Question

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

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

Answer 1

The correct option is C $\frac{1}{4}(h_1-h_2{)}^{2}gA\rho$

If $h$ is the common height when they are connected, by conservation of mass
$\rho A_1{h}_1+\rho A_2{h}_2=\rho h(A_1+A_2)$

$h=(h_1+h_2)/2$ [as $A_1=A_2=A$ given ]

As $\left(h_1/2\right)$ and $h_2/2$ are heights of initial centre of gravity of liquid in two vessels, the initial potential energy of the system

$U_i=\left(h_{1}A\rho \right)g\frac{h_1}{2}+\left(h_{2}A\rho \right)g\frac{h_2}{2}=\rho gA\frac{(h_1^2+h_2^2)}{2}$

When vessels are connected, the height of center of gravity of liquid in each vessel will be $h/2,$

i.e., $\frac{(h_1+h_2)}{4}$ [As $h=(h_1+h_2)/2]$

Final potential energy of the system

$U_f=\left[\frac{2(h_1+h_2)}{2}A\rho \right]g\left(\frac{h_1+h_2}{4}\right)$

$U_f=A\rho g\left[\frac{(h_1+h_2{)}^2)}{4}\right]$

Work done by gravity

$W=U_i-U_f$

$=\frac{1}{4}\rho gA\left[2\right(h_1^2+h_2^2)-(h_1+h_2{)}^2]$

$=\frac{1}{4}\rho gA(h_1-h_2{)}^2$