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Let's solve question 10.

### Problem Interpretation:
We have a U-tube manometer partially filled with water. Oil, which is less dense than water, is then poured into the left arm of the tube. The oil-water interface is situated at the midpoint of the U-tube. Both ends of the tube are open to the air, so the pressures at the top of both arms are equal to atmospheric pressure (\( P_0 \)).

We need to find the relationship between the heights \( h_{\text{agua}} \) (height of the water column) and \( h_{\text{aceite}} \) (height of the oil column).

### Variables and Known Data:
- \( \rho_{\text{agua}} \): Density of water.
- \( \rho_{\text{aceite}} \): Density of oil (less than water).
- \( h_{\text{agua}} \): Height of the water column on the right side.
- \( h_{\text{aceite}} \): Height of the oil column on the left side.

### Approach:
Since both ends are open to the air, the pressure at the top of each arm is equal to \( P_0 \). By balancing the pressures at the oil-water interface point, we can set up an equation to find the relationship between \( h_{\text{agua}} \) and \( h_{\text{aceite}} \).

#### Step-by-Step Solution:
1. **Pressure at the oil-water interface from the left side (oil side):**
   - The pressure at the oil-water interface (from the left) is:
     \[
     P_{\text{interface (left)}} = P_0 + \rho_{\text{aceite}} \cdot g \cdot h_{\text{aceite}}
     \]

2. **Pressure at the oil-water interface from the right side (water side):**
   - The pressure at the oil-water interface (from the right) is:
     \[
     P_{\text{interface (right)}} = P_0 + \rho_{\text{agua}} \cdot g \cdot h_{\text{agua}}
     \]

3. **Equating the pressures at the oil-water interface:**
   \[
   P_0 + \rho_{\text{aceite}} \cdot g \cdot h_{\text{aceite}} = P_0 + \rho_{\text{agua}} \cdot g \cdot h_{\text{agua}}
   \]
   Simplifying, we get:
   \[
   \rho_{\text{aceite}} \cdot h_{\text{aceite}} = \rho_{\text{agua}} \cdot h_{\text{agua}}
   \]

4. **Solving for the ratio of heights:**
   \[
   \frac{h_{\text{aceite}}}{h_{\text{agua}}} = \frac{\rho_{\text{agua}}}{\rho_{\text{aceite}}}
   \]

### Final Answer:
The relationship between the heights \( h_{\text{aceite}} \) and \( h_{\text{agua}} \) is:
\[
\frac{h_{\text{aceite}}}{h_{\text{agua}}} = \frac{\rho_{\text{agua}}}{\rho_{\text{aceite}}}
\]

This shows that the height of the oil column is inversely proportional to the ratio of the densities of water and oil.

Would you like further explanation on any step, or do you have questions about this solution?

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### Related Questions:
1. How would the result change if the manometer were closed at one end?
2. What would happen to the oil-water interface if oil had a density higher than water?
3. How does the presence of air at the top of each arm affect the pressure calculation?
4. If the tube were tilted at an angle, how would you approach the pressure balance?
5. What practical applications involve similar principles of pressure balance in fluids?

### Tip:
When dealing with U-tube manometers, remember that the pressure at any horizontal level in a connected fluid system must be equal on both sides of the U-tube.

Un tubo de manómetro se llena parcialmente con agua. Después se vierte aceite (que no se mezcla con el agua y tiene menor densidad que el agua) en el brazo izquierdo del tubo hasta que la interfaz aceite-agua está en el punto medio del tubo. Ambos brazos están abiertos al aire. Determine la relación entre las alturas h_agua y h_aceite.

Learn how to calculate the height relationship between water and oil columns in a U-tube manometer. This fluid mechanics problem explores concepts such as pressure equilibrium and density ratios, applying Pascal's Principle to determine the ratio of heights h_agua and h_aceite. Suitable for high school physics students.

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