Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.20

Question 10.20:
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).
Solution:
Excess pressure inside the soap bubble is 20 Pa;
Pressure inside the air bubble is 1.06 X 105 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 = 4S / r
= 4 X 2.5 X 10-2 / 5 X 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 X 2.5 X 10-2 / (5 X 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 X 105 + 0.4 X 1.2 X 103 X 9.8 + 10
= 1.06 X 105 Pa
Therefore, the pressure inside the air bubble is 1.06 X 105 Pa

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