Find the difference in excess pressure on the inside and outside of a rain drop, if its diameter changes from 1.003 cm to 1.002 cm by evaporation ? Surface tension of water is `72 xx 10^(-3) Nm^(-1)`.

Explanation

To find the difference in excess pressure on the inside and outside of a raindrop as its diameter changes, we can follow these steps:

### Step 1: Determine the radii of the raindrop
The radius \( R \) of a sphere is half of its diameter \( D \). Given the diameters:
- Initial diameter \( D_1 = 1.003 \, \text{cm} = 1.003 \times 10^{-2} \, \text{m} \)
- Final diameter \( D_2 = 1.002 \, \text{cm} = 1.002 \times 10^{-2} \, \text{m} \)

We can calculate the radii:
\[ R_1 = \frac{D_1}{2} = \frac{1.003 \times 10^{-2}}{2} = 0.5015 \times 10^{-2} \, \text{m} = 5.015 \times 10^{-3} \, \text{m} \]
\[ R_2 = \frac{D_2}{2} = \frac{1.002 \times 10^{-2}}{2} = 0.501 \times 10^{-2} \, \text{m} = 5.01 \times 10^{-3} \, \text{m} \]

### Step 2: Use the formula for excess pressure
The excess pressure \( P \) inside a droplet due to surface tension is given by:
\[ P = \frac{2T}{R} \]
where \( T \) is the surface tension of water, given as \( T = 72 \times 10^{-3} \, \text{N/m} \).

### Step 3: Calculate the excess pressure inside and outside the raindrop
1. **Excess pressure inside the raindrop \( P_i \)**:
\[ P_i = \frac{2T}{R_1} = \frac{2 \times 72 \times 10^{-3}}{5.015 \times 10^{-3}} \]
Calculating this gives:
\[ P_i = \frac{144 \times 10^{-3}}{5.015 \times 10^{-3}} \approx 28.743 \, \text{N/m}^2 \]

2. **Excess pressure outside the raindrop \( P_o \)**:
\[ P_o = \frac{2T}{R_2} = \frac{2 \times 72 \times 10^{-3}}{5.01 \times 10^{-3}} \]
Calculating this gives:
\[ P_o = \frac{144 \times 10^{-3}}{5.01 \times 10^{-3}} \approx 28.714 \, \text{N/m}^2 \]

### Step 4: Calculate the difference in excess pressure
The difference in excess pressure \( \Delta P \) is:
\[ \Delta P = P_o - P_i = 28.714 - 28.743 \approx -0.029 \, \text{N/m}^2 \]

### Conclusion
The difference in excess pressure on the inside and outside of the raindrop is approximately:
\[ \Delta P \approx 0.029 \, \text{N/m}^2 \]