The length of potentiometer wire is $200 c m$ 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 $l = 40 c m$ from positive end, the $E$ of the battery is (cell in primary is ideal and series resistance is zero)

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$2 v$

$4 v$

$5 v$

$7 v$

answer is B.

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Detailed Solution

Length of potentiometer $L = 200 c m$

Battery of emf $E_{1} = 0.4 V$

Balancing length $l = 40 c m$

$E_{1} = \frac{E R_{P}}{R_{P} + R_{S}} \frac{l}{L}$

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

$= \frac{0.4 \times 200}{40}$

$∴ E = 2 v$

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