To determine the depth of the fluid in the cylindrical container, we use the relationship between pressure, fluid density, gravity, and depth:

$P_{\text{absolute}} = P_{\text{atmospheric}} + \rho g h$

Where:

• $P_{\text{absolute}}$ is the absolute pressure ($115 \, \text{kPa} = 115,000 \, \text{Pa}$),
• $P_{\text{atmospheric}}$ is the atmospheric pressure ($1.013 \times 10^5 \, \text{Pa}$),
• $\rho$ is the fluid density ($740 \, \text{kg/m}^3$),
• $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$),
• $h$ is the depth of the fluid (to be determined).

Step 1: Solve for $h$

Rearranging the equation for $h$:

$h = \frac{P_{\text{absolute}} - P_{\text{atmospheric}}}{\rho g}$

Step 2: Substitute known values

$h = \frac{115,000 - 101,300}{740 \cdot 9.8}$

$h = \frac{13,700}{7252}$

$h \approx 1.89 \, \text{m}$

Final Answer:

The depth of the fluid is approximately 1.89 meters.

Would you like a detailed explanation of how to interpret this result or apply it in other contexts?

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Here are 5 related questions for further exploration:

1. What is the gauge pressure in the fluid?
2. How would the depth change if the density of the fluid were increased?
3. What is the force exerted by the fluid on the base of the container?
4. How does atmospheric pressure affect the total pressure calculation?
5. How would the result change if the fluid density were given in $\text{g/cm}^3$?

Tip: Always confirm if you're using absolute pressure or gauge pressure in fluid problems to avoid errors in your calculations.