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(C) The potential gradient $x$ is given by the formula: $x = \frac{E}{R + R_h + r} 	imes \frac{R}{L}$.Given: $E = 2 \ V$,$r = 0 \ \Omega$,$R = 3 \ \O

Vedclass · Accepted Answer

$57$

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(C) The potential gradient $x$ is given by the formula: $x = \frac{E}{R + R_h + r} 	imes \frac{R}{L}$.Given: $E = 2 \ V$,$r = 0 \ \Omega$,$R = 3 \ \Omega$,$L = 10 \ cm$,and $x = 1 \ mV/cm = 10^{-3} \ V/cm$.Substituting the values into the formula:$10^{-3} = \frac{2}{3 + R_h + 0} 	imes \frac{3}{10}$.$10^{-3} = \frac{6}{10(3 + R_h)}$.$10^{-2} = \frac{6}{3 + R_h}$.$3 + R_h = \frac{6}{10^{-2}} = 600$.$R_h = 600 - 3 = 597 \ \Omega$.Wait,re-evaluating the calculation: $x = 1 \ mV/cm = 0.1 \ V/m$. The potential drop across the wire is $V = I 	imes R$. The potential gradient $x = \frac{V}{L} = \frac{E}{R + R_h} 	imes \frac{R}{L}$.$10^{-3} = \frac{2}{3 + R_h} 	imes \frac{3}{10}$.$10^{-3} = \frac{0.6}{3 + R_h}$.$3 + R_h = \frac{0.6}{10^{-3}} = 600$.$R_h = 597 \ \Omega$. Given the options provided,there appears to be a discrepancy in the problem statement or options. However,if we assume the potential gradient is $1 \ mV/mm$ or similar,the calculation changes. Based on the provided solution logic $R_h = 57 \ \Omega$,it implies $10^{-1} = \frac{6}{3 + R_h} \Rightarrow 3 + R_h = 60 \Rightarrow R_h = 57 \ \Omega$. This corresponds to a potential gradient of $10 \ mV/cm$.

$A$ wire of length $10 \ cm$ is connected to a cell of $emf$ $2 \ V$ and negligible internal resistance. The resistance of the wire is $3 \ \Omega$. The value of the resistance required to obtain a potential gradient of $1 \ mV/cm$ is ................ $\Omega$.

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