A thin circular ring of mass $$M$$ and radius $$R$$ is rotating with a constant angular velocity $$2$$ rad s$$^{-1}$$ in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass $$m$$ be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rad s$$^{-1}$$).

A thin circular ring of mass $$M$$ and radius $$R$$ rotates with angular velocity $$\omega_0 = 2$$ rad/s. Two objects each of mass $$m$$ are attached gently to opposite ends of a diameter.

Since the masses are attached gently (no external torque), angular momentum is conserved:
$$L_i = L_f$$
$$I_i \omega_0 = I_f \omega_f$$.

Initial moment of inertia (ring only) is
$$I_i = MR^2$$
and final moment of inertia (ring + two point masses at distance $$R$$) is
$$I_f = MR^2 + 2mR^2 = (M + 2m)R^2$$.

Substituting into the conservation equation,
$$MR^2 \times 2 = (M + 2m)R^2 \times \omega_f$$,
this gives
$$\omega_f = \frac{2M}{M + 2m}$$ rad/s.

The correct answer is Option C.