Q. Two resistances $R_{1}$ and another $R_{2}$ of the same material but twice the length and half the thickness are connected in series with a standard battery $E$ of internal resistance $r.$ The balancing point is
A) $\frac{1}{8l}$
B) $\frac{1}{4l}$
C) $8\ l$
D) $16\ l$
Explanation
(c) If $l$ is the balancing length for $R_{1},$ and $l'$ for $R_{2},$
$\frac{V_{0}}{L_{0}}.l = \frac{ER_{1}}{(r + R_{1} + R_{2})} \Rightarrow \frac{V_{0}}{L_{0}}.l = \frac{ER_{1}}{\left( r_{1} + 9R_{1} \right)}$ as $R_{2} = 8R_{1}$
$\frac{V_{0}}{L_{0}}l^{'} = \frac{E.R_{2}}{(r_{1} + 9R_{1})} = \frac{E.8R_{1}}{(r_{1} + 9R_{1})}$
$\therefore\frac{l^{'}}{l} = \frac{R_{2}}{R_{1}} = \frac{8R_{1}}{R_{1}} \Rightarrow l^{'} = 8\ l$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.