A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of the emf’s is : A. 3:4 B. 5:4 C. 5:1 D. 3:2

The correct answer is $\boxed{\text{C. }5:1}$.

When two cells are connected in series to support one another, the total emf is the sum of the two emfs. The balance point is obtained at a point on the potentiometer wire that

When two cells are connected in series in opposite directions, the total emf is the difference of the two emfs. The balance point is obtained at a point on the potentiometer wire that is proportional to the difference in emf. In this case, the balance point is obtained at 10 cm from the positive end of the wire. This means that the difference in emf is equal to the emf of one cell multiplied by 10/100 = 1/10.

The ratio of the emf’s is therefore 5/10 : 1/10 = 5:1.