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=P_a + hg

Where P₁ =pressure at depth h

P_a=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

V_a be the volume of air bubble at surface of river

Where R_a = radius of bubble at surface of river

So, according to Boyle’s law

P₁ V₁=P_a V_a

R_a=2.210^-3m

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