Q88 full solution............
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I think it's answer is 3/2MR^2 because the inertia os disc about a rotational axis is MR^2/2.
so, we add it by area of cross section
MR^2/2+MR^2
= 3MR^2/2
so, we add it by area of cross section
MR^2/2+MR^2
= 3MR^2/2
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