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(D) For a thin uniform square sheet of side $a$ and mass $m$,the moment of inertia about an axis passing through the center and perpendicular to the p

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$I = \frac{ma^2}{12}$

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(D) For a thin uniform square sheet of side $a$ and mass $m$,the moment of inertia about an axis passing through the center and perpendicular to the plane is $I_z = \frac{ma^2}{6}$.By the perpendicular axis theorem,$I_z = I_x + I_y$,where $I_x$ and $I_y$ are the moments of inertia about the two perpendicular axes passing through the center in the plane of the sheet.Due to symmetry,$I_x = I_y = \frac{ma^2}{12}$.Let $I_d$ be the moment of inertia about a diagonal. By the perpendicular axis theorem applied to the two diagonals (which are also perpendicular to each other and lie in the plane),$I_z = I_{d1} + I_{d2}$.Since the square is symmetric,$I_{d1} = I_{d2} = I$.Thus,$I_z = 2I$,which implies $I = \frac{I_z}{2} = \frac{ma^2/6}{2} = \frac{ma^2}{12}$.Wait,let us re-evaluate: The moment of inertia of a square plate about an axis passing through the center and parallel to a side is $I_{side} = \frac{ma^2}{12}$.The moment of inertia about a diagonal is $I_{diag} = \frac{ma^2}{12}$.Therefore,$I = \frac{ma^2}{12}$.

Consider a thin uniform square sheet made of a rigid material. If its side is $a$,mass is $m$,and the moment of inertia about one of its diagonals is $I$,then:

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