$A B C D$ is a square plate with centre $O$ . The moment of inertia of the plate about the axes $1, 2, 3$ and $4$ are $I_{1}, I_{2}, I_{3} & I_{4}$ respectively. It follows that

I_{1} + I_{2} = I_{3} + I_{4}

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I_{1} + I_{4} = I_{2} + I_{3}

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I_{1} + I_{3} = I_{2} + I_{4}

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I_{1} - I_{3} = I_{2} - I_{4}

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Solution

The correct option is C $I_{1} + I_{3} = I_{2} + I_{4}$
According to theorem of perpendicular axes, moment of inertia of the plate about an axis passing through the center $O$ and perpendicular to the plane is,
$I_{o} = I_{1} + I_{3} = I_{2} + I_{4}$
(Since, diagonals are mutually perpendicular to each other)

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