MHTCET 25 Apr 2025 Shift 2 Paper

Concept:The loss in gravitational potential energy of the falling bob is converted into the translational kinetic energy of the bob and the rotational kinetic energy of the flywheel.Explanation:Initial potential energy of the bob is $mgh$.Since the string unwinds without slipping, the speed of the bob is $v = R\omega$.Kinetic energy of the bob: $\frac{1}{2}mv^2 = \frac{1}{2}mR^2\omega^2$.Moment of inertia of the flywheel, which is a disc of mass $m$ and radius $R$, is $I = \frac{1}{2}mR^2$.Rotational kinetic energy of the flywheel: $\frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{1}{2}mR^2\right)\omega^2 = \frac{1}{4}mR^2\omega^2$.Total kinetic energy $= \frac{1}{2}mR^2\omega^2 + \frac{1}{4}mR^2\omega^2 = \frac{3}{4}mR^2\omega^2$.By conservation of energy: $mgh = \frac{3}{4}mR^2\omega^2$.Cancelling $m$ and solving for $\omega$: $\omega = \frac{2}{R}\sqrt{\frac{gh}{3}}$.Answer:$\omega = \frac{2}{R}\sqrt{\frac{gh}{3}}$, so the correct option is A.