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(B) The moment of inertia of a square plate of mass $m$ and side $a$ about an axis passing through its center of mass and perpendicular to its plane i

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$\frac{2}{3}$

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(B) The moment of inertia of a square plate of mass $m$ and side $a$ about an axis passing through its center of mass and perpendicular to its plane is $I_{cm} = \frac{ma^2}{6}$.Using the parallel axis theorem,$I = I_{cm} + md^2$,where $d$ is the distance from the center of mass to the corner.The distance from the center to a corner is $d = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a}{\sqrt{2}}$.Thus,$I = \frac{ma^2}{6} + m(\frac{a}{\sqrt{2}})^2$.$I = \frac{ma^2}{6} + \frac{ma^2}{2} = \frac{ma^2 + 3ma^2}{6} = \frac{4ma^2}{6} = \frac{2}{3}ma^2$.Comparing this with $\alpha ma^2$,we get $\alpha = \frac{2}{3}$.

$A$ square plate of mass $m$ and side $a$ has a moment of inertia about an axis perpendicular to its plane and passing through one of its corners equal to $\alpha \,ma^2$. Then $\alpha$ is:

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