Density/Specific Gravity Question from AAMC #4

Hi Alex,

 

Sorry to bother you again-but this question from AMCAS MCAT # 4 is driving me insane and I think it's because I'm overthinking it. So, here goes:

 

18) An object with 15 grams mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the approximate specific gravity of the object? (Data: Specific gravity of benzene = 0.7)

 

a) 1.4

b) 1.8

c) 2.1

d) 3.0

 

The answer is (c).
Solution given: The buoyant foce on an object immersed in a fluid is equal to the weight of the fluid displaced by the object (Archimedes' principle). There were 5 g of liquid displaced; thus, the ratio of object mass to fluid mass is 15/5 = 3. The specific gravity of the object (mass per unit volume compared to water) is three times the specific gravity of benzene (3 x 0.7= 2.1) because the volumes of object and displaced liquid are equal.

 

*I don't understand why the ratio of the mass of the object to the mass of the liquid is 3. Where are they getting these numbers? Please explainnn :(

-----------------

When we approach this question, the first thing that should hopefully stick out is that we're dealing a bit with buoyancy.  We put the object in a fluid and the weight changes; what is responsible for that change?  The buoyant force, of course.

So let's think about what forces are acting on the object before it's in the fluid.  We have a force of gravity equal to the mass of the object times the acceleration due to gravity.  That's really the only one we're concerned about in that case.

When we put the object into the fluid, the new net force is less.  The "apparent loss of mass" here is due to the buoyant force, and must be equal to five grams times the acceleration due to gravity (why?  Because if the apparent mass is now 10 grams, then the apparent weight would be 10 times the acceleration due to gravity.  We've lost 5 grams times the acceleration due to gravity, which is completely attributable to the buoyant force).

Remember that, by definition, the buoyant force is equal to the weight of the fluid that's been displaced.  It is also equal to the density of the fluid times the acceleration due to gravity times the submerged volume of the fluid (in this question, since they use the word "immersed," we know the full volume has been submerged).  We care about the volume here, because we can use that to calculate the density of the submerged object.  If we know the submerged object's volume (which must be equal to the volume of displaced fluid, logically), and its mass (provided in the question stem), we can figure out its density and thus figure out its specific gravity.

Our calculations would look like this:

(rho of fluid)(g)(V)=(mass of fluid)(g)

Taking out g on both sides, we have:

(rho of fluid)(V)=(mass of fluid).

Note that this is basically just a reiteration of the density equation.

From there, let's plug in:

(0.7)(V)=5 grams

and V thus equals 5/0.7.

Taking that one step further, we can say that if the volume of the cube must also be 5/0.7, and its mass is 15 grams, then its density is 15/(5/0.7) = 3*0.7 = 2.1.  That gets us to answer choice C.