Solve Gauge Pressure Problem Step-By-Step

[einstein_from_oz]

Can someone please attempt this problem?

Question:
A tall cylinder with a cross-sectional area of 12.0 cm^2 is partially filled with mercury; the surface of the mercury is 5 cm above the bottom of the cylinder. Water is slowly poured in on top of the mercury and the two fluids don't mix. What volume of the water must be added to double the gauge pressure at the bottom of the cylinder.

Can some please write step-by-step solution, not just the answer?

Thank you

 

[vincentchan]

No, we don't write solution for people...

 

[wais]

for providing this problem. Solving gauge pressure problems involves understanding the concepts of fluid pressure and the relationship between pressure and depth. Here is a step-by-step solution to the problem:

Step 1: Understand the given information
In this problem, a tall cylinder with a cross-sectional area of 12.0 cm^2 is partially filled with mercury. The surface of the mercury is 5 cm above the bottom of the cylinder. The remaining space is then filled with water, which does not mix with the mercury. We are asked to find the volume of water that needs to be added in order to double the gauge pressure at the bottom of the cylinder.

Step 2: Identify the known values
We are given the cross-sectional area of the cylinder (12.0 cm^2), the height of the mercury (5 cm), and the fact that the two fluids do not mix. We also know that we need to double the gauge pressure, which means that the final pressure at the bottom of the cylinder will be twice the original pressure.

Step 3: Determine the original gauge pressure
To find the original gauge pressure, we need to first calculate the pressure at the bottom of the cylinder. This can be done using the formula P = ρgh, where P is pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid. In this case, the density of mercury is 13.6 g/cm^3 and the density of water is 1 g/cm^3. Therefore, the pressure at the bottom of the cylinder is P = (13.6 g/cm^3)(9.8 m/s^2)(5 cm) = 666.4 Pa.

Step 4: Determine the final gauge pressure
Since we need to double the gauge pressure, the final gauge pressure will be 2 times the original gauge pressure, which is 2(666.4 Pa) = 1332.8 Pa.

Step 5: Use the gauge pressure formula to find the volume of water needed
The gauge pressure formula is P = ρgh, where P is the gauge pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid. In this case, we know the final gauge pressure (1332.8 Pa), the density of water (1 g/cm^3), and the height of the