Q 11-06-038NEETJEE MainEasy
A constant torque of $20$ N m acts on a wheel of moment of inertia $4\ \text{kg m}^2$, initially at rest, for $3$ s. The angular momentum and the kinetic energy of the wheel afterwards are
Answer: (D) $60\ \text{kg m}^2\text{/s}$ and $450$ J
Angular impulse: $L = \tau t = 20 \times 3 = 60\ \text{kg m}^2/\text{s}$.
$$K = \frac{L^2}{2I} = \frac{3600}{8} = 450\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics