Question

Two identical cylindrical vessels with their bases at same level each contains a liquid of density ρ. The height of the liquid in one vessel is h₁ 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. $\frac{1}{2}(h_1-h_2{)}^{2}gA\rho$
• D. $\frac{1}{4}(h_1-h_2{)}^{2}gA\rho$

Answer 1

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

If h is the common height when they are connected, by conservation of mass
$\rho A_1{h}_1+\rho A_2{h}_2=\rho h(A_1+A_2)$
$h=\frac{(h_1+h_2)}{2}[as{A}_1=A_2=Agiven]$
As $\left(\frac{h_1}{2}\right)$ and $\left(\frac{h_2}{2}\right)$ are heights of initial centre of gravity of liquid in two vessels, the initial potential energy of the system
$U_i=\left(h_{1}A\rho \right)g\frac{h_1}{2}+\left(h_{2}A\rho \right)\frac{h_2}{2}=\rho gA\frac{(h_1^2+h_2^2)}{2}....\left(i\right)$
When vessels are connected the height of centre of gravity of liquid in each vessel will be $\frac{h}{2}$,
i.e. $\frac{(h_1+h_2)}{4}[ash=\frac{(h_1+h_2)}{2}]$
Final potential energy of the system
$U_f=\left[\frac{(h_1+h_2)A\rho }{2}\right]g(\frac{h_1+h_2}{4})=A\rho g\left[\frac{(h_1+h_2{)}^2}{4}\right]....\left(ii\right)$
Work done by gravity
$W=U_i-U_f=\frac{1}{4}\rho gA\left[2\right(h_1^2+h_2^2)-(h_1+h_2{)}^2]$
$=\frac{1}{4}\rho gA(h_1\sim h_2{)}^2$