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A potentiometer has a wire of 100cm length and its resistance is 10 Ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40cm length of potentiometer wire, the value of ‘E’ is

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0.08V

0.4V

0.8V

0.16V

answer is D.

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Detailed Solution

$\begin{array}{l} E^{'} = \frac{E \times R_{P} \times l}{(r + R_{s} + R_{p}) L} \\ E^{'} = \frac{2 \times 10 \times 40}{(0 + 40 + 10) 100}; E^{'} = 0 .16 V \end{array}$

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