Question

Two identical cylindrical vessels, with their bases at the same level, each contain a liquid of density $\rho$. The height of liquid in one vessel in $h_1$ and that in the other is $h_2$. The area of either base is $A$. What is the work done by gravity in equalising the levels when the vessels are interconnected?

• A. $A\rho g{\left(\frac{h_1-h_2}{2}\right)}^2$
• B. $A\rho g{\left(\frac{h_1+h_2}{2}\right)}^2$
• C. $\frac{1}{2}A\rho g{(h_1-h_2)}^2$
• D. None of these

Answer 1

The correct option is A $A\rho g{\left(\frac{h_1-h_2}{2}\right)}^2$

The figure can be shown diagrammatically as shown here.
Let us assume that the height is $h$
So, $h=\left(\frac{h_1+h_2}{2}\right)$
Hence decrease in height in vessel of height $h_1$
$\Delta h=h_1-\left(\frac{h_1+h_2}{2}\right)=\left(\frac{h_1-h_2}{2}\right)$
Mass of liquid would be equal to
$m=\left(\frac{h_1-h_2}{2}\right)\rho A$
Thus we can find the work done equal to
$W=mg\Delta h=\left(\frac{h_1-h_2}{2}\right)\rho A\left(\frac{h_1-h_2}{2}\right)={\left(\frac{h_1-h_2}{2}\right)}^{2}g\rho A$