The moment of inertia of a rod of length L about an axis passing through its center of mass and perpendicular to rod is I. The moment of inertia of hexagonal shape formedby six such rods, about an axis passing through its center of mass and perpendicular to its plane will be

Moment of inertia of rod AB about its center and perpendicular to the length =mL212=I⇒mL2=12INow moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of  hexagon Irod =mL212+mx2[From the theorem of parallel axes]=mL212+m32L2=5mL26Now the moment of inertia of system,Isystem =6×Irod =6×5mL26=5mL2Isystem =5(12I)=60I  As mL2=12I