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Question

A disc has mass 9M . A small disc of radius R/3 is cut from it as shown in the figure. The moment of inertia of remaining part about an axis passing through the centre ′O′ of the disc and perpendicular to the plane of the disc is :


A
8 MR2
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B
4 MR2
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C
409 MR2
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D
379 MR2
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Solution

The correct option is B 4 MR2
Mass of the disc =9M
Mass of the 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+M(2R3)2=118MR2+49MR2=12MR2
Therefore, moment of inertia of the remaining portion of disc about O is
=9MR22MR22=4MR2

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