Assuming the surface tension of rain water to be `72 "dyne"//"cm"`, find the differnce of pressure inside and outside a rain drop of diameter `0.02 cm`. What would this pressure difference amount to , if the drop were to be decreased by evaporation to a diameter of `0.0002 cm` ?

Explanation

To find the difference of pressure inside and outside a raindrop, we can use the formula for pressure difference due to surface tension:

\[ \Delta P = \frac{2S}{R} \]

where:
- \(\Delta P\) is the pressure difference,
- \(S\) is the surface tension,
- \(R\) is the radius of the raindrop.

### Step 1: Calculate the radius of the raindrop

Given the diameter of the raindrop is \(0.02 \, \text{cm}\), we can find the radius:

\[ R = \frac{\text{Diameter}}{2} = \frac{0.02 \, \text{cm}}{2} = 0.01 \, \text{cm} \]

To convert this to meters:

\[ R = 0.01 \, \text{cm} = 0.01 \times 10^{-2} \, \text{m} = 0.0001 \, \text{m} \]

### Step 2: Convert surface tension to SI units

The surface tension \(S\) is given as \(72 \, \text{dyne/cm}\). To convert this to Newtons per meter:

\[ S = 72 \, \text{dyne/cm} = 72 \times 10^{-5} \, \text{N/m} = 72 \times 10^{-3} \, \text{N/m} \]

### Step 3: Calculate the pressure difference for the initial raindrop

Now, we can substitute the values into the pressure difference formula:

\[ \Delta P = \frac{2S}{R} = \frac{2 \times 72 \times 10^{-3}}{0.0001} \]

Calculating this gives:

\[ \Delta P = \frac{144 \times 10^{-3}}{0.0001} = 1440 \, \text{N/m}^2 \]

### Step 4: Calculate the new radius for the evaporated raindrop

If the raindrop evaporates to a diameter of \(0.0002 \, \text{cm}\), the new radius is:

\[ R' = \frac{0.0002 \, \text{cm}}{2} = 0.0001 \, \text{cm} = 0.0001 \times 10^{-2} \, \text{m} = 0.000001 \, \text{m} \]

### Step 5: Calculate the pressure difference for the smaller raindrop

Now we can calculate the pressure difference for the new radius:

\[ \Delta P' = \frac{2S}{R'} = \frac{2 \times 72 \times 10^{-3}}{0.000001} \]

Calculating this gives:

\[ \Delta P' = \frac{144 \times 10^{-3}}{0.000001} = 144000 \, \text{N/m}^2 \]

### Final Results

1. The pressure difference for the raindrop of diameter \(0.02 \, \text{cm}\) is \(1440 \, \text{N/m}^2\).
2. The pressure difference for the raindrop of diameter \(0.0002 \, \text{cm}\) is \(144000 \, \text{N/m}^2\).