5. (15 points) Two identical vessels of height $H$ are connected by a thin tube with a valve $K$ at a height of $0.2 H$. Water is poured into the left vessel, and gasoline into the right vessel. The levels of the liquids are the same and equal to $0.9 H$. Determine the levels of the liquids in the vessels that will be established after opening the valve. The density of water is 1000 kg $/ \mathrm{m}^{3}$, and the density of gasoline is 600 kg $/ \mathrm{m}^{3}$.

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Answer: $h_{\text {left }}=0.69 H ; h_{\text {right }}=H$.

Solution. When the tap is opened, water will begin to flow into the right vessel. Several situations are possible.

First: the water will not reach the tube, and the gasoline will not spill out of the vessel.

The equilibrium condition at the level of the tube: $\rho_{\text {water }} g(0.7 H-x)=\rho_{\text {gasoline }} g(0.7 H+x)$,

where $x$ is how much the water level in the left vessel has decreased.

We get, $x=0.175 H$, which means this situation does not work.

Second: the water does not reach the tube in the right vessel, but part of the gasoline spills out.

The equilibrium condition in this case: $\rho_{\text {water }} g(0.7 H-x)=\rho_{\text {gasoline }} g \cdot 0.8 H$. (1 point)

We get, $x=0.22 H$, which means this situation does not work either.

Third: the water reaches the tube and part of the gasoline spills out. (1 point)

The equilibrium condition in this case: $\rho_{\text {water }} g(0.7 H-x)=\rho_{\text {water }} g(x-0.2 H)+$ $\rho_{\text {gasoline }} g(H-x)$.

We get, $x=\frac{3}{14} H$.

The level in the left vessel: $h_{\text {left }}=0.9 H-\frac{3}{14} H=\frac{24}{35} H \approx 0.69 H$.

The level in the right tube: $h_{\text {right }}=H$.

Note. All possible situations should be considered in the solution.