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

Answer

B

Explanation

Let.s suppose points are A, B, C, D
Total moment of inertia about an axis passing through point O and perpendicular to the plane of paper is
`I_("net") =I_(AB) + I_(BO) +I_(OC) +I_(CD) ....(i)`
Let.s suppose points P, Q, R, S are the middle points of all rods respectively.
Now using parallel axis theorem
`I_(BO)=I_(OC) =(ml^(2))/(12) +(m) ((1)/(2))^(2)`
`I_(BO) =I_(OC) =(ml^(2))/(3)`
Now,
Distance PO = ?

Using pythogoras theorem,
`x= PO = sqrt(l^(2) +(l^(2))/(4)) = (sqrt(51))/(2)`
Now,
`I_(AB)=I_(CD) =(ml^(2))/(12) + m((sqrt(51))/(2))^(2)`
(Using parallel axis theorem)
So, `I_(AB)=I_(CD)=(4)/(3) ml^(2)`
Now putting values in eq. (i)
`I_("net") =(ml^(2))/(3) +(ml^(2))/(3) +(4)/(3)ml^(2) +(4)/(3)ml^(2)`
`I_("net") =(10)/(3)ml^(2)`
So option (2) is correct