Q 11-06-222JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
Moment of inertia of a body about a given axis is $1.5\ \text{kg m}^2$. Initially the body is at rest. In order to produce a rotational kinetic energy of $1200\ \text{J}$, the angular acceleration of $20\ \text{rad/s}^2$ must be applied about the axis for a duration of
Answer: (B) $2\ \text{s}$
$$\frac12I\omega^2 = 1200 \Rightarrow \omega^2 = \frac{2400}{1.5} = 1600 \Rightarrow \omega = 40\ \text{rad/s}$$
Starting from rest, $t = \dfrac\omega\alpha = \dfrac{40}{20} = 2\ \text{s}$.
Solution by Sreeraj P, M.Sc Physics