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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a
I
b
Isin2θ
c
Icos2θ
d
Icos2θ2
answer is A.
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Detailed Solution
Let IZ is the moment of inertia of square plate about the axis which is passing through the centre and perpendicular to the plane.IZ=IAB+IA'B'=ICD+IC'D'[By the theorem of perpendicular axis]IZ=2IAB=2IA'B'=2ICD=2IC'D' [As AB,A′B′ and CD,C′D′ are symmetric axis] Hence ICD=IAB=I