Q 12-03-086JEE MainJEE Main 2024 (27 Jan, Shift 1)Medium
A wire of length $10\ \text{cm}$ and radius $\sqrt7\times10^{-4}\ \text{m}$ is connected across the right gap of a meter bridge. When a resistance of $4.5\ \Omega$ is connected on the left gap by using a resistance box, the balance length is found to be at $60\ \text{cm}$ from the left end. If the resistivity of the wire is $R\times10^{-7}\ \Omega\,\text{m}$, then value of $R$ is:
Answer: (C) 66
Balance condition of the meter bridge:
$$\frac{4.5}{X} = \frac{60}{40} \;\Rightarrow\; X = 3\ \Omega$$
Resistivity of the wire:
$$\rho = \frac{XA}{l} = \frac{3\times\pi\times(\sqrt7\times10^{-4})^2}{0.10} = \frac{3\times\frac{22}{7}\times7\times10^{-8}}{0.10} = 66\times10^{-7}\ \Omega\,\text{m}$$
So $R = 66$.
Solution by Sreeraj P, M.Sc Physics