Question-22 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 balance point at distance ‘x’. The value of ‘x’ is
(A) \frac{5L}{12}
(B) \frac{4L}{15}
(C) \frac{3L}{10}
(D) \frac{2L}{15}
Answer : (C)
Explanation:
If the length of potentiometer is increased by \frac{L}{2}, the balancing length will also increase by the same proportion. Therefore, the balancing length will increase from
\frac{L}{5} to \left(\frac{1}{2}\times\frac{L}{5}\right).
⇒ \frac{L}{5}+\frac{L}{10}=\frac{3L}{10}