Q 11-08-112JEE MainJEE Main 2025 (8 Apr, Shift 2)Easy
A $3\ \text{m}$ long wire of radius $3\ \text{mm}$ shows an extension of $0.1\ \text{mm}$ when loaded vertically by a mass of $50\ \text{kg}$ in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is $P\times10^{11}\ \text{N m}^{-2}$, where the value of $P$ is: (Take $g = 3\pi\ \text{m/s}^2$)
Answer: (A) $5$
$$Y = \frac{FL}{A\Delta L} = \frac{50\times3\pi\times3}{\pi(3\times10^{-3})^2\times0.1\times10^{-3}} = \frac{450}{9\times10^{-10}} = 5\times10^{11}\ \text{N m}^{-2}$$
So $P = 5$.
Solution by Sreeraj P, M.Sc Physics