Question

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

Answer 1

`I_(1)=(ML^(2))/(3)`
`L=2piRimplies R=(L)/(2pi)`
`I_(2)=(1)/(2)MR^(2)=(1)/(2)M((L)/(2pi))^(2)=(ML^(2))/(8pi^(2))`
`(I_(1))/(I_(2))=(8pi^(2))/(3)`