Question

# A thin uniform disc of mass M and radius R has concentric hole of radius r. Find the moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane.

A
12RM(R3+r3)
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B
12M(R2r2)
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C
13M(R2+r2)
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D
12M(R2+r2)
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Solution

## The correct option is D 12M(R2+r2)Mass per unit area of the disc is m=Mπ(R2−r2) Mass of the disc if it was complete (i.e without hole) is M1=m×πR2=Mπ(R2−r2)×πR2 =MR2R2−r2 Mass of the removed portion is M2=m×πr2=Mr2R2−r2 Since the two portions are concentric, the moment of inertia of the given disc about the given axis is I=12M1R2−12M2r2 =12[MR4R2−r2−Mr4R2−r2] =M2[R4−r4R2−r2]=12M(R2+r2)

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