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

Answer

B

Explanation

`I=(mL^(2))/(3), I_(1)=(mR^(2))/(2)`
`(I)/(I_(1))=(mL^(2))/(3)xx(2)/(MR^(2))`
`= (2)/(3)xx(L^(2))/(R^(2))`
`=(2)/(3)xx((2pi R^(2)))/(R^(2)) because L = 2 pi R`
`= (2)/(3)xx (4pi^(2)R^(2))/(R^(2))`
`(I)/(I_(1))=(8pi^(2))/(3)`.