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A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the container the pressure is 116 kPa.

(a) What is the depth of the fluid?
(b) Find the pressure at the bottom of the container after an additional 2.05 X 10^-3 m^3 of this fluid is added to the container. Assume that no fluid spills out of the container.

User Simran
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1 Answer

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Final answer:

To find the depth of the fluid in the cylindrical container, we can use the equation for pressure due to the weight of the fluid. The pressure at the bottom of the container can be calculated by subtracting the atmospheric pressure from the given pressure and then dividing by the density, gravitational acceleration, and cross-sectional area. The depth is found to be 0.0188 m. For part (b), the pressure at the bottom of the container after adding additional fluid can be calculated by adding the weight of the added fluid to the weight of the original fluid, and then dividing by the cross-sectional area. The pressure is found to be 227136.4 Pa.

Step-by-step explanation:

To find the depth of the fluid in the cylindrical container, we can use the equation for pressure due to the weight of the fluid. The pressure at the bottom of the container is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid. The pressure due to the fluid is equal to the weight of the fluid divided by the area of the cross section. We can rearrange the equation to solve for the depth of the fluid:

Depth = (Pressure - Atmospheric Pressure) / (Density x Gravitational Acceleration x Cross Sectional Area)

Substituting the given values:

Depth = (116,000 Pa - 101,325 Pa) / (806 kg/m^3 x 9.8 m/s^2 x (65.2 cm^2 / 10000))

Depth = 0.0188 m

For part (b), we can use the same equation but with the volume of the additional fluid added to the original volume of the container:

Pressure = (Weight of Fluid + Weight of Added Fluid) / Area

Substituting the given values:

Pressure = ((806 kg/m^3 x 9.8 m/s^2 x (0.0652 m^2)) + (806 kg/m^3 x 9.8 m/s^2 x (2.05 x 10^-3 m^3))) / (0.0652 m^2)

Pressure = 227136.4 Pa

User Carlo Field
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