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Let l be the moment of inertia of an uniform square plate about an axis AB that passes through its center and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the center of the plate and makes an angle $θ$ with AB. The moment of inertia of the plate about the axis CD is then equal to

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$I \cos^{2} \frac{θ}{2}$

$I \cos^{2} θ$

$I \sin^{2} θ$

$I$

answer is A.

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Detailed Solution

Let I_Z is the moment of inertia of square plate about the axis which is passing through the centre and perpendicular to the plane.

$I_{Z} = I_{A B} + I_{A' B'} = I_{C D} + I_{C' D'}$

[By the theorem of perpendicular axis]

$\begin{array}{l} I_{Z} = 2 I_{A B} = 2 I_{A' B'} = 2 I_{C D} = 2 I_{C' D'} \\ [As A B, A^{'} B^{'} and C D, C^{'} D^{'} are \\ symmetric axis] \\ Hence I_{C D} = I_{A B} = I \end{array}$

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