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?

{} \\\ \left( A \right)A\rho g{{\left( \dfrac{{{h}_{1}}-{{h}_{2}}}{2} \right)}^{2}} \\\ {} \\\ \left( B \right)A\rho g{{\left( \dfrac{{{h}_{1}}+{{h}_{2}}}{2} \right)}^{2}} \\\ {} \\\ \left( C \right)\dfrac{1}{2}A\rho g{{\left( {{h}_{1}}-{{h}_{2}} \right)}^{2}} \\\ {} \\\ \left( D \right)None\text{ }of\text{ }these \\\ \end{array}$$

Answer 1

At first find the average height of the liquid filled in the vessel by taking the average of both liquid column heights, then find the mass of liquid and apply the formula for work done and put the value of calculated average height and mass.

Formula used: Work done: $W=mg\Delta h$
$m$- mass of liquid
$h$- average height of liquid
$g$- acceleration due to gravity

Complete step by step answer:
Let us assume that the height is $h$
So, $h=\left( \dfrac{{{h}_{1}}+{{h}_{2}}}{2} \right)$
Hence decrease in height in vessel of height ${{h}_{1}}$
$\Delta h={{h}_{1}}-\left( \dfrac{{{h}_{1}}+{{h}_{2}}}{2} \right)=\left( \dfrac{{{h}_{1}}-{{h}_{2}}}{2} \right)$
Mass of liquid would be equal to
$m=\left( \dfrac{{{h}_{1}}-{{h}_{2}}}{2} \right)\rho A$
Thus we can find the work done equal to

& W=mg\Delta h=\left( \dfrac{{{h}_{1}}-{{h}_{2}}}{2} \right)\rho gA\left( \dfrac{{{h}_{1}}-{{h}_{2}}}{2} \right) \\\ & W={{\left( \dfrac{{{h}_{1}}-{{h}_{2}}}{2} \right)}^{2}}g\rho A \\\ \end{aligned}$$ **So, the correct answer is “Option A”.** **Additional Information:** When the body is at rest above the surface of the world at finite height, if some external force is applied on the body, therefore the work performed by the gravity of the world to bring the body to the surface is termed as work done by the gravity. **Note:** We should calculate the average height of both the columns because when we apply the pressure on one then the liquid level rises in another column. So, We have to maintain the average height to calculate the work done. And it should always be noticed that the total mass is often alleged to be concentrated at the middle of the filled part.