91Ó°ÊÓ

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

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

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,$\mathrm{ÒÏ}=\frac{M}{V}$ …(¾±)

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),

$$\begin{gathered} {\mathrm{ÒÏ}}_{fish}=\frac{m_{fish}}{V} \\ =1.08g/c{m}^3 \end{gathered}$$ …(¾±¾±)

$$\begin{gathered} p_w=\frac{m_{fish}}{V+V_a} \\ =1.00g/c{m}^3 \end{gathered}$$ …(¾±¾±¾±)

where,$p_w$is the density of water.

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

role="math" localid="1657254700810" $$\begin{gathered} m_{fish}={\mathrm{ÒÏ}}_{fish}V \\ =\left(1.08g/c{m}^3\right).V \end{gathered}$$ …(¾±±¹)

$$\begin{gathered} m_{fish}={\mathrm{ÒÏ}}_w\left(V+V_a\right) \\ =\left(1.00g/c{m}^3\right).\left(V+V_a\right) \end{gathered}$$ …(±¹)

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

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

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%.