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From a circular disc of radius R and ...

From a circular disc of radius R and 9M , a small disc of mass M and radius `(R )/(3)` is removed concentrically .The moment of inertia of the remaining disc about and axis perpendicular to the plane of the disc and passing through its centre is

A

`(40)/(9) MR^(2)`

B

`MR^(2)`

C

`4MR^(2)`

D

`(4)/(9) MR^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`I= (9M (R)^(2))/(2)- (M ((R)/(3))^(2))/(2)`
`= (9MR^(2))/(2)- (MR^(2))/(18)= (80 MR^(2))/(18)`
`= (40MR^(2))/(9)`
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