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

To address this issue, we will employ the principles of pressure differential and surface tension. The pressure differential stemming from surface tension for an air bubble is quantified by the following equation:

\(P_{\text{inside}} - P_{\text{outside}} = \frac{4T}{r}\)

In this formula, \(T\) represents the liquid's surface tension, and \(r\) denotes the air bubble's radius.

Given that the internal pressure of the bubble exceeds atmospheric pressure by 2100 N/m\(^2\), we can express this as:

\(P_{\text{inside}} - P_{\text{outside}} = 2100 \, \text{N/m}^2\)

Furthermore, with the radius specified as \(r = 0.1 \, \text{cm}\), which converts to 0.001 m, we can integrate these values into the equation:

\(2100 = \frac{4T}{0.001}\) 

Rearranging to solve for \(T\) yields:

\(T = \frac{2100 \times 0.001}{4} = \frac{2.1}{4} = 0.525 \times 0.1 = 0.05 \, \text{N/m}\)

Consequently, the surface tension of the liquid is determined to be \(0.05 \, \text{N/m}\).

The calculated surface tension is 0.05.