JEE Physics 2023 Question with Solution

Given: Initial angular velocity is $\omega_1 = \omega$, initial moment of inertia is $I_1 = I_0$, and final moment of inertia is $I_2 = 3I_0$.

Find: The value of $x$ if the final kinetic energy is $\frac{E}{x}$.

Since there is no external torque, angular momentum is conserved:

$$I_1 \omega_1 = I_2 \omega_2$$

Substituting the given values:

$$I_0 \omega = 3I_0 \omega_2$$

Hence,

$$\omega_2 = \frac{\omega}{3}$$

The initial kinetic energy is:

$$E = \frac{1}{2} I_0 \omega^2$$

The final kinetic energy is:

$$E_f = \frac{1}{2} I_2 \omega_2^2$$

Substituting $I_2 = 3I_0$ and $\omega_2 = \frac{\omega}{3}$:

$$E_f = \frac{1}{2} (3I_0) \left(\frac{\omega}{3}\right)^2$$ $$E_f = \frac{1}{2} \cdot 3I_0 \cdot \frac{\omega^2}{9}$$ $$E_f = \frac{1}{6} I_0 \omega^2$$

Comparing with $E = \frac{1}{2} I_0 \omega^2$:

$$E_f = \frac{E}{3}$$

Therefore, the value of $x$ is $3$.

Given: The moment of inertia becomes three times, from $I_0$ to $3I_0$.

Find: The value of $x$ in $E_f = \frac{E}{x}$.

When angular momentum $L$ is conserved, rotational kinetic energy can be written as:

$$E = \frac{L^2}{2I}$$

So energy is inversely proportional to moment of inertia:

$$E \propto \frac{1}{I}$$

If moment of inertia becomes three times, the kinetic energy becomes one-third:

$$E_f = \frac{E}{3}$$

Therefore, the correct value is $x = 3$.