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 of1.08g/cm3. 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

Expert verified

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

01

The given data

Density of fish,ρ=1.08g/cm3

Density of water,ρ=1.00g/cm3

02

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 ratioVaV+Va.

Formula:

Density of a material,ρ=MV …(i)

03

Calculation for fraction of volume of body of fish

Let the volume of expanded air sacs be Vaand the volume of fish with its air sacs collapsed be V.

Using equation (i),

ρfish=mfishV=1.08g/cm3 …(ii)

pw=mfishV+Va=1.00g/cm3 …(iii)

where,pwis the density of water.

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

role="math" localid="1657254700810" mfish=ρfishV=1.08g/cm3.V …(iv)

mfish=ρwV+Va=1.00g/cm3.V+Va …(v)

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

ρfishV=ρwV+Vaρfishρw=V+VaVV+VaV=1.081.00V+Va=1.08VVa=1.08-1V=0.08V

Substituting the value Vaof, we get

VV+Va=0.081.08=0.074=7.4%

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

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