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 409MR2Mass of the disc = 9MMass of removed portion of disc = MThe moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane isI1=92MR2Now, the moment of inertia of the disc with removed portionI2=12M(R3)2=118MR2Therefore, moment of inertia of the remaining portion of disc about O isI=I1−I2=9MR22−MR218=40MR29

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