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Question

From a circular disc of radius R and mass 9M, a small disc of mass M and radius R3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

A
409MR2
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B
MR2
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C
4MR2
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D
49MR2
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Solution

The correct option is A 409MR2
Mass of the disc = 9M
Mass of removed portion of disc = M
The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is
I1=92MR2
Now, the moment of inertia of the disc with removed portion
I2=12M(R3)2=118MR2
Therefore, moment of inertia of the remaining portion of disc about O is
I=I1I2
=9MR22MR218=40MR29
287356_306094_ans.png

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