A uniform rod of mass 6M and length 6l is bent to make an equilateral hexagon. Its M.I. about an axis passing through the centre of mass and perpendicular to the plane of hexagon is

The moment of inertia (I) of a uniform rod about an axis through its center and perpendicular to its length is given by the formula I = Ml²/12, where M is the mass of the rod and l is the length of the rod.

In this case, the rod is bent into an equilateral hexagon. This means the rod is divided into 6 equal parts, each with mass M and length l.

The moment of inertia of each part about the center of the hexagon can be calculated using the parallel axis theorem, which states that the moment of inertia about any axis parallel to and a distance d away from an axis through the center of mass is given by I = Icm + Md², where Icm is the moment of inertia about the center of mass axis.

For each part of the hexagon, Icm = Ml²/12 and d = l√3/2 (the distance from the center of the hexagon to the midpoint of a side).

So, the moment of inertia of each part about the center of the hexagon is I = Ml²/12 + M(l√3/2)² = Ml²/12 + 3Ml²/4 = 5Ml²/6.

Since there are 6 parts, the total moment of inertia of the hexagon about the center is 6 * 5Ml²/6 = 5Ml².