Q 11-03-082JEE MainJEE Main 2023 (30 Jan, Shift 2)Easy
A stone tied to $180\ \text{cm}$ long string at its end is making $28$ revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is $\dfrac{1936}{x}\ \text{m s}^{-2}$. The value of $x$ is ______. [Take $\pi=\frac{22}{7}$]
Numerical value type. Enter your answer.
Answer: 125
$\omega=2\pi\times\dfrac{28}{60}=\dfrac{44}{15}\ \text{rad s}^{-1}$.
$$a=\omega^2r=\frac{1936}{225}\times1.8=\frac{1936}{125}\ \text{m s}^{-2}$$
So $x=125$.
Solution by Sreeraj P, M.Sc Physics