Question

Please solve this question at earliest.

Answer 1

Given:

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

V' = 2V

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

=> p' = (40/3) p

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

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 will be

P'V' = P V

Where P = hpg + atmospheric pressure

Answer 2

The pressure at the bottom of the lake is = The pressure due to hte air + The pressure due to the water column=P1

The pressure at the top is only the atmosphereic pressure=P2 Initial Vol=V1 and final Vol is =V2

NOW YOU HAVEN'T MENTIONED ABOUT THE TEMP SO I AM CONSIDERING IT TO BE CONSTANT

So from gas eq we get P1V1=P2V2

V2=2V1

Putting the values in the eq 1 we get,

H=10 meter