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A hydrometer consists of a spherical bulb and a cylindrical stem with a cross- sectional area of 0.400 \(\mathrm{cm}^{2}\) (see Fig. 14.13a). The total volume of bulb and stem is \(13.2 \mathrm{cm}^{3} .\) When immersed in water, the hydrometer floats with 8.00 \(\mathrm{cm}\) of the stem above the water surface. When the hydrometer is immersed in an organic fluid, 3.20 \(\mathrm{cm}\) of the stem is above the surface. Find the density of the organic fluid. (Nore: This illustrates the precision of such a hydrometer. Relatively small density differences give rise to relatively large differences in hydrometer readings.)

Short Answer

Expert verified
The density of the organic fluid is approximately 0.8395 g/cm³.

Step by step solution

01

Understand the Problem

We have a hydrometer with a known total volume and a cylindrical stem with a known cross-sectional area. The hydrometer floats differently in two fluids: water and an organic fluid. The goal is to find the density of the organic fluid based on this information.
02

Calculate the Displaced Volume in Water

When the hydrometer floats in water with 8.00 cm of the stem above, the submerged volume is the total volume minus the stem volume above water. The stem volume above is obtained by multiplying the height (8.00 cm) by the cross-sectional area (0.400 cm²). Thus, the volume of the stem above water is:\[ V_{\text{above, water}} = 8.00 \, \mathrm{cm} \times 0.400 \, \mathrm{cm}^2 = 3.20 \, \mathrm{cm}^3 \]The volume submerged in water is:\[ V_{\text{submerged, water}} = 13.2 \, \mathrm{cm}^3 - 3.20 \, \mathrm{cm}^3 = 10.00 \, \mathrm{cm}^3 \]
03

Use Archimedes Principle for Water

The buoyant force on the hydrometer equals the weight of the water displaced. The mass of the displaced water is equal to the volume submerged multiplied by the density of water (assumed to be 1 g/cm³). Therefore, the buoyant force is:\[ F_{\text{buoyant, water}} = 10.00 \, \mathrm{cm}^3 \times 1 \, \mathrm{g/cm}^3 = 10.00 \, \mathrm{g} \]
04

Calculate the Displaced Volume in Organic Fluid

When the hydrometer floats in the organic fluid, 3.20 cm of the stem is above the surface. Thus, the volume of the stem above the fluid is:\[ V_{\text{above, fluid}} = 3.20 \, \mathrm{cm} \times 0.400 \, \mathrm{cm}^2 = 1.28 \, \mathrm{cm}^3 \]The volume submerged in the organic fluid is:\[ V_{\text{submerged, fluid}} = 13.2 \, \mathrm{cm}^3 - 1.28 \, \mathrm{cm}^3 = 11.92 \, \mathrm{cm}^3 \]
05

Use Archimedes Principle for Organic Fluid

Similarly, the buoyant force in the organic fluid is equal to the weight of the organic fluid displaced. The mass of the displaced organic fluid is:\[ F_{\text{buoyant, fluid}} = 11.92 \, \mathrm{cm}^3 \times \rho \]where \(\rho\) is the density of the organic fluid.
06

Equate Buoyant Forces

The buoyant forces when floating in water and the organic fluid are equal since the hydrometer's weight doesn't change. Set both buoyant forces equal:\[ 10.00 = 11.92 \times \rho \]Solve for \(\rho\):\[ \rho = \frac{10.00}{11.92} \approx 0.8395 \, \mathrm{g/cm}^3 \]
07

Verify the Answer

Double-check calculations ensuring each volume and step is correctly computed. Compare calculated density with expected ranges for common organic fluids.

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Key Concepts

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

Density Determination
Density is fundamental in understanding how substances behave in different environments. Determining the density of a liquid involves observing how an object floats within it. A hydrometer, which we're focusing on, is a simple tool used for this purpose. It's designed to help us determine the density of a liquid based on how much of the hydrometer is submerged.
A hydrometer has two key parts:
  • A spherical bulb that holds most of its volume.
  • A cylindrical stem used for measuring the floating level.
By noting the length of the stem either above or below the liquid surface, we can derive the liquid's density. Mathematically, this comes from comparing the volume of liquid displaced (i.e., the submerged volume) to the total volume of the hydrometer. By measuring how much of the stem still floats above the liquid, we use simple arithmetic and the known geometry of the hydrometer to determine what proportion of the entire hydrometer is submerged. This, in turn, allows us to calculate the density of the unknown liquid using the formula for density \[\rho = \frac{\text{mass}}{\text{volume}}\] but by observing buoyancy effects rather than direct mass measurements.
Buoyant Force
Buoyant force is an upward force exerted by a fluid on a submerged object. It's why things float! It's a fundamental concept when discussing why hydrometers work. Simply put, buoyant force is equal to the weight of the fluid that the object displaces. This is why when you place an object in water, the level of the water rises. The part of the object that is submerged is pushing the water out of the way, hence the object experiences an upward push.
The equation for buoyant force is given as:\[ F_{\text{buoyant}} = V_{\text{submerged}} \times \rho_{fluid} \times g \]where:
  • \( V_{\text{submerged}} \) is the submerged volume of the object.
  • \( \rho_{fluid} \) is the density of the fluid.
  • \( g \) is the acceleration due to gravity.
For our hydrometer, understanding the force helps solve the mystery of how deeply it submerges in different fluids. The hydrometer sinks until the upward buoyant force equals the downward gravitational force of its weight.
This principle allows us to explore unknown liquid densities, linking directly to Archimedes' Principle, which we’ll explore next.
Archimedes Principle
Archimedes' Principle is a cornerstone of fluid mechanics. Formulated by the ancient Greek mathematician Archimedes, it explains how bodies are supported in fluids. The principle states that any object submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object.
This means that if you have an object floating in water - like our hydrometer - the weight of the water displaced is exactly equal to the weight of the object. This is a direct result of the balance between gravitational forces and buoyancy.
In simpler terms, Archimedes' Principle helps us understand why objects sink, float, or remain suspended in fluids. By applying this principle to our hydrometer, we can deduce the density of an unknown fluid by measuring how much of the hydrometer stays above the surface and using that to find the volume submerged. This is because the buoyant force acting upward on the hydrometer is equal to the force of gravity acting downward, leading to the formula:\[V_{\text{submerged}} \times \rho = W \]where \( W \) is the weight of the hydrometer. This allows students to not only solve physics problems but also connects them to real-world applications.

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Most popular questions from this chapter

Lift on an Airplane. Air streams horizontally past a small airplane's wings such that the speed is 70.0 \(\mathrm{m} / \mathrm{s}\) over the top surface and 60.0 \(\mathrm{m} / \mathrm{s}\) past the bottom surface. If the plane has a wing area of 16.2 \(\mathrm{m}^{2}\) on the top and on the bottom, what is the net vertical force that the air exerts on the airplane? The density of the air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) .

Exploring Venus. The surface pressure on Venus is 92 arm, and the acceleration due to gravity there is 0.894\(g .\) In a future exploratory mission, an upright cylindrical tank of benzene is sealed at the top but still pressurized at 92 atm just above the benzene. The tank has a diameter of \(1.72 \mathrm{m},\) and the benzene column is 11.50 \(\mathrm{m}\) tall. Ignore any effects due to the very high temperature on Venus. (a) What total force is exerted on the inside surface of the bottom of the tank? (b) What force does the Venusian atmosphere exert on the outside surface of the bottom of the tank ? \((\mathrm{c})\) What total inward force does the atmosphere exert on the vertical walls of the tank?

A barge is in a rectangular lock on a freshwater river. The lock is 60.0 \(\mathrm{m}\) long and 20.0 \(\mathrm{m}\) wide, and the steel doors on each end are closed. With the barge floating in the lock, a \(2.50 \times 10^{6} \mathrm{N}\) load of scrap metal is put onto the barge. The metal has density \(9000 \mathrm{kg} / \mathrm{m}^{3} .\) (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?

A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?

A hunk of aluminum is completely covered with a gold shell to form an ingot of weight 45.0 \(\mathrm{N}\) . When you suspend the ingot from a spring balance and submerge the ingot in water, the balance reads 39.0 \(\mathrm{N}\) . What is the weight of the gold in the shell?

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