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Chapter 14: Problem 79

An object hangs from a spring balance. The balance registers \(30 \mathrm{~N}\) in air, \(20 \mathrm{~N}\) when this object is immersed in water, and \(24 \mathrm{~N}\) when the object is immersed in another liquid of unknown density. What is the density of that other liquid?

Short Answer

Expert verified
The density of the other liquid is approximately 600 kg/m鲁.

Step by step solution

01

Understand the Problem

We are given the weight of an object in air, in water, and in another liquid. We need to find the density of the other liquid using the differences in weight due to buoyancy.
02

Calculate the Weight Loss in Water

The object weighs 30 N in air and 20 N in water, showing a buoyant force of 10 N when submerged in water. This loss of weight corresponds to the weight of the volume of water displaced by the object.
03

Calculate the Volume of the Object

The buoyant force in water is equal to the weight of water displaced by the object, which is 10 N. The density of water is \(1000 \text{ kg/m}^3\), and gravity is \(9.8 \text{ m/s}^2\). Thus, the volume \(V\) can be calculated using the formula: \( \text{Buoyant force} = \text{density} \times V \times g \Rightarrow 10 = 1000 \times V \times 9.8 \Rightarrow V = \frac{10}{1000 \times 9.8} \approx 0.00102 \text{ m}^3.\)
04

Calculate the Weight Loss in the Unknown Liquid

When the object is immersed in the unknown liquid, it weighs 24 N, thus it experiences a buoyant force of \(30 - 24 = 6 \text{ N}.\)
05

Calculate the Density of the Unknown Liquid

Using the volume of the object from Step 3 and the buoyant force in the unknown liquid, we apply the formula \( \text{Buoyant force} = \text{density} \times V \times g.\) Solve for density: \(6 = \text{density} \times 0.00102 \times 9.8\), so \(\text{density} = \frac{6}{0.00102 \times 9.8} \approx 600 \text{ kg/m}^3.\)

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Key Concepts

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

Density
Every form of matter has a property called density. It measures how much mass is contained in a given volume. In simple terms, density tells us how tightly packed the molecules in a substance are. The formula to calculate density is given by:
  • \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
For any given object, the mass is the amount of matter it contains, and it is usually measured in kilograms (kg). Volume is the amount of space the object takes up, measured in cubic meters (m鲁).
In our exercise, the unknown liquid's density was determined by measuring the change in weight of an object when it was submerged in it. This method utilizes the buoyancy that acts on the object, which we'll explore further in the next section.
Archimedes' Principle
Archimedes' Principle is a cornerstone of fluid mechanics and deals with buoyancy. It states that any object submerged in a fluid experiences an upward force equal to the weight of the fluid displaced by the object.
So, if you place an object in water, Archimedes' Principle tells us the fluid will exert a force upwards against the object. This force is called the 'buoyant force.'
In this example, the object lost weight when placed in water and an unknown liquid due to the buoyant force acting upwards.
  • The buoyant force in water was 10 N
  • In the unknown liquid, the buoyant force was 6 N
These changes were crucial in identifying the density of the unknown fluid. Knowing the upward force and using the known density of water allowed us to measure the volume of the object, leading us to calculate the density of the other liquid.
Volume
Volume is a fundamental concept that describes the three-dimensional space occupied by a substance. It is often a pivotal part of calculations in science, including our exercise on buoyancy and density.
In our exercise, the displacement of water by the submerged object revealed its volume. Once immersed in water, the weight loss corresponded to the weight of the water displaced, which was 10 N. By employing the density of water (1000 kg/m鲁) and standard gravity (9.8 m/s虏), we calculated the volume of the object.
The calculation was straightforward:
  • Using the formula: \( \text{Buoyant force} = \text{density} \times V \times g \)
  • From which, Volume \( V = \frac{10}{1000 \times 9.8} \propto 0.00102 \text{ m}^3 \)
This volume then helped us determine the buoyant force and subsequently the density of the unknown liquid.

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Most popular questions from this chapter

A simple open U-tube contains mercury. When \(11.2 \mathrm{~cm}\) of water is poured into the right arm of the tube, how high above its initial level does the mercury rise in the left arm?

A hollow sphere of inner radius \(8.0 \mathrm{~cm}\) and outer radius \(9.0 \mathrm{~cm}\) floats half-submerged in a liquid of density 800 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.

To suck lemonade of density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) up a straw to a maximum height of \(4.0 \mathrm{~cm}\), what minimum gauge pressure (in atmospheres) must you produce in your lungs?

When a pilot takes a tight turn at high speed in a modern fighter airplane, the blood pressure at the brain level decreases, blood no longer perfuses the brain, and the blood in the brain drains. If the heart maintains the (hydrostatic) gauge pressure in the aorta at 120 torr (or mm Hg) when the pilot undergoes a horizontal centripetal acceleration of \(4 g\), what is the blood pressure (in torr) at the brain, \(30 \mathrm{~cm}\) radially inward from the heart? The perfusion in the brain is small enough that the vision switches to black and white and narrows to "tunnel vision" and the pilot can undergo \(g\) -LOC ("g-induced loss of consciousness"). Blood density is \(1.06 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\)

When you cough, you expel air at high speed through the trachea and upper bronchi so that the air will remove excess mucus lining the pathway. You produce the high speed by this procedure: You breathe in a large amount of air, trap it by closing the glottis (the narrow opening in the larynx), increase the air pressure by contracting the lungs, partially collapse the trachea and upper bronchi to narrow the pathway, and then expel the air through the pathway by suddenly reopening the glottis. Assume that during the expulsion the volume flow rate is \(7.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\). What multiple of the speed of sound \(v_{s}(=343 \mathrm{~m} / \mathrm{s})\) is the airspeed through the trachea if the trachea diameter (a) remains its normal value of \(14 \mathrm{~mm}\) and (b) contracts to \(5.2 \mathrm{~mm}\) ?
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