Question

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 1.05 g/cm3. To what fraction of its expanded body volume must the fish inflate the air sacs to reduce its density to that of water?

Answer 1

To reduce its density to that of water, the fish needs to inflate its air sacs to a certain fraction of its expanded body volume.

Step-by-step explanation:

To reduce its density to that of water, the fish needs to inflate its air sacs to a certain fraction of its expanded body volume. We can use the formula:

FV = (m_air_sacs) / (m_air_sacs + m_fish)

where FV is the fractional volume, m_air_sacs is the mass of the air sacs, and m_fish is the mass of the fish. Since density is mass divided by volume, we can use the equation:

d_fish = (m_fish) / (V_fish)

where d_fish is the density of the fish, m_fish is the mass of the fish, and V_fish is the volume of the fish. By combining these two equations, we can solve for the fractional volume:

FV = [1 - (d_water / d_fish)]

Answer 2

V’/V= 1.05

Step-by-step explanation:

The density is defined with the ratio of the mass to the volume, for the fish with the collapsed sack

ρ₁ = m / V

The density of the fish with the bag full of air is

ρ₂ = m / V’

For the fish to float if it exerts its density must be exactly equal to that of the surrounding water

We clear the mass and match

m = ρ1 V = ρ2 V'

ρ₁ V = ρ₂ V’

V ’/ V = ​​ρ₁ / ρ₂

V ’/ V = ​​1.05 / 1

V ’= 1.05 V

This is that the fish should increase its volume by 5%