### 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 × 10^{5} Pa).

**Answer:-**Excess pressure inside the soap bubble is 20 Pa;

Pressure inside the air bubble is 1.06 × 10^{5} 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 × 10^{3} kg/m^{3}

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 × 10^{5} Pa

Acceleration due to gravity, *g* = 9.8 m/s^{2}

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 + *h**ρ*g + *P*’

= 1.01 × 10^{5} + 0.4 × 1.2 × 10^{3} × 9.8 + 10

= 1.06 × 10^{5} Pa

Therefore, the pressure inside the air bubble is 1.06 × 10^{5} Pa.