## 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? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-10 Mechanical Properties of Fluids

### Q.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).

Excess pressure inside the soap bubble is 20 Pa;
Pressure inside the air bubble is 1.06 × 105 Pa
Soap bubble is of radius, r = 5.00 mm = 5 × 10–3 m
Surface tension of the soap solution, = 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 = 4S / r
= 4 × 2.5 × 10-2 / 5 × 10-3
= 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
P‘ = 2S / r
= 2 × 2.5 × 10-2 / (5 × 10-3)
= 10 Pa
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
= 1.01 × 105 + 0.4 × 1.2 × 103 × 9.8 + 10
= 1.06 × 105 Pa
Therefore, the pressure inside the air bubble is 1.06 × 105 Pa.