The diameter of rain-drop is 0.02 cm . If surface tension of water be `72 xx 10^(-3)` newton per metre , then the pressure difference of external and internal surfaces of the drop will be

Options

A. `1.44 xx 10^(4) "dyne" - cm^(-2)`
B. `1.44 xx 10^(4) "newton" - m^(-2)`
C. `1.44 xx 10^(3) "dyne" - cm^(-2)`
D. `1.44 xx 10^(5) "newton" - m^(-2)`

Answer

To solve the problem of finding the pressure difference between the external and internal surfaces of a raindrop, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Diameter of the raindrop (d) = 0.02 cm
- Surface tension of water (γ) = 72 x 10^(-3) N/m

2. **Calculate the Radius:**
- The radius (r) is half of the diameter.
- \( r = \frac{d}{2} = \frac{0.02 \text{ cm}}{2} = 0.01 \text{ cm} \)

3. **Convert Radius to Meters:**
- To convert centimeters to meters, we use the conversion factor \( 1 \text{ cm} = 0.01 \text{ m} \).
- \( r = 0.01 \text{ cm} = 0.01 \times 10^{-2} \text{ m} = 0.0001 \text{ m} \)

4. **Use the Formula for Pressure Difference:**
- The formula for the pressure difference (ΔP) between the inside and outside of a spherical drop is given by:
\[ \Delta P = \frac{2 \gamma}{r} \]
- Substitute the values into the formula:
\[ \Delta P = \frac{2 \times (72 \times 10^{-3} \text{ N/m})}{0.0001 \text{ m}} \]

5. **Calculate the Pressure Difference:**
- Calculate the numerator:
\[ 2 \times 72 \times 10^{-3} = 144 \times 10^{-3} \text{ N/m} \]
- Now divide by the radius:
\[ \Delta P = \frac{144 \times 10^{-3}}{0.0001} = 1440 \text{ N/m}^2 \]

6. **Convert Pressure to Dines per Centimeter Square:**
- Since \( 1 \text{ N/m}^2 = 10^5 \text{ dine/cm}^2 \):
\[ \Delta P = 1440 \times 10^5 \text{ dine/cm}^2 = 1.44 \times 10^8 \text{ dine/cm}^2 \]

### Final Answer:
The pressure difference of the external and internal surfaces of the drop is \( 1.44 \times 10^8 \text{ dine/cm}^2 \).

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