__Question 10.20__:^{–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).

__Solution__:Pressure inside the air bubble is 1.06 X 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 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 +

*h*

*ρ*g +

*P*’

= 1.01 X 10

^{5}+ 0.4 X 1.2 X 10

^{3}X 9.8 + 10

= 1.06 X 10

^{5}Pa

Therefore, the pressure inside the air bubble is 1.06 X 10

^{5}Pa

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