JEE Physics 2025 Question with Solution

Given: Initially, the solid sphere has mass $M$, radius $R$, and angular velocity $\omega_1$. Later, its radius becomes $\frac{R}{2}$ while density remains uniform and shape remains spherical.

Find: The value of $x$ if $\omega_2 = x\omega_1$.

Use conservation of angular momentum:

$$I_1\omega_1 = I_2\omega_2$$

For a solid sphere,

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

Initially,

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

Since density remains constant,

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

When the radius becomes $\frac{R}{2}$, the new mass is $M_2$ and

$$\rho = \frac{M_2}{\frac{4}{3}\pi \left(\frac{R}{2}\right)^3}$$

Equating densities,

$$\frac{M}{\frac{4}{3}\pi R^3} = \frac{M_2}{\frac{4}{3}\pi \left(\frac{R}{2}\right)^3}$$

So,

$$M_2 = \frac{M}{8}$$

Now the new moment of inertia is

$$I_2 = \frac{2}{5}M_2\left(\frac{R}{2}\right)^2$$ $$I_2 = \frac{2}{5}\left(\frac{M}{8}\right)\frac{R^2}{4} = \frac{MR^2}{80}$$

Apply conservation of angular momentum:

$$\frac{2}{5}MR^2\omega_1 = \frac{MR^2}{80}\omega_2$$

Hence,

$$\omega_2 = 32\omega_1$$

Therefore, the value of $x$ is $32$.

Given: The sphere keeps the same shape and uniform density while shrinking from radius $R$ to $\frac{R}{2}$.

Find: The factor relating $\omega_2$ to $\omega_1$.

For constant density, mass scales as volume:

$$M \propto R^3$$

For a solid sphere,

$$I \propto MR^2 \propto R^5$$

So when radius halves,

$$I_2 = I_1\left(\frac{1}{2}\right)^5 = \frac{I_1}{32}$$

With no external torque, angular momentum is conserved:

$$I_1\omega_1 = I_2\omega_2$$

Thus,

$$\omega_2 = \frac{I_1}{I_2}\omega_1 = 32\omega_1$$

Therefore, the value of $x$ is $32$.