/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Two identical containers are ope... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of \(1.00 \mathrm{m},\) one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

Short Answer

Expert verified
The fluid level in the left container is approximately 1.86 meters.

Step by step solution

01

Understanding the Concept

We are dealing with two liquids that are immiscible and different densities, one being water and the other mercury. When the valve is opened, the liquids will seek equilibrium by their weights (hydrostatic pressure).
02

Setting up the Equation

The principle that governs this problem is that, at equilibrium, the pressure at the bottom of both columns of liquid must be the same, because they are connected. We can use the pressure equation: \[ P = ho gh \] where \( P \) is the pressure, \( \rho \) is the density of the liquid, \( g \) is the gravitational acceleration, and \( h \) is the height of the liquid column.
03

Defining the Densities

The density of water is \( \rho_{water} = 1000 \text{ kg/m}^3 \) and the density of mercury is \( \rho_{mercury} = 13600 \text{ kg/m}^3 \). These values are essential for calculating the pressure at the base of each liquid column.
04

Equating the Pressures

At equilibrium, the pressure due to the height of water \( h_w \) must equal the pressure due to the height of mercury \( h_m \) since they are connected at the same horizontal level:\[ \rho_{water} g h_w = \rho_{mercury} g h_m \]Here, \( g \) is the gravitational acceleration, which is the same for both, so they cancel out.
05

Solving for the Fluid Level

Rearrange the equation to solve for \( h_w \):\[ h_w = \frac{\rho_{mercury}}{\rho_{water}} h_m \]Since the initial height is 1.00 m for both, but since they have the same base level initially, at equilibrium, the sum of the heights of water and mercury must remain constant. Let \( h_w = x \) and \( h_m = 2 - x \), then\[ x = \frac{13600}{1000} (2 - x) \]Solve for \( x \).
06

Final Calculation

Solve the previous equation:\[ x = 13.6 (2 - x) \]\[ x = 27.2 - 13.6x \]\[ 14.6x = 27.2 \]\[ x = \frac{27.2}{14.6} \approx 1.86 \text{m} \]
07

Conclusion

The water level in the left container will equilibrate to approximately 1.86 meters when both containers reach equilibrium.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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. This pressure increases with depth as more fluid is piled on top, pushing down with gravitational force. It is described by the formula: - \[ P = \rho gh \]- where:- \( P \) is the hydrostatic pressure,- \( \rho \) is the density of the fluid,- \( g \) is the acceleration due to gravity,- and \( h \) is the height of the fluid column.When dealing with fluid containers connected at the base, like in this exercise, the hydrostatic pressure at the bottom must be equal for equilibrium. This principle directs how the fluids seek to balance themselves once the valve connecting them is opened.
Immiscible Fluids
Immiscible fluids are liquids that do not mix or form a homogeneous mixture when combined. In this case, water and mercury are immiscible. - This means they will retain their distinct layers in the container, even when the valve is opened. The distinct immiscibility is crucial to determine the equilibrium condition. Water's and mercury's distinct densities and their refusal to mix mean that they exert pressure independently. Understanding this separation helps us tailor the pressure equation for calculating the respective heights needed at equilibrium.
Density Comparison
Density is a measure of how much mass is contained within a specific volume. Here, density is symbolized by \( \rho \) in the pressure equation, and it significantly influences how liquids settle when released. - Water has a density of approximately \( 1000 \text{ kg/m}^3 \), while mercury has a much higher density of \( 13600 \text{ kg/m}^3 \). This difference explains why mercury settles lower and water higher when the two are connected. - Denser fluids like mercury exert more pressure over a shorter column height compared to less dense fluids like water. This difference in density directly affects the redistribution of liquid levels until equilibrium is reached.
Pressure Equation
The pressure equation plays an essential role in solving equilibrium problems involving fluids. For the problem at hand, we know that the pressure at the base of each container must be equal when equilibrium is achieved. The fundamental equation to equate is:- \[ \rho_{water} g h_w = \rho_{mercury} g h_m \]Here, \( h_w \) refers to the water column height and \( h_m \) to the mercury column height. By canceling the gravitational constant \( g \), we simplify the comparison to purely dependent on heights and densities of fluids.- This equation illustrates how even with vastly different densities, the respective fluid column heights adjust to maintain equilibrium.Finally, solving the modified equation shows how each fluid adjusts its level in response to opening the valve, achieving the balanced pressure across the system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A cylindrical storage tank has a radius of \(1.22 \mathrm{m} .\) When filled to a height of \(3.71 \mathrm{m},\) it holds \(14300 \mathrm{kg}\) of a liquid industrial solvent. What is the density of the solvent?

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is \(1.80 \times 10^{5}\) Pa. Assuming that the top and bottom surfaces of the cap each have an area of \(4.10 \times 10^{-4} \mathrm{m}^{2}\), obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.

A liquid is flowing through a horizontal pipe whose radius is \(0.0200 \mathrm{m}\). The pipe bends straight upward through a height of \(10.0 \mathrm{m}\) and joins another horizontal pipe whose radius is \(0.0400 \mathrm{m} .\) What volume flow rate will keep the pressures in the two horizontal pipes the same?

Neutron stars consist only of neutrons and have unbelievably high densities. A typical mass and radius for a neutron star might be \(2.7 \times 10^{28} \mathrm{kg}\) and \(1.2 \times 10^{3} \mathrm{m} .\) (a) Find the density of such a star. (b) If a dime \((V=\) \(\left.2.0 \times 10^{-7} \mathrm{m}^{3}\right)\) were made from this material, how much would it weigh (in pounds)?

To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is \(5.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) What fraction of the specimen's apparent volume is solid?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.