Q. Resistance of $100\ cm$ long potentiometer wire is $10\Omega,$ it is connected to a battery $(2\ volt)$ and a resistance $R$ in series. A source of $10\ mV$ gives null point at $40\ cm$ length, then external resistance $R$ is
A) $490\ \Omega$
B) $790\ \Omega$
C) $590\ \Omega$
D) $990\ \Omega$
Explanation
(b) $E = \frac{e}{\left( R + R_{h} + r \right)}.\frac{R}{L} \times l$
$\Rightarrow 10 \times 10^{- 3} = \frac{2}{(10 + R + 0)} \times \frac{10}{1} \times 0.4 \Rightarrow R = 790\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.