A spherical drop of water as `1mm` radius. If the surface tension of the the water is `50xx10^-3(N)/(m)`, then the difference of pressure between inside and outside the spherical drop is:

Options

A. `25 N//m^(2)`
B. `1000 N//m^(2)`
C. `100 N//m^(2)`
D. `50 N//m^(2)`

Explanation

To solve the problem of finding the difference in pressure between the inside and outside of a spherical drop of water, we can use the formula related to surface tension. Here’s a step-by-step solution:

### Step 1: Identify the given values
- **Radius of the drop (r)**: 1 mm = \(1 \times 10^{-3}\) m
- **Surface tension (T)**: \(50 \times 10^{-3}\) N/m

### Step 2: Write the formula for pressure difference
The formula for the difference in pressure (\(\Delta P\)) between the inside and outside of a spherical drop due to surface tension is given by:
\[ \Delta P = \frac{2T}{r} \]

### Step 3: Substitute the values into the formula
Now, we can substitute the values of surface tension and radius into the formula:
\[ \Delta P = \frac{2 \times (50 \times 10^{-3})}{1 \times 10^{-3}} \]

### Step 4: Calculate the numerator
Calculating the numerator:
\[ 2 \times (50 \times 10^{-3}) = 100 \times 10^{-3} \text{ N/m} \]

### Step 5: Divide by the radius
Now, divide the result by the radius:
\[ \Delta P = \frac{100 \times 10^{-3}}{1 \times 10^{-3}} = 100 \text{ N/m}^2 \]

### Step 6: Final result
Thus, the difference in pressure between the inside and outside of the spherical drop is:
\[ \Delta P = 100 \text{ N/m}^2 \]