Two identical cylindrical vessels are kept on the ground and each… - Vidyadip

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

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Answer

$U _{ i }=\left(\rho Sx _{1}\right) g \cdot \frac{ x _{1}}{2}+\left(\rho Sx _{2}\right) g \cdot \frac{ x _{2}}{2}$

$U _{ f }=\left(\rho Sx _{ f }\right) g \cdot \frac{ x _{ f }}{2} \times 2$

By volume conservation

$Sx _{1}+ Sx _{2}= S \left(2 x _{ f }\right)$

$x_{f}=\frac{x_{1}+x_{2}}{2}$

$\Delta U =\rho \operatorname{Sg}\left[\left(\frac{ x _{1}^{2}}{2}+\frac{ x _{2}^{2}}{2}\right)- x _{ f }^{2}\right]$

$=\rho \operatorname{Sg}\left[\frac{ x _{1}^{2}}{2}+\frac{ x _{2}^{2}}{2}-\left(\frac{ x _{1}+ x _{2}}{2}\right)^{2}\right]$

$=\frac{\rho Sg }{2}\left[\frac{ x _{1}^{2}}{2}+\frac{ x _{2}^{2}}{2}- x _{1} x _{2}\right]$

$=\frac{\rho Sg }{4}\left( x _{1}- x _{2}\right)^{2}$

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