Question

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

• A. $A\rho g(h_1-h_2{)}^2$
• B. $A\rho g(h_1+h_2{)}^2$
• C. $A\rho g{\left(\frac{h_1-h_2}{2}\right)}^2$
• D. $A\rho g{\left(\frac{h_1+h_2}{2}\right)}^2$

Answer 1

The correct option is C

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

The work done by gravity equals the change in the potential energy of the system after the vessels are interconnected. We may regard the liquid in each vessel as equivalent to a point mass kept at their respective centres of gravity. Remembering that the mass of the liquid is given by $\left(Ah\rho \right)$ and the PE of a mass at a height h in earth’s gravity is mgh, we have

Total PE at start $=\left(A{h}_1\rho \right)g\frac{h_1}{2}+\left(A{h}_2\rho \right)g\frac{h_2}{2}\frac{A\rho g}{2}(h_1^2+h_2^2)$
After the vessels are connected, the height of liquid in each vessel is $\frac{(h_1+h_2)}{2}.$

$\begin{array}{cc}\text{Hence}& \text{PE after connection}=\left\{A\rho \left(\frac{h_1+h_2}{1}\right)g\left(\frac{h_1+h_2}{2}\right)\right\}\\ & =\frac{A\rho g}{4}(h_1+h_2{)}^2\\ \text{Change in PE}& \frac{A\rho g}{4}\left\{\right(h_1+h_2{)}^2-2(h_1^2+h_2^2)\}\\ & -\frac{A\rho g}{4}(h_1-h_2{)}^2\\ & =-A\rho g{\left(\frac{h_1-h_2}{1}\right)}^2\end{array}$

This must be equal to the work done ‘by’ gravity on the liquid. Thus the work done ‘by’ gravity is
$A\rho g{\left(\frac{h_1-h_2}{1}\right)}^2$

Hence the correct choice is (c).