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Question

# A ring having linear mass density 5 kg/m and radius 0.5 m is rotated about the axis passing through it's centre and perpendicular to it's plane. Find the moment of inertia (MI) of the ring about the required axis.

A
0.5 kg-m2
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B
1 kg-m2
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C
2.5π kg-m2
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D
1.25π kg-m2
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Solution

## The correct option is D 1.25π kg-m2Mass of ring,(m) = linear mass density × circumference r=0.5 m ⇒m=5×(2πr)=5×(2×π×0.5) ∴m=5π kg MI of ring about the required axis YY′ as shown in figure is: I=mr2 I=(5π)(0.5)2 ∴I=1.25π kg-m2 So, moment of inertia of ring about the axis YY′ is 1.25π kg-m2

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