V$. The potential gradient $k_1$ for the original wire of length $L$ is $k_1 = \frac{V}{L}$.The balance point is at $l_1 = \frac{L}{5}$,so $E = k_1 l_1 = \frac{V}{L} \cdot \frac{L}{5} = \frac{V}{5}$.When the length of the wire is increased by $\frac{L}{2}$,the new length is $L' = L + \frac{L}{2} = \frac{3L}{2}$.The potential difference $V$ across the wire remains the same (assuming the driver circuit is unchanged). The new potential gradient $k_2$ is $k_2 = \frac{V}{L'} = \frac{V}{3L/2} = \frac{2V}{3L}$.For the same cell $E$,the new balance point $x$ is given by $E = k_2 x$.Substituting the values: $\frac{V}{5} = \left( \frac{2V}{3L} \right) x$.Solving for $x$: $x = \frac{V}{5} \cdot \frac{3L}{2V} = \frac{3L}{10}$.

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,the same cell will give a balance point at distance $x$. The value of $x$ is

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