Calculating Apparent Volume and Density: A Geologist's Guide

[tmkgemini]

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 solid part of the specimen has a density of 5.13×103kg/m3. What fraction of the specimen's apparent volume is solid?

I tried setting up a ratio of densities but it didn't work at all... please help!



I also need help with this one:

A geologist finds that a moon rock whose mass is 9.22kg has an apparent mass of 5.58kg when submerged in water. What is the density of the rock?

I tried using the equation W=pvg and it didn't get me the right answer.

 

[dextercioby]

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 solid part of the specimen has a density of 5.13×103kg/m3. What fraction of the specimen's apparent volume is solid?

I tried setting up a ratio of densities but it didn't work at all... please help!



I also need help with this one:

A geologist finds that a moon rock whose mass is 9.22kg has an apparent mass of 5.58kg when submerged in water. What is the density of the rock?

I tried using the equation W=pvg and it didn't get me the right answer.

The first problem is more complicated.The solid has a given volume which is not apparent at all.It's the same volume whether it's sunk in water,air or any other fluid.Incomplessible.For air it doesn't really hold,but let's give credit to Archimedes.
I hope to be familiar with the notations used:
$$G_{apparent}^{water}=mg-F_{Archimedes} ^{water}$$
$$G_{apparent}^{air}=mg-F_{Archimedes}^{air}$$
$$G_{apparent}^{air}=2G_{apparent}^{water}$$ ($)
$$\rho_{real}^{body}=...?$$
$$F_{Archimedes}^{air}=V_{body} \rho_{air} g$$
$$F_{Archimedes}^{water}=V_{body} \rho_{water} g$$
$$G_{apparent}^{air}= V_{body}g(\rho_{real}^{body}-\rho^{air})$$
$$G_{apparent}^{water}=V_{body}g(\rho_{real}^{body}-\rho^{water})$$


From the last 2 equations and the relation ($) u'll be given that:
$$\rho_{real}^{body}=2\rho^{water}-\rho^{air}$$ ($$)

$$\rho_{real}^{body}=\frac{\rho^{air}V_{inside}^{air}+\rho_{exterior}^{solid}V_{exterior}^{solid}}{V_{body}}$$

Daniel.

 

[marlon]

second question :

5.58*g = -9.22*g + pVg where p is 1000kg/m³ (waterdensity)... and solve for V...

V is the volume of the object...

density : 9.22/V

regards
marlon, just of the top of my head...lol

 

[dextercioby]

From the last equation,u'll need the quantity:
$$\frac{V_{exterior}^{solid}}{V_{body}}$$.
Make some more simple manipulations of the forumulas and u'll finnally be having:
$$\frac{V_{exterior}^{solid}}{V_{body}}= \frac{\rho_{real}^{body} -\rho_{air}}{\rho_{solid}-\rho_{air}}$$
,which could be put in the more "familiar" way:
$$\frac{V_{exterior}^{solid}}{V_{body}}=2\frac{\rho_{water}-\rho_{air}}{\rho_{solid}-\rho_{air}}$$
,which rounds roughly to 0,4.

1.Apply the same kind of logics assuming the stone was filled with water instead of air.What would change in the prior arguments?? Would the final number differ?
2.Use the same kind of technics to solve the second problem and to check Marlon's guessing...( ).

Daniel.

 

[Angie913]

I don't have this problem, but I don't understand number one and what you did. Could you explain it better? I'm not sure what G is and all of that. Thanks!

 

[dextercioby]

I don't have this problem, but I don't understand number one and what you did. Could you explain it better? I'm not sure what G is and all of that. Thanks!

G is the apparent gravity force.mg is the real gravity force,and F_{Archimedes} is the buoyant force.
I think it's clear enough.