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(C) Let $V$ be the $EMF$ of the primary cell and $R_0$ be the resistance of the potentiometer wire of length $L_1 = 10 \ m$. The current in the primar

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$2.75$

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(C) Let $V$ be the $EMF$ of the primary cell and $R_0$ be the resistance of the potentiometer wire of length $L_1 = 10 \ m$. The current in the primary circuit is $I = \frac{V}{R_0}$.The potential gradient $k_1$ is given by $k_1 = \frac{V}{L_1} = \frac{V}{10}$.The null point is obtained at $l_1 = 2.5 \ m$,so the $EMF$ of the secondary cell is $E = k_1 	imes l_1 = \frac{V}{10} 	imes 2.5 = 0.25V$.When the length of the wire is increased to $L_2 = 10 + 1 = 11 \ m$,the new resistance is $R_0' = \frac{11}{10}R_0$. The new current in the primary circuit is $I' = \frac{V}{R_0'} = \frac{V}{1.1 R_0} = \frac{I}{1.1}$.The new potential gradient $k_2$ is $k_2 = \frac{I' 	imes R_0'}{L_2} = \frac{V}{L_2} = \frac{V}{11}$.For the same secondary cell,the new null point $l_2$ satisfies $E = k_2 	imes l_2$.$0.25V = \frac{V}{11} 	imes l_2$.$l_2 = 0.25 	imes 11 = 2.75 \ m$.

$A$ cell in a secondary circuit gives a null deflection for $2.5 \ m$ length of a potentiometer wire having a total length of $10 \ m$. If the length of the potentiometer wire is increased by $1 \ m$ without changing the cell in the primary circuit,the new position of the null point is: (in $m$)

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