AAMC 4 #18: Fluid Mechanics

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

UTPHS

Full Member
10+ Year Member
Joined
Aug 1, 2012
Messages
14
Reaction score
0
An object with 15 grams is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the specific gravity of the object? The specific gravity of benzene is 0.7.

A) 1.4
B) 1.8
C) 2.1
D) 3.0

Actual answer: C

My logic:

Apparent weight = mg - Fb.
=> 10 = 15 - Fb => Fb = 5
=> Fb = pvg = p(object)vg / p(benzene)vg
=> p(object)vg = Fb*p(benzene)vg (1)

sg of benzene = p(benzene)vg / p(water)vg
=> p(water)vg = p(benzene)vg / sg of benzene (2)

(1) / (2) => p(object)/p(water) = [Fb*p(benzene)] / [p(benzene)/sg of benzene] (note: vg/vg and p(benzene)/p(benzene) cancels out)

=> Fb*(sg of benzene) = 5*(0.7) = 3.5

While my logic did not get me to any of the answers, can someone explain what was incorrect about my logic?

Members don't see this ad.
 
I think you made your mistake in setting Fb to be 5. The problem says that it seems as though it has lost a mass of 5 grams. When setting up any force, you know that a force is a mass (in this case .005 kg) times an acceleration (in this case, g = about 10), Fb is .005 times 10 or .05.

Thus you get:

Fb = (.005 kg)(g) = .05 = pvg = .7vg
v = .005/.7
mass of actual object = .015 kg
m/v = density of object =.015/(.005/.7) = (.015*.7)/.005 = 3*.7 = 2.1
 
2 things: as someone else mentioned, "apparent loss of mass of 5 grams" means that 5 grams of fluid was displaced by the object. "immersed" is another keyword here. this tells us that the object is fully submerged, thus the volume of the fluid displaced = the volume of the object.

here is another way to look at it:

(1) V_benzene displaced = V_object

rewrite equation (1) using definition of density: ρ = m / V (i.e., V = m / `):

m_benzene / ρ_benzene = m_object / ρ_object

rearrange the equation to put ρ_object on one side:

ρ_object = (m_object submerged / m_benzene displaced) ρ_benzene
ρ_object = (15g/5g) ρ_benzene
ρ_object = 3 ρ_benzene

divide each side by ρ_H2O to get SG of object and benzene:

SG_object = 3 SG_benzene = 3 * 0.7 = 2.1

also, do a sanity check. does this make sense for something that is not floating in benzene? why?
 
Last edited:
I think you made your mistake in setting Fb to be 5. The problem says that it seems as though it has lost a mass of 5 grams. When setting up any force, you know that a force is a mass (in this case .005 kg) times an acceleration (in this case, g = about 10), Fb is .005 times 10 or .05.

Thus you get:

Fb = (.005 kg)(g) = .05 = pvg = .7vg
v = .005/.7
mass of actual object = .015 kg
m/v = density of object =.015/(.005/.7) = (.015*.7)/.005 = 3*.7 = 2.1

For this method, wouldn't you also have to divide by the density of water to get sg? Like sg = density of object (m/v) / density of water

2 things: as someone else mentioned, "apparent loss of mass of 5 grams" means that 5 grams of fluid was displaced by the object. "immersed" is another keyword here. this tells us that the object is fully submerged, thus the volume of the fluid displaced = the volume of the object.

here is another way to look at it:

(1) V_benzene displaced = V_object

rewrite equation (1) using definition of density: ρ = m / V (i.e., V = m / `):

m_benzene / ρ_benzene = m_object / ρ_object

rearrange the equation to put ρ_object on one side:

ρ_object = (m_object submerged / m_benzene displaced) ρ_benzene
ρ_object = (15g/5g) ρ_benzene
ρ_object = 3 ρ_benzene

divide each side by ρ_H2O to get SG of object and benzene:

SG_object = 3 SG_benzene = 3 * 0.7 = 2.1

also, do a sanity check. does this make sense for something that is not floating in benzene? why?

When you divided by the density of water which value would you use? I'm seeing p=1000 kg/m^3 and 1 g/cm^3. I'm guess you used the 1g/cm^3 but would 1000 kg/m^3 have worked?
 
Members don't see this ad :)
for my method, you don't need that information.

when i wrote "divide each side by ρ_H2O..." that was just to show you that ρ_sample/ρ_H2O = SG_sample. this makes life easier, since the problem asks for SG_object, and you are given SG_benzene. 3 is just a factor.
 
Last edited:
What is the point of doing all the math? If the object is 15 grams and the amount of benzene displaced only equates to 5 grams, then the object is 3 times as dense as benzene.
 
What is the point of doing all the math? If the object is 15 grams and the amount of benzene displaced only equates to 5 grams, then the object is 3 times as dense as benzene.

That is how I solved it too.

On the MCAT you don't want to be setting up 10 lines of math to solve a problem.
 
What is the point of doing all the math? If the object is 15 grams and the amount of benzene displaced only equates to 5 grams, then the object is 3 times as dense as benzene.

When you do that, are you getting it from the eq:

Vdisp / V = p(benzene) / p(obj) => 5/15 = p(benzene) / p(obj)

So it leads to, p(obj) = 3*p(benzene) = 3*(0.7)*(pH2O), since sg = p(benzene)/p(H2O)

In that case, does that mean you are able to equate volume with the mass displaced?
 
Top