Physics · Horizontal Circular Motion

JEE Main 2024 — 5 April, Shift 2 — Question 39

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

Option A:
zero
Option B:
$16 \pi^{2} \mathrm{~ms}^{-2}$
Correct
Option C:
$4 \pi^{2} \mathrm{~ms}^{-2}$
Option D:
$57600 \pi^{2} \mathrm{~ms}^{-2}$

Answer: B

Step-by-step solution

. Given : $\mathrm{R}=9 \mathrm{~m}$ ,

120 revolution in 3 min

$\omega=\frac{120 \mathrm{Re} \mathrm{v} .}{3 \mathrm{~min} .}=\frac{120 \times 2 \pi \mathrm{rad}}{3 \times 60 \mathrm{sec}}=\frac{4 \pi}{3} \mathrm{rad} / \mathrm{s}$

$a_{\text {centripetal }}=\omega^{2} R=\left(\frac{4 \pi}{3}\right)^{2} \times 9=16 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Physics
Chapter: Horizontal Circular Motion
Topic: Dynamics of circular motion