Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are `0.465Nm^-1, 0.03Nm^-1 and 0.076Nm^-1` respectively.

Explanation

To solve the problem of finding the excess pressure inside a drop of mercury, a soap bubble, and an air bubble, we will use the formulas related to excess pressure due to surface tension.

### Step-by-Step Solution:

1. **Excess Pressure in a Drop of Mercury:**
- Given:
- Radius of mercury drop, \( R = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Surface tension of mercury, \( S = 0.465 \, \text{N/m} \)
- Formula for excess pressure inside a drop:
\[ P = \frac{2S}{R} \]
- Substituting the values:
\[ P = \frac{2 \times 0.465}{2 \times 10^{-3}} = \frac{0.93}{0.002} = 465 \, \text{N/m}^2 \]

2. **Excess Pressure in a Soap Bubble:**
- Given:
- Radius of soap bubble, \( R = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
- Surface tension of soap solution, \( S = 0.03 \, \text{N/m} \)
- Formula for excess pressure inside a soap bubble:
\[ P = \frac{4S}{R} \]
- Substituting the values:
\[ P = \frac{4 \times 0.03}{4 \times 10^{-3}} = \frac{0.12}{0.004} = 30 \, \text{N/m}^2 \]

3. **Excess Pressure in an Air Bubble in Water:**
- Given:
- Radius of air bubble, \( R = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
- Surface tension of water, \( S = 0.076 \, \text{N/m} \)
- Formula for excess pressure inside an air bubble:
\[ P = \frac{2S}{R} \]
- Substituting the values:
\[ P = \frac{2 \times 0.076}{4 \times 10^{-3}} = \frac{0.152}{0.004} = 38 \, \text{N/m}^2 \]

### Summary of Results:
- Excess pressure inside the mercury drop: **465 N/m²**
- Excess pressure inside the soap bubble: **30 N/m²**
- Excess pressure inside the air bubble: **38 N/m²**