The resistance of 10, m long potentiometer wi | vedclass

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(C) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V_{wire}}{L}$.Given: Length of wire $L

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(C) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V_{wire}}{L}$.Given: Length of wire $L = 10\, m$,Resistance per unit length $\lambda = 1\,\Omega/m$,so total resistance of wire $R = 10\,\Omega$.$e.m.f.$ of the cell $E = 2.2\, V$.Let the resistance taken from the resistance box be $R_h$.The current $I$ in the circuit is $I = \frac{E}{R + R_h} = \frac{2.2}{10 + R_h}$.The potential drop across the wire is $V_{wire} = I 	imes R = \frac{2.2 	imes 10}{10 + R_h}$.The potential gradient $x = \frac{V_{wire}}{L} = \frac{2.2 	imes 10}{(10 + R_h) 	imes 10} = \frac{2.2}{10 + R_h}$.We are given $x = 2.2\, mV/m = 2.2 	imes 10^{-3}\, V/m$.Equating the two: $2.2 	imes 10^{-3} = \frac{2.2}{10 + R_h}$.$10 + R_h = \frac{2.2}{2.2 	imes 10^{-3}} = 10^3 = 1000$.$R_h = 1000 - 10 = 990\,\Omega$.

The resistance of $10\, m$ long potentiometer wire is $1\,\Omega/m$. $A$ cell of $e.m.f.$ $2.2\, V$ and a high resistance box are connected in series to this wire. The value of resistance taken from the resistance box for getting a potential gradient of $2.2\, mV/m$ will be ............... $\Omega$.

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