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?

Answer 1

Step 1: Write the given data.

Two identical cylindrical vessels with density $\rho$.

$h_1$is the height of liquid in one vessel.

$h_2$ is the height of the other vessel.

$A$ is the area of the base.

Step 2: Write the work done formula.

$\mathit{W}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{g}\mathbf{∆}\mathit{h}$

Where,

$m$ is the mass of the liquid.

$∆h$is the height of the liquid.

Step 3: To find the height of the liquid.

Let us assume that the height is $h$

$A$ is an area of cross-section.

The height of the liquid is given by: $\mathit{h}\mathbf{=}\frac{\left(h_1+h_2\right)}{\mathbf{2}}$

Hence decrease in height in a vessel height $h_1$is given by,
$\begin{array}{rcl}∆h& =& h_1=\frac{\left(h_1+h_2\right)}{2}\\ ∆h& =& \frac{2h_1-h_1-h_2}{2}\\ ∆h& =& \frac{h_1-h_2}{2}\end{array}$

Step 4: Find the mass of the liquid.

Mass of liquid would be equal to $\mathit{m}\mathbf{=}\mathbf{∆}\mathit{h}\mathit{\rho }\mathit{A}$

Substitute $∆h$in the above formula,

$\mathit{m}\mathbf{=}\left(\frac{h_1-h_2}{2}\right)\mathit{\rho }\mathit{A}$

Step 5: Solve for work done.

$\mathit{W}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{g}\mathbf{∆}\mathit{h}$

$m$ is the mass of the liquid.

$∆h$ is the height of the liquid.

Replace the above data in the work done formula,

$\begin{array}{rcl}W& =& \left(\frac{h_1+h_2}{2}\right)g\rho A\left(\frac{h_1-h_2}{2}\right)\\ W& =& {\left(\frac{h_1-h_2}{2}\right)}^{2}g\rho A\end{array}$

Therefore the work done by gravity in equalising the levels when the vessels are interconnected is ${\left(\frac{h_1-h_2}{2}\right)}^{\mathbf{2}}\mathit{g}\mathit{\rho }\mathit{A}$.