In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.

Solution

Physics sir · 3 steps

Same wire, same gradient: emf is proportional to length, so the 3 : 2 ratio carries over to the lengths.

1. Why emf tracks length. At balance no current flows through the cell, so its emf equals the potential drop across the balancing length, which is kl (k is the potential gradient). Both cells use the same wire and the same driver circuit, so k is the same.

   E₁/E₂ = l₁/l₂ = 3/2

2. Find the second length. The first cell balances at l₁ = 75 cm. So l₂ = 2/3 × 75 = 50 cm. The second cell has the smaller emf, so it balances on a shorter length.
3. So. The difference is 75 - 50 = 25 cm.

The common slip: 37.5. Flipping the ratio to l₂ = 3/2 × 75 = 112.5 cm gives a difference of 37.5. The first cell has the larger emf, so the second length must be shorter.

The idea to keep. Same wire and same driver cell means the same potential gradient, so E ∝ l. Check which cell is bigger before you scale the length up or down.

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