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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 emf's is

1. 5 : 1
2. 5 : 4
3. 3 : 4
4. 3 : 2

Correct answer: 5 : 4

Solution

When the cells are connected in series supporting each other, the effective emf is E1 + E2, and the balance point is at 50 cm. When connected in opposition, the effective emf is E1 - E2, and the balance point is at 10 cm. Using the proportionality of emf to balance length, (E1 + E2)/(E1 - E2) = 50/10 = 5. Solving gives E1/E2 = 5/4.

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