Q76 / 90 · JEE Main 2021 (26 Aug Shift 2) · Physics

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 $\frac{24}{π} \times 10^{2} N m^{- 2}$ . What is the minimum radius of the wire ?
( take $g = 10 {m s}^{- 2})$

Solution

Tension in the string $= \frac{2 m_{1} m_{2}}{(m_{1} + m_{2})} g$

$= \frac{2 \times 3 \times 5 \times 10}{8} = \frac{75}{2}$

Stress $= \frac{T}{A}$

$\frac{24}{π} \times 10^{2} = \frac{75}{2 (π r^{2})}$

$r^{2} = \frac{75}{48} \times 10^{- 2}$

$r^{2} = 0.015625$

$r = 0.125 m = 12.5 cm$

Hence, the correct answer is option: 'D'.