1. Use conservation of angular momentum

Since the two small spheres are attached gently and symmetrically at opposite ends of the rim, there is no external torque about the axis of the disc.

So, $I_i\omega_1 = I_f\omega_f$ where:

• $I_i$ = initial moment of inertia of the disc
• $I_f$ = final moment of inertia of disc + two spheres

2. Initial moment of inertia of the disc

For a circular disc about its central axis, $I_i = \frac{1}{2}MR^2$

3. Final moment of inertia after attaching two spheres

Each small sphere of mass $m$ is attached at the edge, i.e. at distance $R$ from the center.

Moment of inertia of one sphere about the disc axis: $mR^2$

For two spheres: $2mR^2$

Hence, $I_f = \frac{1}{2}MR^2 + 2mR^2$

4. Apply conservation of angular momentum

$\frac{1}{2}MR^2\,\omega_1 = \left(\frac{1}{2}MR^2 + 2mR^2\right)\omega_f$

Cancel $R^2$: $\frac{1}{2}M\omega_1 = \left(\frac{1}{2}M + 2m\right)\omega_f$

Multiply numerator and denominator by $2$: $\omega_f = \frac{M}{M+4m}\,\omega_1$

5. Check options

This matches: $\boxed{\left(\frac{M}{M+4m}\right)\omega_1}$ which is Option C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.