Find the pressure difference of an air bubble of radius 2 cm formed 20 cm below an open water surface and atmospheric pr…

Explanation

The pressure difference inside an air bubble submerged in water is due to two contributions:

1. Due to surface tension: \(\Delta {P}_{\text{surface tension}}=\frac{2T}{r}\)​, where \(T\) is the surface tension and \(r\) is the radius of the bubble.
2. Due to water column: \(\Delta {P}_{\text{water}}=\rho gh\), where \(\rho\) is the density of water, \(g\) is gravitational acceleration, and \(h\) is the depth of the bubble below the surface.

Step 1: Calculate Pressure Difference Due to Surface Tension

\(\Delta {P}_{\text{surface tension}}=\frac{2T}{r}\)​

Given:

\(T=70\times 1{0}^{−3}\text{ }\text{N/m},\ r=2\text{ }\text{cm}=0.02\text{ }\text{m}\),

\(\Delta {P}_{\text{surface tension}}=\frac{2\times 70\times 1{0}^{−3}}{0.02}\),

​ \(\Delta {P}_{\text{surface tension}}=7\text{ }{\text{N/m}}^{2}\)

Step 2: Calculate Pressure Difference Due to Water Column

\(\Delta {P}_{\text{water}}=\rho gh\)

Given:

\(\rho =1000\text{ }{\text{kg/m}}^{3},\ g=9.8\text{ }{\text{m/s}}^{2},\ h=20\text{ }\text{cm}=0.2\text{ }\text{m}\)

\(\Delta {P}_{\text{water}}=1000\times 9.8\times 0.2\)

\(\Delta {P}_{\text{water}}=1960\text{ }{\text{N/m}}^{2}\)

Step 3: Total Pressure Difference

The total pressure difference is the sum of the two contributions:

\(\Delta P=\Delta {P}_{\text{surface tension}}+\Delta {P}_{\text{water}}\)

\(\Delta P=7+1960=1967\text{ }{\text{N/m}}^{2}\)

Convert to standard form:

\(\Delta P=0.01967\times 1{0}^{5}\text{ }{\text{N/m}}^{2}\text{≈}0.02\times 1{0}^{5}\text{ }{\text{N/m}}^{2}\)

Final Answer:

\((b)\text{ }0.02\times 1{0}^{5}\text{ }{\text{N/m}}^{2}\)​