Question

A bubble of radius 2 mm is formed from soap solution of surface tention

Answer 1

Work is Force times distance.

So its pretty obvious that the work done in expanding a soap bubble is equal to the surface tension by the area, because this gives the right units:

E = N * m = N/m * m^2

E = 0.06 N/m * xm^2

Where x is the change in surface area.

So work out the area of 2 cm bubble in square metres, same for the 5 cm bubble, subtract them to find the change in surface area, then multiply by 0.06

Answer 2

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, 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 × 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.

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