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A potentiometer circuit has wire AB of length 50 cm and resistance 10 ohm. A battery of emf 4 V (negligible internal resistance) is connected in series with resistors R1 = 15 ohm and R2 = 5 ohm. When only the main circuit key K1 is closed and the secondary key K2 is open, the null point is found at 31.25 cm from end A. Find the potential gradient (in V/cm) of the potentiometer wire AB under these conditions.

1. 0.04 V/cm
2. 0.032 V/cm
3. 0.08 V/cm
4. 0.02 V/cm

Correct answer: 0.032 V/cm

Solution

Standard potentiometer setup: the rheostat R1 = 15 ohm is in series with wire AB (10 ohm) in the main circuit. R2 = 5 ohm is in the secondary (cell under test) branch controlled by K2. When K2 is open, secondary branch is disconnected. Main circuit current: I = 4/(15+10) = 4/25 = 0.16 A. Potential drop across AB = 0.16 * 10 = 1.6 V. Potential gradient = 1.6/50 = 0.032 V/cm. Cross-check null point: emf of secondary cell = 0.032 * 31.25 = 1.0 V (a typical standard cell value — consistent).

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