A spherical drop of water has 2.5 mm radius. If the surface tension of water is `70 xx 10^(-3)` N `m^(-1)`, then the excess pressure inside the drop is

Options

A. 70 N `m^(-2)`
B. 56 N `m^(-2)`
C. 35 N `m^(-2)`
D. 48 N `m^(-2)`

Answer

To find the excess pressure inside a spherical drop of water, we can use the formula:

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

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

### Step 1: Identify the given values
- Radius of the drop, \(R = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m}\)
- Surface tension, \(S = 70 \times 10^{-3} \, \text{N/m}\)

### Step 2: Substitute the values into the formula
We substitute the values of \(S\) and \(R\) into the formula for excess pressure:

\[ \Delta P = \frac{2 \times (70 \times 10^{-3})}{2.5 \times 10^{-3}} \]

### 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:

\[ \Delta P = \frac{140 \times 10^{-3}}{2.5 \times 10^{-3}} \]

### Step 4: Perform the division
Now, we can simplify the fraction:

\[ \Delta P = \frac{140}{2.5} \, \text{N/m}^2 \]

Calculating \( \frac{140}{2.5} \):

\[ \Delta P = 56 \, \text{N/m}^2 \]

### Final Answer
The excess pressure inside the drop is:

\[ \Delta P = 56 \, \text{N/m}^2 \]