Given:

• Initial moment of inertia $= I$
• Initial angular velocity $= \omega$
• Initial kinetic energy $= E$

When the person stretches out his hands:

• The moment of inertia becomes triple. $I' = 3I$

Apply conservation of angular momentum Since no external torque acts on the system, $I\omega = I'\omega'$ $I\omega = 3I\omega'$ $\omega' = \frac{\omega}{3}$

Initial rotational kinetic energy $E = \frac{1}{2}I\omega^2$

Final rotational kinetic energy $E' = \frac{1}{2}I'\omega'^2$ Substituting $I' = 3I$ and $\omega' = \frac{\omega}{3}$: $E' = \frac{1}{2}(3I)\left(\frac{\omega}{3}\right)^2$ $E' = \frac{1}{2}(3I)\cdot \frac{\omega^2}{9}$ $E' = \frac{1}{2}I\omega^2 \cdot \frac{1}{3}$ $E' = \frac{E}{3}$

Therefore, $\frac{E}{x} = \frac{E}{3} \implies x = 3$