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The length of potentiometer wire is 200 cm and the emf of standard cell in primary circuit is E volts. It is employed to a battery of emf 0.4 v the balance point is obtained at l =40cm from positive end, the E of the battery is (cell in primary is ideal and series resistance is zero)

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answer is B.

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Detailed Solution

$\begin{array}{l} E^{'} = \frac{E \times R_{P} \times l}{(r + R_{s} + R_{p}) L} \\ 0 .4 = \frac{E \times R_{p} \times 40}{(0 + 0 + R_{p}) (200)}; E = 2 V \end{array}$

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