In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of $40$ cm from A. If a $10\ \Omega$ resistor is connected in series with $R_1$, the null point shifts by $10$ cm. The resistance that should be connected in parallel with $(R_1 + 10)\ \Omega$ such that the null point shifts back to its initial position is
Answer: (C) $60\ \Omega$
Initially $\dfrac{R_1}{R_2} = \dfrac{40}{60}$. Increasing $R_1$ moves the null point away from A, to $50$ cm:
$$\frac{R_1 + 10}{R_2} = \frac{50}{50} \Rightarrow R_1 + 10 = R_2$$
With $R_1 = \dfrac23R_2$: $R_2 = 30\ \Omega$, $R_1 = 20\ \Omega$.
To restore the null point, $(R_1 + 10) = 30\ \Omega$ in parallel with $X$ must equal $20\ \Omega$:
$$\frac{30X}{30 + X} = 20 \Rightarrow X = 60\ \Omega$$
Solution by Sreeraj P, M.Sc Physics