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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 7HWe 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=8H−H = 7H

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