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. `(h_1 -h_2) grho`
B. `(h_1 -h_2) gArho`
C. `1/2 (h_1-h_2)^2 gArho`
D. `1/4 (h_1-h_2)^2 gArho`

Answer

D

Explanation

`hrarr` Common height
Conservation of mass –
`rhoA_(1) h_1 +rhoA_(2) h_2 = rhoh(A_1+A_2)`
`implies h = ((h_1+h_2)/(2)) " "` [As `A_1=A_2 =A` ]
As `(h_1/2) rarr` initial C.G. `" "` [centre of gravity]
Similarly, `(h_2/2) rarr` initial C.G. `" "` [centre of gravity]
The initial potential energy
`U_i=(h_(1)Arho)g""(h_1)/2+(h_(2)Arho)gh_2/2=rhogA ((h_1^2+h_2^2))/(2)`
when vessels are connected –
`h/2 rarrC` Centre of gravity
i.e. `(((h_1+h_2))/4)` [ as `h=(h_1+h_2)//2`]
Potential energy –
`U_f = [((h_1+h_2)/2)Arho]g((h_1+h_2)/4)=Arhog[((h_1+h_2)^(2))/(4)]`
Work done by gravity –
`W=U_i-U_f=1/4 .rhogA[2h_1^2+2h_2^2-h_1^2-h_2^2-2h_1h_2]`
`W=1/4 rhogA(h_1-h_2)^2`