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(C) For a meter bridge,the balancing condition is $\frac{R_1}{R_2} = \frac{l}{100-l}$.Given $l = 40\,cm$,so $\frac{R_1}{R_2} = \frac{40}{60} = \frac{2

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$60$

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(C) For a meter bridge,the balancing condition is $\frac{R_1}{R_2} = \frac{l}{100-l}$.Given $l = 40\,cm$,so $\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \implies R_2 = 1.5 R_1$ .....$(i)$When $10\,\Omega$ is connected in series with $R_1$,the new resistance is $(R_1 + 10)\,\Omega$. The null point shifts by $10\,cm$. Since $R_1$ increases,the null point shifts towards $B$,so the new length $l' = 40 + 10 = 50\,cm$.Thus,$\frac{R_1 + 10}{R_2} = \frac{50}{50} = 1 \implies R_1 + 10 = R_2$ .....$(ii)$Substituting $(i)$ into $(ii)$: $R_1 + 10 = 1.5 R_1 \implies 0.5 R_1 = 10 \implies R_1 = 20\,\Omega$ and $R_2 = 30\,\Omega$.Now,we need to connect a resistance $R$ in parallel with $(R_1 + 10) = 30\,\Omega$ such that the null point returns to $40\,cm$ (i.e.,ratio remains $\frac{2}{3}$).Let the equivalent resistance be $R_{eq} = \frac{30R}{30+R}$.Then $\frac{R_{eq}}{R_2} = \frac{2}{3} \implies \frac{R_{eq}}{30} = \frac{2}{3} \implies R_{eq} = 20\,\Omega$.So,$\frac{30R}{30+R} = 20 \implies 30R = 600 + 20R \implies 10R = 600 \implies R = 60\,\Omega$.

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