Question

A potentiometer wire of length $L$ and resistance $10\Omega$ is connected in series with a battery of e.m.f. $2.5V$ and a resistance in its primary circuit. The null point corresponding to a cell of e.m.f. $1V$ is obtained at a distance $L/2$. If the resistance in the primary circuit is doubled then the position of new null point will be

• A. $0.4L$
• B. $0.5L$
• C. $0.6L$
• D. $0.8L$

Answer 1

The correct option is C $0.6L$

Initial current in the circuit $=\frac{2.5V}{R+10\Omega }$
Potential difference across $\frac{L}{2}$ or $5\Omega =\frac{2.5}{R+10}\times 5=1V$
because a cell of $1volt$ balances.
From this $R=2.5\Omega$
$\therefore$ When $R$ is doubled,
$i=\frac{2.5}{5+10}A$ If this balances at a length $x,\frac{2.5}{15}\times R^{\prime }=1V$ where $R^{\prime }$ is the resistance corresponding to the length $x$.
$\Rightarrow R^{\prime }=\frac{15\times 1}{5/2}=6\Omega$ i.e., $x=\frac{L\times 6}{10}=0.6L$.