A ring of mass M and radius R is rotating about its axis with angular velocity $$\omega$$. Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be :

Initially, the ring has mass $$M$$ and radius $$R$$, rotating about its axis with angular velocity $$\omega$$. The moment of inertia of the ring about its axis is $$I_i = M R^2$$. The initial kinetic energy is given by:

$$K_i = \frac{1}{2} I_i \omega^2 = \frac{1}{2} (M R^2) \omega^2$$

When two identical bodies, each of mass $$m$$, are gently attached at the ends of a diameter, they are at a distance $$R$$ from the axis. The moment of inertia of each body is $$m R^2$$, so the total added moment of inertia is $$2m R^2$$. The final moment of inertia becomes:

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

Since the attachment is done gently (no external torque), angular momentum is conserved. Initial angular momentum $$L_i = I_i \omega = M R^2 \omega$$. Let the final angular velocity be $$\omega_f$$. Final angular momentum $$L_f = I_f \omega_f = R^2 (M + 2m) \omega_f$$. Setting $$L_i = L_f$$:

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

Canceling $$R^2$$ (assuming $$R \neq 0$$):

$$M \omega = (M + 2m) \omega_f$$

Solving for $$\omega_f$$:

$$\omega_f = \frac{M \omega}{M + 2m}$$

The final kinetic energy is:

$$K_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \left[ R^2 (M + 2m) \right] \left( \frac{M \omega}{M + 2m} \right)^2$$

Simplifying:

$$K_f = \frac{1}{2} R^2 (M + 2m) \cdot \frac{M^2 \omega^2}{(M + 2m)^2} = \frac{1}{2} R^2 \cdot \frac{M^2 \omega^2}{M + 2m}$$

The loss in kinetic energy is $$\Delta K = K_i - K_f$$:

$$\Delta K = \frac{1}{2} M R^2 \omega^2 - \frac{1}{2} R^2 \cdot \frac{M^2 \omega^2}{M + 2m}$$

Factor out $$\frac{1}{2} R^2 \omega^2$$:

$$\Delta K = \frac{1}{2} R^2 \omega^2 \left( M - \frac{M^2}{M + 2m} \right)$$

Combine the terms inside the parentheses:

$$M - \frac{M^2}{M + 2m} = \frac{M(M + 2m) - M^2}{M + 2m} = \frac{M^2 + 2mM - M^2}{M + 2m} = \frac{2mM}{M + 2m}$$

Substitute back:

$$\Delta K = \frac{1}{2} R^2 \omega^2 \cdot \frac{2mM}{M + 2m} = \frac{1}{2} \cdot 2 \cdot \frac{mM}{M + 2m} R^2 \omega^2 = \frac{Mm}{M + 2m} \omega^2 R^2$$

Thus, the kinetic energy loss is $$\frac{Mm}{(M+2m)} \omega^2 R^2$$. Comparing with the options, this matches option C.

Hence, the correct answer is Option C.