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Question:

When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes 5r/4. Taking the atmospheric pressure to be equal to 10m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

8.7m

9.5m

10.5m

11.2m

Solution:

Pressure at bottom (P1) = Patm + ρgh + 4T/R1__________(1)
Pressure at top (P2) = Patm + 4T/R2____________(2)
Given
R1 = r
R2 = 5r/4
So P1V1 = P2V2
(P1)(4/3πr³)=(P2)(4/3)(125r³/64)
Dividing (1) and (2)
P1/P2 = (Patm + ρgh + 4T/r) / (Patm + 4T/(5r/4))
125/64 = (ρg(10) + ρgh) / (ρg(10))
640 + 64h = 1250
On solving we get
h = 9.5m