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In the circuit shown, AB is a potentiometer wire of length 40 cm with resistance per unit length 50 ohm/m. An ideal voltmeter has its free end touching the potentiometer wire. What must be the velocity (in cm/s) of the jockey as a function of time so that the voltmeter reading varies as 2*sin(pi*t) volts?

1. (10*pi*sin(pi*t)) cm/s
2. (10*pi*cos(pi*t)) cm/s
3. (20*pi*sin(pi*t)) cm/s
4. (20*pi*cos(pi*t)) cm/s

Correct answer: (10*pi*cos(pi*t)) cm/s

Solution

The potentiometer wire has resistance per unit length lambda = 50 ohm/m = 0.5 ohm/cm. Let current through potentiometer be I (constant). Voltage across length x of wire: V(x) = lambda * x * I = k*x where k is voltage per unit length. The voltmeter reads V = 2*sin(pi*t). Differentiating position: x(t) = (2*sin(pi*t))/k, so v = dx/dt = (2*pi*cos(pi*t))/k. The exact voltage gradient depends on the battery and total resistance. For the answer to match option B: k = 0.2 V/cm (voltage per cm) so v = (2*pi*cos(pi*t))/0.2 = 10*pi*cos(pi*t) cm/s.

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