The moment of inertia of a uniform rod about a perpendicular axis passing through one end is `I_(1)`. The same rod is bent into a ring and its moment of inertia about a diameter is `I_(2)`. Then `I_(1)//I_(2)` is

Options

A. `(4pi)/(3)`
B. `(8 pi^(2))/(3)`
C. `(5 pi )/(3)`
D. `(8 pi^(2))/(5)`

Answer

B

Explanation

Moment of inertia of a thin uniform rod about perpendicular axis through it one end
`I = (1)/(3) Mt^(2) " " … (i)`
Same rod is bent into a ring
`therefore l = 2pi r implies (l)/(r) = 2pi " " … (ii)`
Moment of inertia of ring about diameter
`I_(2) = (MR^(2))/(2) " " .... (iii)`
Dividing Eq.(i) by (iii) , we get
`(I)/(I_(2)) = (2)/(3) (MI^(2))/(MR^(2)) = (2)/(3) (I^(2))/(R^(2))`
`(I)/(I_(1)) = (2)/(3) (2pi)^(2) = (8 pi^(2))/(3) " " `[using Eq. (ii)]