In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length-of null point increases by 40 cm. The value of E is x/10 V. The value of x is ____________.

Solution

Physics sir · 3 steps

In a potentiometer, emf is proportional to balancing length, so a simple ratio gives E.

1. Why a ratio works. The wire has a fixed potential drop per unit length, k. At the null point no current flows through the cell, so its emf equals the drop across the balancing length: E = kℓ.
2. New balancing length. The first cell balances at ℓ₁ = 60 cm. The new length is 40 cm more, so ℓ₂ = 60 + 40 = 100 cm.

   E/1.5 = ℓ₂/ℓ₁ = 100/60 ⇒ E = 1.5 × 100/60 = 2.5 V

3. So. E = 2.5 V = 25/10 V, so x = 25.

The common slip: 10. Taking the new balancing length as 40 cm instead of 60 + 40 = 100 cm gives E = 1 V, so x = 10.

The idea to keep. With the same driver circuit, the potential gradient stays fixed, so emfs compare directly as balancing lengths: E₁/E₂ = ℓ₁/ℓ₂. Read carefully whether a length is given or a change in length.

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