Problem 1 - Fundamentals of Physics

Question

A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density the same as that of the water. Suppose that with its air sacs collapsed, a fish has a density of \(1.08 \mathrm{~g} / \mathrm{cm}^{3} .\) To what fraction of its expanded body volume must the fish inflate the air sacs to reduce its density to that of water?

Step-by-Step Solution

Answer

The fish must inflate 7.41% of its body volume with air to match water density.

Step 1: Understanding the Problem

The fish needs to adjust its density from 1.08 g/cm³ to 1.00 g/cm³ (the density of water) by inflating its air sacs. Our goal is to find the fraction of the fish's volume needs to be occupied by air to achieve this.

Step 2: Setting Up the Equation

Let the initial volume of the fish (without inflated sacs) be represented as \( V_f \), and the final volume with inflated air sacs as \( V_t \). The mass of fish is \( m = 1.08 \times V_f \). After inflating, the fish's density will be \( \frac{1.08 \times V_f}{V_t} = 1.00 \).

Step 3: Calculating the Final Volume

Rearrange the equation to solve for \( V_t \): \[ \frac{1.08 \times V_f}{V_t} = 1.00 \] Multiply both sides by \( V_t \) and divide by 1.00 to find: \[ V_t = 1.08 \times V_f \].

Step 4: Determining Volume Fraction of Air Sacs

The volume of the air sacs is \( V_t - V_f \). The fraction of total volume that must be air is \( \frac{V_t - V_f}{V_t} = \frac{1.08 \times V_f - V_f}{1.08 \times V_f} = \frac{0.08 \times V_f}{1.08 \times V_f} \). This simplifies to \( \frac{0.08}{1.08} \).

Step 5: Final Calculation

Calculate \( \frac{0.08}{1.08} = 0.0741 \). Thus, the fish must expand its body volume by about 7.41% with air to achieve neutral buoyancy.

Key Concepts

Density

Density is a measure of how much mass exists in a specific volume. When talking about buoyancy, density plays a crucial role in determining whether an object sinks or floats in a fluid. For our fish to float in water, its density must match the density of water, which is typically 1.00 g/cm³.

In more practical terms, if an object is denser than the fluid it is in, it will tend to sink. Conversely, if it is less dense, it will float. Density is calculated using the formula:

• Density = Mass / Volume

This relationship helps us understand how the fish can adjust its density by changing its volume, using air sacs. If its density is too high, it will adjust by taking in air to increase its volume without changing its mass, effectively lowering its density to achieve neutral buoyancy.

Volume

Volume is the space that an object occupies and is essential in understanding how buoyancy works. In the context of our fish, the fish's volume changes when it inflates or deflates its air sacs.

For the fish to achieve neutral buoyancy, it needs to increase its volume to reduce its density. This concept is beautifully demonstrated with Archimedes' principle, which states that a body will float or be neutrally buoyant when the weight of the fluid displaced is equal to the body's weight. Hence, by increasing its volume through inflation of the air sacs, the fish displaces enough water to equal its weight.

From a mathematical perspective, the volume change is managed by the formula:

• Final Volume = Initial Volume × (1 + Volume Fraction of Air Sacs)

This fraction helps in determining the expansion required for the fish to become buoyant.

Neutral Buoyancy

Neutral buoyancy is the point at which an object's density is equal to the fluid's density, resulting in a state where it neither sinks nor floats. This is the goal for the fish, as being neutrally buoyant allows it to conserve energy while maintaining its position in the water column.

Achieving neutral buoyancy requires precise adjustments to the fish's overall density, primarily done through changes in body volume. In our exercise, the fish starts with a density higher than water at 1.08 g/cm³ and must adjust to equal water's density of 1.00 g/cm³. By manipulating its internal air sacs, it changes its volume fraction, achieving a balance whereby its density equals that of the water.

• Percisely controlled air inflation helps in achieving that perfect balance.
• Neutral buoyancy means the fish can hover at its preferred depth effortlessly.

Air Sacs

Air sacs are specialized structures in fish that help them control their buoyancy. These sacs can inflate and deflate to adjust the fish's volume and density.

In our exercise, air sacs play a vital role. Initially, when the fish's air sacs are deflated, its density is 1.08 g/cm³. By inflating the air sacs, the fish increases its total volume, which subsequently reduces its overall density. The formula that helps calculate this process is:

• Volume Fraction of Air Sacs = (Increase in Volume / Total Volume)

Once inflated optimally to occupy about 7.41% of its total expanded volume, the fish's density matches that of the water, achieving neutral buoyancy.

These air sacs also help the fish in ascending or descending through water by simple adjustments, showcasing nature's efficiency in animal survival mechanisms.