Ncert Exercises & Solutions

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 × $10^{5}$ Pa).

Given,

The radius of the soap bubble , r = 5.00 mm = 5 × 10^–3 m

The surface tension of the soap solution, S = 2.50 × 10^–2 Nm^–1

Hence, the excess pressure inside the soap bubble is:

$∆ P_{s} = \frac{4 S}{r} = \frac{4 \times 2.5 \times 10^{- 2}}{5 \times 10^{- 3}} = 20 Pa$

Radius of the air bubble, r = 5 mm = 5 × 10^–3 m

The excess pressure inside the air bubble is:

$∆ P_{a} = \frac{2 S}{r} = \frac{2 \times 2.5 \times 10^{- 2}}{5 \times 10^{- 3}} = 10 Pa$

the depth of the air bubble, h = 40 cm = 0.4 m

The relative density of the soap solution = 1.20

Density of the soap solution, ρ = 1.2 × 10³ kg/m³

Acceleration due to gravity, g = 9.8 m/s²

$At a depth of 0.4 m, the total pressure inside the air bubble = Atmospheric pressure + hρg + ∆ P_{a} = 1.01 \times 10^{5} + 0.4 \times 1.2 \times 10^{3} \times 9.8 + 10 = 1.057 \times 10^{3} Pa = 1.06 \times 10^{5} Pa$

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