
Icomplete =2MR2Iremoved I=2MR2=(4M)(2R)221+(4M)(2R)2=323MR2−323MR2=3213MR2
JEE Main 2025 — Physics Mechanics
A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above. 
Held on 22 Jan 2025 · Verified 6 Jul 2026.
327MR2
329MR2
3217MR2
3213MR2
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