91Ó°ÊÓ

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$g$.

$F=\mathrm{ÒÏ}Vg$

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.

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

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

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

Here, ${\mathrm{ÒÏ}}_f$is the density of the fish, ${\mathrm{ÒÏ}}_{\text{air}}$is the density of air, ${\mathrm{ÒÏ}}_w$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

$g$,

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

Divide the above equation by $V$

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

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

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

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

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

$\left({\mathrm{ÒÏ}}_f−{\mathrm{ÒÏ}}_w\right)=\left({\mathrm{ÒÏ}}_w−{\mathrm{ÒÏ}}_{\text{air}}\right)X$

$X=\frac{\left({\mathrm{ÒÏ}}_f−{\mathrm{ÒÏ}}_w\right)}{\left({\mathrm{ÒÏ}}_w−{\mathrm{ÒÏ}}_{\text{air}}\right)}$

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

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

$=0.0801$

$=8.01\mathrm{\%}$