The length of the potentiometer wire is L. A cell of emf ‘e’ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. Assuming there is no contribution in resistance from any part of the circuit except the potentiometer wire, the balancing length for the same cell if length of the wire is increased by $\frac{L}{2}$

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$\frac{4 L}{15}$

$\frac{3 L}{10}$

$\frac{L}{5}$

$\frac{2 L}{15}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} balancing equation for potentiometer e & = & \frac{E}{R_{p}} \frac{R_{p}}{L} \frac{L}{5} - - - (1) \\ when length increases to \frac{3 L}{2}, e & = & \frac{E}{3 L / 2} l - - - - - (2) \\ equating l & = & \frac{3 L}{10} \end{array}$

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