In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, (ε ₁)/(ε ₂) is :

Solution

Physics sir · 4 steps

Trace which cells sit between A and the jockey: first only ε₁, then both cells in series.

1. Read the circuit. Point A joins the left end of ε₁. Point (1) is the junction between the two cells, and point (2) is the far end of ε₂. Both cells face the same way, so their emfs add.
2. Balance condition. At balance, the emf in the galvanometer branch equals the potential drop across AC, which is proportional to the length: ε = kl.

   ε₁ = k(250), ε₁ + ε₂ = k(400)

3. Find each emf. Subtracting gives ε₂ = k(400 - 250) = k(150). Then ε₁/ε₂ = 250/150.
4. So. ε₁/ε₂ = 5/3, which is option (a).

The idea to keep. In a potentiometer, the balancing length measures whatever emf lies between the wire's end and the galvanometer. Find out which cells are in that path before you write any ratio.

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