The potentiometer wire AB shown is 40 cm long of resistance $50 Ω / m$ , free end of an ideal voltmeter is touching the potentiometer wire at the middle initially. What should be the velocity of the jockey as a function of time so that reading in voltmeter is varying with time as $(2 \sin π t)$ .

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$10 π \cos π tcm / s$

$20 π \sin π tcm / s$

$20 π \cos π tcm / s$

$10 π \sin π tcm / s$

answer is D.

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Detailed Solution

$\frac{V_{1}}{V_{2}} = \frac{R_{1}}{R_{2}} = \frac{x}{40 - x};$

$V_{1} = \frac{4}{x + 40 - x} \cdot x = \frac{x}{10};$

$V_{A} - V_{Q} = 2; V_{A} - V_{P} = x / 10$

$V_{Q} - V_{P} = \frac{x}{10} - 2;$

$\Rightarrow \frac{x}{10} - 2 = 2 \sin π t;$

$\frac{d x}{dt} = 20 \cos π \cdot π$

$\Rightarrow V = 20 π \cos π tcm / s$

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