/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A uniform lead sphere and a unif... [FREE SOLUTION] | 91Ó°ÊÓ

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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 radius of the aluminum sphere to the radius of the lead sphere is approximately 1.53.

Step by step solution

01

Recall the formula for the volume of a sphere.

The formula for the volume of a sphere is given by \( V = \frac{4}{3} \pi \ r^{3} \) where \( r \) is the radius.
02

Express the mass of a sphere.

The mass of a sphere is given by \( m = \rho V \), where \( \rho \) is the density of the material and \( V \) is the volume of the sphere. Therefore, using the definition of \( V \), we have \( m = \rho \frac{4}{3} \pi \ r^{3} \).
03

Set up and solve the ratio.

Given that the masses of the two spheres are equal, the ratio of the radii cubed of the aluminum sphere (\( r_{A} \)) to the lead sphere (\( r_{L} \)) is equal to the inverse ratio of their densities:\( \frac{r_{A}^{3}}{r_{L}^{3}} = \frac{\rho_{L}}{\rho_{A}} \). To find \( \frac{r_{A}}{r_{L}} \), we take the cube root of \( \frac{\rho_{L}}{\rho_{A}} \). Using known values for the densities of lead (11340 kg/m³) and aluminium(2700 kg/m³), we get \( \frac{r_{A}}{r_{L}} = \sqrt[3]{\frac{11340}{2700}} \).

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Key Concepts

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

Sphere Volume
To understand the concept of sphere volume, imagine inflating a balloon. As it gets larger, the volume inside increases. Mathematically, the volume \( V \) of a sphere is calculated using the formula:
  • \( V = \frac{4}{3} \pi r^3 \)
Here, \( r \) represents the radius of the sphere, which is the distance from its center to any point on its surface. The constant \( \pi \) is approximately 3.14159, though it's used in its exact form in calculations when possible.
The cubed radius \( r^3 \) signifies that the volume grows immensely with even small increases in radius, as the radius is multiplied by itself twice. Therefore, if you double the radius, the volume increases by a factor of eight \((2^3 = 8)\). Remember, the volume captured by a sphere depends heavily on its radius, a core principle that applies to our exercise problem.
Density Relation
Density is a measure of how much mass occupies a particular volume. In simpler terms, it's essentially how "heavy" something feels for its size. This is calculated with the formula:
  • \( \rho = \frac{m}{V} \)
Here, \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume.
In the given exercise problem, the concept of density is essential as it relates directly to both materials' properties: lead and aluminum. Although both spheres have the same mass, their densities are different, impacting their respective volumes, and consequently, their radii. Since lead is denser than aluminum, an equivalent mass of lead will occupy less volume than the same mass of aluminum.
This relationship explains why in our solution, the ratio of the radii is inversely proportional to the ratio of the densities. Understanding density helps us grasp why lighter materials take up more space and heavier ones take up less.
Cube Root Calculation
Cube root calculations are about finding a number that, when multiplied by itself three times, equals another number. It's like reversing the process of cubing a number.
In the context of the problem, we need to find the cube root to determine the ratio of radii given the radii are related by their cubed values. Let's see this relationship mathematically:
  • \( \frac{r_A^3}{r_L^3} = \frac{\rho_L}{\rho_A} \)
  • \( \frac{r_A}{r_L} = \sqrt[3]{\frac{\rho_L}{\rho_A}} \)
In our problem, the cube root is taken of the density ratio \( \frac{11340}{2700} \), simplifying the calculation to find the radius ratio. The cube root is a necessary step here because the initial equation gives us a relation in terms of cube powers, and finding the cube root simplifies that relation to direct radii. Remember, the cube root calculation translates abstract mathematical ratios into something tangible like the radius, accommodating easier understanding of practical problems.

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Most popular questions from this chapter

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