a rod of mass m and length l lies in the axis, with its centre at the origin. what is its moment of inertia about the line x=y?

dear reshab ,

first take elemental part dx anywhere on the rod & let the distance of this part from origin is x ..

angle bw rod & axis is @ = 45^o (in this case)

perpendicular distance bw that part & axis is r = xsin@

mass of this element = m/l (dx)

dI (inertia of this small part) = (dm) r²

                                        = (m/l) (xsin@)² dx

I = (msin²@/l) x²dx     lim from -l/2 to l/2

after integrating & putting limit we get

I = ml² / 24