$I_{1}, I_{2}, I_{3}, I_{4}$ are respectively the moments of inertia of a thin square plate ABCD of uniform thickness about axes 1,2,3 and 4 which are in the plane of the plate. The moment of inertia about an axis passing through the centre O and perpendicular to the plane of the plate is

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$2 (I_{3} + I_{4})$

$I_{1} + I_{3}$

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

$2 (I_{1} + I_{2})$

answer is C.

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Detailed Solution

Because of symmetry of axes 1 and 2, $I_{1} = I_{2} .$ Similarly, $I_{3} = I_{4}$ From perpendicular axes theorem, it follows that the moment of inertia of the plate about an axis passing through the centre and perpendicular to the plane of the plate is

$I = I_{1} + I_{2} = I_{3} + I_{4} = 2 I_{1} = 2 I_{3} (∵ I_{1} = I_{2}, I_{3} = I_{4}) or I_{1} = I_{3}$