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Aluminum cylinder in alcohol

  1. Oct 29, 2006 #1
    An aluminum cylinder weighs 1.01 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.88 × 10-5 m3. The apparent weight of the cylinder when completely submerged is 0.778 N. What is the specific gravity of the alcohol?

    m(cylinder)=.10296 kg
    .10296kg/3.88*10^5 => 2653.608kg/m^3 (avg density)
    SG= 2653.608/1000=> 2.65

    What did I do wrong? Thanks
  2. jcsd
  3. Oct 29, 2006 #2

    Doc Al

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    Staff: Mentor

    You found the specific gravity of the cylinder, not the alcohol. Hint: Buoyant force is involved.
  4. Oct 29, 2006 #3
    Doc Al,

    So I can find the mass of the alcohol, then the avg density, divide by 1000 and get the alcohol's SG?

    Also, how do I relate Buoyant force to this problem? Thanks again
  5. Oct 29, 2006 #4


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    Staff Emeritus
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    The force applied upwards by the alcohol is equal to the weight of the displaced alcohol
    Last edited: Oct 30, 2006
  6. Oct 30, 2006 #5

    Doc Al

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    Staff: Mentor

    That'll work.

    The buoyant force equals the weight of the displaced fluid (alcohol). You can find the buoyant force by comparing the actual weight of the cylinder to its apparent weight when submerged in the alcohol.
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