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Question

Figure represents half of an annular disc of radii R1 and R2. The mass of this half disc is M. The moment of inertia of this disc about an axis passing through centre of full disc and perpendicular to the plane of disc is
149240.png

A
M2(R22R21)
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B
M(R22+R21)
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C
M2(R22+R21)
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D
M2(R42+R41)(R22R21)
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Solution

The correct option is D M2(R22+R21)
Taking an element of thickness dr at a distance r from the center ,
we have dI=dmr2 where dm=2Mπ(R22R21)πrdr
Hence , I=R2R12MR22R21r3dr
Hence , I=M2(R22+R21)
191586_149240_ans.png

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