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(C) In a meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ and $Q$ are resistances in the two gaps and $l$ is

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$20$

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(C) In a meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ and $Q$ are resistances in the two gaps and $l$ is the balance length.Case $1$: $R_1$ is in series with $10 \,\Omega$ in one gap,and $R_2$ is in the other gap. The balance point is at $50 \, cm$.$\frac{R_1 + 10}{R_2} = \frac{50}{100-50} = \frac{50}{50} = 1$$\Rightarrow R_2 = R_1 + 10$  --- (Equation $1$)Case $2$: The $10 \,\Omega$ resistance is removed,so only $R_1$ remains in the first gap. The balance point shifts to $40 \, cm$.$\frac{R_1}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$$\Rightarrow 3R_1 = 2R_2$ --- (Equation $2$)Substitute Equation $1$ into Equation $2$:$3R_1 = 2(R_1 + 10)$$3R_1 = 2R_1 + 20$$R_1 = 20 \,\Omega$.

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