Q 11-03-063JEE MainJEE Main 2024 (5 Apr, Shift 2)Easy
A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9\ \text{m}$ and completes 120 revolutions in 3 minutes. The magnitude of the centripetal acceleration of the monkey is (in $\text{m/s}^2$)
Answer: (D) $16\pi^2\ \text{m s}^{-2}$
Frequency: $f = \dfrac{120}{180} = \dfrac23\ \text{Hz}$, so $\omega = 2\pi f = \dfrac{4\pi}{3}\ \text{rad/s}$.
$$a = \omega^2r = \frac{16\pi^2}{9}\times9 = 16\pi^2\ \text{m s}^{-2}$$
Solution by Sreeraj P, M.Sc Physics