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Question

Moment of Inertia (MI) of four bodies, having same mass and radius, are reported as -

I1=MI of thin circular ring about its diameter.

I2=MI of circular disc about an axis perpendicular to the disc and passing through the centre.

I3=MI of solid cylinder about its axis.

I4=MI of solid sphere about its diameter.

Then :

A
I1=I2=I3<I4
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B
I1+I2=I3+52I4
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C
I1+I3<I2+I4
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D
I1=I2=I3>I4
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Solution

The correct option is D I1=I2=I3>I4
I1=MI of thin circular ring about its diameter.
I1=MR22=0.5MR2

I2=MI of circular disc about an axis perpendicular to the disc and passing through the centre.
I2=MR22=0.5MR2

I3=MI of solid cylinder about its axis.
I3=MR22=0.5MR2

I4=MI of solid sphere about its diameter.
I4=2MR25=0.4MR2

I1=I2=I3>I4

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