Question

A liquid of density $\rho$ is filled in two identical cylindrical vessels with base area $A$ and height of the liquid in one vessel is $h_1$ and that in the other vessel is $h_2$ . If both the vessels are connected with their base at the same level then calculate the work done by gravity in equalizing the levels.

• A. $\left(h_1-h_2\right)g\rho$
• B. $\left(h_1-h_2\right)gA\rho$
• C. $\frac{1}{2}{\left(h_1-h_2\right)}^{2}gA\rho$
• D. $\frac{1}{4}{\left(h_1-h_2\right)}^{2}gA\rho$

Answer 1

Answer: (D)

Here $h$ is the average of $h_1$ and $h_2$
So $h=\frac{h_1+h_2}{2}$
And
$\Delta h=h_1-h=h_1-\left(\frac{h_1+h_2}{2}\right)=\left(\frac{h_1-h_2}{2}\right)$
Mass of liquid displaced is
$\rho =\frac{m}{V}m=\rho Vm=\rho \times A\times \Delta hm=\rho \times A\times \left(\frac{h_1-h_2}{2}\right)$
where $\rho =$ density of liquid
$A=$ area of vessel
$V=$ volume of liquid displaced
$W=-\Delta UW=mg\Delta hW=\rho gA\times \left(\frac{h_1-h_2}{2}\right)\times \left(\frac{h_1-h_2}{2}\right)W=\frac{\rho gA}{4}{\left(h_1-h_2\right)}^2$