A thin rod of length 4l, mass 4 m is bent at the point as shown in the figure. What is the moment of inertia of the rod about the axis passing through O and perpendicular to the plane of the paper?

Options

A. `(Ml^(2))/3`
B. `(10Ml^(2))/3`
C. `(Ml^(2))/12`
D. `(Ml^(2))/24`

Answer

B

Explanation

Total `MI=I_(1)+I_(2)+I_(3)+I_(4)`
`=2I_(1)+2I_(2)`
`=2(I_(1)+I_(2))[I_(1)=I_(3),I_(1)=I_(4)]`
now `I_(2)=I_(3)=(Ml^(2))/3`
using parallel axis theorem, we have
`I=I_(CM)+Mx^(2)` and `x=sqrt(l^(2)+(l/2)^(2))`
`I_(1)=I_(4)=(Ml^(2))/12+M[sqrt(l^(2)+(l/2)^(2))]`
putting all values we get
`I=(10 Ml^(2))/3`