A spherical drop of water has `1mm` radius. If the surface tension of water is `70xx10^(-3)N//m`, then the difference of pressure between inside and outside of the spherical drop is:

Options

A. `35 N m^(-2)`
B. `70 N m^(-2)`
C. `140 N m^(-2)`
D. Zero

Answer

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

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

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

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

### Step 2: Substitute the values into the formula
Now we substitute the values of \(T\) and \(R\) into the pressure difference formula:

\[ \Delta P = \frac{2 \times (70 \times 10^{-3})}{1 \times 10^{-3}} \]

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

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

Now substituting this back into the equation:

\[ \Delta P = \frac{140 \times 10^{-3}}{1 \times 10^{-3}} = 140 \text{ N/m}^2 \]

### Step 4: Final result
Thus, the difference of pressure between the inside and outside of the spherical drop is:

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

### Summary
The difference of pressure between the inside and outside of the spherical drop is \(140 \text{ N/m}^2\).

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