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

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(D) Let the potential difference across the wire be $V$. The potential gradient $k$ is given by $k = \frac{V}{L_{total}}$.For the first case,the balan

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$\frac{3 L}{10}$

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(D) Let the potential difference across the wire be $V$. The potential gradient $k$ is given by $k = \frac{V}{L_{total}}$.For the first case,the balancing length $l_1 = \frac{L}{5}$. The e.m.f. $E$ is given by $E = k_1 l_1 = \frac{V}{L} \cdot \frac{L}{5} = \frac{V}{5}$.When the length of the wire is increased by $\frac{L}{2}$,the new total length becomes $L' = L + \frac{L}{2} = \frac{3L}{2}$.The new potential gradient is $k_2 = \frac{V}{L'} = \frac{V}{3L/2} = \frac{2V}{3L}$.Let the new balancing length be $l_2$. Then $E = k_2 l_2$.Equating the two expressions for $E$: $\frac{V}{5} = \frac{2V}{3L} \cdot l_2$.Solving for $l_2$: $l_2 = \frac{V}{5} \cdot \frac{3L}{2V} = \frac{3L}{10}$.

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 $\frac{L}{2}$,at what distance will the same cell give a balance point?

Similar Questions

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