Q. A torque of $30\ N$-$m$ is applied on a $5\ kg$ wheel whose moment of inertia is $2kg - m^{2}$ for $10\ sec$. The angle covered by the wheel in $10\ sec$ will be
A) $750\ rad$.
B) $1500\ rad$.
C) $3000\ rad$.
D) $6000\ rad$.
Explanation
(a) $\alpha = \frac{\tau}{I} = \frac{30}{2} = 15\ rad/s^{2}$
$\because\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 0 + \frac{1}{2} \times (15) \times (10)^{2} = 750\ rad$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.