Question

The average density of the body of a fish is $1080\mathrm{kg}/{\mathrm{m}}^3$. To keep from sinking, a fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air. By what percent must the fish increase its volume to be neutrally buoyant in fresh water? The density of air at ${20}^{\circ }\mathrm{C}\text{is}1.19\mathrm{kg}/{\mathrm{m}}^3$.

Short answer

The percentage of increase in its volume of fish is $8.01\mathrm{\%}$

Step-by-step solution

Step :1 Introduction 

The density of water discharged reflects the biological body's volume. The sum of the format's density, volume, and buoyant force is determined., and acceleration due to gravity$g$.

$F=\rho Vg$

A fish's internal air bladder helps it to control its buoyancy.

By extending or contracting the size of its bladder, the fish may simply move up and down, increasing or decreasing its soaring strength.

Step :2 Explanation 

According to Archimedes' Principle, the drifting energy is proportional to the displacement caused.

We have the following equation if the fish is neutrally buoyant:

${\rho }_{f}Vg+{\rho }_{air}{V}_{et}g={\rho }_{w}g\left(V+V_{et}\right)$

Here, ${\rho }_f$is the density of the fish, ${\rho }_{\text{air}}$is the density of air, ${\rho }_w$the water is probably viscosity, $V$is the fish's actual quantity, $V_{ext}$is the additional volume created by inflating the air bladder, and $g$is the acceleration due to gravity.

On the left side of the equation is the weight of the fish with an expanded air bladder.

On the right hand side we can see the weight of the water it displaced.

By multiplying the previous equation by

$g$,

${\rho }_{f}V+{\rho }_{air}{V}_{ext}={\rho }_w\left(V+V_{ext}\right)$

Divide the above equation by $V$

${\rho }_f+{\rho }_{\text{air}}\frac{V_{ext}}{V}={\rho }_w+{\rho }_w\frac{V_{ext}}{V}\dots \dots \left(1\right)$

Let $X$be the proportion change in the amount required for the fish to be primarily water, then:

$X=\frac{V_{ext}}{V}$

Step :3 Substitution

Substitute equation (2) into (1), we get

${\rho }_f+{\rho }_{air}X={\rho }_w+{\rho }_{w}X$

$\left({\rho }_f-{\rho }_w\right)=\left({\rho }_w-{\rho }_{\text{air}}\right)X$

$X=\frac{\left({\rho }_f-{\rho }_w\right)}{\left({\rho }_w-{\rho }_{\text{air}}\right)}$

Substitute $1080\mathrm{kg}/{\mathrm{m}}^3\text{for}{\rho }_f,1000\mathrm{kg}/{\mathrm{m}}^3\text{for}{\rho }_{\mathrm{w}}\text{and}1.19\mathrm{kg}/{\mathrm{m}}^3\text{for}{\rho }_{\text{air}}$

$X=\frac{\left(1080\mathrm{kg}/{\mathrm{m}}^3-1000\mathrm{kg}/{\mathrm{m}}^3\right)}{\left(1000\mathrm{kg}/{\mathrm{m}}^3-1.19\mathrm{kg}/{\mathrm{m}}^3\right)}$

$=0.0801$

$=8.01\mathrm{\%}$