Two cylindrical vessels of the same type contain a liquid of density rho. The bases of both the vessels lie on the same horizontal plane. The depth of the liquid in the left vessel is h_(1) and that in the right vessel is h_(2). The area of cross-section of the base of each vessel is A. If the vessels are connected by a tube, then how much is the work done by gravity to equalize the levels of the liquid in the vessels (suppose h_(1)gth_(2))?

Two cylindrical vessels of the same type contain a liquid of density r

1.	Two cylindrical vessels of the same type contain a liquid of density rho. The bases of both the vessels lie on the same horizontal plane. The depth of the liquid in the left vessel is h_(1) and that in the right vessel is h_(2). The area of cross-section of the base of each vessel is A. If the vessels are connected by a tube, then how much is the work done by gravity to equalize the levels of the liquid in the vessels (suppose h_(1)gth_(2))?
Answer» SOLUTION :When the vessels are connected, the level of the liquid in the left-hand VESSEL will fall and that in the right-hand one will go up. The initial difference in the liquid levels in the vessles = `h_(1)-h_(2)`. When the levels in both the vessels become the same, the level of liquid in the left-hand vessel will fall by `1/2(h_(1)-h_(2))` and that in the right-hand vessel will rise by `1/2(h_(1)-h_(2))`. MASS of this liquid = `1/2(h_(1)-h_(2))Arho`. The centre of gravity of this liquid moves up against gravity by `1/2(h_(1)-h_(2))`. `therefore` Increase in potential ENERGY of this liquid = work DONE against gravity `1/2(h_(1)-h_(2))Arhoxxgxx1/2(h_(1)-h_(2))=1/4Arhog(h_(1)-h_(2))^(2)`.