Question

pls solve this question guys..need your help..!!!​

Answer 1

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Moment of Inertia  of a thin  rod about the axis perpendicular to its length and passing through its center 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 center of mass is MR^2   i.e

I2 = MR^2

(a)/(b) = I1/I2 = (Ml^2/12 )/Ml^2/4π^2  

                            = π^2/3