An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0 × 105 Pa and density of water = 1000 kg m–3.

Given


Radius of bubble at the bottom of deep river R1=2.0mm=2.0m


Depth of the river h=3.3m


Density of water = 1000 kg m–3


We know that


Pressure at depth inside a fluid is related to atmospheric pressure by relation


P1=Pa + hg


Where P1 =pressure at depth h


Pa=atmospheric pressure=1.0 × 105 Pa


g=acceleration due to gravity=9.8ms-2


=density of fluid.


So,


P1=1.0105+3.310009.8=1.32105Pa


Temperature is same for both water at the bottom and the water at the surface. So, we can apply Boyle’s law which says that PV=constant, when temperature is constant.


Let V1 be the volume of air bubble at bottom of deep river



Va be the volume of air bubble at surface of river



Where Ra = radius of bubble at surface of river


So, according to Boyle’s law


P1 V1=Pa Va






Ra=2.210-3m


Radius of the air bubble at the surface of river is 2.210-3m.


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