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>I4I1=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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