What is the difference of pressure between the inside and outside of a spherical drop of water of radius 1mm? Surface tension of water `= 7.2 xx 10^(-2)Nm^(-1)`

Explanation

To find the difference of pressure between the inside and outside of a spherical drop of water, we can use the formula for excess pressure in a liquid drop:

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

where:
- \(\Delta P\) is the difference of pressure,
- \(\sigma\) is the surface tension of the liquid,
- \(R\) is the radius of the drop.

### Step 1: Identify the given values
- Surface tension of water, \(\sigma = 7.2 \times 10^{-2} \, \text{N/m}\)
- Radius of the water drop, \(R = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\)

### Step 2: Substitute the values into the formula
Using the formula for excess pressure:

\[ \Delta P = \frac{2 \times \sigma}{R} \]

Substituting the known values:

\[ \Delta P = \frac{2 \times (7.2 \times 10^{-2} \, \text{N/m})}{1 \times 10^{-3} \, \text{m}} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ 2 \times (7.2 \times 10^{-2}) = 14.4 \times 10^{-2} \, \text{N/m} \]

### Step 4: Divide by the radius
Now, divide by the radius:

\[ \Delta P = \frac{14.4 \times 10^{-2} \, \text{N/m}}{1 \times 10^{-3} \, \text{m}} = 14.4 \times 10^{1} \, \text{N/m}^2 \]

### Step 5: Simplify the result
This simplifies to:

\[ \Delta P = 144 \, \text{N/m}^2 \]

### Final Answer
The difference of pressure between the inside and outside of the spherical drop of water is:

\[ \Delta P = 144 \, \text{N/m}^2 \]

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