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

Explanation

Moment of inertia of a system about an axis 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_(1))[as I_(B_(1))=I_(B_(2))]`
`:. I_(B_(1))=I_(B_(2))=(I_(Z))/2=(ML^(2))/12`