ID=2mr2
Moment of inertia of removed portion about the axis.
Iremoved=21(16m)(16r2)+(16m)(169r2)(by parallel axis theorem)
=512mr2+18mr2
=51219mr2
Iremaining=2mr2−51219mr2
=512237mr2
A circular hole of radius 4R is made in a thin uniform disc having mass and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is-

Held on 9 Apr 2017 · Verified 6 Jul 2026.
256219MR2
512237MR2
256197MR2
51219MR2
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