Q 12-03-076JEE MainJEE Main 2024 (1 Feb, Shift 2)Medium
In a metre bridge, when a resistance in the left gap is $2\ \Omega$ and an unknown resistance is in the right gap, the balance length is found to be $40\ \text{cm}$. On shunting the unknown resistance with $2\ \Omega$, the balance length changes by
Answer: (A) $22.5\ \text{cm}$
$$\frac{2}{X} = \frac{40}{60} \Rightarrow X = 3\ \Omega$$
Shunted: $X' = \dfrac{3\times2}{3 + 2} = 1.2\ \Omega$.
$$\frac{2}{1.2} = \frac{l}{100 - l} \Rightarrow 5(100 - l) = 3l \Rightarrow l = 62.5\ \text{cm}$$
Change in balance length $= 62.5 - 40 = 22.5\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics