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 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension an?

Answer 1

change in pressure of bubble in air

Δp = (2T/R)×2=4T/R

where

T-temperature

R-radius

at bottom surface

p1=pa+pgh+us/r

p2-pa=us/(5r/4)=(16s/5r)

also p1v1=p2v2

p1.(4π/3)r∧3=p2.4π/3*(125/64)r∧3

p1=(125/64)p2 or p1/p2=125/64

∴ pa+ pgh +4s/r/pa+16s/5r=125/64

ignore surface tension

∴ pa+pgh/pa=125/64 or 1 +pgh/pa=125/64

∴ pgh/pa=61/64 or pgh= (61/64)pg*10

h=9.5m