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

Answer

A

Explanation

We know that, radius of ring, `R=(L)/(2pi)" …(i)"`
Moment of inertia of thin uniform rod.
`l=(ML^(2))/(12)" …(ii)"`
and same rod is bent into a ring, then its moment of inertia,
`l'=(1)/(2)MR^(2)" [From Eq. (i)]"`
`l'=(1)/(2)(ML^(2))/(4pi^(2))=(ML^(2))/(8pi^(2))" ...(iii)"`
On dividing Eq. (ii) by Eq. (iii), we get
`(l)/(l')=(ML^(2))/(12)xx(8pi^(2))/(ML^(2))=(8pi^(2))/(1)=(2pi^(2))/(3)`