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(D) Given that the rotational kinetic energies are equal:$(K.E.)_A = (K.E.)_B$$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2$$\frac{I_1}{I_2

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$3 I_1$

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(D) Given that the rotational kinetic energies are equal:$(K.E.)_A = (K.E.)_B$$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2$$\frac{I_1}{I_2} = \frac{\omega_2^2}{\omega_1^2} \implies \frac{\omega_2}{\omega_1} = \sqrt{\frac{I_1}{I_2}} \quad ...(i)$Also,rotational kinetic energy is given by $K.E. = \frac{L^2}{2I}$. Since kinetic energies are equal:$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$$\frac{I_2}{I_1} = \frac{L_2^2}{L_1^2}$Given the ratio of angular momenta $\frac{L_1}{L_2} = \frac{1}{\sqrt{3}}$,we have $\frac{L_2}{L_1} = \sqrt{3}$.Therefore,$\frac{I_2}{I_1} = (\sqrt{3})^2 = 3$.Thus,$I_2 = 3 I_1$.

Two bodies $A$ and $B$ have their moments of inertia $I_1$ and $I_2$ respectively about their axis of rotation. If their kinetic energies of rotation are equal and their angular momenta $L_1$ and $L_2$ respectively are in the ratio $1 : \sqrt{3}$,then $I_2$ will be

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