Question

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

The ratio of the aluminum sphere's radius to the lead sphere's radius is approximately 1.61.

Step-by-step solution

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.

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 \( \rho_{lead} V_{lead} = \rho_{aluminum} V_{aluminum} \).

Express Volumes in Terms of Radii

Using the sphere volume formula, express the volumes for lead and aluminum spheres:- \( V_{lead} = \frac{4}{3}\pi r_{lead}^3 \)- \( V_{aluminum} = \frac{4}{3}\pi r_{aluminum}^3 \)

Set Up the Mass Equation

Since the masses are the same, set their equations equal:\( \rho_{lead} \cdot \frac{4}{3} \pi r_{lead}^3 = \rho_{aluminum} \cdot \frac{4}{3} \pi r_{aluminum}^3 \).

Simplify the Equation

Cancel out the common terms \( \frac{4}{3}\pi \):\( \rho_{lead} r_{lead}^3 = \rho_{aluminum} r_{aluminum}^3 \).

Solve for the Radius Ratio

Rearrange the simplified equation to find the radius ratio:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = \frac{\rho_{lead}}{\rho_{aluminum}} \).

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:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = 4.2 \).

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 \( \frac{r_{aluminum}}{r_{lead}} \approx 1.61 \).

Key concepts

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.

Calculating this volume is crucial, especially when comparing the volumes of spheres made from different materials, such as in physics problems and engineering applications.

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.

When the mass remains constant, the volume and the density are inversely related. This means, if the volume increases, the density decreases, assuming the mass doesn't change. In our problem, both spheres made of different materials have the same mass, but differing volumes due to their material's differing densities.

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.

This comparison enabled us to find the radius ratio of aluminum to lead spheres, resulting in \( \frac{r_{aluminum}}{r_{lead}} \approx 1.61 \). This ratio shows that the aluminum sphere needs a larger radius to reach the same mass as the smaller lead sphere due to its lower density.