Question

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d$. The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x_1$ and in the other, $x_2$. When both cylinders are connected through a pipe of negligible volume, very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:

• A. $gdS(x_2^2+x_1^2)$
• B. $gdS(x_2+x_1{)}^2$
• C. $\frac{3}{4}gdS(x_2-x_1{)}^2$
• D. $\frac{1}{4}gdS(x_2-x_1{)}^2$

Answer 1

The correct option is D $\frac{1}{4}gdS(x_2-x_1{)}^2$


Initial potential energy,

$U_i=\left(\rho S{x}_1\right)g.\frac{x_1}{2}+\left(\rho S{x}_2\right)g.\frac{x_2}{2}$

$=\frac{\rho Sg{x}_1^2}{2}+\frac{\rho Sg{x}_2^2}{2}$

Final potential energy,
$U_f=\left(\rho S{x}_f\right)g.\frac{x_f}{2}\times 2=\frac{\rho Sg{x}_f^2}{2}$

By volume conservation,
$S{x}_1+S{x}_2=S\left(2x_f\right)$

$x_f=\frac{x_1+x_2}{2}$

When the valve is opened, change in potential energy will occur till the water level become same, on both the sides.

Now, $\Delta U=U_i-U_f$

$\Delta U=\rho Sg[(\frac{x_1^2}{2}+\frac{x_2^2}{2})-x_f^2]$

$=\rho Sg[\frac{x_1^2}{2}+\frac{x_2^2}{2}-{\left(\frac{x_1+x_2}{2}\right)}^2]$

$=\frac{\rho Sg}{2}[\frac{x_1^2}{2}+\frac{x_2^2}{2}-x_1{x}_2]$

$=\frac{\rho Sg}{4}(x_1-x_2{)}^2$

Hence, $\left(D\right)$ is the correct answer.