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Question

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 centre of mass and perpendicular to its plane is

A
MR22
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B
MR26
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C
MR23
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D
2MR23
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Solution

The correct option is D 2MR23
Given, surface mass density
σ=kr2
So, mass of the disc can be calculated by considering small element of area 2rπdr on it and then integrating it for complete disc, i.e,

dm=σdA=σ×2πrdr
dm=M=R0(kr2)2πrdr
M=2πkR44=12πkR4

Moment of inertia about the axis of the disc,
I=dI=dmr2=σdAr2
=R0kr2(2πrdr)r2
I=2πkR0r5dr=2πkR66=πkR63....(ii)
From Eqs. (i) and (ii), we get
I=23MR2

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