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

Options

A. `(Ml^(2))/(6)`
B. `(Ml^(2))/(12)`
C. `(Ml^(2))/(3)`
D. `(Ml^(2))/(4)`

Answer

B

Explanation

Moment of inertia of system about an axes which is perpendicular to plane of rods and passing through the common centre of rods
`I_(z)=(Ml^(2))/(12)+(Ml^(2))/(12)=(Ml^(2))/(6)`
Again from perpendicular axes theorem `I_(z)=I_(B_(1))+I_(B_(2))=2I_(B_(1))=2I_(B_(2))=(Ml^(2))/(6)["As "I_(B_(1))=I_(B_(2))]`
`therefore I_(B_(1))=I_(B_(2))=(Ml^(2))/(12)`.