Q 11-04-087JEE MainJEE Main 2024 (29 Jan, Shift 2)Medium
A stone of mass $900\ \text{g}$ is tied to a string and moved in a vertical circle of radius $1\ \text{m}$ making $10\ \text{rpm}$. The tension in the string, when the stone is at the lowest point, is (if $\pi^2 = 9.8$ and $g = 9.8\ \text{m s}^{-2}$):
Answer: (B) $9.8\ \text{N}$
$\omega = \dfrac{2\pi\times10}{60} = \dfrac\pi3\ \text{rad s}^{-1}$, so $\omega^2 r = \dfrac{\pi^2}{9} = \dfrac{9.8}{9}\ \text{m s}^{-2}$.
At the lowest point:
$$T = m(g + \omega^2r) = 0.9\left(9.8 + \frac{9.8}{9}\right) = 0.9\times9.8\times\frac{10}{9} = 9.8\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics