Two uniform identicla rods each of mass M and length l are joined to form a cross as shown in figure. Find the momet of inertia of the cross about a bisector as shown doted in the figure

Options

A. `(mI^(2))/6`
B. `(mI^(2))/3`
C. `(mI^(2))/(12)`
D. `(2mI^(2))/3`

Answer

C

Explanation

Moment of inertia of the combined rods about an axis passing through their point of intersection O and perpendicular to the plane of the rods is
`I_(0) = (mI^(2))/(12) + (mI^(2))/12 = (mI^(2))/6`
Now xy and are two mutually perpendicular axes passing through O lying in the plane of two rods. Then by symmetry.

`I_(XY) = I_(X'Y') =I` (Say)
and `I_(XY) + I_(X'Y') = I_(0)` (from theorem of perpendicular axes).
or `2I = (mI^(2))/6` So, `I=(mI^(2))/12`