The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20Ω, the null point on it is found to be a 1000 cm. The resistance of whole wire is :

Solution

Physics sir · 3 steps

At the null point no current flows through the cell, so its 20 Ω internal resistance plays no part.

1. Null point. At balance, the galvanometer reads zero, so no current is drawn from the cell. With no current there is no drop across its internal resistance, and the full emf of 5 V balances the wire.
2. Potential gradient. The wire carries I = 60 mA = 0.06 A, so the drop across the whole wire is IR. The drop per unit length is IR/L, and the balancing length l gives the emf.

   ε = IR/L l ⇒ R = (ε L)/(I l) = (5 × 1200)/(0.06 × 1000)

3. So. R = 6000/60 = 100 Ω, which is option (d).

The idea to keep. A potentiometer measures emf because it draws no current at balance. Internal resistance only matters when current flows through the cell, so ignore it at the null point.

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