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

Answer

B

Explanation

Moment of inertia of a thin uniform rod about perpendicular axis through it one end
`l=(1)/(3)Ml^(2)" …(i)"`
Same rod is bent into a ring, `l=2piR rArr (l)/(R)=2pi" …(ii)"`
Moment of inertia of ring about diameter, `l_(1)=(MR^(2))/(2)" ...(iii)"`
Dividing Eq. (i) by (iii), we get
`(l)/(l_(2))=(2)/(3)(Ml^(2))/(MR^(2))=(2)/(3)(l^(2))/(R^(2))`
`(l)/(l_(1))=(2)/(3)(2pi)^(2)=(8pi^(2))/(3)" [Using Eq. (ii)]"`