The `M.I.` of a rod about an axis through its center and perpendicular to it is `I_(0)`. The rod is bent in the middle so that the two halves make an angle `theta`. The moment of inertia of the bent rod about the same axis would be

Options

A. `I_(0)sin^(2)theta`
B. `I_(0)cos^(2)theta`
C. `I_(0)`
D. `(I_(0))/(2)`

Explanation

`I_(A)=(ML^(2))/(12)=I_(0)`
Now rod is bent at `A`

`M.I.` of half rod (`1`) about `A` and `bot^(ar)` to the plane
`I_(1)=(1)/(3)(M)/(2)((L)/(2))^(2)=(ML^(2))/(24)=I_(2)`
`I_(A)=I_(1)+I_(2)=(ML^(2))/(12)=I_(0)`
`M.I.` will remain the same for any value of `theta`.