Q 11-06-092JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
A thin circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. If another disc of same dimensions but of mass $M/2$ is placed gently on the first disc co-axially, then the new angular velocity of the system is:
Answer: (C) $\dfrac23\omega$
No external torque acts, so angular momentum is conserved:
$$\frac12MR^2\omega = \left(\frac12MR^2 + \frac12\cdot\frac M2R^2\right)\omega' = \frac34MR^2\omega'$$
$$\omega' = \frac23\omega$$
Solution by Sreeraj P, M.Sc Physics