MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Rotational kinetic energy depends on the moment of inertia and the square of angular velocity.Explanation:Initial kinetic energy is given by $K = \frac{1}{2} I \omega^2$.Here, $I$ is the initial moment of inertia and $\omega$ is the initial angular velocity.When the dancer stretches his arms, the new moment of inertia becomes $I' = 3I$.The new angular velocity becomes $\omega' = \frac{\omega}{3}$.The new rotational kinetic energy is $K' = \frac{1}{2} I' \omega'^2$.Substituting the new values, we get $K' = \frac{1}{2}(3I)\left(\frac{\omega}{3}\right)^2$.Simplifying, $K' = \frac{1}{2}(3I)\frac{\omega^2}{9}$.This becomes $K' = \frac{1}{3}\left(\frac{1}{2} I \omega^2\right)$.Since $\frac{1}{2} I \omega^2 = K$, we have $K' = \frac{K}{3}$.Therefore, the kinetic energy of the dancer after stretching his arms is $\frac{K}{3}$.Answer:$\frac{K}{3}$Correct option: B.