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$\mathbf{1}\mathbf{.}\mathbf{08}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$. To what fraction of its expanded body volume must the fish inflate the air sacs to reduce its density to that of water?

Short answer

The fraction of its expanded body volume that the fish must inflate the air sacs to reduce its density to that of water is 7.4%

Step-by-step solution

The given data

Density of fish,$\mathrm{\rho }=1.08\mathrm{g}/{\mathrm{cm}}^3$

Density of water,$\mathrm{\rho }=1.00\mathrm{g}/{\mathrm{cm}}^3$

Understanding the concept of density

The density is inversely proportional to that of the volume of a body. Using this formula of density and volume, we can write two expressions for the mass of fish in terms of the density of water and fish. Equating these two equations, we can find the required ratio$\frac{{\mathbf{V}}_{\mathbf{a}}}{\mathbf{V}\mathbf{+}{\mathbf{V}}_{\mathbf{a}}}$.

Formula:

Density of a material,$\rho =\frac{M}{V}$ …(i)

Calculation for fraction of volume of body of fish

Let the volume of expanded air sacs be $V_a$and the volume of fish with its air sacs collapsed be V.

Using equation (i),

$$\begin{gathered} {\rho }_{fish}=\frac{m_{fish}}{V} \\ =1.08g/c{m}^3 \end{gathered}$$ …(ii)

$$\begin{gathered} p_w=\frac{m_{fish}}{V+V_a} \\ =1.00g/c{m}^3 \end{gathered}$$ …(iii)

where,$p_w$is the density of water.

Using equation (ii) & (iii), we can write

role="math" localid="1657254700810" $$\begin{gathered} m_{fish}={\rho }_{fish}V \\ =\left(1.08g/c{m}^3\right).V \end{gathered}$$ …(iv)

$$\begin{gathered} m_{fish}={\rho }_w\left(V+V_a\right) \\ =\left(1.00g/c{m}^3\right).\left(V+V_a\right) \end{gathered}$$ …(v)

Again, by comparing equations (iv) and (v), we get

$$\begin{gathered} {\rho }_{fish}V={\rho }_w\left(V+V_a\right) \\ \frac{{\rho }_{fish}}{{\rho }_w}=\frac{\left(V+V_a\right)}{V} \\ \frac{\left(V+V_a\right)}{V}=\frac{1.08}{1.00} \\ V+V_a=1.08V \\ {V}_a=\left(1.08-1\right)V \\ =0.08V \end{gathered}$$

Substituting the value $V_a$of, we get

$$\begin{gathered} \frac{V}{V+V_a}=\frac{0.08}{1.08} \\ =0.074 \\ =7.4\% \end{gathered}$$

Hence, the fraction of volume of the body of fish inside water is 7.4%.