Exams › JEE Main › Physics
A potentiometer wire PQ of length 1 m is driven by a standard cell E1. A second cell E2 of EMF 1.02 V is connected in series with a resistance r and a switch S. With S open, the balance (null) point is located 49 cm from end Q. Find the potential gradient along the wire.
- 0.02 V/cm
- 0.04 V/cm
- 0.01 V/cm
- 0.03 V/cm
Correct answer: 0.02 V/cm
Solution
At null, the EMF 1.02 V equals the potential drop over the balancing length measured from P. The gradient is EMF divided by that length.
Related JEE Main Physics questions
⚔️ Practice JEE Main Physics free + battle 1v1 →