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A potentiometer wire has length 4m and resistance $8 Ω$ . The resistance that must be connected in series with the wire and an accumulator of e.m.f 2V, so as to get a potential gradient 1mV per cm on the wire is

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$44 Ω$

$40 Ω$

$48 Ω$

$32 Ω$

answer is D.

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Detailed Solution

$x = \frac{E_{B} R_{p}}{(R_{P} + R_{s} + r) L}$

$0.1 = \frac{2 \times 8}{(8 + R + 0) 4}$

$R = 32 Ω$

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