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?

Options

A. `2 rho Ag(h_(2)-h_(1))^(2)`
B. `rho Ag(h_(2)-h_(1))^(2)`
C. `(1)/(2) rho Ag(h_(2)-h_(1))^(2)`
D. `(1)/(4) rhoAg(h_(2)-h_(1))^(2)`

Answer

D

Explanation

Common height after they are connected can be determined by equating the volume. Hence
`(A_(1)+A_(2))h=A_(1)h_(1)+A_(2)h_(2) " " (A_(1)=A_(2)=A)`
`therefore " " h=(h_(1)+h_(2))/(2)`
Work done by gravity `= - Delta U = U_(i)-U_(f)`
`={Ah_(1) rho. g. (h_(1))/(2)+Ah_(2)rho.g.(h_(2))/(2)}-A(h_(1)+h_(2))rho. g. ((h_(1)+h_(2)))/(4)`
`=(rho Ag)/(4)(h_(2)-h_(1))^(2)`