Question

a long rod ABC of mass m and length l has two particles of masses m and 2m attached to it as shown in the fig. the system is initially in the horizontal position the work done it keep it verticalily with A is at the bottom is

Answer 1

Here when we make this rod verticle with A at its bottom , C will raise to top along with the mass 2m

I have attached the diagrams of before and after the rotation condition

Work done = change in potential energy

Assuming the horizontal condition to be a reference and potential energy is condidered zero at the reference. 
So change in potential energy of the 
2m mass
= 2mgl
m mass
=mgl/2

and for the rod lets assume that whole mass is centered at centere of mass at l/2
so change in potential energy
=mgl/2

net change in potential energy
= 2mgl+mgl/2+mgl/2
=3mgl

Answer 2

the work done is mgl is the right answer