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Magazine Chapter 14: 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 radii is approximately 1.613.
Step by step solution
01
Understand the Problem
We are given that a lead sphere and an aluminum sphere have the same mass and we need to find the ratio of their radii. The volume and mass relationship is essential here since both materials have different densities.
02
Recall the Formula for Volume
The volume \( V \) of a sphere is computed using the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere.
03
Relate Volume to Mass and Density
The mass \( m \) of a substance is given by \( m = \text{density} \times \text{volume} = \rho V \). For both spheres, the mass can be expressed as \( m = \rho \frac{4}{3} \pi r^3 \).
04
Determine Equations for Each Sphere
For the lead sphere with density \( \rho_{Lead} \) and radius \( r_{Lead} \), the mass equation is: \( m = \rho_{Lead} \frac{4}{3} \pi r_{Lead}^3 \). For the aluminum sphere with density \( \rho_{Al} \) and radius \( r_{Al} \): \( m = \rho_{Al} \frac{4}{3} \pi r_{Al}^3 \).
05
Set Equations for Mass Equal
Since both spheres have the same mass, their equations can be set equal: \( \rho_{Lead} \frac{4}{3} \pi r_{Lead}^3 = \rho_{Al} \frac{4}{3} \pi r_{Al}^3 \).
06
Simplify the Equation
Cancel \( \frac{4}{3} \pi \) from both sides, leaving: \( \rho_{Lead} r_{Lead}^3 = \rho_{Al} r_{Al}^3 \). Thus, \( r_{Al}^3 = \frac{\rho_{Lead}}{\rho_{Al}} r_{Lead}^3 \).
07
Solve for the Radius Ratio
Taking the cube root on both sides, we find \( r_{Al} = \left(\frac{\rho_{Lead}}{\rho_{Al}}\right)^{1/3} r_{Lead} \). Therefore, the ratio \( \frac{r_{Al}}{r_{Lead}} = \left(\frac{\rho_{Lead}}{\rho_{Al}}\right)^{1/3} \).
08
Substitute Density Values
Substitute the known densities: \( \rho_{Lead} = 11.34 \, \text{g/cm}^3 \) and \( \rho_{Al} = 2.70 \, \text{g/cm}^3 \) into the formula: \[ \frac{r_{Al}}{r_{Lead}} = \left(\frac{11.34}{2.70}\right)^{1/3} \].
09
Calculate the Ratio
Compute the ratio using the values: \[ \frac{r_{Al}}{r_{Lead}} = \left(\frac{11.34}{2.70}\right)^{1/3} = \left(4.2\right)^{1/3} \]. This gives us \( \frac{r_{Al}}{r_{Lead}} \approx 1.613 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sphere volume formulas
Understanding the volume of a sphere is vital when solving problems involving spheres. The formula for calculating the volume of a sphere involves finding the cube of its radius, which is then multiplied by the constant factor. The formula is: \( V = \frac{4}{3} \pi r^3 \), where \( V \) stands for volume and \( r \) represents the radius of the sphere.
This equation shows that the volume of a sphere increases significantly as its radius increases.
Knowing the way the volume correlates to the radius is crucial when comparing spheres with different properties, such as in determining which has a bigger enclosed space given similar or different masses.
This equation shows that the volume of a sphere increases significantly as its radius increases.
Knowing the way the volume correlates to the radius is crucial when comparing spheres with different properties, such as in determining which has a bigger enclosed space given similar or different masses.
Mass and density relationship
In physics, understanding the relationship between mass, volume, and density is fundamental.
The mass of an object is the product of its density and volume, written as \( m = \rho \cdot V \). Here, \( m \) represents mass, \( \rho \) is density, and \( V \) represents volume.
When dealing with spheres made from different materials like lead and aluminum, their densities will affect the volume and the sphere's characteristics.
Given that both spheres in our problem have the same mass, we can use density to determine the differences in size, specifically their radius.
Differences in density cause variations in volume, which helps determine how materials distribute across the mass.
The mass of an object is the product of its density and volume, written as \( m = \rho \cdot V \). Here, \( m \) represents mass, \( \rho \) is density, and \( V \) represents volume.
When dealing with spheres made from different materials like lead and aluminum, their densities will affect the volume and the sphere's characteristics.
Given that both spheres in our problem have the same mass, we can use density to determine the differences in size, specifically their radius.
Differences in density cause variations in volume, which helps determine how materials distribute across the mass.
Cube root calculations
The cube root of a number is a value that, when multiplied by itself twice, returns the original number.
In mathematics, the cube root is expressed as \( x^{1/3} \).
To solve the relationship between the radii of the aluminum and lead spheres, as explained, we resolve the equation \( r_{Al}^3 = \frac{\rho_{Lead}}{\rho_{Al}} r_{Lead}^3 \).
Finding the cube root on both sides, we derive \( r_{Al} = \left(\frac{\rho_{Lead}}{\rho_{Al}}\right)^{1/3} r_{Lead} \).
This calculation helps us in finding the proportion between different radii.
In mathematics, the cube root is expressed as \( x^{1/3} \).
To solve the relationship between the radii of the aluminum and lead spheres, as explained, we resolve the equation \( r_{Al}^3 = \frac{\rho_{Lead}}{\rho_{Al}} r_{Lead}^3 \).
Finding the cube root on both sides, we derive \( r_{Al} = \left(\frac{\rho_{Lead}}{\rho_{Al}}\right)^{1/3} r_{Lead} \).
This calculation helps us in finding the proportion between different radii.
- Calculators often have a cube root function to make this calculation easier.
- Understanding cube roots is essential in physics where power relationships occur.
