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(A) The rotational kinetic energy is given by $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.Since there are

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$3$

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(A) The rotational kinetic energy is given by $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.Since there are no external torques acting on the system,the angular momentum $L$ remains constant.Therefore,the kinetic energy is inversely proportional to the moment of inertia: $K \propto \frac{1}{I}$.Let $I_i$ and $K_i = E$ be the initial moment of inertia and initial kinetic energy,respectively.Let $I_f$ and $K_f$ be the final moment of inertia and final kinetic energy,respectively.Given that $I_f = 3I_i$,we have:$\frac{K_f}{K_i} = \frac{I_i}{I_f} = \frac{I_i}{3I_i} = \frac{1}{3}$.Thus,$K_f = \frac{K_i}{3} = \frac{E}{3}$.Comparing this with $\frac{E}{x}$,we get $x = 3$.

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