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Question

# From a circular disc of radius R and mass 9M, a small disc of radius R3 is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc passing through O is

A
4MR2
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B
409MR2
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C
40MR2
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D
379MR2
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Solution

## The correct option is A 4MR2Solution:- (A) 4MR2As we know that the moment of inertia of complete disc about a perpendicular axis passing through centre O is given as-I=12MR2Whereas M and R are the mass and radius of the disc respectively.Given that the mass of disc is 9M and radius is R.Therefore,I1=12×9M×R2=92MR2Now, mass of cut out disc of radius R3-m=9MπR2×π(R3)2⇒m=MNow using the theorem of parallel axis, the moment of inertia of cut out disc about the perpendicular axis passing through cenre O is,I2=12×M×(R3)2M(2R3)2⇒I2=12MR2Therefore, the moment of inertia of residue disc is-I=I1−I2⇒I=92MR2−12MR2⇒I=4MR2Hence the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc passing through O is 4MR2.

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