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(A) The potential gradient along the potentiometer wire $PQ$ is given by the ratio of the measured $e.m.f.$ to the balancing length.Potential gradient

Vedclass · Accepted Answer

$995$

Vedclass · Answer

(A) The potential gradient along the potentiometer wire $PQ$ is given by the ratio of the measured $e.m.f.$ to the balancing length.Potential gradient $= \frac{6.00 \, mV}{0.600 \, m} = 10 \, mV/m$.Since the total length of the potentiometer wire $PQ$ is $1 \, m$,the total potential difference across $PQ$ is $V_{PQ} = 10 \, mV/m 	imes 1 \, m = 10 \, mV = 0.01 \, V$.The current $I$ flowing through the potentiometer wire $PQ$ is $I = \frac{V_{PQ}}{R_{PQ}} = \frac{0.01 \, V}{5 \, \Omega} = 0.002 \, A = 2 \, mA$.Applying Ohm's law to the primary circuit,the total voltage of the driver cell is the sum of the voltage drops across $R$ and $PQ$: $E = I(R + R_{PQ})$.$2 \, V = 0.002 \, A 	imes (R + 5 \, \Omega)$.$R + 5 \, \Omega = \frac{2 \, V}{0.002 \, A} = 1000 \, \Omega$.$R = 1000 \, \Omega - 5 \, \Omega = 995 \, \Omega$.

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