If I1 is the MI of a thin rod about and axis perpendicular to its length and passing through its centre of mass and I2 is the moment of inertia of the ring formed by bending the rod what is ratio of I1:I2?

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)

Plz Approve!

Approved