A thin circular ring having mass M and radius r rotates about its...

Initial moment of inertia of the ring is $I_1 = M r^2$.

Initial angular speed is $\omega_1 = \omega$.

When two particles each of mass $m$ are attached at diametrically opposite points on the ring, the new moment of inertia becomes $I_2 = M r^2 + 2m r^2 = (M + 2m) r^2$.

Since no external torque acts on the system, the angular momentum is conserved.

$I_1 \omega_1 = I_2 \omega_2$

$M r^2 \omega = (M + 2m) r^2 \omega_2$

$\omega_2 = \dfrac{M \omega}{M + 2m}$