A uniform thin rod of 120 cm length and 1600 g mass is bent as shown in the figure. The moment of inertia of the bent rod about and axis passing through the point 'O' and perpendicular to the plane of the paper is . . . . . . `kg -m^(2)` .

Options

A. 0 . 084
B. 0 . 360
C. 0 . 018
D. 0 . 120

Answer

A

Explanation

There are four rods each of mass m = 400 g
` = (400)/(1000)`
= 0 . 4 kg
and each having a length of 1 = 30 cm
` = (30)/(100) m `
= 0 . 3 m
Now, out of therefore two sets of two rods have same m,I.
So, we can claculate moment of inertia as follows
`h^(2) = I^(2) - ((l)/(2))^(2)`
`=(3)/(4) l^(2)`
M.I system ` = (MI "of" I) xx 2 + (M . I "of" II) xx2`
`= (ml^(2))/(3) xx2 + 2 ((ml^(2))/(12) + mh^(2))`
`= (2)/(3) ml^(2) + 2((ml^(2))/(12)+ (3)/(4) ml^(2))`
`= (2)/(3) ml^(2) + 2ml^(2) [(1)/(12)+(3)/(4)]`
`= (2)/(3) ml^(2) + 2ml^(2) xx (10)/(12)`
`= (2)/(3)ml^(2)`
`= (7)/(3) xx 0 . 4 xx 0 . 3 xx . 0 3`
` = 0 . 084 kg -m^(2)`