The correct answer is (b)
Explanation:
Moment of inertia of the ring \(I=M r^{2}\)
Angular Momentum \(=1 \omega\)
When the masses are attached, the moment of Inertia I'= \(M r^{2}+2 m r^{2}\) \(=(M+2 m) r^{2}\)
Let the new angular speed be w'. So the angular momentum \(=l^{\prime} \omega^{\prime} .\)
Since the angular momentum is conserved. I' \(\omega\) ' \(=I \omega\)
\(\rightarrow \omega^{\prime}=\mid \omega / I^{\prime}=\omega M r^{2} /(M+2 m) r^{2}=\omega M /(M+2 m)\)