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(B) Let the additional resistance be $R$.The total resistance of the circuit is $R_{total} = R + R_{wire} = R + 3\, \Omega$.The current flowing throug

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$57$

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(B) Let the additional resistance be $R$.The total resistance of the circuit is $R_{total} = R + R_{wire} = R + 3\, \Omega$.The current flowing through the circuit is $I = \frac{E}{R_{total}} = \frac{2}{R + 3}$.The potential gradient $\phi$ is given as $1\, mV/cm = 10^{-3}\, V/cm = 0.1\, V/m$. Since the wire length is $100\, cm = 1\, m$,the total potential drop across the wire is $V_{wire} = \phi 	imes L = 10^{-3}\, V/cm 	imes 100\, cm = 0.1\, V$.Also,$V_{wire} = I 	imes R_{wire} = \left( \frac{2}{R + 3} ight) 	imes 3$.Equating the two expressions for $V_{wire}$:$0.1 = \frac{6}{R + 3}$.$R + 3 = \frac{6}{0.1} = 60$.$R = 60 - 3 = 57\, \Omega$.

$A$ wire of length $100\, cm$ is connected to a cell of emf $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential difference of $1\, mV/cm$ is ............. $\Omega$.

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