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In a potentiometer circuit, 10 identical cells each of emf E and internal resistance r are connected in series in the primary circuit. A cell of emf E0 and internal resistance r0 is balanced at length l1 on the potentiometer wire. When the polarity of two cells in the primary circuit is reversed, the same cell of emf E0 is balanced at length l2. Find the ratio l1/l2.

1. 3/5
2. 5/3
3. 1/2
4. 2/1

Correct answer: 5/3

Solution

The balance length in a potentiometer is proportional to the potential gradient, which is proportional to the total emf of the primary circuit (assuming the wire and external resistance are fixed). Initially: total emf = 10E. After reversing 2 cells: net emf = (10-2)*E - 2*E =... more carefully: if 2 cells are reversed, 8 cells contribute +E each and 2 cells contribute -E each, so net = 8E - 2E = 6E. l1/l2 = 10E/6E = 5/3.

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