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
8MR2
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B
4MR2
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C
409MR2
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D
379MR2
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Solution
The correct option is B4MR2 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 =9MR22−MR22=4MR2