Fraction of the fish’s total volume.

Question

Suppose a fish has an average density of 1030 kg/m³, compared to the density of the surrounding water, 1000 k/m³. One way the fish can keep from slowly sinking is by using an air bladder (the density of air is 1.2 kg/m³). What fraction of the fish’s total volume must be air in order for the fish to be neutrally buoyant (the buoyant force is equal and opposite to the weight). Assume that the volume V of the fish’s tissue is fixed, so in order to increase the volume U of the air bladder, the total volume of the fish V + U must increase.

Answer 1

Let

f = 1030 kg/m³ fish density,

w = 1000 hg/m³ water density,

a = 1.2 kg/m³ be air density,

g = 9.81 m/s² be gravitational acceleration.

For neutral buoyancy, the weight of the fish (including its air filled bladder)

must be the same as the buoyancy force of water.

The total mass of the fish together with its bladder is

Vf + Ua

The buoyancy force on an object of volume V+U is

(V+U)wg

So the equation is

(Vf + Ua)g = (V+U)wg

Manipulating it into giving us U/V:

Vf + Ua = Vw + Uw

U/V = (f-w)/(w-a)

Substituting actual numbers

U/V = 30/998.8 ≈ 0.03

So the fish must add about 3% of its original volume in the form of air.

Notice that this result is independent of the gravitational acceleration g.

So if you take your fish in its water bowl to the moon, it will be just as buoyant

as on the Earth.