Two cylindrical vessels of cross-sectional areas A and 3A are connected at their bottoms by a thin pipe with a closed valve. The narrower vessel contains a liquid of density up to a height 5h , and the wider vessel contains the same liquid up to a height h . When the valve is opened, the liquid levels equalize. The work done by the force of gravity during this process is

Step-by-step solution

Let the final equilibrium height of the liquid in both vessels be h_f . By conservation of volume: A(5h) + 3A(h) = (A + 3A)h_f 8Ah = 4A h_f h_f = 2h The volume of liquid that flows from the narrower vessel to the wider vessel is: V = A(5h) - A(2h) = 3Ah The mass of this transferred liquid is m = 3 Ah . In the narrower vessel, this liquid was originally located between heights 2h and 5h . Its initial centre of mass height is: y_i = 2h + 5h 2 = 3.5h In the wider vessel, this liquid occupies the space between heights