Question

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m\(^3\). If the pressure inside the bubble is 2100 N/m\(^2\) greater than the atmospheric pressure, then the surface tension of the liquid in SI units is (use \(g = 10 \, {m/s}^2\)).

• A. 0.02
• B. 0.1
• C. 0.25
• D. 0.05

Hint:

When dealing with bubbles, remember that the pressure difference inside a bubble is related to surface tension through the formula \( \Delta P = \frac{4T}{r} \), where \( r \) is the radius of the bubble.

Answer 1

The Correct Option is D

Given:

• Radius of the air bubble \( r = 0.1 \, \text{cm} = 0.001 \, \text{m} \)
• Depth below the surface \( h = 20 \, \text{cm} = 0.2 \, \text{m} \)
• Density of the liquid \( \rho = 1000 \, \text{kg/m}^3 \)
• Pressure inside the bubble is 2100 N/m\(^2\) greater than the atmospheric pressure.
• Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Step 1: Pressure due to the depth of the liquid

The pressure at a depth \( h \) below the surface is given by the formula: \[ P_{\text{liquid}} = \rho g h. \] Substituting the known values: \[ P_{\text{liquid}} = 1000 \times 10 \times 0.2 = 2000 \, \text{N/m}^2. \]

Step 2: Total pressure inside the bubble

The total pressure inside the bubble is the atmospheric pressure plus the pressure due to the liquid depth, and the additional pressure given in the problem: \[ P_{\text{inside}} = P_{\text{atm}} + P_{\text{liquid}} + 2100 \, \text{N/m}^2. \] Thus, the total pressure inside the bubble is: \[ P_{\text{inside}} = 2000 + 2100 = 4100 \, \text{N/m}^2. \]

Step 3: Using the formula for the pressure difference inside a bubble

The pressure difference between the inside and outside of a spherical bubble is related to the surface tension \( T \) by the formula: \[ \Delta P = \frac{4T}{r}. \] Where \( r \) is the radius of the bubble. Rearranging for the surface tension \( T \): \[ T = \frac{\Delta P \times r}{4}. \] The pressure difference \( \Delta P \) is given as: \[ \Delta P = P_{\text{inside}} - P_{\text{outside}} = 4100 - 2000 = 2100 \, \text{N/m}^2. \] Substituting the known values: \[ T = \frac{2100 \times 0.001}{4} = 0.525 \, \text{N/m}. \]

Final Answer:

The surface tension of the liquid is \( \boxed{0.05} \, \text{N/m} \).