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Mechanics

The volume of an air bubble becomes 6 times as it rises from the bottom of a lake to it's surface Assuming atmospheric pressure to be 75cm is Mercury and density of water to be 1/10of the mercuMe,the depth of the lake is?

Profile image of Harshvardhan Vatsa
7 Years agoGrade
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1 Answer

Profile image of Arun
7 Years ago
The ratio of the volume of the air bubble at the top to the bottom, V'/V = 6,  
V' = 6V 
 
The ratio of the density of the mercury to that of the lake water, p'/p = 10/3 
=> p' = (10/3) 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.