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Q.

A potentiometer wire is 5m long and a potential difference of 6V is maintained between its ends. Find the emf of a cell which balances against length of 180cm of the potentiometer wire.

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Detailed Solution

$\begin{array}{l} l_{1} = 5 m, E_{1} = 6 V; l_{2} = 180 cm = 1 .8 m \\ \frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow E_{2} = E_{1} \frac{l_{2}}{l_{1}} = \frac{6 \times 1 .8}{5} \\ E_{2} = 2 .16 volt. \end{array}$

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