MCAT Physics Question — Floating Objects
 by
 Jan 20, 2017
 MCAT Question of the Day
A. 1/20
B. 1/2
C. The cork will completely submerge because of the mercury’s greater cohesiveness than water.
D. 5/13
Click for Explanation
Since the cork is floating, it must be displacing water (and mercury!) equal to its own mass. That lets us write the following:
m_{water} = m_{mercury}
Since ρ = m / V, we can rearrange the density equation to get:
m = ρV
Thus:
ρ_{water}V_{water }= ρ_{mercury}V_{mercury}
_{ }We’re told that the cork floats with half of its volume submerged. So the cork is displacing 5 cm^{3} of water.
(1 g/cm^{3})(5cm^{3}) = (13 g/cm^{3})(V_{mercury})
V_{mercury} = 5/13 cm^{3}
The cork is 10 cm^{3}, which means that 5/13 cm^{3} represents a tiny fraction of the cork’s volume. Among the answer choices, only (a) is anywhere close.
Alternatively, you can solve this question using the ratio of the specific gravity of the cork to the specific gravity of mercury. Since water has a specific gravity of 1 and half of the cork’s volume is submerged when in water, the cork must have a specific gravity of 0.5. The ratio of this value to the S.G. of mercury (13) gives us the fraction of the cork’s volume that is submerged: 0.5/13 = 1/26, closest to choice A.
A. 1/20th, correct
B. 1/2, incorrect, this answer reflects no change in submersion, even though mercury is far more dense than water
C. incorrect, although mercury has greater cohesiveness than water, this does not affect the buoyancy of objects floating in mercury
D. 5/13, incorrect, this answer reflects using 5/13 cm^{3}, 0.38, as the portion submerged rather than comparing to the total volume of 10 cm^{3}
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