Q. A potentiometer wire of length $L$ and resistance $10\ \Omega$ is connected in series with a battery of e.m.f. $2.5\ V$ and a resistance in its primary circuit. The null point corresponding to a cell of e.m.f. $1V$ is obtained at a distance $\frac{L}{2}.$ If the resistance in the primary circuit is doubled then the position of new null point will be
A) $0.4\ L$
B) $0.5\ L$
C) $0.6\ L$
D) $0.8\ L$
Explanation
(c) $I = \frac{E}{R_{T}} = \frac{2.5}{10 + R}$ and $V = I.R = \frac{2.5 \times 10}{10 + R} = \frac{25}{10 + R}$
$x = \frac{V}{L} = \frac{25}{(10 + R)L}$
$E = x.l_{1}$
$\Rightarrow 1 = \frac{25}{(10 + R)L} \times \frac{L}{2} \Rightarrow 25 = 20 + 2R$
$\Rightarrow 2R = 5 \Rightarrow R = \frac{5}{2}$
$\therefore$ Now the resistance is doubled
$R^{1} = \frac{5}{2} \times 2 = 5\Omega$
$\therefore x = \frac{25}{(10 + 5)L} = \frac{25}{15.L} = \frac{5}{3L}$
$E = x.l_{2}$
$\Rightarrow 1 = \frac{5}{3L}.l_{2} \Rightarrow l_{2} = \frac{3L}{2} = 0.6L$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.