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

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2 V

4 V

7 V

5 V

answer is B.

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Detailed Solution

$E_{1} = i ρ l \Rightarrow E_{1} = i (R / L) l$

$\Rightarrow \frac{E_{1}}{l} = \frac{E}{L} (l)$

$\Rightarrow E = 2 V$

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