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Aamc #5 q:49

Discussion in 'MCAT: Medical College Admissions Test' started by bubblegum123, Aug 6, 2011.

  1. bubblegum123

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    Hey SDNers,

    I have come across question 49 on AAMC practice test 5 which I am having difficulty understanding the concept and problem set up. The explanation provided from AAMC is vague while that of Kaplan is too long.
    It would be helpful if anyone can help me understand how to set this problem up. Plus this question was similar to a problem in AAMC #4 question 18. Any help would be apprecaited!

    The question is:

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

    a)0.3
    b)0.9
    c)1.7
    d) 2.3

    ans is c.

    AAMC 4, question 18:

    An object of 15 grams is immersed in benzene and suffers an apparent weight loss of 5 grams. What is the apprx specfic gravity of the object? (specifc gravity benzene 0.7)

    a)1.4
    b)1.8
    c)2.1
    d)3.0

    ans c
     
  2. MAP2

    2+ Year Member

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    I really like the way Ek explained this. If the object is "totally immersed", then we can assume it isn't floating, and so it displaces its volume. We know the buoyant force is Fb = (density)(Volume)(g). For the Q from AAMC five, we can see that if the Buoyant force is greater by a factor of 12/5, then since the V is the same and g is constant, the density must be the value that changes (the V is constant since the object displaces its volume, and obviously its volume is constant). So, to fit the equation, the density of the unknown must be 12/5 x 0.7, which comes out to 1.7, or answer c. The idea behind the question from AAMC 4 is that the object displaces it volume and that the 5g "loss of mass" (just another way to tell you the buoyant force) is due to a loss of 5g of liquid. So, if within the same volume, the object weighs 15g, while the liquid weighs 5g, then the object must be three times as dense, by Rho = m/V. Hope this makes sense!!
     
  3. Bumbl3b33

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    B/b=densityofB/densityofb (basically, buoyant forces are proportional)

    w/B=densityofobject/densityoffluid
    (for future reference, w/B=densityofobject/densityoffluid=Vsubmerged/Vtotal incase you ever run into those "how much of hte object is seen above the water" questions) all of these are derived from weight of object = densityofobject*volumesubmerged*g and Buoyant force=densityoffluid*V*g and setting them equal to each other. Try it yourself, you'll remember it better that way! :)

    Hope that helps!
     

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