Physics!!!!!! A fish maintains its depth in fresh water by adjusting the

Please hep... i keep getting 0.12, but its wrong...!!!

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 1.12 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?

Thx!

The mass of the air is insignificant with respect to the weight of the fish. So we can write.

Mo = Me

where Mo is the mass of the fish before expanding and Me is its mass after expanding.

Mo = 1.12g/cc * Vo

The density of water is about 1.0 g/cc

Me = 1.0/cc * Ve

So

1.12g/cc * Vo = 1.0/cc * Ve

Note that

Vo/Ve = 1.0/1.12

We are looking for
(Ve - Vo)/Ve = 1 - Vo/Ve

1 - 1.0/1.12 = 0.107 = 10.7%

Lets assume the unexpanded fish has a volume of 1cm^3. It would have a mass of 1.12g.

So the expanded volume would be:
V = m/D
V = 1.12g / (1g/cm^3)
V = 1.12cm^3

The difference in size is the size of the air sacs: 1.12cm^3 - 1.0cm^3 => 0.12cm^3

Now, if it wanted the fraction of the *unexpanded* volume, you'd have 0.12. But it wants the fraction of the expanded volume.

0.12cm^3 / 1.12cm^3 => 0.107

0.12 as a fraction is 12/100, or 3/25ths.

who cares catch the fish and fry it up instead

.12 isn't a fraction, a fraction would be 3/25