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Moment of Inertia of a thin rod about the axis perpendicular to its length and passing through its centre of mass is Ml^2/12 i.e
I1 = Ml^2/12.........................................(a)
Now, rod of length ''l'' is formed into a circular ring say Radius ''R''
then
2πR = l
R = l/2π
Moment of Inertia of the ring passing through the centre of mass is MR^2 i.e
I2 = MR^2
OR
I2 = Ml^2/4π^2...................................................(b)
(a)/(b) = I1/I2 = (Ml^2/12 )/Ml^2/4π^2
= π^2/3
or 9.8/3 or 98/30 = 49/15
For more approximatio treat π^2 = 10
thus, I1/I2 = 10/3 or 10 : 3
HERE π IS (Pie)
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