# Physics Question (from AAMC)

#### Tobi222

Hi all,

Can anyone tell me how to solve for specific gravity problem? I don't recall learning specific gravity when I took physics but maybe I did and just forgot. Anyway, the problem from the Physics qpack states:

An object that is totally immersed in benzene (specific gravity = 0.7) is subject to a buoyancy force of 5 N. When the same object is totally immersed in an unknown liquid, the buoyancy force is 12 N. What is the approximate specific gravity of the unknown liquid?

The answer was 1.7 and according to their explanation, their equation was (12/5) x 0.7

I know that buoyant force is density x volume x gravity and that (upon looking it up) specific gravity is the ratio of the unknown's density to water but I'm just confused as to how they got about to that equation.

Thank you!

#### Matthew9Thirtyfive

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For starters, specific gravity is how the density of an object relates to the density of water. The formula is s.g. = density of the object/density of water. So something with a specific gravity of 0.7 is less dense than water.

But anyway, you actually don't really need to know that to answer this. This sort of question is just a relative fraction question.

You have two sets of the same value, and they must equal each other since they refer to the same object. So when immersed in a fluid with a specific gravity of 0.7, it has a buoyant force of 5 N. You can write that as 0.7/5. Now, you have another liquid of unknown density, which we'll call p, where it experiences a buoyant force of 12 N. So we can write that as p/12. Now, since it is the same object in both fluids, you can set them equal to each other: 0.7/5 = p/12. Solve for p, and you get p = (12*0.7)/5 = 1.7.

#### mk536

Vol×gravity×density(material)/spec. gravity= vol×gravity×density(h2o)

The right side is a constant for a given size object. So we know that the buoyant force of an object in a fluid divided by the fluid's specific gravity will be a constant regardless of the material. Therefore:
5/0.7=12/spec. gravity

Solving for the spec. gravity then yields the correct answer.

#### Tobi222

For starters, specific gravity is how the density of an object relates to the density of water. The formula is s.g. = density of the object/density of water. So something with a specific gravity of 0.7 is less dense than water.

But anyway, you actually don't really need to know that to answer this. This sort of question is just a relative fraction question.

You have two sets of the same value, and they must equal each other since they refer to the same object. So when immersed in a fluid with a specific gravity of 0.7, it has a buoyant force of 5 N. You can write that as 0.7/5. Now, you have another liquid of unknown density, which we'll call p, where it experiences a buoyant force of 12 N. So we can write that as p/12. Now, since it is the same object in both fluids, you can set them equal to each other: 0.7/5 = p/12. Solve for p, and you get p = (12*0.7)/5 = 1.7.

This makes so much more sense! Thank you!!

1 user

#### Tobi222

Vol×gravity×density(material)/spec. gravity= vol×gravity×density(h2o)

The right side is a constant for a given size object. So we know that the buoyant force of an object in a fluid divided by the fluid's specific gravity will be a constant regardless of the material. Therefore:
5/0.7=12/spec. gravity

Solving for the spec. gravity then yields the correct answer.

I see now! Thank you!!

1 user