# AAMC 4 buoyancy q

#### chiddler

##### Full Member
10+ Year Member
AAMC 4 Discrete # 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? (Sp. gravity of benzene = 0.7)

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

I lucked out with this. I'd really like if somebody goes over my work because my answer has the wrong number of 0's!

My work:

15 g = 0.15 N
5g = 0.05 N

0.05 N pushing is the buoyant force.

Fb = 0.005 = V*rho(benzene)*g = V * 700*10

V = 0.005 / 7000

rounding to hell, I put the 10's aside and work with the important numbers:

5/7 *10^-6 ~ a bit more than half, say, 0.7 * 10^-6 = 6 * 10^-7 meters cubed

Density = mass / volume = 0.015 / 7*10^-7 = 1.5 / 7 * 10^6 = 2.1* 10^6

*10^6!!! it should be *10^3!

#### MrNeuro

##### Full Member
7+ Year Member
AAMC 4 Discrete # 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? (Sp. gravity of benzene = 0.7)

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

I lucked out with this. I'd really like if somebody goes over my work because my answer has the wrong number of 0's!

My work:

15 g = 0.15 N
5g = 0.05 N

0.05 N pushing is the buoyant force.

Fb = 0.005 = V*rho(benzene)*g = V * 700*10

V = 0.005 / 7000

rounding to hell, I put the 10's aside and work with the important numbers:

5/7 *10^-6 ~ a bit more than half, say, 0.7 * 10^-6 = 6 * 10^-7 meters cubed

Density = mass / volume = 0.015 / 7*10^-7 = 1.5 / 7 * 10^6 = 2.1* 10^6

*10^6!!! it should be *10^3!

this is how i set it up

B = .005*g = pbenzeneVobjg
W= .015*g = pobjVobjg

Vobj=.005/pbenzene
Vobj=.015/pobj

p benzene =0.7

.005/0.7=.015/pobj

pobj = 21/10 = 2.1

#### kasho11

##### Full Member
I think it's much better for this problem to think about it conceptually. It suffers an apparent loss of 1/3rd of its mass so it must be 3 times as dense. Similarly if it lost 1/2 its mass it would be twice as dense.

To solve for it mathematically you start with SG benzene = p benzene / p water

rearrange to, p water = p benzene / SG benzene

p of benzene = mass of benzene / volume

p water = mass of benzene / SG benzene x volume

SG object = p object / p water, plugging in the equation for p water we get

SG object = p object / mass of benzene / SG benzene x volume, rearrange to get

SG object = p object x SG benzene x volume / mass of benzene

plug in p object = mass of object / volume to get

SG object = mass of object x SG benzene x volume / mass of benzene x volume

Volume cancels out, you are left with SG object = mass of object x SG benzene / mass of benzene

plugging in 15g for mass of object, 0.7 for SG benzene, and 5g for mass of benzene you get

15g x 0.7 / 5g = 3 x 0.7 = 2.1 - no need to convert to kilograms or deal with any other units because they cancel out for specific gravity.

So a bit more work than you need if you just recognize that because only 1/3rd of it is displaced by benzene, it must be 3 times as dense.

#### chiddler

##### Full Member
10+ Year Member
see with the solutions you guys posted, to me it seems like you're just writing your first equations out of nowhere.

i reasoned that there is buoyant force reducing apparent mass. buoyant force has volume in the equation. i have mass. i need to find volume to find density.

how did you guys start your solutions? what was the thought process?

#### kasho11

##### Full Member
To be honest I got the right answer on the test but I didn't really do it the right away, it ended up working but it wasn't quite right. I figured out the proper way to figure it out after. Well the way I did it was actually right, it just didn't seem right. I set 0.7/1 = x/3 because 1/3rd of the density is displaced.

You are given specific gravity of benzene = 0.7. Specific gravity is a formula you should know from content review. It is simply the density of whatever divided by the density of water.

When you are given a problem that involves object submerged in water, think specific gravity and density. If you see something involving acceleration or sinking or equilibrium start to think buoyant force.

#### pfaction

##### Full Member
10+ Year Member
I think typical indian asked this and I answered by setting pobj to each other.

#### nindra

##### Full Member
The way I did it was by writing a ratio between the object's weight and the liquid's weight. The equation is obviously Force equalling the product of density, volume and grav acceleration. Since volume and grav acceleration are the same for both equations in the ratio, you just cancel them out.
Therefore the ratio is basically down to that of the density ratio equalling the ratio of the two forces. Thus, its basically mg/mg, and cancelling g, its 15g: 5g, which equals 3:1. Thus the object's density is 3 (0.7) = 2.1

#### typicalindian

##### Full Member
5+ Year Member
7+ Year Member
I think typical indian asked this and I answered by setting pobj to each other.

yeah I did. When I did this question I didn't even bother writing down a formula or calculating anything. I solved it conceptually which is why I was a little confused when you did it numerically

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