Correct Answer is: (d) Wh/4
The liquid levels will equalize at Q and R. This is equivalent to the section PQ moving down to the position RS.
Mass of section PQ = ρAh/ 2.
The centre of mass of this section moves down through a distance h/2.
Loss in potential energy
= (1/2 ρAh) g. (h/2) = 1/2 (ρAgh) h = 1/4 Wh
= work done by gravity forces on the liquid.
