Physics · Rotational Dynamics

JEE Main 2024 — 8 April, Shift 1 — Question 38

A clock has $75 \mathrm{~cm}, 60 \mathrm{~cm}$ long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel x distance more than the tip of minute hand. The value of $x$ in meter is nearly (Take $\pi=3.14$ ) :

Option A:
139.4
Correct
Option B:
140.5
Option C:
220
Option D:
118.9

Answer: A

Step-by-step solution

$\quad \mathrm{x}_{\min }=\pi \times \mathrm{r}_{\text {min }}$

=\pi \times \frac{60}{100} \mathrm{~m}

$\mathrm{x}_{\text {second }}=30 \times 2 \pi \times \mathrm{r}_{\text {second }}$

=30 \times 2 \pi \times \frac{75}{100}

$\mathrm{x}=\mathrm{x}_{\text {second }}-\mathrm{x}_{\text {min }}$ $=139.4 \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Kinematics of General Motion of a Rigid Body