Question

the volume of air bubble is doubled as it rises from the bottom of lake to its surface. the atmospheric presure is 75 cm of the mercury and the ratio of the density of mercury to that of lake water is 40/3. what is the depth of of the lake ?

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 75cm 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  

 

So, we can now find the value of h.

Answer 2

Answer:

10m

Explanation:

2P0 = P0 + dgh  P0 = dgh   P0 = 75 cm mercury [Atmospheric pressure]