AB and CD are two identical rods each of length l and mass m joined to form a cross. The moment of inertia of these two rods about a bisector (XY) of angle between the rods is

Options

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

Answer

A

Explanation

Moment of inertia of rods AB and CD about an axis passing through their point of intersection O and perpendicular to the plane of the rods is
`I_(O) = (ml^(2))/(12) + (ml^(2))/(12) = (ml^(2))/(6)`
In figure , XY and XY are two mutually perpendicular axes passing through O lying in the plane of the two rods . By symmetry `I_(XY) = I_(X. Y.) = I` , say
From the theorem of perpendicular axes `I_(XY) + I_(X. Y.) = I_(O)`
or `2I = (ml^2)/(6) or I = (ml^2)/(12) `