Question

Two identical cylindrical vessel with their bases at the same level each contain a liquid of density `rho`. The height of the liquid in one vessel is `h_(1)` and in the other is `h_(2)` the area of either base is A. What is the work done by gravity is equalising the levels when the two vessels are connected?

Answer 1

Correct Answer - `(gArho)/(4)(h_(t) - h_(2))^(2)`
Let `h` be level in equilibrium. Equating the volumes, we have
`Ah_(1) + Ah_(2) = 2Ah`
`:. H = ((h_(1) + h_(2))/(2))`
Work done by gravity `= U_(i) - U_(f)`
Work `W = (m_(1)gh_(1)/(2) + m_(2)gh_(2)/(2)) - (m_(1) + m_(2))g h/(2)`
`= (Ah_(1)rhogh_(1))/(2) + (Ah_(2)rhogh_(2))/(2) - [Ah_(1) - Ah_(2)rho + Ah_(2)rholg((h_(1) + h_(2))/(4))`
Simplifying this, we get
`W = (rhoAg)/(4)(h_(1) - h_(2))^(2)`