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 $1.08g/c{m}^2$ . To what fraction of its expanded body volume must the fish inflate the air so as to reduce its density to that of water ?

Answer 1

Given,

density of the inflated fish equal to density of water

Density of fish,${\rho }_f=1.08g/c{m}^3$

Density of water, ${\rho }_w=1g/c{m}^3$

Volume of fish when collapsed, $v$

Volume of air added by fish into bone, $v_{af}$

Volume of inflated body $v+v_{af}$

From conservation of mass of fish

${\rho }_{f}v={\rho }_w(v+v_{af})$

$1.08v=1\times (v+v_{af})$

$\frac{v_{af}}{v}=0.08\Rightarrow \frac{v}{v_{af}}=12.5$

Add 1 on both side

$\frac{v}{v_{af}}+1=12.5+1$

$\frac{v+v_{af}}{v_{af}}=13.5$

Fraction of volume of air added to the volume of inflated body

$\frac{v_{af}}{v+v_{af}}=0.074=7.4percent$