What is the difference in pressure between the inside and outside of a spherical drop of radius 2mm Surface tension of water `=70 xx 10^(-3) "Nm"^(-1).`

Answer

To find the difference in pressure between the inside and outside of a spherical drop, we can use the formula derived from the principles of fluid mechanics, specifically for a soap bubble or a spherical drop. The formula is:

\[ \Delta P = P' - P_0 = \frac{2S}{R} \]

where:
- \(\Delta P\) is the difference in pressure,
- \(P'\) is the pressure inside the drop,
- \(P_0\) is the pressure outside the drop,
- \(S\) is the surface tension of the liquid,
- \(R\) is the radius of the drop.

### Step 1: Identify the given values
- Surface tension, \(S = 70 \times 10^{-3} \, \text{N/m}\)
- Radius of the drop, \(R = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\)

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

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

Substituting the values of \(S\) and \(R\):

\[ \Delta P = \frac{2 \times (70 \times 10^{-3})}{2 \times 10^{-3}} \]

### Step 3: Simplify the equation
Calculating the numerator:

\[ 2 \times (70 \times 10^{-3}) = 140 \times 10^{-3} \, \text{N/m} \]

Now substituting this back into the equation:

\[ \Delta P = \frac{140 \times 10^{-3}}{2 \times 10^{-3}} \]

### Step 4: Perform the division
Now, divide \(140 \times 10^{-3}\) by \(2 \times 10^{-3}\):

\[ \Delta P = \frac{140}{2} = 70 \, \text{N/m}^2 \]

### Step 5: Final result
Thus, the difference in pressure between the inside and outside of the spherical drop is:

\[ \Delta P = 70 \, \text{N/m}^2 \]