Q. The length of a wire of a potentiometer is 100cm, and the emf of its stand and cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5$\Omega$. If the balance point is obtained at $l = 30$cm from the positive end, the emf of the battery is
A) $\frac{30E}{100.5}$
B) $\frac{30E}{100 - 0.5}$
C) $\frac{30(E - 0.5i)}{100},\ Where\ i\ is\ the\ current\ in\ the$
$\ potentiometer\ wire.$
D) $\frac{30E}{100}$
Explanation
(d) $V \propto l$
$\Longrightarrow \frac{V}{E} = \frac{l}{L}$
Where, $l$= balance point
L= length of potentiometer wire.
Or $V = \frac{l}{L}E$
$or\ \ V = \frac{30 \times E}{100} = \frac{30}{100}E$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.