Math Problem Statement

Archimedes supposedly was asked to determine whether a crown made for the king consisted of pure gold. According to legend, he solved this problem by weighing the crown first in air and then in water. If the scale read 7.96 N when the crown was in air and 6.70 N when it was in water, determine the density of the crown.

Give your answer in units of kg/m3, accurate to the nearest 10 kg/m3 (this being the maxium accuracy possible from the variables given).

Your Answer:

Solution

We can solve this problem using Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. Here's the step-by-step method to find the density of the crown.

Step 1: Calculate the buoyant force

The buoyant force FbF_b is the difference in weight of the crown in air and in water: Fb=WairWwater=7.96N6.70N=1.26NF_b = W_{air} - W_{water} = 7.96\,\text{N} - 6.70\,\text{N} = 1.26\,\text{N}

Step 2: Relating buoyant force to volume

The buoyant force is also equal to the weight of the water displaced by the crown, which is: Fb=ρwatergVF_b = \rho_{water} \cdot g \cdot V where:

  • ρwater=1000kg/m3\rho_{water} = 1000\,\text{kg/m}^3 is the density of water,
  • g=9.8m/s2g = 9.8\,\text{m/s}^2 is the acceleration due to gravity,
  • VV is the volume of the crown.

Rearranging this equation to solve for VV: V=Fbρwaterg=1.26N1000kg/m39.8m/s2=1.2857×104m3V = \frac{F_b}{\rho_{water} \cdot g} = \frac{1.26\,\text{N}}{1000\,\text{kg/m}^3 \cdot 9.8\,\text{m/s}^2} = 1.2857 \times 10^{-4}\,\text{m}^3

Step 3: Find the mass of the crown

We can find the mass of the crown using its weight in air: Wair=mgm=Wairg=7.96N9.8m/s2=0.812kgW_{air} = m \cdot g \quad \Rightarrow \quad m = \frac{W_{air}}{g} = \frac{7.96\,\text{N}}{9.8\,\text{m/s}^2} = 0.812\,\text{kg}

Step 4: Calculate the density of the crown

Now, using the mass and the volume of the crown, we can find the density ρcrown\rho_{crown}: ρcrown=mV=0.812kg1.2857×104m3=6316.8kg/m3\rho_{crown} = \frac{m}{V} = \frac{0.812\,\text{kg}}{1.2857 \times 10^{-4}\,\text{m}^3} = 6316.8\,\text{kg/m}^3

Step 5: Round the answer

Rounding to the nearest 10 kg/m3\text{kg/m}^3: ρcrown6320kg/m3\rho_{crown} \approx 6320\,\text{kg/m}^3

Thus, the density of the crown is 6320 kg/m³.

Would you like further details or have any questions? Here are some related questions to explore this concept:

  1. How does Archimedes' principle help determine an object's density?
  2. What is the significance of the buoyant force in measuring densities?
  3. How would the result change if the crown were submerged in a fluid other than water?
  4. How can you distinguish between pure gold and an alloy using density?
  5. How would the problem change if the crown weighed 10 N in air?

Tip: When solving problems involving buoyancy, always remember that the buoyant force equals the weight of the displaced fluid.

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Math Problem Analysis

Mathematical Concepts

Physics
Buoyancy
Density

Formulas

Buoyant force: F_b = W_air - W_water
Buoyant force and volume relation: F_b = ρ_water * g * V
Density: ρ_crown = m / V
Weight: W = m * g

Theorems

Archimedes' Principle

Suitable Grade Level

Grades 9-12