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

Answer

B

Explanation

For a rod of mass M and length L, the M.I. about a perpendicular axis passing through one end is `I_(1)=(ML^(2))/(3)`
When it is bent to form a ring, then
`L=2piR" "therefore" "R=(L)/(2pi)`
`therefore " The M.I. of the ring about its diameter is"`
`I_(2)=(MR^(2))/(2)=(M.L^(2))/(4pi^(2).2)=(ML^(2))/(8pi^(2))`
`therefore" "(I_(1))/(I_(2))=(ML^(2))/(3)xx(8pi^(2))/(ML^(2))=(8)/(3)pi^(2)`