From a square disc of uniform thickness and mass Ma circular disc of maximum possible area is cut. If the moment of inertia of the circle with the axis of rotation at the centre and
(Mr^2)\2
perpendicular to the plane of the circle is
then the side of the square is
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Mass removed m=M/4
radius of mass removed=r=R/2
M.I of whole disc about its center=MR2/2
M.I of mass removed about its center=mr2/2=MR2/32
By parallel axis theorem,
M.I of mass removed about center of disc=mr2/2+mr2=3MR2/32
M.I of remaining mass about center of disc=MR2/2−3MR2/32
=13MR2/32
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