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A thin disc of mass M and radius R has m...

A thin disc of mass M and radius R has mass per unit area `sigma( r) =kx^(2)` where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:

A

`(MR^(2))/(3)`

B

`(2MR^(2))/(3)`

C

`(MR^(2))/(6)`

D

`(MR^(2))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`dI=dmr^(2)`
`I=int_(0)^(R)Kr^(2)2pirdr.r^(2)=K2pi.(R^(6))/(6)=(KpiR^(6))/(3)`
`int_(0)^(m)dm=int_(0)^(R)Kr^(2)2pirdr`
`m=2piK.(R^(4))/(4)=(piKR^(4))/(2)`
`I=(2)/(3)mR^(2)`.
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