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A hollow spherical aluminum shell floats exactly half submerged in water. The density of aluminum is 2710 kg/m³ and that of water is 1000 kg/m³. Draw a diagram and define all relevant variables. What is the ratio of the outer to inner radius of the spherical shell?

A hollow spherical aluminum shell floats exactly half submerged in water. The density of aluminum is 2710 kg/m³ and that of water is 1000 kg/m³. Draw a diagram and define all relevant variables. What is the ratio of the outer to inner radius of the spherical shell?

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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### Problem Statement A hollow spherical aluminum shell floats exactly half submerged in water. The density of aluminum is \( 2710 \, \text{kg/m}^3 \) and that of water is \( 1000 \, \text{kg/m}^3 \). 1. Draw a diagram and define all relevant variables. 2. What is the ratio of the outer to inner radius of the spherical shell? ### Explanation This problem involves understanding buoyancy and density. The shell floats half-submerged because its average density equals that of water. The task is to consider the shell's geometry and apply the concept of buoyancy to find the desired ratio of radii. **Variables Explanation:** - Let \( R \) be the outer radius of the shell. - Let \( r \) be the inner radius of the shell. - The volume of the sphere is determined by the formula \( V = \frac{4}{3}\pi R^3 \). - The volume of the hollow part is \( \frac{4}{3}\pi r^3 \). **Relevant Equations:** - Archimedes' principle of buoyancy: The buoyant force equals the weight of the displaced fluid. - Density relation: Mass = Density × Volume. ### Diagram (Hypothetical Explanation) Imagine a sphere with an outer radius \( R \) floating such that its center is at the surface of the water. The inner hollow portion has a radius \( r \). The thickness of the shell is \( R - r \). This setup allows us to understand how the shell’s volume relates to its buoyancy and the density of the materials involved. By solving algebraically, you can find the ratio \( \frac{R}{r} \) to maintain balance between the buoyant force and gravitational force on the half-submerged sphere.
Transcribed Image Text:### Problem Statement A hollow spherical aluminum shell floats exactly half submerged in water. The density of aluminum is \( 2710 \, \text{kg/m}^3 \) and that of water is \( 1000 \, \text{kg/m}^3 \). 1. Draw a diagram and define all relevant variables. 2. What is the ratio of the outer to inner radius of the spherical shell? ### Explanation This problem involves understanding buoyancy and density. The shell floats half-submerged because its average density equals that of water. The task is to consider the shell's geometry and apply the concept of buoyancy to find the desired ratio of radii. **Variables Explanation:** - Let \( R \) be the outer radius of the shell. - Let \( r \) be the inner radius of the shell. - The volume of the sphere is determined by the formula \( V = \frac{4}{3}\pi R^3 \). - The volume of the hollow part is \( \frac{4}{3}\pi r^3 \). **Relevant Equations:** - Archimedes' principle of buoyancy: The buoyant force equals the weight of the displaced fluid. - Density relation: Mass = Density × Volume. ### Diagram (Hypothetical Explanation) Imagine a sphere with an outer radius \( R \) floating such that its center is at the surface of the water. The inner hollow portion has a radius \( r \). The thickness of the shell is \( R - r \). This setup allows us to understand how the shell’s volume relates to its buoyancy and the density of the materials involved. By solving algebraically, you can find the ratio \( \frac{R}{r} \) to maintain balance between the buoyant force and gravitational force on the half-submerged sphere.
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