By parallel axis theorem,
Mass of smaller disc =πR29Mπ(3R)2=M
MI of smaller disc =2M(3R)2+M(32R)2
MI=18MR2+94MR2
MI=2MR2
MI=MI(Bigger)−MI(Smaller)
=29MR2−2MR2
=4MR2
JEE Main 2018 — Physics Mechanics
From a uniform circular disc of radius R and mass 9 M, a small disc of radius 3R is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:

Held on 8 Apr 2018 · Verified 6 Jul 2026.
937MR2
4MR2
940MR2
10MR2
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