From a uniform circular disc of radius R and mass 9M, a small disc of radius R3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is
A
10MR2
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B
379MR2
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C
4MR2
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D
409MR2
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Solution
The correct option is C4MR2
[Using negative mass concept] 1=9MR22−⎡⎢
⎢
⎢⎣M(R3)22+M(2R3)2⎤⎥
⎥
⎥⎦=MR2[92−118−49]=4MR2