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(C) Given: $I_1 = I$ and $I_2 = 2I$.Rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.Since the rotational kinetic energies are equal

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$\sqrt{2}:1$

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(C) Given: $I_1 = I$ and $I_2 = 2I$.Rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.Since the rotational kinetic energies are equal,we have $K_1 = K_2$.Therefore,$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2$.Substituting the values: $\frac{1}{2} (I) \omega_1^2 = \frac{1}{2} (2I) \omega_2^2$.Simplifying the equation: $\omega_1^2 = 2 \omega_2^2$.Taking the square root on both sides: $\frac{\omega_1}{\omega_2} = \sqrt{\frac{2}{1}} = \frac{\sqrt{2}}{1}$.Thus,the ratio of their angular velocities is $\sqrt{2}:1$.

The moments of inertia of two bodies about their axes of rotation are $I$ and $2I$ respectively. If their rotational kinetic energies are equal,what is the ratio of their angular velocities?

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