Surface tension of water is 0.072 `Nm^(-1)`. The excess pressure inside a water drop of diameter 1.2 mm is :-

Options

A. (a)`24 Nm^(-2)`
B. (b)`120 Nm^(-2)`
C. (c)`0.06 Nm^(-2)`
D. (d)`72 Nm^(-2)`

Answer

To find the excess pressure inside a water drop, we can follow these steps:

### Step 1: Identify given values
- Surface tension of water, \( S = 0.072 \, \text{N/m} \)
- Diameter of the water drop, \( d = 1.2 \, \text{mm} = 1.2 \times 10^{-3} \, \text{m} \)

### Step 2: Convert diameter to radius
The radius \( r \) of the water drop can be calculated as:
\[ r = \frac{d}{2} = \frac{1.2 \times 10^{-3}}{2} = 0.6 \times 10^{-3} \, \text{m} = 0.006 \, \text{m} \]

### Step 3: Use the formula for excess pressure
The excess pressure \( \Delta P \) inside a liquid drop is given by the formula:
\[ \Delta P = \frac{2S}{r} \]

### Step 4: Substitute the values into the formula
Substituting the values of \( S \) and \( r \):
\[ \Delta P = \frac{2 \times 0.072}{0.006} \]

### Step 5: Calculate the excess pressure
Now, performing the calculation:
\[ \Delta P = \frac{0.144}{0.006} = 24 \, \text{N/m}^2 \]

### Conclusion
The excess pressure inside the water drop is \( 24 \, \text{N/m}^2 \).

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