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Question

An air bubble rises from the bottom of a lake and its radius is doubled on reaching the surface. If the temperature is constant the depth of the lake is (Pressure is 1 atm above the lake)

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Solution

Let,
P1= Pressure under water at asked depth.
P2= Pressure at surface of water = atmosph P.
Let,
V1= Volume of bubble at original depth.
V2= Volume of bubble at surface at water.
Let original radius of bubble =r
new radius =2r
Now, P1V1=P2V2 (for constant temperature)
P1(43πr3)=P2(43π(2r)3)
P1(43πr3)=P2(43π.8r3)
P1=8P2
But, P2=1 atm
P1=7 atm=P1+hρg
1 atm+h(ρg)=8atm
h(ρg)=7atm=P
h=70 m
[10m column of water produces 1atm of pr]


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