1. Moment of inertia of the sphere

For a solid sphere of mass $M$ and radius $R$, about a diameter:

$I = \frac{2}{5}MR^2$

Initially,

$I_1 = \frac{2}{5}M_1R^2$

with angular velocity $\omega_1$.

2. What remains constant?

The sphere loses mass uniformly and its shape remains spherical. Since the lost mass leaves symmetrically, there is no external torque on the sphere.

Hence, angular momentum is conserved:

$I_1\omega_1 = I_2\omega_2$

3. Relation between mass and radius

Uniform density is maintained, so

$M = \rho \cdot \frac{4}{3}\pi r^3$

Thus mass is proportional to $r^3$:

$M \propto r^3$

When radius becomes $R/2$,

$M_2 = M_1\left(\frac{R/2}{R}\right)^3 = M_1\left(\frac{1}{2}\right)^3 = \frac{M_1}{8}$

4. Final moment of inertia

At radius $R/2$,

$I_2 = \frac{2}{5}M_2\left(\frac{R}{2}\right)^2$

Substitute $M_2 = M_1/8$:

$I_2 = \frac{2}{5}\cdot \frac{M_1}{8} \cdot \frac{R^2}{4}$

$I_2 = \frac{2}{5}M_1R^2 \cdot \frac{1}{32}$

So,

$I_2 = \frac{I_1}{32}$

5. Use conservation of angular momentum

$I_1\omega_1 = I_2\omega_2$

$I_1\omega_1 = \frac{I_1}{32}\omega_2$

Cancelling $I_1$:

$\omega_2 = 32\omega_1$

Thus,

$x = 32$

6. Comparison with stored answer

Derived answer: $32$

Stored correct answer: $32$

They match.