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In this problem, we are dealing with a U-shaped manometer, which contains three different fluids: engine oil, water, and fluid "X". The objective is to determine the specific gravity of fluid X.

### Data from the diagram:
- Height of engine oil: \( h_{\text{engine oil}} = 10 \, \text{cm} \) (left side) and \( 9 \, \text{cm} \) (right side)
- Height of water: \( h_{\text{water}} = 7 \, \text{cm} \) (left side) and \( 5 \, \text{cm} \) (right side)
- Height of fluid X: \( h_{\text{X}} = 4 \, \text{cm} \) (left side) and \( 6 \, \text{cm} \) (right side)

### Key Concepts:
- Since both ends are open to atmospheric pressure, the pressures at the same horizontal level in the fluid must be equal. This principle is used to establish the balance of pressures exerted by the fluids in both arms.
- The pressure at a point in a liquid column is given by:
  \[
  P = \rho g h
  \]
  where \( \rho \) is the fluid density, \( g \) is the gravitational acceleration, and \( h \) is the height of the fluid column.

- The relationship between the densities of the fluids and their heights must be balanced at the same depth. Let the densities be as follows:
  - \( \rho_{\text{engine oil}} \) = \( 0.85 \, \rho_{\text{water}} \)
  - \( \rho_{\text{water}} \) = \( 1 \, \rho_{\text{water}} \)
  - \( \rho_X \) is the unknown density of fluid X.

### Applying pressure balance:
At the same horizontal level where fluid X is located, the pressures on both sides of the manometer must be equal:

1. **Left side:**
   \[
   P_{\text{left}} = \rho_{\text{engine oil}} g h_{\text{engine oil}} + \rho_{\text{water}} g h_{\text{water}} + \rho_X g h_X
   \]
   \[
   P_{\text{left}} = (0.85 \, \rho_{\text{water}}) g (10 \, \text{cm}) + (1 \, \rho_{\text{water}}) g (7 \, \text{cm}) + \rho_X g (4 \, \text{cm})
   \]

2. **Right side:**
   \[
   P_{\text{right}} = \rho_{\text{engine oil}} g h_{\text{engine oil}} + \rho_{\text{water}} g h_{\text{water}} + \rho_X g h_X
   \]
   \[
   P_{\text{right}} = (0.85 \, \rho_{\text{water}}) g (9 \, \text{cm}) + (1 \, \rho_{\text{water}}) g (5 \, \text{cm}) + \rho_X g (6 \, \text{cm})
   \]

Since \( g \) and \( \rho_{\text{water}} \) cancel out, we get the equation:

\[
(0.85 \times 10) + 7 + 4 \times S_X = (0.85 \times 9) + 5 + 6 \times S_X
\]

Where \( S_X = \frac{\rho_X}{\rho_{\text{water}}} \) is the specific gravity of fluid X.

Let's solve this equation.The specific gravity of fluid X is approximately 1.43.

Would you like more details or have any questions about the solution?

Here are some related questions to explore:
1. How would the solution change if the density of engine oil were different?
2. What if the manometer had a different fluid instead of water in the middle section?
3. How does the manometer work if only two fluids are present?
4. Can we determine the density of the unknown fluid in \( \text{kg/m}^3 \) if we know the density of water?
5. What would happen to the fluid levels if the atmospheric pressure were increased?

**Tip:** When using manometers, remember that the heights of the fluids are inversely proportional to their densities for the same pressure levels.

In the provided diagram, both ends of a U-shaped manometer are exposed to atmospheric pressure. Given the observed fluid levels in the two arms, estimate the specific gravity of the fluid labeled 'X.'

Learn how to estimate the specific gravity of an unknown fluid using a U-shaped manometer exposed to atmospheric pressure. This problem involves key fluid mechanics concepts such as pressure balance and the specific gravity formula. Suitable for advanced high school or early college students.

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