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Chapter 13: Problem 63

A glass tube is bent into the form of a U. A \(50.0-\mathrm{cm}\) height of olive oil in one arm is found to balance \(46.0 \mathrm{~cm}\) of water in the other. What is the density of the olive oil?

Short Answer

Expert verified
The density of the olive oil is \(920 \mathrm{~kg/m^3}\).

Step by step solution

01

Understanding the Problem

We need to find the density of olive oil given that a height of olive oil balances a different height of water in a U-shaped tube. We know the height of olive oil is 50.0 cm and the height of water is 46.0 cm.
02

Using the Concept of Hydrostatic Equilibrium

In a U-tube, the pressure exerted by the liquids must be equal at the same level. This means that the pressure due to olive oil should equal the pressure due to water at their respective heights.
03

Applying the Hydrostatic Pressure Formula

The formula for hydrostatic pressure is given by \(P = \rho gh\), where \(\rho\) is the density of the liquid, \(g\) is the gravitational acceleration (\(9.8 \mathrm{~m/s^2}\)), and \(h\) is the height of the liquid column.
04

Writing the Pressure Equilibrium Equation

For the olive oil and water in the U-tube to balance, their pressures must be equal: \(\rho_{oil} gh_{oil} = \rho_{water} gh_{water}\). Given \(h_{oil} = 50.0 \text{ cm} = 0.5 \text{ m}\) and \(h_{water} = 46.0 \text{ cm} = 0.46 \text{ m}\), and knowing that the density of water \(\rho_{water}\) is \(1000 \mathrm{~kg/m^3}\).
05

Solving for Olive Oil Density

We eliminate \(g\) from both sides of the equation and rearrange to solve for \(\rho_{oil}\): \(\rho_{oil} = \rho_{water} \frac{h_{water}}{h_{oil}}\). Substituting the values we have: \(\rho_{oil} = 1000 \cdot \frac{0.46}{0.5}\).
06

Calculating the Result

Perform the calculation: \(\rho_{oil} = 1000 \cdot 0.92 = 920 \mathrm{~kg/m^3}\). Thus, the density of the olive oil is \(920 \mathrm{~kg/m^3}\).

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Key Concepts

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

Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It's crucial in understanding how different liquids balance in a system, such as a U-shaped tube. A fundamental formula for calculating hydrostatic pressure is:
  • \( P = \rho gh \)
where:
  • \( P \) is the pressure in pascals (Pa).
  • \( \rho \) is the fluid density in kilograms per cubic meter (kg/m鲁).
  • \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏).
  • \( h \) is the height of the fluid column in meters (m).
Understanding this concept is essential for calculating and comparing the forces exerted by different fluids in any system. In the U-tube scenario, both sides of the tube reach a state where the pressures caused by the liquid columns equalize.
U-tube Equilibrium
A U-tube is a practical apparatus for demonstrating fluid equilibrium. When two different liquids are placed in a U-shaped tube, they will settle such that the pressure exerted by the liquid columns at the same height level remains equal. This is known as U-tube equilibrium. This principle is used to compare the densities of the liquids.
In our example, olive oil and water in the U-tube reach equilibrium when the height of 50.0 cm for olive oil equals the pressure exerted by 46.0 cm height of water. The equilibrium signifies that the forces exerted by the liquids below the surface are balanced, illustrating their densities' relationship.
Fluid Density Comparison
Comparing the densities of different fluids can be straightforward using instruments like a U-tube. Since pressure also depends on fluid density, different colored liquids will balance at varying heights within the tube. This balance point allows us to determine or compare densities.
In our scenario, despite the different heights, the pressures due to the water and olive oil are equal at equilibrium set by the differing column heights. Using the equation for pressure equilibrium:
  • \( \rho_{oil} gh_{oil} = \rho_{water} gh_{water} \)
and solving for the unknown density, you can see how density is directly connected with the height of the liquid when pressures are balanced. This illustrates why olive oil, with a density of 920 kg/m鲁, requires a greater height than water to exert the same pressure at equilibrium.
Physics Problem Solving
In physics, problem-solving often involves breaking complex problems into manageable steps, as demonstrated in calculating the density of oleaginous substances. The process begins with understanding the problem and what is being asked. Next, applying relevant physical principles, such as hydrostatic pressure, helps frame the approach.
For our U-tube example, you:
  • Identify equilibrium conditions and ensure understanding of the forces at play.
  • Apply the hydrostatic pressure formula to set up an equation representing the balance in forces.
  • Use logical reasoning to solve for the unknown variable, in our case, which is the density of olive oil.
  • Perform any necessary calculations, ensuring units are consistent for correct results.
Each step builds on the previous, illustrating a clear strategy to tackle similar physics-based problems efficiently.

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Most popular questions from this chapter

What fraction of the volume of a piece of quartz \(\left(\rho=2.65 \mathrm{~g} / \mathrm{cm}^{3}\right)\) will be submerged when it is floating in a container of mercury \((\rho\) \(\left.=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\)

When a submarine dives to a depth of \(120 \mathrm{~m}\), to how large a total pressure is its exterior surface subjected? The density of seawater is about \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\) \(P=\) Atmospheric pressure \(+\) Pressure of wate $$ \begin{array}{l} =1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\rho g h=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\left(1030 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(120 \mathrm{~m}) \\ =1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+12.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=13.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.31 \mathrm{MPa} \end{array} $$

What must be the volume \(V\) of a \(5.0\) -kg balloon filled with helium \(\left(\rho_{\mathrm{He}}=0.178 \mathrm{~kg} / \mathrm{m}^{3}\right)\) if it is to lift a 30 -kg load? Use \(\rho_{\text {air }}=\) \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\) The buoyant force, \(V \rho_{\text {airg }}\), must lift the weight of the balloon, its load, and the helium within it: $$ V \rho_{\text {air }} g=(35 \mathrm{~kg})(g)+V \rho_{\mathrm{He}} g $$ which gives $$ V=\frac{35 \mathrm{~kg}}{\rho_{\text {air }}-\rho_{\mathrm{He}}}=\frac{35 \mathrm{~kg}}{1.11 \mathrm{~kg} / \mathrm{m}^{3}}=32 \mathrm{~m}^{3} $$

A foam plastic \(\left(\rho_{p}=0.58 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is to be used as a life preserver. What volume of plastic must be used if it is to keep 20 percent (by volume) of an 80 -kg man above water in a lake? The average density of the man is \(1.04 \mathrm{~g} / \mathrm{cm}^{3}\). Keep in mind that a density of \(1 \mathrm{~g} / \mathrm{cm}^{3}\) equals \(1000 \mathrm{~kg} / \mathrm{m}^{3}\). At equilibrium where subscripts \(m, w\), and \(p\) refer to man, water, and plastic, respectively. But \(\rho_{m} V_{m}=80 \mathrm{~kg}\) and so \(V_{m}=(80 / 1040) \mathrm{m}^{3}\). Substitution gives \(\left[(1000-580) \mathrm{kg} / \mathrm{m}^{3}\right] V_{p}=\left[(1040-800) \mathrm{kg} / \mathrm{m}^{3}\right]\left[(80 / 1040) \mathrm{m}^{3}\right]\) from which \(V_{p}=0.044 \mathrm{~m}^{3}\).

The area of a piston of a force pump is \(8.0 \mathrm{~cm}^{2}\). What force must be applied to the piston to raise oil \(\left(\rho=0.78 \mathrm{~g} / \mathrm{cm}^{2}\right)\) to a height of \(6.0 \mathrm{~m}\) ? Assume the upper end of the oil is open to the atmosphere.
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