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

Options

A. `(1)/(2)(h_(1)-h_(2))^(2)Arhog`
B. `(1)/(2) (h_(1)+h_(2))^(2)Arhog`
C. `(1)/(2) (h_(1)^(2)-h_(2)^(2))^(2)Arhog`
D. `(1)/(4) (h_(1)-h_(2))^(2)Arhog`

Answer

D

Explanation

Let `h=`final height of liquid
Equanting the volume we have
`Ah_(1)+Ah_(2)=A(2h)`
`:. h=(h_(1)+h_(2))/(2)`
`W_(mg)=-Delta U=U_(i)-U_(f)`
`= [(Ah_(1)rhog.(h_(1))/(2))+(Ah_(2)rhog.(h_(2))/(2))]`
`-[2A((h_(1)+h_(2))/2)rho((h_(1)+h_(2))/4)]`
Simplifyig this expression we get the result.