Q 12-03-126JEE MainJEE Main 2023 (24 Jan, Shift 1)Easy
A hollow cylindrical conductor has length of $3.14\ \text{m}$, while its inner and outer diameters are $4\ \text{mm}$ and $8\ \text{mm}$ respectively. The resistance of the conductor is $n\times10^{-3}\ \Omega$. If the resistivity of the material is $2.4\times10^{-8}\ \Omega\,\text{m}$, the value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 2
Cross-section $A=\pi(r_2^2-r_1^2)=\pi(16-4)\times10^{-6}=12\pi\times10^{-6}\ \text{m}^2$.
$$R=\frac{\rho l}{A}=\frac{2.4\times10^{-8}\times3.14}{12\times3.14\times10^{-6}}=2\times10^{-3}\ \Omega$$
Solution by Sreeraj P, M.Sc Physics