Understanding Gauge Pressure in a Rain Drop

This question involves calculating the gauge pressure inside a small spherical liquid drop due to surface tension. Surface tension is a property of liquids that allows them to resist an external force, acting like a stretched elastic membrane on the surface. For a spherical liquid drop, this surface tension creates an inward pull, resulting in a higher pressure inside the drop compared to the outside. This excess pressure is known as gauge pressure.

Formula for Gauge Pressure

The excess pressure ($ \Delta P $) inside a spherical drop due to surface tension ($ \gamma $) is given by the Laplace pressure formula for a single interface:

$$ \Delta P = \frac{2 \gamma}{r} $$

Where:

• $ \Delta P $ is the gauge pressure (the difference between the inside pressure and the outside pressure).
• $ \gamma $ is the surface tension of the liquid-air interface.
• $ r $ is the radius of the spherical drop.

Given Data

From the question, we have the following information:

• Surface tension ($ \gamma $) = 0.073 N/m
• Diameter of the rain drop ($ d $) = 1 mm

Step-by-Step Calculation

To find the gauge pressure, we first need to determine the radius of the rain drop and ensure all units are consistent.

1. Calculate the Radius: The radius is half of the diameter.

   $$ r = \frac{d}{2} $$

   $$ r = \frac{1 \, \text{mm}}{2} = 0.5 \, \text{mm} $$

2. Convert Units: Since the surface tension is given in Newtons per meter (N/m), we need to convert the radius from millimeters (mm) to meters (m).

   $$ 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} $$

   Therefore,

   $$ r = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} $$

3. Apply the Formula: Now, substitute the values of $ \gamma $ and $ r $ into the gauge pressure formula.

   $$ \Delta P = \frac{2 \gamma}{r} $$

   $$ \Delta P = \frac{2 \times 0.073 \, \text{N/m}}{0.5 \times 10^{-3} \, \text{m}} $$

4. Calculate the Result:

   $$ \Delta P = \frac{0.146 \, \text{N/m}}{0.5 \times 10^{-3} \, \text{m}} $$

   $$ \Delta P = \frac{0.146}{0.5} \times 10^3 \, \text{N/m}^2 $$

   $$ \Delta P = 0.292 \times 10^3 \, \text{N/m}^2 $$

   $$ \Delta P = 292 \, \text{N/m}^2 $$

Conclusion

The calculated gauge pressure inside the rain drop is 292 N/m². This value corresponds to one of the given options.