Archimedes purportedly used his principle to verify that the king's crown was pure gold by weighing the crown submerged in water. Suppose the crown's actual weight was \(25.0 \mathrm{N}\). What would be its apparent weight if it were made of (a) pure gold and (b) \(75 \%\) gold and \(25 \%\) silver, by volume? The densities of gold, silver, and water are \(19.3 \mathrm{g} / \mathrm{cm}^{3}, 10.5 \mathrm{g} / \mathrm{cm}^{3},\) and \(1.00 \mathrm{g} / \mathrm{cm}^{3},\) respectively.

Short Answer

Expert verified
The apparent weight if the crown were made of pure gold is \(23.71 \mathrm{N}\) and if it was made of 75% gold and 25% silver, it would be \(20.91 \mathrm{N}\).

Step by step solution

01

Find the volume of the crown

The crown's weight, as provided by the statement, is \(25.0 \mathrm{N}\). On Earth, the acceleration due to gravity is roughly \(9.8 \mathrm{m/s^2}\). Hence, using the equation \[Force = mass \cdot acceleration\], rearranges to: \[mass = \frac{Force}{acceleration}\].Substituting the given values, we find the mass of the crown is \(\frac{25.0 \mathrm{N}}{9.8 \mathrm{m/s^2}}\) which equals approximately \(2.55 \mathrm{kg}\). Convert the mass into grams to match the density units and it is equal to \(2550 \mathrm{g}\). npw, we will use the density of gold which is \(19.3 \mathrm{g/cm^3}\). By rearranging the density equation \[density = \frac{mass}{volume}\], to find the volume, \[volume = \frac{mass}{density}\]. Therefore, the volume of the crown is \(\frac{2550 \mathrm{g}}{19.3 \mathrm{g/cm^3}}\) which equals approximately \(132 \mathrm{cm^3}\).
02

Calculate the weight of water displaced

The weight of the water displaced by the crown when it is fully immersed in water is found using the density of water, which is \(1.00 \mathrm{g/cm^3}\), and the acceleration due to gravity. The displaced water's mass is equal to the volume of the crown times the density of water, so \[mass = volume \cdot density = 132 \mathrm{cm^3} \cdot 1.00 \mathrm{g/cm^3} = 132 \mathrm{g}\], which is equal to \(0.132 \mathrm{kg}\). The weight of this displaced water is then found using the equation \[weight = mass \cdot gravity = 0.132 \mathrm{kg} \cdot 9.8 \mathrm{m/s^2} = 1.29 \mathrm{N}\].
03

Calculate the apparent weight of the crown made of pure gold

With the actual weight of the crown and the weight of the water it displaces known, the apparent weight of the crown when it is submerged in water can be calculated. This is found by subtracting the weight of the displaced water from the actual weight of the crown: \[apparent weight = actual weight - weight of displaced water = 25.0 \mathrm{N} - 1.29 \mathrm{N} = 23.71 \mathrm{N}\].
04

Calculate the apparent weight of the crown made of 75% gold and 25% silver

Assuming the crown is made up of 75% gold and 25% silver by volume, the density of the crown now is \[(0.75 \times 19.3 \mathrm{g/cm^3}) + (0.25 \times 10.5 \mathrm{g/cm^3}) = 17.2 \mathrm{g/cm^3}\]. So, the mass of the crown now is \[mass = volume \cdot density = 132 \mathrm{cm^3} \cdot 17.2 \mathrm{g/cm^3} = 2270 \mathrm{g} = 2.27 \mathrm{kg}\]. The weight of the crown is:\[weight = mass \cdot gravity = 2.27 \mathrm{kg} \cdot 9.8 \mathrm{m/s^2} = 22.2 \mathrm{N}\].However, the buoyant force (weight of displaced water) remains the same as in Step 2. Hence, the apparent weight of the crown when submerged in water is: \[apparent weight = actual weight - weight of displaced water = 22.2 \mathrm{N} - 1.29 \mathrm{N} = 20.91 \mathrm{N}\].

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 Vaia!

Key Concepts

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

Buoyant Force
When an object is submerged in a fluid, it experiences an upward force called buoyant force. This is the core of Archimedes' Principle, which states that the buoyant force on a submerged object is equal to the weight of the fluid that the object displaces. In the case of the crown, when it is submerged in water, it displaces a volume of water equal to its own volume, and this displaced water has weight. The crown will therefore feel a lighter force acting on it, which is what we refer to as a reduction in the object's apparent weight.

Understanding Through Experimentation

One way to truly understand the concept is to perform a simple experiment. Take a solid object, measure its weight in the air and then measure it again while it's fully submerged in water. The difference you observe is due to the buoyant force.
Density
Density is a measure of how compact the mass in a substance or object is. It is defined as mass per unit volume and is a critical factor in determining whether an object will sink or float in a fluid. In our exercise involving the crown, knowing the density of gold, silver, and water allows us to calculate the volume of the crown, given its mass. This is because the crown's mass, distributed over the volume it occupies, gives us its density, which should be consistent with that of pure gold. Different densities of materials can drastically change the apparent weight of the object when submerged.

The Secret Ingredient

When it comes to density, it's like the secret ingredient that defines the character of a substance. For example, packing more mass into the same volume will increase the density, just like adding more seasoning makes the flavor of a dish more intense.
Apparent Weight
Apparent weight refers to the weight of an object as measured when it is submerged in a fluid, such as water. The concept of apparent weight is directly tied to the buoyant force. As Archimedes' Principle tells us, the object will feel an upward buoyant force equal to the weight of the fluid it displaces. Thus, the object seems lighter and its weight, as measured while submerged, is reduced. This is what we call the apparent weight.

Real-World Application

Understanding apparent weight is crucial in many fields like shipbuilding, where calculating the buoyancy of a vessel is necessary to ensure it can float and carry a designated load without sinking. Apparent weight helps us grasp how submerged objects interact with the fluid around them, reinforcing concepts like buoyancy and density with tangible consequences.

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 spherical rubber balloon with mass \(0.85 \mathrm{g}\) and diameter \(30 \mathrm{cm}\) is filled with helium (density \(0.18 \mathrm{kg} / \mathrm{m}^{3}\) ). How many 1.0 -g paper clips can you hang from the balloon before it loses buoyancy?

Barometric pressure in the eye of a hurricane is 0.91 atm \((27.2\) in. of mercury). How does the level of the ocean surface under the eye compare with the level under a distant fair-weather region where the pressure is 1.0 atm?

Density and pressure in Earth's atmosphere are proportional: \(\rho=p / h_{0} g,\) where \(h_{0}=8.2 \mathrm{km}\) is a constant and \(g\) is the gravitational acceleration. (a) Integrate Equation 15.2 for this case to show that atmospheric pressure as a function of height \(h\) above the surface is given by \(p=p_{0} e^{-h / l h_{0}},\) where \(p_{0}\) is the surface pressure. (b) At what height will the pressure have dropped to half its surface value?

You're a private investigator assisting a large food manufacturer in tracking down counterfeit salad dressing. The genuine dressing is by volume one part vinegar (density \(1.0 \mathrm{g} / \mathrm{cm}^{3}\) ) to three parts olive oil (density \(0.92 \mathrm{g} / \mathrm{cm}^{3}\) ). The counterfeit dressing is diluted with water (density \(1.0 \mathrm{g} / \mathrm{cm}^{3}\) ). You measure the density of a dressing sample and find it to be \(0.97 \mathrm{g} / \mathrm{cm}^{3} .\) Has the dressing been altered?

Water flows through a 2.5 -cm-diameter pipe at \(1.8 \mathrm{m} / \mathrm{s}\). If the pipe narrows to 2.0 -cm diameter, what's the flow speed in the constriction?

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.

Sign-up for free