A potentiometer wire has a length of 100 cm | vedclass

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(D) In a potentiometer,the potential drop across a length $\ell$ of the wire is given by $V = k\ell$,where $k$ is the potential gradient.Given the tot

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$\frac{30E}{100}$

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(D) In a potentiometer,the potential drop across a length $\ell$ of the wire is given by $V = k\ell$,where $k$ is the potential gradient.Given the total length $L = 100 \ cm$ and the total $emf \ E$ applied across the wire,the potential gradient is $k = \frac{E}{L} = \frac{E}{100}$.At the balance point,no current flows through the secondary circuit (the circuit containing the battery with $emf \ E_0$). Therefore,the internal resistance of the battery does not affect the potential difference measured at the balance point.The $emf \ E_0$ is equal to the potential drop across the length $\ell = 30 \ cm$.Thus,$E_0 = k 	imes \ell = \left(\frac{E}{100}ight) 	imes 30 = \frac{30E}{100}$.

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