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A tall cylinder with a cross-sectional area \(12.0 \mathrm{~cm}^{2}\) is partially filled with mercury; the surface of the mercury is \(8.00 \mathrm{~cm}\) 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 water must be added to double the gauge pressure at the bottom of the cylinder?

Short Answer

Expert verified
Upon calculation, volume of water to be added can be calculated based on the calculated height and given cross-sectional area. Convert the volume answer in \( m^3 \) to \( cm^3 \) to meet the given unit standard.

Step by step solution

01

Understanding Pressure in Fluids

Pressure in a fluid column is given by the equation \( P = \rho g h \), where \( P \) is the pressure, \( \rho \) is the density of the fluid, \( g \) is acceleration due to gravity and \( h \) is the height of the fluid column. Initially, the pressure at the bottom of the cylinder is due to the mercury alone.
02

Calculating Initial Pressure

Pressure due to the mercury can be calculated using the given values: cross-sectional area \( A = 12.0 \mathrm{~cm}^{2} \), height of mercury \( h_{mercury} = 8.00 \mathrm{~cm} \), and the density of mercury \( \rho_{mercury} = 13600 \mathrm{kg/m^3} \). Remember to convert all measurements to SI units. This gives \( P_{mercury} = \rho_{mercury} \cdot g \cdot h_{mercury} \)
03

Doubling the Pressure with Water

Since we want to double the pressure at the bottom of the cylinder, the added pressure from the water must be equal to the initial pressure of the mercury. Using the equation for pressure in a fluid column with the density of water \( \rho_{water} = 1000 \mathrm{kg/m^3} \), we can solve for the height of the water column needed: \( h_{water} = P_{mercury} / (\rho_{water} \cdot g) \)
04

Calculating Volume of Water

Finally, knowing the height of the water column and the cross-sectional area of the cylinder, we can calculate the volume of water needed using the equation \( V = A \cdot h \), where \( A \) is the area and \( h \) is the height.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure in Fluids
Pressure in fluids is a fundamental concept in fluid mechanics. It refers to the force exerted by a fluid per unit area. In a static fluid, the pressure at any point depends on the depth of the point from the fluid's surface and the fluid's density. This is often described by the formula:
  • \( P = \rho g h \)
Here, \( P \) represents the pressure, \( \rho \) is the fluid density, \( g \) is the gravitational acceleration (approximately \( 9.81 \mathrm{~m/s}^2 \)), and \( h \) is the height of the fluid column above the point where the pressure is being measured.

Understanding the distribution of pressure within fluids helps in predicting how fluids will behave when contained or when interacting with different objects. This knowledge is crucial for solving problems related to hydraulic systems, atmospheric science, and engineering designs involving fluids.
Density of Fluids
Density is another key concept when discussing fluids. It is defined as the mass of the fluid per unit volume. The formula to calculate density is:
  • \( \rho = \frac{m}{V} \)
where \( \rho \) denotes the density, \( m \) the mass, and \( V \) the volume of the fluid.

In our exercise, we use the density of mercury and water to determine the pressure exerted by each fluid. Mercury has a higher density (\( 13600 \mathrm{~kg/m^3} \)) compared to water (\( 1000 \mathrm{~kg/m^3} \)). This disparity in density significantly affects how pressure is calculated and perceived in various contexts. Higher density fluids exert more pressure at a given depth compared to lower density fluids.
Gauge Pressure
Gauge pressure is the pressure relative to the atmospheric pressure. In many situations, it's more practical to measure gauge pressure rather than absolute pressure, especially because many sensors and instruments naturally read gauge pressure. The relationship between gauge pressure \( P_g \) and absolute pressure \( P_a \) is:
  • \( P_g = P_a - P_0 \)
where \( P_0 \) is the atmospheric pressure at the reference level.

In the given exercise, we're asked to double the gauge pressure at the bottom of the cylinder. This implies that we need to calculate the additional pressure exerted by the water being added above the mercury, without considering atmospheric pressure. Knowing the gauge pressure helps in applications where only the overpressure beyond the atmosphere matters, such as tire pressure readings.
Column of Fluid
The column of fluid concept addresses how the height or depth of a fluid contributes to the pressure exerted at a given point within a container. Typically, the greater the height (or depth), the higher the pressure at any given point beneath it. This is because pressure in a fluid increases with depth due to the weight of the fluid above it.

In the exercise, the column includes both mercury and the water being added. Initially, the pressure calculation involves just the mercury column, but adding water increases the total height of the fluid column, thereby affecting the overall pressure at the bottom of the cylinder. Understanding the column of fluid concept is essential for contexts ranging from simple immersion problems to complex engineering applications involving large fluid reservoirs.
Hydrostatics
Hydrostatics is the branch of fluid mechanics that studies fluids at rest. It involves understanding pressure distribution in a fluid column and how different fluid densities affect this distribution. The key principle governing hydrostatics is that fluid pressure increases with depth due to the weight of the fluid above.

This exercise exemplifies hydrostatic principles as it explores how fluid layers, mercury and water, combine their pressures. The integration of these pressures results in the overall pressure at the bottom of the container. Hydrostatics helps in predicting the behavior of fluid in static conditions, such as in dams or pressure calculations for submerged objects, ensuring safety and efficiency in related engineering designs.

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Most popular questions from this chapter

On another planet that you are exploring, a large tank is open to the atmosphere and contains ethanol. A horizontal pipe of cross sectional area \(9.0 \times 10^{-4} \mathrm{~m}^{2}\) has one end inserted into the tank just above the bottom of the tank. The other end of the pipe is open to the atmosphere. The viscosity of the ethanol can be neglected. You measure the volume flow rate of the ethanol from the tank as a function of the depth \(h\) of the ethanol in the tank. If you graph the volume flow rate squared as a function of \(h,\) your data lie close to a straight line that has slope \(1.94 \times 10^{-5} \mathrm{~m}^{5} / \mathrm{s}^{2} .\) What is the value of \(g,\) the acceleration of a free-falling object at the surface of the planet?

A large, cylindrical water tank with diameter \(3.00 \mathrm{~m}\) is on a platform \(2.00 \mathrm{~m}\) above the ground. The vertical tank is open to the air and the depth of the water in the tank is \(2.00 \mathrm{~m}\). There is a hole with diameter \(0.500 \mathrm{~cm}\) in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole. (a) When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank? (b) How long does it take you to collect 1.00 gal of water in the bucket? Based on your answer in part (a), is it reasonable to ignore the change in the depth of the water in the tank as 1.00 gal of water flows out?

The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid's density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform - so the hole is \(50.0 \mathrm{~cm}\) above the ground. The table gives your measurements of the horizontal distance \(R\) that the initially horizontal stream of liquid pouring out of the tank travels before it strikes the ground and the gauge pressure \(p_{\mathrm{g}}\) of the air in the tank. $$\begin{array}{l|lllll}p_{\mathrm{g}}(\mathrm{atm}) & 0.50 & 1.00 & 2.00 & 3.00 & 4.00 \\\\\hline \boldsymbol{R}(\mathrm{m}) & 5.4 & 6.5 & 8.2 & 9.7 & 10.9\end{array}$$ (a) Graph \(R^{2}\) as a function of \(p_{\mathrm{g}}\). Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height \(h\) (in meters) of the liquid in the tank and the density of the liquid (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). Use \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) Assume that the liquid is nonviscous and that the hole is small enough compared to the tank's diameter so that the change in \(h\) during the measurements is very small.

A cubical block of density \(\rho_{\mathrm{B}}\) and with sides of length \(L\) floats in a liquid of greater density \(\rho_{\mathrm{L}}\). (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \(\rho_{\mathrm{W}}\) ) and does not mix with it. If water is poured on the surface of that liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L, \rho_{\mathrm{B}}, \rho_{\mathrm{L}},\) and \(\rho_{\mathrm{W}}\). (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and \(L=10.0 \mathrm{~cm}\).

A large plastic cylinder with mass \(30.0 \mathrm{~kg}\) and density \(370 \mathrm{~kg} / \mathrm{m}^{3}\) is in the water of a lake. A light vertical cable runs between the bottom of the cylinder and the bottom of the lake and holds the cylinder so that \(30.0 \%\) of its volume is above the surface of the water. What is the tension in the cable?

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