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Vinil7 [7]
3 years ago
6

A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density th

e same as that of the water. Suppose that with its air sacs collapsed, a fish has a density of 1.18 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?
Biology
1 answer:
Volgvan3 years ago
5 0

Answer:

0.153

Explanation:

We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.

So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.

ρVg = ρ'V'g + ρ"V"g

ρV = ρ'V'g + ρ"V"

Its new body volume = volume of water displaced, V = V' + V"

ρ(V' + V") = ρ'V' + ρ"V"

ρV' + ρV" = ρ'V' + ρ"V"

ρV' - ρ"V'  = ρ'V" - ρV"

(ρ - ρ")V'  = (ρ' - ρ)V"

V'/V" = (ρ - ρ")/(ρ' - ρ)

= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)

= (0.9988 g/cm³ ÷ 0.18 g/cm³)

V'/V" = 5.55

Since V = V' + V"

V' = V - V"

(V - V")/V" = 5.55

V/V" - V"/V" = 5.55

V/V" - 1 = 5.55

V/V" = 5.55 + 1

V/V" = 6.55

V"/V = 1/6.55

V"/V = 0.153

So, the fish must inflate its air sacs to 0.153 of its expanded body volume

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