A potentiometer wire PQ of 1 m length is connected to a standard cell E₁. Another cell E₂ of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm from Q. The potential gradient in the potentiometer wire is :

Solution

Physics sir · 3 steps

The cell is tied to end P, so measure the balancing length from P: 100 − 49 = 51 cm.

1. Where to measure. In the figure, E₂ is connected to end P, and the jockey J taps the wire. The balancing length is the wire between P and J, so it is 100 - 49 = 51 cm, not 49 cm.
2. Role of r. At the null point the galvanometer reads zero, so no current flows through E₂ or through r. With S open, the voltage across the wire length PJ is exactly the emf of E₂, which is 1.02 V.

   k = E₂/l = (1.02 V)/(51 cm) = 0.02 V cm⁻¹

3. So. The potential gradient along the wire is 0.02 V/cm, which is option (c).

The idea to keep. In a potentiometer, the balancing length always starts from the end where the cell is connected. At balance no current flows through the cell, so a series resistance makes no difference.

Answer from the NTA final answer key. Solution by PrepWiser.