Question

Two identical cylindrical vessels with their bases at the same level each
contain a liquid of density $1.30\times 10kg/m^3$. The area of each base is $4.00c{m}^2$,but in one vessel the liquid height is $0.854m$and in the other it is $1.560m$. Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

Answer 1

when the levels are the same, the height of the liquid is $h=(h_1+h_2)/2$, where $h_1$ and $h_2$ are the originals heights. Suppose $h_1$ is greater than $h_2$. The final situation can then be achieved by taking liquid with volume $A(h_1-h)$ and mass $\rho A(h_1-h)$, in the first vessel, and lowering it a distance $h-h_2$. The work done by the force of gravity is $W=\rho A(h_1-h)g(h-h_2)$. We substitute $h=(h_1+h_2)/2$ to obtain
$W=\frac{1}{4}\rho gA(h_1-h_2{)}^2=\frac{1}{4}(1.30\times {10}^{3}kg/m^3\left)\right(9.80m/s^2\left)\right(4.00\times {10}^{-4}{m}^2\left)\right(1.56m-0.854m{)}^2=0.635J$