Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A wheel with moment of inertia $2 \mathrm{~kg} \mathrm{~m}^2$ and rotating at $\mathrm{50 \: rpm}$ is brought to rest. Find the torque required to bring it to rest in $\mathrm{1 \mathrm{~min}}$

Option: 1
$\frac{\pi}{6}$

Option: 2
$\frac{\pi}{12}$

Option: 3
$\frac{\pi}{15}$

Option: 4
$\frac{\pi}{18}$

Answers (1)

best_answer

Given,

$\mathrm{I=2 \mathrm{~kg}-\mathrm{m}^2}$ , $\mathrm{\omega_0=50 \times \frac{\pi}{30}}$

$\mathrm{ t=\operatorname{60\: sec} }$

$\mathrm{\omega_{0}=\frac{5\pi}{3},\omega=0}$

$\mathrm{ \omega=\omega_0-\alpha t}$

$\uparrow$ for retardation

$\mathrm{ 0=\frac{5 \pi}{3}-\alpha .60 }$

$\mathrm{ \alpha=\frac{5 \pi}{3 \times 60}=\frac{\pi}{36} \mathrm{rad} / \mathrm{sec}^2}$

$\mathrm{ Torque \: \tau=I \cdot \alpha}$

$\mathrm{ \tau=2 \times \frac{\pi}{36}=\frac{\pi}{18} \mathrm{~N-m} }$

Hence option 4 is correct.

Posted by

Irshad Anwar

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET 2026: Odisha aspirant dies by suicide fearing failure

anam-khan

NEET 2026 Answer Key (Out) LIVE: Challenge window link at neet.nta.nic.in soon; result date, cutoff

anam-khan

NEET UG Answer Key 2026 (OUT) LIVE: NTA provisional keys out neet.nta.ac.in; challenge window opens soon

Latest Question