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Question

From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is


A
4MR2
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B
404MR2
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C
10MR2
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D
379MR2
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Solution

The correct option is A 4MR2
Let σ be the mass per unit area. The total mass of the disc = σ×πR2=9M.

M = Mass of the circular disc cut
= σ×π(R3)2
= σ×πR29=M
Let us consider the above system as a complete disc of mass 9M and negative mass M superimposed on it.
Moment of inertia (I1) of the complete disc = 9MR2/2 about an axis passing through O and perpendicular to the plane of the disc.
MI of the cutout portion about an axis passing through O' and perpendicular to the plane of disc is 12×M×(R3)2
Therefore, MI (I2) of the cutout portion about an axis passing through O and perpendicular to the plane of disc is [12×M×(R3)2+M×(2R3)2]
[Using perpendicular axis theorem]
Therefore, the total MI of the system about an axis passing through O and perpendicular to the plane of the disc is
I=I1+I2
=129MR2[12×M×(R3)2+M×(2R3)2]2
=129MR2MR2[118+49]

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