A man standing on a turn-table is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is $75 \%$ of that with outstretched arms, then his rotational kinetic energy will

Let $I_1$ and $\omega_1$ represent the initial moment of inertia and angular velocity, with $I_2$ and $\omega_2$ the final values after the man folds his arms. The moment of inertia decreases to $I_2 = \frac{3}{4}I_1$.

Angular momentum conservation gives $I_1\omega_1 = I_2\omega_2$. Substituting yields $I_1\omega_1 = \frac{3}{4}I_1\omega_2$, so $\omega_2 = \frac{4}{3}\omega_1$.

The rotational kinetic energy increases from $KE_1 = \frac{1}{2}I_1\omega_1^2$ to $KE_2 = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}\left(\frac{3}{4}I_1\right)\left(\frac{4}{3}\omega_1\right)^2 = \frac{1}{2}I_1\omega_1^2 \cdot \frac{4}{3} = \frac{4}{3}KE_1$.

The kinetic energy increases by $\Delta KE = KE_2 - KE_1 = \frac{1}{3}KE_1$, representing a $\frac{1}{3} \times 100\% \approx 33.3\%$ increase.

The rotational kinetic energy increases by approximately $33.3\%$.