A thin circular ring of mass $$m$$ and radius $$R$$ is rotating about its axis with a constant angular velocity $$\omega$$. Two objects each of mass $$M$$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $$\omega' =$$

The external torque about the axis is zero, so angular momentum is conserved.

For the thin ring,  $$I_i=mR^2$$

Hence, initial angular momentum is  $$L_i=I_i\omega=mR^2\omega$$

When two masses $$M$$ are attached at opposite ends of the diameter, each mass is at distance $$R$$ from the axis.

Therefore, the final moment of inertia is  $$I_f=mR^2+MR^2+MR^2$$

$$I_f=(m+2M)R^2$$

If the new angular velocity is $$\omega'$$, then  $$L_f=I_f\omega'=(m+2M)R^2\omega'$$

By conservation of angular momentum,  $$L_i=L_f$$

$$mR^2\omega=(m+2M)R^2\omega'$$

Cancelling $$R^2$$,  $$\omega'=\frac{m\omega}{m+2M}$$

Therefore,  $${\omega'=\frac{\omega m}{m+2M}}$$

Hence, the correct option is A.