Q. The figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 2.5 V cell. The balance point of the cell in the open circuit is 75cm. When a resistor of 10$\ \Omega$is used in the external circuit of the cell, the balance point shifts to 65 cm length of potentiometer wire. Then the internal resistance of the cell is
A) 2.5$\ \Omega$
B) 2.0$\ \Omega$
C) 1.54$\ \Omega$
D) 1.0$\ \Omega$
Explanation
(c) $r = R\left( \frac{l_{1}}{l_{2}} - 1 \right) = 10\ \left( \frac{75}{65} - 1 \right)$
$\ \ \ = 10 \times 0.0154 = 1.54\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.