Chapter 12: Problem 5
A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?
Short Answer
Expert verified
The ratio of the aluminum sphere's radius to the lead sphere's radius is approximately 1.61.
Step by step solution
01
Understand the Problem
We are given two spheres, one made of lead and the other made of aluminum, both having equal mass. We need to find the ratio of their radii.
02
Identify Relevant Formulas
The volume of a sphere is given by the formula \( V = \frac{4}{3}\pi r^3 \). Mass is equal to density times volume, given by \( m = \rho V \). Since the masses are equal, we have \( \rho_{lead} V_{lead} = \rho_{aluminum} V_{aluminum} \).
03
Express Volumes in Terms of Radii
Using the sphere volume formula, express the volumes for lead and aluminum spheres:- \( V_{lead} = \frac{4}{3}\pi r_{lead}^3 \)- \( V_{aluminum} = \frac{4}{3}\pi r_{aluminum}^3 \)
04
Set Up the Mass Equation
Since the masses are the same, set their equations equal:\( \rho_{lead} \cdot \frac{4}{3} \pi r_{lead}^3 = \rho_{aluminum} \cdot \frac{4}{3} \pi r_{aluminum}^3 \).
05
Simplify the Equation
Cancel out the common terms \( \frac{4}{3}\pi \):\( \rho_{lead} r_{lead}^3 = \rho_{aluminum} r_{aluminum}^3 \).
06
Solve for the Radius Ratio
Rearrange the simplified equation to find the radius ratio:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = \frac{\rho_{lead}}{\rho_{aluminum}} \).
07
Calculate the Ratio Using Densities
Use the known densities of lead \( \rho_{lead} \approx 11340 \, \text{kg/m}^3 \) and aluminum \( \rho_{aluminum} \approx 2700 \, \text{kg/m}^3 \) to calculate:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = \frac{11340}{2700} \).Simplify to find:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = 4.2 \).
08
Find the Final Radius Ratio
Take the cube root to find the final ratio:\( \frac{r_{aluminum}}{r_{lead}} = \sqrt[3]{4.2} \). Using approximation, this gives \( \frac{r_{aluminum}}{r_{lead}} \approx 1.61 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sphere Volume Calculation
To understand the calculation of a sphere's volume, imagine a perfectly round object like a basketball. The volume of this sphere is the space it occupies, which can be mathematically described using a specific formula. The formula to calculate the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \), where \( V \) represents volume and \( r \) is the radius of the sphere. This formula is derived considering that a sphere is a symmetrical three-dimensional shape.
- \( \pi \) is a constant, approximately equal to 3.14159.
- \( r^3 \) means we multiply the radius by itself twice, showing how the volume increases rapidly with the growth of radius.
Density and Mass Relationship
Density is a measure of how much mass is contained in a given volume. It's like understanding how 'heavy' something feels for its size. The formula to represent this relationship is \( \rho = \frac{m}{V} \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.
- This relationship is pivotal in physics as it helps to identify how compact a material is.
- A higher density means more mass in a smaller volume, whereas a lower density implies lighter mass in the same volume.
Material Density Comparison
When comparing two materials, such as lead and aluminum, their densities serve as a critical point of comparison. Lead is denser than aluminum, meaning for the same mass, lead takes up less space than aluminum. Here, we used the densities of lead \( \rho_{lead} \approx 11340 \text{ kg/m}^3 \) and aluminum \( \rho_{aluminum} \approx 2700 \text{ kg/m}^3 \).
- The higher the density, the smaller the volume for the same mass.
- The equation \( \rho_{lead} V_{lead} = \rho_{aluminum} V_{aluminum} \) signifies equal masses but differing volumes.