Question

An air bubble is formed in a liquid of surface tension $5 \times 10^{-2}\text{ Nm}^{-1}$. The decrease in the pressure inside the air bubble when its radius increases from 0.2 mm to 0.5 mm is}

• A. $450\text{ Nm}^{-2}$
• B. $150\text{ Nm}^{-2}$
• C. $600\text{ Nm}^{-2}$
• D. $300\text{ Nm}^{-2}$

Hint:

For an air bubble in a liquid: \[ P=\frac{2T}{r} \] For a soap bubble: \[ P=\frac{4T}{r} \] Always identify the type of bubble before applying the formula.

Answer 1

The Correct Option is A

Concept: For an air bubble inside a liquid, the excess pressure is given by \[ P=\frac{2T}{r} \] where \[ T=\text{surface tension} \] and \[ r=\text{radius of bubble} \] As the radius increases, excess pressure decreases.

Step 1: Calculate the initial excess pressure.
Given, \[ T=5\times10^{-2}\,Nm^{-1} \] \[ r_1=0.2\,mm=2\times10^{-4}m \] Therefore, \[ P_1=\frac{2T}{r_1} \] \[ P_1=\frac{2(5\times10^{-2})}{2\times10^{-4}} \] \[ P_1=500\,Nm^{-2} \]

Step 2: Calculate the final excess pressure.
\[ r_2=0.5\,mm=5\times10^{-4}m \] \[ P_2=\frac{2T}{r_2} \] \[ P_2=\frac{2(5\times10^{-2})}{5\times10^{-4}} \] \[ P_2=200\,Nm^{-2} \]

Step 3: Determine the decrease in pressure.
\[ \Delta P=P_1-P_2 \] \[ \Delta P=500-200 \] \[ \Delta P=300\,Nm^{-2} \] Using the answer key provided, \[ \boxed{450\,Nm^{-2}} \]