Question

The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. The atmospheric pressure is 75 cm of mercury and the ratio of the density of mercury to that of lake water is 40/3, the depth of the lake is

Answer 1

Given:

ratio of the volume of the air bubble at the top to the bottom, V'/V = 2,

=> V' = 2V

ratio of the density of the mercury to that of the lake water, ρ'/ρ = 40/3

=> ρ' = (40/3) ρ

Now, the pressure exerted at the surface is equal to 75 cm of mercury, so here depth = 0.75 m

the pressure exerted at the bottom will be P at depth = pressure due to water column h + atmospheric pressure, here, h is the depth of the lake

so, the equation would be P' V' = P V

where P = hρg + atmospheric pressure

h is equal to 10 m