Question

A constant torque of $ 3.14\,Nm $ is exerted on a pivoted wheel. If the angular acceleration of the wheel is $ 4\pi \,rad/s^{2}, $ then the moment of Inertia of the wheel is

• A. $ 0.25\,kg-m^{2} $
• B. $ 2.5\,kg-m^{2} $
• C. $ 4.5\,kg-m^{2} $
• D. $ 25\,kg-m^{2} $

Answer 1

The Correct Option is A

Torque, $ \tau =I\omega $
Moment of inertia, $ I=\frac{\tau }{\omega } $
$ =\frac{3.14}{4\times \pi } $
$ =\frac{3.14}{4\times 3.14}=\frac{1}{4} $
$ =0.25\,kg-m^{2} $