Question

If i1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and i2 is the moment of inertia of the ring about an axis perpendicular to plane of ring and passing through its centre formed by bending the rod, then: options

Answer 1

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

or 9.8/3 or 98/30 = 49/15  

For more approximation treat π^2 = 10  

thus, I1/I2 = 10/3 or 10 : 3