Q 11-08-087JEE MainJEE Main 2021 (26 Aug, Shift 2)Medium
Two blocks of masses $3$ kg and $5$ kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{\pi}\times10^2$ N m$^{-2}$. What is the minimum radius of the wire? (take $g = 10$ m s$^{-2}$)
Answer: (D) $12.5$ cm
Tension in the wire:
$$T = \frac{2m_1m_2g}{m_1 + m_2} = \frac{2\times3\times5\times10}{8} = 37.5\ \text{N}$$
The wire must not break: $\dfrac{T}{\pi r^2} \le \dfrac{24}{\pi}\times10^2$
$$r^2 \ge \frac{37.5}{2400} = 0.015625 \Rightarrow r \ge 0.125\ \text{m} = 12.5\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics