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From a circular disc of radius R and mas...

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

A

`4 MR^(2)`

B

`40/9 MR^(2)`

C

`40 MR^(2)`

D

`37/9 MR^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Mass per unit area of disc
`=(9M)/(piR^(2))`
Mass of removed portion of disc
`=(9M)/(piR^(2)) xx pi(R/3)^(3) = M`
Moment of inertia of removed portion about an axis passing through center of disc and perpendicular to the plane of disc, using theorem of parallel axes is
`I_(1)=M/2(R/3)^(2)+1/2(MR^(2))=4MR^(2)`
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