A diver ascends quickly to the surface from the bottom of a lake of depth H metre. During this period, he neither exhales nor inhales air. Assuming constant temperature, what would be the fractional i

Question

A diver ascends quickly to the surface from the bottom of a lake of depth H metre. During this period, he neither exhales nor inhales air. Assuming constant temperature, what would be the fractional increase in volume of his lungs? The atmospheric pressure is 9H metre of water.

TestHub · Accepted Answer

At the bottom P₁ = 9H + H = 10H m of water; at the surface P₂ = 9H m. Boyle's law: V₂ = V₁ · 10H/9H = 10V₁/9. Fractional increase = 10/9 - 1 = 1/9. Matches (d).

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