Angular Momentum & its Conservation question from MHT CET 2024 (4 May, Shift 2)

Answer: (b)

1. No external torque, so angular momentum is conserved: $I_1\omega = (I_1+I_2)\omega'$.
2. $I_1 = \frac{1}{2}MR^2$, $I_2 = \frac{1}{2}\left(\frac{M}{3}\right)R^2$, so $I_1+I_2 = \frac{1}{2}\cdot\frac{4M}{3}R^2$.
3. $\omega' = \frac{M}{4M/3}\,\omega = \frac{3}{4}\omega$
4. So the answer is (b) $\frac{3}{4}\omega$.