5. (15 points) Two identical vessels of height $H$ are connected by a thin tube with a valve $K$ at a height of $0.15 H$. 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/m³, and the density of gasoline is $600 \mathrm{kg} / \mathrm{m}^{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-09.jpg?height=320&width=465&top_left_y=1248&top_left_x=881)
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 tube level:

$\rho_{\text {water }} g(0.75 H-x)=\rho_{\text {gasoline }} g(0.75 H+x)$,

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

We get $x=0.1875 H$, 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: $\rho_{\text {water }} g(0.75 H-x)=\rho_{\text {gasoline }} g \cdot 0.75 H$.

Third: the water reaches the tube and part of the gasoline spills out. (1 point) The equilibrium condition in this case:

$\rho_{\text {water }} g(0.75 H-x)=\rho_{\text {water }} g(x-0.15 H)+\rho_{\text {gasoline }} g(H-x)$.

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.