Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.4k views
in Physics by (86.6k points)
closed by
A wheel of moment of inertia `2.0 xx 10^3 kgm^2` is rotating at uniform angular speed of `4 rads^-1`. What is the torque required to stop it in one second.
A. `0.5 xx 10^3 Nm`
B. `8.0 xx 10^3 Nm`
C. `2.0 xx 10^3 Nm`
D. none of these

1 Answer

0 votes
by (79.4k points)
selected by
 
Best answer
Correct Answer - B
(b) `prop = (omega)/(t) = (4)/(1) = 4 rad//s^2`
`tau = I prop = 2 xx 10^3 xx 4 = 8 xx 10^3 N`.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...