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Chapter 15: Problem 110

A geode is a hollow rock with a solid shell and an airfilled interior. Suppose a particular geode weighs twice as much in air as it does when completely submerged in water. If the density of the solid part of the geode is \(2500 \mathrm{kg} / \mathrm{m}^{3}\), what fraction of the geode's volume is hollow?

Short Answer

Expert verified
The hollow fraction of the geode is 0.2 or 20%.

Step by step solution

01

Understand the Problem Statement

We are asked to find the fraction of a geode's volume that is hollow. The geode weighs twice as much in air as in water. This suggests we need to consider buoyancy and density aspects to solve the problem.
02

Apply Archimedes' Principle

According to Archimedes' principle, the buoyant force on the geode when submerged equals the weight of the water displaced by it. Let the volume of the whole geode be \(V\) and the hollow part be \(V_h\), then the solid part is \(V_s = V - V_h\).
03

Express Weight Equations

The weight of the geode in air is \(W_{air} = \rho_s V_s g\), where \(\rho_s = 2500 \ \text{kg/m}^3\) is the density of the solid, and \(g\) is the gravitational acceleration. In water, the geode's weight is \(W_{water} = \rho_s V_s g - \rho_w V g\), where \(\rho_w = 1000 \ \text{kg/m}^3\) is the density of water.
04

Set Up the Weight Equation

We know that \(W_{air} = 2 W_{water}\). Substitute the expressions from Step 3: \(\rho_s V_s g = 2 (\rho_s V_s g - \rho_w V g)\).
05

Simplify the Equation

Simplify the equation to find the relationship between volumes: \(\rho_s V_s = 2 \rho_s V_s - 2 \rho_w V\). This simplifies to \(\rho_s V_s = 2 \rho_w V\).
06

Relate Volumes of Solid and Whole Geode

Substitute \(V_s = V - V_h\), giving us \(\rho_s (V - V_h) = 2 \rho_w V\). Solving gives \(V_h = V - \frac{2\rho_w V}{\rho_s}\).
07

Find the Fraction of the Hollow Portion

The fraction of the geode's volume that is hollow is \(\frac{V_h}{V} = 1 - \frac{2\rho_w}{\rho_s}\). Substitute \(\rho_s = 2500 \ \text{kg/m}^3\) and \(\rho_w = 1000 \ \text{kg/m}^3\), yielding the fraction \(\frac{V_h}{V} = 1 - \frac{2 \times 1000}{2500} = 1 - 0.8 = 0.2\).

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Key Concepts

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

Understanding Buoyancy
Buoyancy is the upward force that a fluid exerts on an object placed in it. When a geode is submerged in water, it experiences this buoyant force, which stems from Archimedes' principle. This principle states that the buoyant force on an object, submerged in a fluid, is equal to the weight of the fluid displaced by the object.

In the exercise above, when the geode is submerged, it appears to weigh less than it does in air. This indicates that the buoyant force is acting upwards, opposing the weight of the geode. Understanding buoyancy helps explain how objects float or sink based on their density relative to the fluid.

Key points to remember about buoyancy:
  • Objects that weigh less than the fluid they displace will float.
  • The ability for an object to float is related to its average density compared to the density of the fluid.
  • Buoyant force depends only on the volume of fluid displaced, not the weight of the object.
Exploring Density
Density is a fundamental property that relates mass to volume, usually expressed as \ \( \rho = \frac{m}{V} \ \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume. In our example, the density of the solid portion of the geode is given as 2500 kg/m鲁.

This tells us how compact the material is. When we compare densities, the relative density plays a crucial role in determining whether an object like a geode will float or sink.

  • In the exercise, the density of water is given as 1000 kg/m鲁.
  • The density of the solid part of the geode is greater than that of water, indicating it would normally sink. However, the hollow interior changes its overall density.
  • This is why Archimedes' principle can be applied to understand how much of the geode's volume is solid and how much is hollow.
Volume Calculation Basics
The exercise involves calculating volumes, which are key to solving weight and buoyancy problems. The geode's total volume comprises its solid and hollow parts, and understanding how to calculate these volumes helps find the fraction of the geode that is hollow.

The volume calculation allows us to apply Archimedes' principle effectively. Knowing the geode weighs more in air than in water suggests that air contributes to its volume but not its weight, as air is less dense than water or the solid geode material. Points to consider with volume:
  • The volume of the geode when submerged determines the amount of water displaced, which determines the buoyant force.
  • Using the given densities, we can set up equations that relate the total volume, solid volume, and weight of the geode in different conditions.
  • The calculated volumes help us determine what fraction, \( \frac{V_h}{V} \), is hollow, allowing us to conclude that 20% of the geode's volume is hollow based on the exercise's final equation.

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