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(B) The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^{2}$.Since the kinetic energies are equal, we have $I_{1} \omega_{1}^{2} =

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$\sqrt{2}: \sqrt{2}: 1$

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(B) The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^{2}$.Since the kinetic energies are equal, we have $I_{1} \omega_{1}^{2} = I_{2} \omega_{2}^{2} = I_{3} \omega_{3}^{2}$, which implies $\omega \propto \frac{1}{\sqrt{I}}$.For a square plate of side $a$ and mass $M$:$1$. For axis $1$ (passing through the center and parallel to sides), $I_{1} = \frac{Ma^{2}}{12}$.$2$. For axis $2$ (passing through the center and parallel to the diagonal), $I_{2} = \frac{Ma^{2}}{12}$.$3$. For axis $3$ (passing through the center and perpendicular to the plane of the plate), by the perpendicular axis theorem, $I_{3} = I_{x} + I_{y} = \frac{Ma^{2}}{12} + \frac{Ma^{2}}{12} = \frac{Ma^{2}}{6}$.Thus, the moments of inertia are in the ratio $I_{1}: I_{2}: I_{3} = \frac{1}{12}: \frac{1}{12}: \frac{1}{6} = 1: 1: 2$.The ratio of angular speeds is $\omega_{1}: \omega_{2}: \omega_{3} = \frac{1}{\sqrt{I_{1}}}: \frac{1}{\sqrt{I_{2}}}: \frac{1}{\sqrt{I_{3}}} = \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{2}} = 1: 1: \frac{1}{\sqrt{2}} = \sqrt{2}: \sqrt{2}: 1$.

Three identical square plates rotate about the axes shown in the figure in such a way that their kinetic energies are equal. Each of the rotation axes passes through the centre of the square. Then the ratio of angular speeds $\omega_{1}: \omega_{2}: \omega_{3}$ is

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