Archimede's Principle Question?

  • Thread starter haflanagan
  • Start date
  • #1

Homework Statement



To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is 5.0 x 10^3 kg/m3. What fraction of the specimen’s apparent volume is solid?

Homework Equations



Buoyant force = weight of water displaced
Weight = density * volume * g

The Attempt at a Solution



I don't even know...
Pr=density rock
Pa=density air
Pw=density water
Vs=volume solid
Vh=volume hollow

I thought that maybe:
Weight of rock in air = Pr*vs*g+Pa*vh*g
Therefore, weight of rock in water = (Pr*vs*g+Pa*vh*g)/2

And that's as far as I got. And I don't even think that's right. I'm so lost.
 

Answers and Replies

  • #2
jbriggs444
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You are given the density of rock as part of the problem description. So you do know ρr

The density of air is low enough that by comparison to rock or water, it is negligible. So you can safely assume that ρa = 0.

The density of water is a well known constant. One kilogram per liter. So you know ρw

Your formula for the weight of the rock in air is technically correct. But the problem is talking about the apparent weight of the rock in air. That would be the true weight minus buoyancy. Can you write a formula for the apparent weight of the rock in air?

Your formula for the apparent weight of the rock in water is based on the givens of the problem -- it's half of the apparent weight in air. But there is also another formula for the apparent weight of the rock in water...
 

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