Q 11-04-129JEE MainJEE Main 2022 (24 Jun, Shift 1)Easy
A boy ties a stone of mass $100\ \text{g}$ to the end of a $2\ \text{m}$ long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of $80\ \text{N}$. If the maximum speed with which the stone can revolve is $\dfrac{K}{\pi}\ \text{rev min}^{-1}$, the value of $K$ is: (Assume the string is massless and un-stretchable)
Answer: (C) 600
The tension provides the centripetal force: $T = m\omega^2 r$.
$$80 = 0.1\times\omega^2\times 2\ \Rightarrow\ \omega = 20\ \text{rad s}^{-1}$$
In revolutions per minute: $\dfrac{20}{2\pi}\times 60 = \dfrac{600}{\pi}\ \text{rev min}^{-1}$, so $K = 600$.
Solution by Sreeraj P, M.Sc Physics