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A potentiometer wire AB of length 1 m has a uniform resistance of 10 ohm/m and is fed by a battery of emf 10 V (assume the full emf appears across the wire). Two cells of emf 2 V and 4 V, joined in series so that their emfs add up, are connected in the galvanometer branch. Their internal resistances are 1 ohm and 5 ohm. If the galvanometer shows no deflection when its sliding contact is at point P, how far is P from end A?

1. 20 cm
2. 40 cm
3. 60 cm
4. 80 cm

Correct answer: 60 cm

Solution

Total wire resistance = 10 ohm/m * 1 m = 10 ohm. With 10 V across it and no extra series resistance, the potential gradient is k = 10 V / 1 m = 10 V/m. The two cells are series-aiding, so the emf to be balanced is 2 + 4 = 6 V. Setting k * L = 6 V gives the balancing length, which is independent of the internal resistances because the galvanometer branch carries no current at null.

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