For the given uniform square lamina ABCD,whos | vedclass

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(D) Let the side of the square be $a$. By the theorem of perpendicular axes,the moment of inertia about an axis passing through the center $O$ and per

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$I_{AC} = I_{EF}$

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(D) Let the side of the square be $a$. By the theorem of perpendicular axes,the moment of inertia about an axis passing through the center $O$ and perpendicular to the plane of the lamina is $I_z = I_x + I_y$. For a square lamina,the moment of inertia about any axis passing through the center and lying in the plane of the lamina is the same. Let $I_{EF}$ be the moment of inertia about the axis $EF$ (passing through the midpoints of sides $AB$ and $CD$). By symmetry,$I_{EF} = I_{GH}$ where $GH$ is the axis passing through the midpoints of sides $AD$ and $BC$. Thus,$I_z = I_{EF} + I_{GH} = 2I_{EF}$. Now,consider the diagonal $AC$. The moment of inertia about the diagonal $AC$ is $I_{AC}$. By symmetry,the moment of inertia about the other diagonal $BD$ is $I_{BD} = I_{AC}$. Since the diagonals are also perpendicular axes in the plane of the lamina,$I_z = I_{AC} + I_{BD} = 2I_{AC}$. Equating the two expressions for $I_z$,we get $2I_{EF} = 2I_{AC}$,which implies $I_{AC} = I_{EF}$.

For the given uniform square lamina $ABCD$,whose centre is $O$,

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