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 $\frac{L}{2}$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{4 L}{15}$

$\frac{3 L}{10}$

$\frac{L}{5}$

$\frac{2 L}{15}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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