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An air bubble doubles in volume as it rises from the bottom of a lake to the surface. The atmospheric pressure equals a column of mercury of height H, and the density of mercury is n times that of the lake water. Neglecting surface tension, find the depth of the lake.
- nH
- nH/2
- 2nH
- H/n
Correct answer: nH
Solution
By Boyle's law, P_bottom * V = P_surface * (2V), so P_bottom = 2 P_surface. The surface pressure is atmospheric = H of mercury = n*H of water (since mercury is n times denser). The bottom pressure = atmospheric + the water column of depth d = nH + d (measured in water column). Setting nH + d = 2(nH) gives d = nH. So the lake depth is nH (in units of water column, i.e. depth = nH).
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