A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius 9 ~m and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (. in m / s^2 ) :

Correct answer: D. 16 π^2 ~ms^ -2 aligned & Given : R=9 m, & 120 revolution in 3 min & ω= 120 Rev . 3 min . = 120 × 2 π rad 3 × 60 sec =(4 π)/(3) rad / s & a_ centripetal =ω^2 R=((4 π)/(3))^2 × 9=16 π^2 m / s^2 aligned

JEE Main 20245 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Evening Shift
Subject: Physics
Chapter: Motion in Two Dimensions

JEE Main 2024 · 5 April · Evening Shift · Physics · Motion in Two Dimensions

A man carrying a monkey on his shoulder does cycling smoothly on a...

A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius

9 \mathrm{~m}

and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is

\left(\right. in \mathrm{m} / \mathrm{s}^2

) :

Options

A $57600 \pi^2 \mathrm{~ms}^{-2}$
BZero
C $4 \pi^2 \mathrm{~ms}^{-2}$
D $16 \pi^2 \mathrm{~ms}^{-2}$

Correct answer

D. $16 \pi^2 \mathrm{~ms}^{-2}$

Step-by-step solution

\begin{aligned} & \text{ Given : } \mathrm{R}=9 \mathrm{ m}, \\ & 120 \text{ revolution\text{ in } } 3 \mathrm{ min} \\ & \omega=\frac{120 \mathrm{Rev} .}{3 \mathrm{ min} .}=\frac{120 \times 2 \pi \mathrm{rad}}{3 \times 60 \mathrm{sec}}=\frac{4 \pi}{3} \mathrm{rad} / \mathrm{s} \\ & \mathrm{a}_{\text{centripetal }}=\omega^2 \mathrm{R}=\left(\frac{4 \pi}{3}\right)^2 \times 9=16 \pi^2 \mathrm{ m} / \mathrm{s}^2\end{aligned}

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