Question

The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface if the atmospheric pressure is hm of mercury and the density of mercury is and time start of lake water find the depth of the lake

Answer 1

Answer:

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

Explanation:

Iam thinking this is the right answer