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(D) Let the resistance of the potentiometer wire be $R$.The current $i$ flowing through the potentiometer wire is given by $i = \frac{4}{R + 780}$.The

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(D) Let the resistance of the potentiometer wire be $R$.The current $i$ flowing through the potentiometer wire is given by $i = \frac{4}{R + 780}$.The potential difference across the entire potentiometer wire $AB$ is $V_{AB} = iR = \frac{4R}{R + 780}$.The potential gradient along the wire is $k = \frac{V_{AB}}{L} = \frac{4R}{(R + 780) 	imes 300}$.The null point is obtained at a length $l = 60\,cm$ for a cell of emf $E = 20\,mV = 20 	imes 10^{-3}\,V$.At the null point,the potential difference across the length $l$ is equal to the emf of the cell:$E = k 	imes l$$20 	imes 10^{-3} = \left( \frac{4R}{(R + 780) 	imes 300} ight) 	imes 60$Simplifying the equation:$20 	imes 10^{-3} = \frac{4R 	imes 60}{(R + 780) 	imes 300}$$0.02 = \frac{4R}{5(R + 780)}$$0.1(R + 780) = 4R$$0.1R + 78 = 4R$$3.9R = 78$$R = \frac{78}{3.9} = 20\,\Omega$.

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