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Question

When a bubble containing ideal gas rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of a column of water height H, then the depth of lake is:

A
H
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B
2H
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C
7H
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D
8H
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Solution

The correct option is C 7H
We know, atmospheric pressure Po ​= ρgH
where ρ is the density of water.

Assuming the depth of the lake be d, the pressure at this depth will be, P = Po+ρgd = ρg(H+d)

Since, the temperature is constant, we can assume that the process takes place isothermally.
So, applying the Boyle's Law

P1V1=P2V2...(i)
where , 1 denotes the water at depth d and 2 denotes the surface of water.

So, P1=ρg(d+H)
at surface, it will be atmospheric pressure i.e P2=ρgH

Volume V = 43πr3
on doubling the radius, the volume will increase to 8 times
i.e V2=8V1
put up the values in equation (i)

Substituting the values, in Boyle's Law,
P1=ρg(d+H)V=ρgH×8V

d=8HH = 7H

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