# Specific Gravity problem from AAMC practice test 4

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#### [email protected]

##### Member
15+ Year Member

An object with 15 grams 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 but I really had trouble understanding how to solve this. This is AAMCs answer The buoyant force on an object immersed in a fluid displaced by the object (Archimedes principle)¸ There were 5 g of liquid displaced, thus, the ratio of object mass to fluid is 15/5 grams. 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 the object and displaced liquid are equal.

I am confused. 5 grams of benzene which has a specific gravity of 0.7 was displaced by an object, right? Thus, the apparent weight of the 15 grams object is 10 grams in fluid. Immersed could mean submerged or floating, right? The MCAT was being vague but since all of the answers are greater than 0.7 we can assume that the object is submerged, right? Thus, this object is displacing its volume in benzene and this volume displaced it equal to 5 grams of benzene, right? Is that why they are doing a ratio of the object mass to the fluid mass 15/5=3. It is saying that the specific gravity of the object is 3 times as great as benzene since it sank and weighs more than the fluid that it displaced. Therefore, that is why they multiplied 3 x 0.7= 2.1 because the object is more dense and the specific gravity of the object is 3 times that of benzene.

Did I figure this out correctly? What did AAMC mean when they said because the volumes of the object and displaced liquid are equal? Were they trying to explain the reason the object is more dense is because it weighs 15grams but only displaced 5grams of liquid?

I am going to redo all the EK 1001s on this subject but does anyone know of more practice problems on this subject. For some reason, I have been having issues in the past on this particular subject and I want to be able to answer them all correctly from now on. Also helpful tactics would be great.

Thank you!

#### Ibn Alnafis MD

##### Full Member
10+ Year Member
An object with 15 grams 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 but I really had trouble understanding how to solve this. This is AAMCs answer The buoyant force on an object immersed in a fluid displaced by the object (Archimedes principle)¸ There were 5 g of liquid displaced, thus, the ratio of object mass to fluid is 15/5 grams. 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 the object and displaced liquid are equal.

I am confused. 5 grams of benzene which has a specific gravity of 0.7 was displaced by an object, right? Thus, the apparent weight of the 15 grams object is 10 grams in fluid. Immersed could mean submerged or floating, right? The MCAT was being vague but since all of the answers are greater than 0.7 we can assume that the object is submerged, right? Thus, this object is displacing its volume in benzene and this volume displaced it equal to 5 grams of benzene, right? Is that why they are doing a ratio of the object mass to the fluid mass 15/5=3. It is saying that the specific gravity of the object is 3 times as great as benzene since it sank and weighs more than the fluid that it displaced. Therefore, that is why they multiplied 3 x 0.7= 2.1 because the object is more dense and the specific gravity of the object is 3 times that of benzene.

Did I figure this out correctly? What did AAMC mean when they said because the volumes of the object and displaced liquid are equal? Were they trying to explain the reason the object is more dense is because it weighs 15grams but only displaced 5grams of liquid?

I am going to redo all the EK 1001s on this subject but does anyone know of more practice problems on this subject. For some reason, I have been having issues in the past on this particular subject and I want to be able to answer them all correctly from now on. Also helpful tactics would be great.

Thank you!

I struggled with this concept too, but after reading and doing some practice I developed some understanding of it.

The way I would approach this problem is by first deciding whether the object is floating or submerged. It's obvious from the question, and from the answer choices, that the object sinks in benzene. Therefore its specific gravity (density) must be higher than that of benzene, but by how much?

We know that a submerged object displaces fluid of a volume that is identical to its volume. The question stem mentions that the apparent weight loss was 5g, and that is the mass of fluid displaced. Now we know that the fluid displaced, which has the same volume of the object, is weighs only 1/3 of the object's weight. Therefore, the object must have a density of 3 times the fluid. 3*.07 = 2.1

#### Francium87

##### Full Member
5+ Year Member
What did AAMC mean when they said because the volumes of the object and displaced liquid are equal? Were they trying to explain the reason the object is more dense is because it weighs 15grams but only displaced 5grams of liquid?

"the volumes of the object and displaced liquid are equal" means the object is completly sinking in liquid benzene therefore the two volumes (the object & the displaced liquid) are the same

#### [email protected]

##### Member
15+ Year Member
I struggled with this concept too, but after reading and doing some practice I developed some understanding of it.

The way I would approach this problem is by first deciding whether the object is floating or submerged. It's obvious from the question, and from the answer choices, that the object sinks in benzene. Therefore its specific gravity (density) must be higher than that of benzene, but by how much?

We know that a submerged object displaces fluid of a volume that is identical to its volume. The question stem mentions that the apparent weight loss was 5g, and that is the mass of fluid displaced. Now we know that the fluid displaced, which has the same volume of the object, is weighs only 1/3 of the object's weight. Therefore, the object must have a density of 3 times the fluid. 3*.07 = 2.1

Yeah. This was causing me some issues when I was learning it the first time. Are you familiar with the formula to solve the equation? Is it W(object)/W(fluid)=Specific density of object/specific density of fluid. :-/ I would like to see how these equations work.

#### Ibn Alnafis MD

##### Full Member
10+ Year Member
F(net) = F(gravity) - F(buoyancy)
0.1N = 0.15N - Fb
Fb = 0.05N

The buoyancy force is equal to the weight of the fluid displaced, so weight of displaced fluid is .05N while the weight of the object is 0.15N. Both, the object and the fluid displaced have the same volume, but the object is 3 times as heavy. Therefore, the object must have a density that is 3 times that of the fluid.

#### [email protected]

##### Member
15+ Year Member
F(net) = F(gravity) - F(buoyancy)
0.1N = 0.15N - Fb
Fb = 0.05N

The buoyancy force is equal to the weight of the fluid displaced, so weight of displaced fluid is .05N while the weight of the object is 0.15N. Both, the object and the fluid displaced have the same volume, but the object is 3 times as heavy. Therefore, the object must have a density that is 3 times that of the fluid.

Where did you get F(net)=0.1N? Don't we already know the Fb=0.5N since the apparent weight loss is 5 grams?

Apparent weight loss=weight of fluid displaced

Apparent weight=Real weight - Apparent weight loss

Am I right?

Then we can use ratios Weight of object/Weight of fluid=S.G of object/S.G of fluid

Right?

I think I am getting it now.

#### Ibn Alnafis MD

##### Full Member
10+ Year Member
Where did you get F(net)=0.1N? Don't we already know the Fb=0.5N since the apparent weight loss is 5 grams?

Apparent weight loss=weight of fluid displaced

Apparent weight=Real weight - Apparent weight loss

Am I right?

Then we can use ratios Weight of object/Weight of fluid=S.G of object/S.G of fluid

Right?

I think I am getting it now.

Yes, you are correct.

In regards to Fnet, it is the weight of the object in the fluid, or in other words it is the apparent weight. Fb is not 0.5N, it is 5g*1kg/1000g * 10m/s62 = 0.05N. Similarly, the apparent weight is 0.1N. The real weight of the object is 0.15N.

On a side note, this problem could be solved much quicker with some common sense. If we are told that the object suffers a weight loss equals to 1/3 of its weight, they we should know that its density is 3 times that of fluid its submerged in. In other words, the resistance to the gravitational force is 1/3 that of the gravitational force.