Two cylindrical vessels of equal cross-sectional area A contain water up to heights `h_(1)` and `h_(2)`. The vessels are interconnected so that the levels in them are equal.Calculate the work done by the force of gravity during the process. The density of water is p.

Options

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

Answer

D

Explanation

If h is the common height when they are connected, by conservation of mass
`rhoA_1h_1rhoA_2h_2=rhoh(A_1+A_2)`
`h=((h_1+h_2))/(2)`
As `((h_1)/(2))` and `((h_2)/(2))` are heights of initial centre of gravity of liquid in two vassels, the initial potential energy of the system
`U_i=(h_1Arho)g(h_1)/(2)+(h_2Arho)(h_2)/(2)=rhogA((h_1^2+h_2^2))/(2)`
When vessels are connected the height of centre of gravity of liquid in each vassel will be `(b)/(2)`,
i.e. `((h_1+h_2))/(4)` [as `h=((h_1+h_2))/(2))`]
`U_F=[((h_1+h_2))/(2)Arho]g((h_1+h_2)/(4))`
`=Arhog[((h_1+h_2)^2)/(4)`
`W=U_i-U_f=(1)/(4)rhogA[2(h_1^2+h_2^2)-(h_1+h_2)^2]`
`=(1)/(4)rhogA(h_1approxh_2)^2`