Q 11-08-092JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
The elastic limit of brass is $379\ \text{MPa}$. The minimum diameter of a brass rod if it is to support a $400\ \text{N}$ load without exceeding its elastic limit will be
Answer: (C) $1.16\ \text{mm}$
The stress must not exceed $379\times10^6\ \text{Pa}$:
$$A = \frac{400}{379\times10^6} = 1.055\times10^{-6}\ \text{m}^2$$
$$d = \sqrt{\frac{4A}{\pi}} = \sqrt{1.34\times10^{-6}} \approx 1.16\times10^{-3}\ \text{m} = 1.16\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics