A vertical tube \(1.0 \mathrm{cm}\) in diameter and open at the top contains \(\left.5.0 \mathrm{g} \text { of oil (density } 0.82 \mathrm{g} / \mathrm{cm}^{3}\right)\) floating on \(5.0 \mathrm{g}\) of water. Find the gauge pressure (a) at the oil-water interface and (b) at the bottom.

Short Answer

Expert verified
The gauge pressures are approximately \(62.8 \, Pa\) at the oil-water interface and \(125.5 \, Pa\) at the bottom of the tube.

Step by step solution

01

Calculation of heights

Initially, the height of oil and water in the tube is calculated using their masses and densities. Since the mass of a substance is the product of its volume and density, the volume can be found as \(V = \frac{m}{p}\), where \(m\) is the mass and \(p\) is the density. Using the given diametre of the tube \(D = 1.0 \, cm\), the volume of a cylinder can be calculated as \(V = \pi r^2 h\) where \(r\) is the radius and \(h\) is the height. Equating these gives \(h = \frac{m}{p \pi r^2}\). After plugging in the values, the height of oil \(h_o = \frac{5.0 \, g}{0.82 \, g/cm^3 \cdot \pi \cdot (0.5 \, cm)^2} \approx 7.8 \, cm\) and the height of water \(h_w = \frac{5.0 \, g}{1.0 \, g/cm^3 \cdot \pi \cdot (0.5 \, cm)^2} \approx 6.4 \, cm\) are found.
02

Calculation of gauge pressure at oil-water interface

For the pressure calculation, the formula used is \(P = pgh\), where \(g\) is the gravity acceleration. At the oil-water interface, there is only the column of oil. By substituting the values into the formula \(P_o = 0.82 \, g/cm^3 \cdot 9.8 \, m/s^2 \cdot 7.8 \, cm \approx 62.8 \, Pa\) - the gauge pressure at the oil-water interface is found.
03

Calculation of gauge pressure at the bottom of the tube

At the bottom of the tube, there is the pressure due to both oil and water column. So the total pressure \(P_t\) is the sum \(P_o + P_w\), where \(P_w\) is the pressure due to the water column. Therefore, \(P_w = 1.0 \, g/cm^3 \cdot 9.8 \, m/s^2 \cdot 6.4 \, cm \approx 62.7 \, Pa\), then \(P_t = P_o + P_w \approx 62.8 \, Pa + 62.7 \, Pa \approx 125.5 \, Pa\) is the gauge pressure at the bottom.

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Key Concepts

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

Fluid Mechanics
In the world of physics, fluid mechanics is the study of how fluids (liquids, gases, and plasmas) behave under various forces. When we delve into fluid mechanics, we're mainly concerned with understanding concepts such as fluid flow, its forces, and its effects on surroundings.

Fluid mechanics is divided into two primary branches: fluid statics, which deals with fluids at rest, and fluid dynamics, which considers the effects of forces on fluid motion. Our exercise involves understanding principles from fluid statics, specifically the hydrostatic pressure exerted by a fluid at rest.

Understanding the behavior of fluid pressure is essential not only for academic purposes but also in various engineering fields. For example, civil engineers need to know about fluid mechanics when designing water supply systems, dams, and bridges, while mechanical engineers might apply these principles in heating and cooling systems.
Density and Pressure
Density, denoted by the symbol \(p\) (rho), is a fundamental concept connecting mass \(m\) and volume \(V\). It's defined as mass per unit volume \(p = m/V\). In the context of our exercise, we calculate the volume of oil and water present in a cylinder, and from there, figure out the height of each liquid.

Pressure, on the other hand, is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. In fluids, pressure can be caused by gravity, external forces, or fluid acceleration. The formula \(P = pgh\) relates the pressure exerted by a static fluid to its density \(p\), the gravitational acceleration \(g\), and the height \(h\) of the fluid column.

In terms of fluids at rest, this is known as hydrostatic pressure. When we're calculating the pressure like in our exercise, we ignore atmospheric pressure and thus refer to it as gauge pressure. Gauge pressure is practical for problems where only the pressure due to the fluid is significant.
Hydrostatics
Hydrostatics is the branch of fluid statics that deals with fluids at rest. An essential concept within hydrostatics is that the pressure at a point within a fluid at rest does not depend on the direction; it is the same in all directions.

This isotropic property of fluid pressure results in the widely used hydrostatic equation \(P = pgh\), which we used in our exercise to calculate the gauge pressure at different levels within the tube. Here we considered the pressures due to two different fluid columns: oil and water.

The gauge pressure provides us insights about the force that the fluid would exert on any object placed at that depth without considering the atmospheric pressure. This concept is crucial in many practical applications, such as determining the force on dam walls, calculating the buoyancy of ships, and even in medical applications like blood pressure measurement.

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Most popular questions from this chapter

A spherical rubber balloon with mass \(0.85 \mathrm{g}\) and diameter \(30 \mathrm{cm}\) is filled with helium (density \(0.18 \mathrm{kg} / \mathrm{m}^{3}\) ). How many 1.0 -g paper clips can you hang from the balloon before it loses buoyancy?

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