Buoyancy and Archimedes Principle, volume ratio/density question

AI Thread Summary
A geode weighs twice as much in air as when submerged in water, with a solid density of 3100 kg/m^3. The discussion focuses on applying Archimedes' principle to determine the volume fraction of the geode that is hollow. The weight of the geode in air is expressed as Vρ_s(1-φ)g, while the weight of the water displaced is Vρ_wg. By setting up the equation based on the weight ratio, the fraction of the geode's volume that is solid is calculated to be approximately 0.645, leading to a hollow volume fraction of about 0.355. The calculations emphasize the importance of correctly applying density values and understanding buoyancy principles.
phoebz
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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!
 
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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?
 
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?
 
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I tried putting in my answer for Vair/Vsolid as 0.55 and it was wrong, so I guess I've messed up somewhere
 
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phoebz said:
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?
 
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.
 
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