/*! 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 31 Determination of the density of ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determination of the density of a fluid has many important applications. A car battery contains sulfuric acid, for which density is a measure of concentration. For the battery to function properly, the density must be inside a range specified by the manufacturer. Similarly, the effectiveness of antifreeze in your car's engine coolant depends on the density of the mixture (usually ethylene glycol and water). When you donate blood to a blood bank, its screening includes determination of the density of the blood, since higher density correlates with higher hemoglobin content. A hydrometer is an instrument used to determine liquid density. A simple one is sketched in Figure P14.31. The bulb of a syringe is squeezed and released to let the atmosphere lift a sample of the liquid of interest into a tube containing a calibrated rod of known density. The rod, of length \(L\) and average density \(\rho_{0},\) floats partially immersed in the liquid of density \(\rho . \mathrm{A}\) length \(h\) of the rod protrudes above the surface of the liquid. Show that the density of the liquid is given by $$\rho=\frac{\rho_{0} L}{L-h}$$

Short Answer

Expert verified
The density of the fluid, \(\rho\), is determined to be \(\rho = \frac{\rho_0 \cdot L}{L - h}\), where \(\rho_0\) is the density of the rod, \(L\) is the total length of the rod, and \(h\) is the length of the rod extending out of the fluid.

Step by step solution

01

Clarify Variables

Set \(\rho_0\) as the density of the rod, \(L\) as the length of the rod and \(h\) as the length of the rod that is not submerged in the fluid. We aim to find \(\rho\), the density of the fluid.
02

Apply Archimedes' Principle

Archimedes’ principle states that an object immersed in a fluid experiences a buoyant force equal to the weight of the fluid it displaces. Here, the rod is balancing because the weight of the rod is balanced by the buoyant force of the fluid it displaces. Therefore, we can write: Weight of rod = Weight of displaced fluid or \(\rho_0 \cdot V_{rod} \cdot g = \rho \cdot V_{fluid} \cdot g\). Where \(V_{rod}\) and \(V_{fluid}\) are the volumes of the rod submerged and fluid displaced, respectively, and \(g\) is the acceleration due to gravity.
03

Express Volumes in Terms of Lengths

Assuming the rod is cylindrical, the volume can be expressed as length times area: \(V_{rod} = A \cdot (L - h)\) and \(V_{fluid} = A \cdot (L - h)\), where \(A\) is the cross-sectional area of the rod and \(h\) is the length of the rod not submerged in fluid.
04

Substitute Volumes into Equation

Substituting into our balance of weights gives: \(\rho_0 \cdot A \cdot (L - h) \cdot g = \rho \cdot A \cdot (L - h) \cdot g\). Note that \(A\) & \(g\) are common to both sides of this equation and will cancel each other out.
05

Solve for Density of Fluid

Solving our equation for \(\rho\) gives: \(\rho = \frac{\rho_0 \cdot L}{L - h}\)

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Key Concepts

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

Archimedes' Principle
Archimedes' principle is a fundamental concept in fluid mechanics used to determine the buoyant force on an object submerged in a fluid. It states that the buoyant force exerted on an object is equal to the weight of the fluid that the object displaces. This means when an object is placed in a fluid, it will be pushed up by a force equal to the gravitational force on the volume of fluid that has been moved aside.

For example, consider a solid rod in a liquid, as in the problem we're examining. The rod displaces an amount of liquid equal to the volume of the rod that is submerged. According to Archimedes' principle, the fluid's force pushing upward on the rod (the buoyant force) must equal the weight of the fluid that would occupy the submerged volume of the rod if the rod were not there. It's this principle that allows us to solve for the density of the fluid in our problem.
Buoyant Force
The concept of buoyant force is closely tied to Archimedes' principle. Buoyant force is the net upward force exerted by a fluid on a submerged or partially submerged object. The key point to remember is that the magnitude of this force depends on the volume of the displaced fluid. For objects that are less dense than the fluid, the buoyant force can exceed the object's weight, allowing it to float. For denser objects, the buoyant force is not enough to counteract gravity fully, and they sink.

When an object is floating partially submerged, like the rod in our textbook problem, the buoyant force is balanced by the weight of the object. This balance is what we leverage to find the density of the fluid; the rod's submerged volume is directly related to the fluid's density, enabling us to calculate it using the measurements provided in the problem.
Hydrometer
A hydrometer is an instrument designed to measure the density of fluids based on the concept of buoyancy. The device typically consists of a calibrated rod that floats in the fluid, and the level at which the rod floats is indicative of the fluid's density.

The working principle is straightforward: when a hydrometer is placed in a fluid, it sinks until the buoyant force is equal to the gravitational force acting on it. The lower the density of the fluid, the further the hydrometer sinks, and vice versa. Hydrometers often have a scale on the rod, which corresponds to the density of the fluid. Based on the portion of the scale that is above the surface, one can related the rod's density and the submerged length to calculate the fluid density, as with the problem we are discussing.
Physics Problem Solving
Physics problem solving often involves applying fundamental principles to deduce unknown quantities from known ones. In our example, the process involved clarifying variables, applying Archimedes' principle, relating volumes to easily measured lengths, and then algebraically solving for the desired density of the fluid.

A systematic approach is essential in physics problem-solving: define the problem, identify the principles and equations that apply, manipulate the equations to isolate the desired quantity, and then compute the solution based on given information. Such an approach, grounded in the basic concepts of physics like those discussed here, can help tackle a wide variety of physics problems effectively and efficiently.

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Most popular questions from this chapter

A \(50.0-\mathrm{kg}\) woman balances on one hecl of a pair of highheeled shoes. If the heel is circular and has a radius of \(0.500 \mathrm{cm},\) what pressure does she exert on the floor?

A swimming pool has dimensions \(30.0 \mathrm{m} \times 10.0 \mathrm{m}\) and a flat bottom. When the pool is filled to a depth of \(2.00 \mathrm{m}\) with fresh water, what is the force caused by the water on the bottom? On each end? On each side?

The weight of a rectangular block of low-density material is 15.0 N. With a thin string, the center of the horizontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When \(25.0 \%\) of the block's volume is submerged, the tension in the string is \(10.0 \mathrm{N}\) (a) Sketch a free-body diagram for the block, showing all forces acting on it. (b) Find the buoyant force on the block. (c) Oil of density \(800 \mathrm{kg} / \mathrm{m}^{3}\) is now steadily added to the beaker, forming a layer above the water and surrounding the block. The oil exerts forces on each of the four side walls of the block that the oil touches. What are the directions of these forces? (d) What happens to the string tension as the oil is added? Explain how the oil has this Effect on the string tension. (c) The string breaks when its tension reaches \(60.0 \mathrm{N}\). At this moment, \(25.0 \%\) of the block's volume is still below the water line; what additional fraction of the block's volume is below the top surface of the oil? (f) After the string breaks, the block comes to a new equilibrium position in the beaker. It is now in contact only with the oil. What fraction of the block's volume is submerged?

(a) Calculate the absolute pressure at an ocean depth of \(1000 \mathrm{m} .\) Assume the density of seawater is \(1024 \mathrm{kg} / \mathrm{m}^{3}\) and that the air above exerts a pressure of \(101.3 \mathrm{kPa}\) (b) At this depth, what force must the frame around a circular submarine porthole having a diameter of \(30.0 \mathrm{cm}\) exert to counterbalance the force exerted by the water?

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. (a) Suppose that a horizontal wind blows in streamline flow with a speed of \(11.2 \mathrm{m} / \mathrm{s}\) outside a large pane of plate glass with dimensions \(4.00 \mathrm{m} \times 1.50 \mathrm{m}\) Assume the density of the air to be uniform at \(1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? (b) What If? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is \(22.4 \mathrm{m} / \mathrm{s}\), twice as high.

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