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Question

A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in figure. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is:
20474_bd0a4f96f7304585afc7b01ffeab0f39.PNG

A
1532MR2
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B
716MR2
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C
1332MR2
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D
38MR2
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Solution

The correct option is C 1332MR2
Let I be the Moment of inertia of the disc about O without removal of the circular portion.
Moment of inertia of remaining portion Iremainder=IIcutoff
I=12MR2 (using the formula for MI of a solid disc about the centre and perpendicular to the plane of the disc).
Radius of the cut-off disc is Rcutoff=R2
By using parallel axis theorem, Icutoff=IO+(Mcutoff)d2
where IO is the Moment of inertia of cut-off disc about the axis passing through O' (which is cm) and d is the distance between the centres O and O'.
Mass of cut-off portion Mcutoff=πρ(R2(R2)2)=π×ρ×34R2
Moment of inertia of cut-off disc Icutoff=12Mcutoff×(R2)2=12πρ×34R2×(R2)2=3ρπ32R4
Iremainder=12πρR2×R23πρ32R4=1332πρR2×R2=1332×(πρR2)×R2=1332MR2
142346_20474_ans_2d6ca47d1c61409dab85d79df01edce7.PNG

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