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| ### Statement |
| > Historical Note: This exact problem was featured at the 2015 International Zhautykov Olympiad (IZhO). It was the first question of Problem 2 and was awarded 2 points. |
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| $5.5.19.$ [Insert the problem statement] |
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| ### Solution |
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| > **Historical Note:** This exact problem was featured at the **2015 International Zhautykov Olympiad (IZhO)**. It was the first question of Problem 2 and was awarded 2 points. |
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| ### Statement |
| ### Statement |
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| $5.5.19.$ In a cylindrical vessel of height $H$, a thin-walled tube of length $\ell$ is inserted vertically through the lid, slightly not reaching the bottom of the vessel. The connection between the lid, the vessel, and the tube is airtight. Liquid is poured into the vessel through the tube. Find the height of the liquid level from the bottom of the vessel when the tube is completely filled with liquid. Atmospheric pressure is $P_0$, and the density of the liquid is $\rho$. |
| $5.5.19.$ In a cylindrical vessel of height $H$, a thin-walled tube of length $\ell$ is inserted vertically through the lid, slightly not reaching the bottom of the vessel. The connection between the lid, the vessel, and the tube is airtight. Liquid is poured into the vessel through the tube. Find the height of the liquid level from the bottom of the vessel when the tube is completely filled with liquid. Atmospheric pressure is $P_0$, and the density of the liquid is $\rho$. |
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| ### Solution |
| ### Solution |
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| Let $S$ be the cross-sectional area of the cylindrical vessel, and $S_0$ be the cross-sectional area of the thin-walled tube. According to the problem setup, the tube is thin, which implies $S_0 \ll S$. Because of this condition, we can neglect the initial volume of air inside the tube and assume that all the trapped air is initially located entirely within the main volume of the vessel. |
| Let $S$ be the cross-sectional area of the cylindrical vessel, and $S_0$ be the cross-sectional area of the thin-walled tube. According to the problem setup, the tube is thin, which implies $S_0 \ll S$. Because of this condition, we can neglect the initial volume of air inside the tube and assume that all the trapped air is initially located entirely within the main volume of the vessel. |
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| **Initial State:** |
| Initial State: |
| Before the liquid is poured, the pressure of the air inside the vessel is equal to the atmospheric pressure: |
| Before the liquid is poured, the pressure of the air inside the vessel is equal to the atmospheric pressure: |
| $$P_1 = P_0$$ |
| $$P_1 = P_0$$ |
| The initial volume of this air is: |
| The initial volume of this air is: |
| $$V_1 = S \cdot H$$ |
| $$V_1 = S \cdot H$$ |
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| **Final State:** |
| Final State: |
| When the tube is completely filled with liquid (the liquid column in the tube has height $\ell$), let the height of the liquid level in the vessel be $h$. |
| When the tube is completely filled with liquid (the liquid column in the tube has height $\ell$), let the height of the liquid level in the vessel be $h$. |
| The trapped air now occupies a smaller volume: |
| The trapped air now occupies a smaller volume: |
| $$V_2 = S \cdot (H - h)$$ |
| $$V_2 = S \cdot (H - h)$$ |
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| To find the new pressure of the trapped air ($P_2$), we use the hydrostatic pressure balance. The pressure at the very bottom of the vessel must be the same everywhere. | | To find the new pressure of the trapped air ($P_2$), we use the hydrostatic pressure balance. The pressure at the very bottom of the vessel must be the same everywhere. |
| Looking from the tube's side, the pressure at the bottom is atmospheric pressure plus the hydrostatic pressure of the liquid column of length $\ell$: | | Looking from the tube's side, the pressure at the bottom is atmospheric pressure plus the hydrostatic pressure of the liquid column of length $\ell$: |
| $$P_{\text{bottom}} = P_0 + \rho g \ell$$ | | $$P_{\text{bottom}} = P_0 + \rho g \ell$$ |
| Looking from the trapped air's side, the pressure at the bottom is the gas pressure $P_2$ plus the hydrostatic pressure of the liquid layer of height $h$: | | Looking from the trapped air's side, the pressure at the bottom is the gas pressure $P_2$ plus the hydrostatic pressure of the liquid layer of height $h$: |
| $$P_{\text{bottom}} = P_2 + \rho g h$$ | | $$P_{\text{bottom}} = P_2 + \rho g h$$ |
| Equating these two expressions, we find the pressure of the trapped gas: | | Equating these two expressions, we find the pressure of the trapped gas: |
| $$P_0 + \rho g \ell = P_2 + \rho g h \implies P_2 = P_0 + \rho g (\ell - h)$$ | | $$P_0 + \rho g \ell = P_2 + \rho g h \implies P_2 = P_0 + \rho g (\ell - h)$$ |
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| Assuming the process is isothermal ($T = \text{const}$), we apply Boyle's Law for the trapped air: | | Assuming the process is isothermal ($T = \text{const}$), we apply Boyle's Law for the trapped air: |
| $$P_1 V_1 = P_2 V_2$$ | | $$P_1 V_1 = P_2 V_2$$ |
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| Substitute the expressions for pressures and volumes: | | Substitute the expressions for pressures and volumes: |
| $$P_0 \cdot S \cdot H = [P_0 + \rho g (\ell - h)] \cdot S \cdot (H - h)$$ | | $$P_0 \cdot S \cdot H = [P_0 + \rho g (\ell - h)] \cdot S \cdot (H - h)$$ |
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| We can cancel the cross-sectional area $S$ from both sides: | | We can cancel the cross-sectional area $S$ from both sides: |
| $$P_0 H = (P_0 + \rho g \ell - \rho g h)(H - h)$$ | | $$P_0 H = (P_0 + \rho g \ell - \rho g h)(H - h)$$ |
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| Now, let's expand the brackets: | | Now, let's expand the brackets: |
| $$P_0 H = P_0 H - P_0 h + \rho g \ell H - \rho g \ell h - \rho g h H + \rho g h^2$$ | | $$P_0 H = P_0 H - P_0 h + \rho g \ell H - \rho g \ell h - \rho g h H + \rho g h^2$$ |
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| Cancel $P_0 H$ from both sides and rearrange the equation into a standard quadratic form with respect to $h$: | | Cancel $P_0 H$ from both sides and rearrange the equation into a standard quadratic form with respect to $h$: |
| $$\rho g h^2 - h(P_0 + \rho g \ell + \rho g H) + \rho g \ell H = 0$$ | | $$\rho g h^2 - h(P_0 + \rho g \ell + \rho g H) + \rho g \ell H = 0$$ |
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| Divide the entire equation by $\rho g$: | | Divide the entire equation by $\rho g$: |
| $$h^2 - h \left( \ell + H + \frac{P_0}{\rho g} \right) + \ell H = 0$$ | | $$h^2 - h \left( \ell + H + \frac{P_0}{\rho g} \right) + \ell H = 0$$ |
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| Solving this quadratic equation for $h$, we get two mathematical roots: | | Solving this quadratic equation for $h$, we get two mathematical roots: |
| $$h = \frac{1}{2} \left[ \ell + H + \frac{P_0}{\rho g} \pm \sqrt{ \left( \ell + H + \frac{P_0}{\rho g} \right)^2 - 4\ell H } \right]$$ | | $$h = \frac{1}{2} \left[ \ell + H + \frac{P_0}{\rho g} \pm \sqrt{ \left( \ell + H + \frac{P_0}{\rho g} \right)^2 - 4\ell H } \right]$$ |
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| Since the height of the liquid $h$ must be physically less than the height of the vessel $H$, taking the "$+$" sign would result in an unphysical root ($h > H$). Therefore, the only valid physical solution corresponds to the "$-$" sign. | | Since the height of the liquid $h$ must be physically less than the height of the vessel $H$, taking the "$+$" sign would result in an unphysical root ($h > H$). Therefore, the only valid physical solution corresponds to the "$-$" sign. |
