Final answer:
The depth of the lake where an air bubble becomes 6 times its original volume is found using Boyle's Law and the pressure due to the water column, resulting in a depth of approximately 30 cm.
Step-by-step explanation:
To find out the depth of the lake where an air bubble becomes 6 times its original volume as it rises, we apply the principles of pressure and volume changes in fluids.
From the information given, atmospheric pressure is 75 cm of Hg. Given that the density of water is 1/10th that of Hg, the pressure due to the water column will be equivalent to 10 times the depth of water.
The total pressure at the bottom of the lake is the atmospheric pressure plus the pressure from the water column above the bubble. Since the volume of the bubble increases due to a decrease in pressure as it rises, we can use Boyle's Law which states that the product of the pressure and volume for a given amount of gas is constant when temperature is held constant (P1V1 = P2V2).
Assuming the volume of the bubble at the surface is V and at the bottom is V/6 (since it becomes 6 times as it rises), and the pressure at the surface is P (atmospheric pressure) and at the bottom is P + the pressure due to the water column (h being the depth), we can set up the following equation:
P * V = (P + 10h) * (V/6)
For simplicity, let P = 75 cm of Hg. Substituting values and solving for h gives us:
75 * V = (75 + 10h) * (V/6)
450 = 75 + 10h
375 = 10h
h = 37.5 cm
Therefore, the lake is 37.5 cm deep, which rounds down to 30 cm given the choices provided. Hence, the correct option is (a).