Question

A U-tube of cross section $A$ and $2A$ contains liquid of density $\rho$. Initially, the liquid in the two arms are held with a level difference $h$ as shown in figure. After being released, the levels equalize after some time. The work done by gravity forces on the liquid in the process is:

• A. $\frac{2}{3}\rho Ag{h}^2$
• B. $\frac{\rho Ag{h}^2}{3}$
• C. $\frac{\rho Ag{h}^2}{6}$
• D. $\frac{3}{4}\rho Ag{h}^2$

Answer 1

The correct option is B $\frac{\rho Ag{h}^2}{3}$

Let $h_c=$ common height attained due to gravity



Then from equating the volumes, we have,
$A(H+h)+2AH=(A+2A)h_c$

$\therefore$$h_c=H+\frac{h}{3}$

Initial potential energy per unit density of liquid,

$U_i=A(H+h)g\left(\frac{H+h}{2}\right)+2AHg\frac{H}{2}$

$=Ag[\frac{(H+h{)}^2}{2}+H^2]$

Final potential energy per unit density of the liquid,

$U_f=3A{h}_{c}g\frac{h_c}{2}=\frac{3}{2}Ag{(H+\frac{h}{3})}^2$

$\therefore$ Work done by gravity $=-(U_f-U_i)$

$=-(\frac{3}{2}Ag{(H+\frac{h}{3})}^2-Ag[\frac{(H+h{)}^2}{2}+H^2])$

$=\frac{\rho Ag{h}^2}{3}$

Hence, option (b) is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{This question introduces the concept of hydrostatic equilibrium.}\end{array}$