Question

Two identical cylindrical vessels with their bases at the same level; each contain a liquid of density $\rho$. The area of both is S, but in one vessel the height of liquid in one vessel is ${h_1}$ and in the other is ${h_2}$. The work done when both cylinders are connected, by gravity in equalizing the levels is
A. $\dfrac{1}{4}g\rho S{\left( {{h_2} - {h_1}} \right)^2}$
B. $g\rho S{\left( {{h_2} - {h_1}} \right)^2}$
C. $g\rho S\left( {{h_2} - {h_1}} \right)$
D. $\dfrac{1}{4}g\rho S\left( {{h_2} - {h_1}} \right)$

Answer 1

Consider that the liquids in both the cylinders will reach some equilibrium height h. The work done by both the cylinders in reaching that height will be equal to the work done by gravity. Gravitational force will balance the level of the fluids. Find this work done on each cylinder, add them for the final answer.

Complete step by step answer:
Fluids have a property that they reach at an equilibrium height irrespective of the volume, area or cross section of their container. In this case too, the liquid will reach some height ‘h’ when the equilibrium is obtained. We have to find the work done by both the cylinders. This work done will be in the form of change in potential energy as follows:
Let us consider that ${h_2} > {h_1}$ so the level of water in ${h_2}$ will decrease and the level of water in other vessels will increase.
Work done by first cylinder and second cylinder will be:
${W_{cylinder1}} = S\rho g\dfrac{{{h_1}}}{2}({h_1} - h)$
${W_{cylinder2}} = S\rho g\dfrac{{{h_2}}}{2}({h_2} - h)$
But this work is done by gravity, so the total work done by gravity will be:
${W_{gravity}} = {W_{cylinder1}} + {W_{cylinder2}}$
$\Rightarrow {W_{gravity}} = S\rho g\dfrac{{{h_1}}}{2}({h_1} - h) + S\rho g\dfrac{{{h_2}}}{2}({h_2} - h)$
Also, we know that, the volume of water must be conserved, this implies:
$h = \dfrac{{{h_1} + {h_2}}}{2}$
Substituting this value in above equation, we have
${W_{gravity}} = S\rho g\dfrac{{{h_1}}}{2}({h_1} - \left( {\dfrac{{{h_1} + {h_2}}}{2}} \right)) + S\rho g\dfrac{{{h_2}}}{2}({h_2} - \left( {\dfrac{{{h_1} + {h_2}}}{2}} \right))$
$\Rightarrow {W_{gravity}} = S\rho g{\left( {\dfrac{{{h_1} - {h_2}}}{2}} \right)^2}$
Therefore, the work done when both cylinders are connected, by gravity in equalizing the levels is $\dfrac{1}{4}S\rho g{\left( {{h_1} - {h_2}} \right)^2}$

So, the correct answer is “Option A”.

Note:
We need to consider that one height is greater than another as it is not mentioned in the question. If both the heights were equal, work done by gravity will be zero as there will be no movement in the fluid. Mass can also be expressed in terms of density and volume.A liquid can have a free surface at which the pressure is constant and equal to the gas or vapor pressure over it. An essential difference between liquids and gases lies in their compressibility.