Let $I_{1}, I_{2}$ and $I_{3}$ be the moment of inertia of a uniform square plate about axes AOC, $x D x^{'}$ and $y B y^{'}$ respectively as shown in the figure. The moments of inertia of the plate $I_{1} : I_{2} : I_{3}$ are in the ratio.

1 : \frac{1}{7} : \frac{1}{7}

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1 : \frac{12}{7} : \frac{12}{7}

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1 : \frac{7}{12} : \frac{7}{12}

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1 : 7 : 7

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Solution

The correct option is D $1 : 7 : 7$
$I_{A O C} = I_{1} I_{y B y^{'}} = I_{2} I_{x o x^{'}} = I_{3}$
Assuming mass and length to be $m, l$
$I_{1} = \frac{1}{12} m c^{2} I_{2} = I_{3}$
By Symmetry
$I_{2} = \frac{1}{12} m l^{2} + m (\frac{l}{\sqrt{2}})^{2} = \frac{1}{12} m l^{2} + \frac{m l^{2}}{2} = \frac{7 m l^{2}}{2} ∴ I_{1} : I_{2} : I_{3} = 1 : 7 : 7$

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