If `I_(1)` is the moment of inertia of a thin' rod about an axis perpendicular to its length and passing through the centre of mass and `I_(2)` is the moment of inertia of the ring formed by bending this rod in the form of a ring, then `(I_1)/(I_2)` is

Options

A. `1 : 1`
B. `pi^2 : 3`
C. `pi : 4`
D. `3 : 5`

Answer

B

Explanation

Let M and L be the mass and the length of the rod respectively .
`therefore I_(1) = (1)/(12) ML^(2) " " because L = 2pi r or r = (L)/(2pi)`
`therefore I_(2) = Mr^(2) = M ((L)/(2pi))^(2)`
Thus , `(I_(1))/(I_(2)) = (pi^2)/(3)`