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(B) The rotational kinetic energy 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_

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$\sqrt 2 : \sqrt 2 : 1$

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(B) The rotational kinetic energy 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 mass $m$ and side $l$:$1$. For axis $1$ (in the plane,passing through the center parallel to a side),the moment of inertia is $I_1 = \frac{ml^2}{12}$.$2$. For axis $2$ (in the plane,passing through the center along the diagonal),the moment of inertia is $I_2 = \frac{ml^2}{12}$.$3$. For axis $3$ (perpendicular to the plane,passing through the center),the moment of inertia is $I_3 = I_1 + I_2 = \frac{ml^2}{12} + \frac{ml^2}{12} = \frac{ml^2}{6}$.Thus,the ratio of angular speeds is $\omega_1 : \omega_2 : \omega_3 = \frac{1}{\sqrt{I_1}} : \frac{1}{\sqrt{I_2}} : \frac{1}{\sqrt{I_3}} = \frac{1}{\sqrt{ml^2/12}} : \frac{1}{\sqrt{ml^2/12}} : \frac{1}{\sqrt{ml^2/6}} = 1 : 1 : \frac{1}{\sqrt{1/2}} = 1 : 1 : \sqrt{2}$.Wait,checking the options,the correct ratio is $1 : 1 : \frac{1}{\sqrt{2}}$ which is equivalent to $\sqrt{2} : \sqrt{2} : 1$.

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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