Let I_1, I_2 and I_3 be the moment of inertia | vedclass

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(D) Let the mass of the square plate be $M$ and the side length be $L$.$1$. The axis $AOC$ passes through the center of the square and lies in the pla

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$1 : 7 : 7$

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(D) Let the mass of the square plate be $M$ and the side length be $L$.$1$. The axis $AOC$ passes through the center of the square and lies in the plane of the plate along the diagonal. The moment of inertia of a square plate about its diagonal is $I_1 = \frac{1}{12} M L^2$.$2$. The axes $xDx'$ and $yBy'$ are parallel to the diagonal $AOC$ and pass through the corners $D$ and $B$ respectively.$3$. The perpendicular distance between the diagonal $AOC$ and the parallel axes $xDx'$ or $yBy'$ is $d = \frac{L}{\sqrt{2}}$.$4$. Using the parallel axis theorem,$I = I_{cm} + M d^2$,where $I_{cm} = I_1 = \frac{1}{12} M L^2$.$5$. Thus,$I_2 = I_3 = \frac{1}{12} M L^2 + M \left( \frac{L}{\sqrt{2}} \right)^2 = \frac{1}{12} M L^2 + \frac{1}{2} M L^2 = \left( \frac{1 + 6}{12} \right) M L^2 = \frac{7}{12} M L^2$.$6$. The ratio $I_1 : I_2 : I_3 = \frac{1}{12} M L^2 : \frac{7}{12} M L^2 : \frac{7}{12} M L^2 = 1 : 7 : 7$.

Let $I_1, I_2$ and $I_3$ be the moment of inertia of a uniform square plate of mass $M$ and side length $L$ about axes $AOC, xDx'$ and $yBy'$ respectively as shown in the figure. The ratio of the moments of inertia $I_1 : I_2 : I_3$ is:

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