I feel like these kind of problems are supposed to screw with you, they give you only three numbers and you are supposed to calculate S.G which requires knowing the density of both the object and the fluid? After derivation of multiple formulas (pretty much all of the formulas relative to fluids and solids) I find out that it comes down to simple ratios.
So if you go ahead and think about the situation you know that the ball is dropped, inferring that it has a w = F = ma, you also know that the EtOH is exerting a force on the ball as well. From basics you should know that the Net F = 0. Thus, w - F1 - F2 = 0.
Due to a series of derivation of formulas you eventually get:
1. F = B
2. ma = pVg
The tricky part is that you might think ohh well you can cancel out both a & g, which actually is plausible, leading to
m = pV
but then you may bring the V over and divide m giving you
m/V = p (Leads to) p = p
this is where you say f*ck and realize well I only know mass of the object, but neither the mass of the fluid or the volume of either, wtf do I do with S.G? You can prevent all this by plugging in what you have from the question in #2
3. w = ma = (12)(10) = 120N
Now the next part is really the whole point of the question, its the stump that the MCAT wants you to overcome, bc the rest is easy. Pretty much elementary stuff. Like I said the Net F = 0. You already know that the F of the Ball is 120N, but what is the F of the fluid? If the ball is originally 12g when dropped and then 8g when in EtOH, from simple deduction you will know that if the object still weighs 8g that means that EtOH is exerting 4g on the ball. 12g - 8g = 4g. With that being said, if the force of the fluid on the object is (4g)(10), it is also the buoyant force.
If we know:
1.F1 = B = 40N
2. w = ma = 120N
Then how is the Net Force = 0? Well alot of us at this point are a little constricted on time and really dont care much about understanding the concepts, but the quickest way to get to the answer. Well if you are really paying attention when you are doing content review or want to have a slight chance of doing remotely well on the MCAT you need to realize questions are not simply plug and chug. You may however become so familiar with equations that you can. Now what accounts for the remaining 80N?!?!?! Well the Force the ball exerts doesnt just vanish, when it is on EtOH it is still exerting Force. Therefore, F2 = ma = (8)(10) = 80N.
w - F1 - F2 = 120 - 40 - 80 = 0N.
Although, you may think that was irrelevant to the question, it is quite critical that you see how all this is intertwined. When I derived the formulas you should have noticed the preliminary derivations that (ill do them again for the lazy people who wont scroll up):
1. F = B
2. ma = pVg ( a & g cancel)
3. m = pV (bring the V over)
4. m/V = p
5. p = p
Now, if you take:
1. F/B = ma/pVg = p/p
This is where you go ahhhhhh I see. The weight of the dropped object (not the floating one) divided by the buoyant force = the density of the object/the density of fluid. Thus:
1.120N/40N = p/(.8)
2. 3 = p/(.8)
3. 3(.8) = p
4. p = 2.4
It does come down to ratios and I realize another user had shown how to do this with just a few sentences, but the explanation wasnt that in-depth, which is what we need. Not shortcuts. Its the familiarization of all the concepts together that will get you far, not the so called "tips and tricks". This isnt the ACT, this is the most important test of your life. So treat it like it is. As insignificant as this question may be.... it is well worth your time learning how this questions works to help your understanding overall. I hope this helps, and I hope I didnt come off to abrasive or offensive. I only mean well and I want everyone to do their best. Good Luck.
