The moment of inertia of a thin square plate | vedclass

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(D) According to the theorem of perpendicular axes,the moment of inertia $I_z$ of a planar body about an axis perpendicular to its plane and passing t

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All of the above

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(D) According to the theorem of perpendicular axes,the moment of inertia $I_z$ of a planar body about an axis perpendicular to its plane and passing through a point $O$ is equal to the sum of the moments of inertia about two mutually perpendicular axes in the plane of the body,both passing through $O$.In the given figure,axes $1$ and $2$ are the diagonals of the square,which are mutually perpendicular. Thus,$I_0 = I_1 + I_2$.Axes $3$ and $4$ are lines passing through the centre $O$ and parallel to the sides of the square. These are also mutually perpendicular. Thus,$I_0 = I_3 + I_4$.Furthermore,due to the symmetry of the square,the moment of inertia about any axis passing through the centre and lying in the plane of the plate is the same if the angle of the axis is the same relative to the sides. Specifically,for a square plate,$I_1 = I_2 = I_3 = I_4 = \frac{ML^2}{12}$.Therefore,$I_0 = I_1 + I_2 = I_3 + I_4 = I_1 + I_3$ (since $I_1 = I_3$).Thus,all the given options are correct.

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