The depth of water at which air bubble of radius 0.4 mm remains in equilibrium is `(T_("water")=72xx10^(-3)N//m)`

Options

A. `3.67cm`
B. `3.67m`
C. `6.37cm`
D. `5.32cm`

Explanation

To find the depth of water at which an air bubble of radius 0.4 mm remains in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Excess Pressure**:
The excess pressure (ΔP) inside a bubble is given by the formula:
\[ \Delta P = \frac{2T}{R} \]
where \( T \) is the surface tension of the liquid and \( R \) is the radius of the bubble.

2. **Given Values**:
- Radius of the bubble, \( R = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \)
- Surface tension of water, \( T = 72 \times 10^{-3} \, \text{N/m} \)

3. **Calculate the Excess Pressure**:
Substitute the values into the excess pressure formula:
\[ \Delta P = \frac{2 \times (72 \times 10^{-3})}{0.4 \times 10^{-3}} \]

4. **Perform the Calculation**:
- Calculate the numerator:
\[ 2 \times (72 \times 10^{-3}) = 144 \times 10^{-3} \, \text{N/m} \]
- Now calculate the excess pressure:
\[ \Delta P = \frac{144 \times 10^{-3}}{0.4 \times 10^{-3}} = \frac{144}{0.4} \, \text{N/m}^2 = 360 \, \text{N/m}^2 \]

5. **Relate Excess Pressure to Depth**:
The pressure at a depth \( H \) in a fluid is given by:
\[ P = \rho g H \]
where \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)), and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

6. **Set Excess Pressure Equal to Hydrostatic Pressure**:
For the bubble to be in equilibrium, the excess pressure must equal the hydrostatic pressure:
\[ \Delta P = \rho g H \]
Thus,
\[ 360 = 1000 \times 9.81 \times H \]

7. **Solve for Depth \( H \)**:
Rearranging gives:
\[ H = \frac{360}{1000 \times 9.81} \]
Calculate \( H \):
\[ H = \frac{360}{9810} \approx 0.03667 \, \text{m} = 36.67 \, \text{mm} = 3.67 \, \text{cm} \]

### Final Answer:
The depth of water at which the air bubble remains in equilibrium is approximately **3.67 cm**.