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

From a circular disc of radius R and mass 9 M , 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

`(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
Initial moment of inertia of the complete disc about an axis passing through its centre and perpendicular to its plane is
`I = (1)/(2) mass xx (radius)^(2) = (1)/(2)xx 9M xx R^(2) = (9)/(2)MR^(2)`
moment of inertia of disc removed, about an axis passing through its centre and perpendicular to
its plane is `I' = (1)/(2)M ((R )/(3))^(2) = (MR^(2))/(18)`
moment of inertia of the remaining portion of the disc `= I - I'`
`= (9)/(2)MR^(2) - (MR^(2))/(18) = (80)/(18)MR^(2) = (40MR^(2))/(9)`
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