Buoyancy and Archimedes Principle, volume ratio/density question

[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!

 

[Chestermiller]

Neglect the weight and density of the air. Let V be the volume of the geode, and let [itex]\varphi[/itex] be the fraction that is solid. In terms of V, [itex]\varphi[/itex], and ρ_s, 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 ρ_w? In terms of these parameters, what is the weight of the geode under water?

 

[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?

 

[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

 

[haruspex]

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

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

 

[Chestermiller]

If the volume of the geode is V, the density of the rock is ρ_s, and the volume fraction of air is [itex]\phi[/itex], then the

weight of the geode in air = [itex]V\rho_s(1-\phi)g[/itex]

If the geode is fully submerged, the the

weight of water displaced = [itex]V\rho_wg[/itex]

From Archimedes principle

weight of geode in water = [itex]V\rho_s(1-\phi)g[/itex]-[itex]V\rho_wg[/itex]

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

[tex]\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}[/tex]

This ratio is equal to 2. So, solve for [itex]\phi[/itex].