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 heigth 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

If h is the common height when they are connected, by conservation of mass
`rho A_(1)h_(1)+rho A_(2)h_(2)=rho h(A_(1)+A_(2))`
`h=(h_(1)+h_(2))//2 " "`[as `A_(1)=A_(2)=A` given]
As (h/2) and (h/2) are heights of initial centre of gravity of liquid in two vessels., the initial potential energy of the system
`U_(i)=(h_(1)A rho)g.(h_(1))/(2)+(h_(2)A rho)(h_(2))/(2)=rho gA((h_(1)^(2)+h_(2)^(2)))/(2) " "` ....(i)
When vessels are connected the height of centre of gravity of liquid in each vessel will be h/2,
i.e. `((h_(1)+h_(2)))/(4) ["as "h=(h_(1)+h_(2))//2]`
Final potential energy of the system
`U_(F)=[((h_(1)+h_(2)))/(2)A rho]g ((h_(1)+h_(2))/(4))`
`= A rho g[((h_(1)+h_(2))^(2))/(4)] " "` ......(ii)
Work done by gravity
`W = U_(i)-U_(f)=(1)/(4)rho gA[2(h_(1)^(2)+h_(2)^(2))-(h_(1)+h_(2))^(2)]`
`= (1)/(4)rho gA (h_(1)~ h_(2))^(2)`