NCERT Solutions
Class 11 Physics
Mechanical Properties of Fluids
Q.9.20
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is (2.50 × 10–2) N m–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
Given:
Excess pressure inside the soap bubble = 20 Pa;
Soap bubble is of radius, r = 5.00 mm = (5 × 10–3) m
Surface tension of the soap solution, S = (2.50 × 10–2) Nm–1
Relative density of the soap solution = 1.20
Density of the soap solution, ρ = (1.2 × 103) kg/m3
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = (5 × 10–3) m
1 atmospheric pressure = (1.01 × 105) Pa
Acceleration due to gravity, g = 9.8 m/s2
Hence, the excess pressure inside the soap bubble is given by the relation:
P= 
= 
=20Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
P’= 
= 
=10Pa
Therefore, the excess pressure inside the air bubble is 10 Pa.
At a depth of 0.4 m, the total pressure inside the air bubble
= (Atmospheric pressure + hρg + P’)
= (1.01x105+0.4x1.2x103x9.8+10)
= (1.057x105) Pa
= (1.06x105) Pa
Therefore, the pressure inside the air bubble is (1.06x105) Pa.