Q.

A thin disc of mass M and radius R has mass per unit area σ(r)=kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its center of mass and perpendicular to its plane is

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a

MR22

b

MR23

c

2MR23

d

MR26

answer is B.

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Detailed Solution

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M=oRkr22πrdr=πkR42

k=2MπR4

I=dI=r2dm=kr2(2πrdr)r2

=πkR63=π2MπR4×R63=23MR2

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