Question
Let 'l' be the moment of inertia of an uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre 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
Answer: Option A
:
A
Let Izis the moment of inertia of square plate about the axis which is passing through the centreand 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
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:
A
Let Izis the moment of inertia of square plate about the axis which is passing through the centreand 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
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