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The meter bridge shown is in balance position with $\frac{P}{Q} = \frac{l_{1}}{l_{2}}$ . If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, what will be balance condition?

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$yes, \frac{P}{\bar{Q}} = \frac{I_{1}}{I_{2}}$

$yes, \frac{P}{Q} = \frac{l_{2} - l_{1}}{l_{2} + l_{1}}$

no, no null point

$yes, \frac{P}{Q} = \frac{I_{2}}{l_{1}}$

answer is A.

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Detailed Solution

In balanced bridge (initially)
$\frac{P}{Q} = \frac{l_{1}}{l_{2}}$

In balanced bridge (finally)
$\begin{array}{l} \frac{P}{l_{1}} = \frac{Q}{l_{2}} \\ \Rightarrow \frac{P}{Q} = \frac{l_{1}}{l_{2}} \end{array}$

On interchanging galvanometer and battery positions, the balance condition remains unchanged.

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