The moment of inertia of a rod about an axis through its centre and perpendicular to it, is `(1)/(12)ML^(2)` (where, M is the mass and L is length of the rod). The rod is bent in the middle, so that two halves make an angle of `60^(@)`. The moment of inertia of the bent rod about the same axis would be

Options

A. `(1)/(48)Ml^(2)`
B. `(1)/(12)Ml^(2)`
C. `(1)/(24)Ml^(2)`
D. `(Ml^(2))/(8 sqrt(3))`

Answer

B

Explanation

Given M.I. of rod about its centre of mass and perpendicular to it is
`I=(1)/(12)Ml^(2)`
Let l. = moment of inertia of bent rod
`I.=(1)/(3)(M)/(2)((l)/(2))^(2)+(1)/(3)(M)/(2)((l)/(2))^(2)`
`=(1)/(12)((M)/(2)l^(2))+(1)/(12)((M)/(2)l^(2))`
`=(2)/(12)Ml^(2)((1)/(2))`
`=(1)/(12)Ml^(2)`