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(A) The potential gradient $k$ of a potentiometer wire is given by $k = \frac{V}{L_{total}}$,where $V$ is the potential difference across the wire and

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$\frac{L}{2}$ from positive end

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(A) The potential gradient $k$ of a potentiometer wire is given by $k = \frac{V}{L_{total}}$,where $V$ is the potential difference across the wire and $L_{total}$ is the total length of the wire.For a cell of emf $E$,the balancing length $l$ is given by $E = k \cdot l = \frac{V}{L} \cdot l$.Initially,$l_1 = \frac{L}{3}$,so $E = \frac{V}{L} \cdot \frac{L}{3} = \frac{V}{3}$.When the length of the wire is increased by $50 \%$,the new length $L' = L + 0.5L = 1.5L = \frac{3L}{2}$.The potential difference $V$ across the wire remains the same as the source voltage is constant.The new potential gradient $k' = \frac{V}{L'} = \frac{V}{1.5L} = \frac{V}{1.5L}$.For the same cell $E$,the new balancing length $l_2$ is $E = k' \cdot l_2$.Substituting the values: $\frac{V}{3} = \frac{V}{1.5L} \cdot l_2$.Solving for $l_2$: $l_2 = \frac{1.5L}{3} = 0.5L = \frac{L}{2}$.

$A$ potentiometer wire has length $L$. For a given cell of emf $E$,the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of the potentiometer wire is increased by $50 \%$,then for the same cell,the balance point is obtained at length

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