Calculate the difference in excess pressure inside a spherical rain drop if its radius changes from 0.05 cm to 0.0005 cm by evaporation. Surface tension of water is `"72 dyne cm"^(-1)`.

Answer

To calculate the difference in excess pressure inside a spherical raindrop as its radius changes from 0.05 cm to 0.0005 cm, we can use the formula for excess pressure inside a spherical droplet, which is given by:

\[ P = \frac{4T}{R} \]

where \( P \) is the excess pressure, \( T \) is the surface tension, and \( R \) is the radius of the droplet.

### Step-by-Step Solution:

1. **Convert Surface Tension to SI Units:**
The surface tension \( T \) is given as 72 dyne/cm. We need to convert this to Newton/meter.
\[ T = 72 \, \text{dyne/cm} = 72 \times 10^{-5} \, \text{N/m} = 0.072 \, \text{N/m} \]

2. **Convert Radii to SI Units:**
The radii are given in centimeters and need to be converted to meters.
- For \( R_1 = 0.05 \, \text{cm} \):
\[ R_1 = 0.05 \times 10^{-2} \, \text{m} = 0.0005 \, \text{m} \]
- For \( R_2 = 0.0005 \, \text{cm} \):
\[ R_2 = 0.0005 \times 10^{-2} \, \text{m} = 0.0000005 \, \text{m} \]

3. **Calculate Excess Pressure for Each Radius:**
- For \( R_1 \):
\[ P_1 = \frac{4T}{R_1} = \frac{4 \times 0.072}{0.0005} = \frac{0.288}{0.0005} = 576 \, \text{N/m}^2 \]
- For \( R_2 \):
\[ P_2 = \frac{4T}{R_2} = \frac{4 \times 0.072}{0.0000005} = \frac{0.288}{0.0000005} = 576000 \, \text{N/m}^2 \]

4. **Calculate the Difference in Excess Pressure:**
Now, we find the difference in excess pressure:
\[ \Delta P = P_2 - P_1 = 576000 - 576 = 575424 \, \text{N/m}^2 \]

### Final Answer:
The difference in excess pressure inside the spherical raindrop as its radius changes from 0.05 cm to 0.0005 cm is \( 575424 \, \text{N/m}^2 \).