Fish maintaining position in water

[Zorodius]

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 1.08 g/cm^3. To what fraction of its expanded body volume must the fish inflate the air sacs to reduce its density to that of water?

Sorry, I'm spacing out entirely on how to set up this problem. Could someone give me a push in the right direction?

It sounds like we're looking for $\frac{V_A}{V_f + V_A}$, where Va is the volume of the air inside the fish and Vf is the volume of the fish.

We know that the fish, once inflated, has a density of $\rho_w$, the density of the water, which gives us:

$$\frac{m_f + m_A}{V_f + V_A} = \rho_w$$

Where Mf is the mass of the fish, Ma is the mass of the air, Vf is the volume of the fish, Va is the volume of the air, and pw is the density of the water.

I can only solve for $\frac{V_A}{V_f + V_A}$ in terms of $\rho_w, \rho_f, \rho_A,$ and $\frac{V_f}{V_f + V_A}$, which doesn't do me any good since that last thing is an unknown for which I have no helpful substitutions.

Can I get a hint on where to go with this?

 

[Zorodius]

I tried solving the problem this way:

Let x be the amount (between 0 and 1) of the inflated fish that is filled with air, then 1 - x is the amount of non-air fish and:

$$\rho_f(1-x)+\rho_A x = \rho_w$$

$$\rho_f - \rho_f x + \rho_A x = \rho_w$$

$$x = \frac{\rho_w - \rho_f}{\rho_A - \rho_f}$$

I tried evaluating this using the density of fresh water (998 kg/m^3), the density of the fish (1080 kg/m^3), and the density of air (1.21 kg/m^3), and came up with 0.076. The book's answer is 0.074. That is different enough to make me question whether I have done the problem correctly. Have I?

 

[krab]

998kg/m^3? I always thought 1000kg/m^3. Also, on this scale, density of air is negligible. So I get 1-(1/1.08)=0.074.

 

[Zorodius]

Thanks for the reply!