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

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 :

TestHub · Accepted Answer

Initial height of liquid in container's of same cross section are $x_{1}$ and $x_{2}$ respectively. Now value is opened find loss in potential energy when water level be become same

loss in $PE = U_{i} - U_{f}$

$= [ρ (A) x_{1} \frac{x_{1}}{2} + {ρAx}_{2} \frac{x_{2}}{2}] g - [ρA (\frac{x_{1} + x_{2}}{2}) \times (\frac{x_{1} + x_{2}}{4}) \times 2] g$

$= ρAg [\frac{x_{1}^{2}}{2} + \frac{x_{2}^{2}}{2} - \frac{{(x_{1} + x_{1})}^{2}}{4}] = \frac{ρAg {(x_{1} - x_{2})}^{2}}{4}$

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