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The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air. By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius.

Answer :

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To solve this problem it is necessary to apply the concept related to density and its definition with respect to mass and volume.

Density can be expressed as

[tex]\rho = \frac{m}{V}[/tex]

Where,

m = mass

V = Volume

The average density of fish air system is equal to the neutral bouyancy when the water displaced by the fish air system can be expressed as

[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]

Where,

[tex]m_a[/tex]= Mass of air

[tex]m_f =[/tex] Mass of fish

[tex]V_a[/tex]= Volume of air

[tex]V_f[/tex]= Volume of fish

As we know that m = \rho V  we can replacing the equivalent value for the mass fro fish and air, then

[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]

[tex]\rho_{b} = \frac{\rho_a V_a+\rho_f V_f}{V_a+V_f}[/tex]

[tex]\rho_{b}(V_a+V_f)=\rho_a V_a+\rho_f V_f[/tex]

[tex]\rho_{b} V_a+\rho_{b}V_f = \rho_a V_a+\rho_f V_f[/tex]

[tex]\rho_{b} V_a- \rho_a V_a=\rho_f V_f-\rho_{b}V_f[/tex]

[tex]V_a(\rho_{b}-\rho_a)=V_f(\rho_f-\rho_{b})[/tex]

[tex]\frac{V_a}{\rho_f}=\frac{(\rho_f-\rho_{b})}{(\rho_{b}-\rho_a)}[/tex]

Replacing with our values we have that

[tex]\frac{V_a}{\rho_f}=\frac{(1000-1080)}{(1.2-1000)}[/tex]

[tex]\frac{V_a}{\rho_f}=0.08[/tex]

[tex]\frac{V_a}{\rho_f}=8\%[/tex]

Therefore the percentage of Volume is increased by 8%

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