Q. Two cells A and B are connected in the secondary circuit of a potentiometer one at a time and the balancing length are respectively 400 cm and 440 cm. The emf of the cell A is 1.08V. The emf of the second cell B is volt is
A) 1.08
B) 1.188
C) 11.88
D) 12.8
Explanation
(b) $\ \ \ \ \ \ \frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Longrightarrow \frac{1.08}{E_{2}} = \frac{400}{440}$
$\Longrightarrow E_{2} = \frac{440 \times 1.08}{400}\mathbf{=}1.188V.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.