# Buoyancy and Archimedes Principle, volume ratio/density question

1. Jan 30, 2013

### phoebz

A geode is a hollow rock with a solid shell and an air-filled interior. Suppose a particular geode weighs twice as much in air as it does when completely submerged in water. If the density of the solid part of the geode is 3100 km/m^3 , what fraction of the geode's volume is hollow?

The density of air is 1.20kg/m^3 and density of water is 1000, and I have been trying to use the equation Fb (force of buoyancy) = W (weight of the object)

(I use the symbol ρ for density)
I have:
Fb=w
ρ_water(volume)g=ρ_geode(volume)g2

and Vair/Vgeode as my unknown... I'm confused on what densities to use.. and if I'm even on the right track.

Any help would be appreciated!

2. Jan 30, 2013

### Staff: Mentor

Neglect the weight and density of the air. Let V be the volume of the geode, and let $\varphi$ be the fraction that is solid. In terms of V, $\varphi$, and ρs, how much does the geode weight in air. If the geode is fully submerged under water, what is the weight of the water displaced in terms of the volume V and density of water ρw? In terms of these parameters, what is the weight of the geode under water?

3. Feb 9, 2013

### phoebz

Okay, for the weight in the air I have: W =Vρsφg

The volume of the water displaced would be the volume of the geode...which we aren't given :s.

But I know the geodes weight under water needs to be multiplied by two to equal the weight of the geode in the air so:

Vρsφg = 2(Vρwg)

solve for φ?

I get φ= 0.645, so then minus that from 1 and I get 0.355... is that about right?

Last edited: Feb 9, 2013
4. Feb 9, 2013

### phoebz

I tried putting in my answer for Vair/Vsolid as 0.55 and it was wrong, so I guess I've messed up somewhere

Last edited: Feb 9, 2013
5. Feb 9, 2013

### haruspex

But according to your previous post you calculated it as .355. Which answer did you give?

6. Feb 9, 2013

### Staff: Mentor

If the volume of the geode is V, the density of the rock is ρs, and the volume fraction of air is $\phi$, then the

weight of the geode in air = $V\rho_s(1-\phi)g$

If the geode is fully submerged, the the

weight of water displaced = $V\rho_wg$

From Archimedes principle

weight of geode in water = $V\rho_s(1-\phi)g$-$V\rho_wg$

The ratio of the geode weight in air to the geode weight in water is:

$$\frac{V\rho_s(1-\phi)g}{V\rho_s(1-\phi)g-V\rho_wg}=\frac{\rho_s(1-\phi)}{\rho_s(1-\phi)-\rho_w}$$

This ratio is equal to 2. So, solve for $\phi$.