AAMC 4 CBT question - fluids discrete

[typicalindian]

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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

I got the right answer when I took the test but I can't seem to conceptually understand this. During the test my thought process was just "well 15/5 is 3...and 3*.7=2.1." Which was right, but that clearly is a poor thought process. There was no passage associated with the question.

 

[pfaction]

Submerged: Wobj/FBuoy = pobj/pliq.

You got it perfectly down.
15(10) / 5(10) => 15/5 = 3
pObj/pLiq: X/0.7
3/1 = X/0.7
X=2.1.


**** I read AAMC4 as 9, so now I'm getting one right in Physics 🙂 taking AAMC4 tomorrow.

Hi, editing this so I can splurge out facts and stuff in white so I can increase my memory and potentially get corrected on things I don't know!
Archemides:

Buoyant force = weight of fluid displaced from object.
When an object is floating, FBuoy = Fwtobject.
pObj x VObject x (grav) = pLiq x VSubObj x (grav)
==>
pObj/pLiq = Vsub/Vobj
Weight of Object = Weight of Fluid

BUT! What happens in subjerged? Vsub=Vobj. Complete. FBuoy is constant at any and all points of depth. And, potentialy since the object is submerged, there's a net force downwards (FGrav>FBuoy) so acceleration can come into play. So if FBuoy is constant at all points, and Fwobj is greater....
WObj/FBuoy = pObj/pLiq
pObj/pLiq can be specific gravity if pLiq is water, or if another spg is given.

If the object isn't moving at all, this must mean Fnet is 0. So, mg=FBuoy! Static equilibrium. I've had TPR questions where a balloon was tied to the bottom of a pool and it asked me to find the tension on the rope, you had to remember Fbuoy and Mg were equal and opposite or some stuff.

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