JEE Physics 2023 Question with Solution

Given: The Earth's volume becomes $\frac{1}{64}$ of the original value while mass remains the same. The initial period of rotation is $24$ hours.

Find: The value of $x$ if the new period is $\frac{24}{x}$ hours.

Use conservation of angular momentum. The solution states that the moment of inertia changes with radius, so a decrease in radius increases angular velocity.

Since volume is proportional to $R^3$,

$$\left(\frac{R_2}{R_1}\right)^3 = \frac{1}{64}$$

Hence,

$$\frac{R_2}{R_1} = \frac{1}{4}$$

So,

$$R_2 = \frac{R_1}{4}$$

From the given working,

$$\frac{2}{5} M R^2 \omega_1^2 = \frac{2}{5} M \left(\frac{R}{4}\right)^2 \omega_2^2$$

Cancel $M$ and $\frac{2}{5}$:

$$\frac{\omega_1}{\omega_2} = \left(\frac{R}{R/4}\right)^2 = \frac{1}{16}$$

The relationship between angular velocity and time period is:

$$\frac{\omega_1}{\omega_2} = \frac{T_2}{T_1}$$

Substituting $\frac{\omega_1}{\omega_2} = \frac{1}{16}$ and $T_1 = 24$:

$$\frac{1}{16} = \frac{T_2}{24}$$

Therefore,

$$T_2 = \frac{24}{16}$$

Comparing with $T_2 = \frac{24}{x}$, we get $x = 16$.

Therefore, the value of $x$ is $16$.

Given: Volume becomes $\frac{1}{64}$ of the original value.

Find: The new rotational period in the form $\frac{24}{x}$.

For a spherical body of constant mass, moment of inertia is proportional to $R^2$. Since the new volume is $\frac{1}{64}$ times the original,

$$V \propto R^3$$

So the radius becomes

$$R_2 = \frac{R_1}{4}$$

Thus the new moment of inertia becomes

$$I_2 = I_1 \left(\frac{1}{4}\right)^2 = \frac{I_1}{16}$$

By conservation of angular momentum,

$$I_1 \omega_1 = I_2 \omega_2$$

Hence,

$$\omega_2 = 16\omega_1$$

Since $T \propto \frac{1}{\omega}$,

$$T_2 = \frac{T_1}{16} = \frac{24}{16}$$

Therefore, in the form $\frac{24}{x}$, we get $x = 16$.