A soap bubble, having radius of 1 mm, is blown from a detergent solution having radius of 1 mm is blown from a detergent solution having a surface tension of `2.5xx10^(-2)N//m`. The pressure inside the bubble equals at a point `Z_(0)` below the free surface of water in a container. Taking `g=10 m//s^(2)`, density of water `=10^(3) kg//m^(3)`, the value of `Z_(0)` is :

Options

A. 0.5cm
B. 100 cm
C. 10 cm
D. 1 cm

Explanation

To solve the problem, we need to find the depth \( Z_0 \) below the surface of water where the pressure inside the soap bubble equals the pressure at that depth.

### Step-by-Step Solution:

1. **Understand the pressure inside the soap bubble**:
The pressure inside a soap bubble is given by the formula:
\[ P_{\text{inside}} = P_0 + \frac{4T}{R} \]
where:
- \( P_0 \) is the atmospheric pressure,
- \( T \) is the surface tension of the soap solution,
- \( R \) is the radius of the bubble.

2. **Identify the given values**:
- Radius of the soap bubble, \( R = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Surface tension, \( T = 2.5 \times 10^{-2} \, \text{N/m} \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

3. **Pressure at depth \( Z_0 \)**:
The pressure at a depth \( Z_0 \) in a fluid is given by:
\[ P_{\text{outside}} = P_0 + \rho g Z_0 \]

4. **Set the pressures equal**:
According to the problem, the pressure inside the bubble equals the pressure at depth \( Z_0 \):
\[ P_0 + \frac{4T}{R} = P_0 + \rho g Z_0 \]
By canceling \( P_0 \) from both sides, we get:
\[ \frac{4T}{R} = \rho g Z_0 \]

5. **Rearranging for \( Z_0 \)**:
We can rearrange the equation to solve for \( Z_0 \):
\[ Z_0 = \frac{4T}{\rho g R} \]

6. **Substituting the values**:
Now substitute the known values into the equation:
\[ Z_0 = \frac{4 \times (2.5 \times 10^{-2})}{(10^3) \times (10) \times (1 \times 10^{-3})} \]

7. **Calculating \( Z_0 \)**:
\[ Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^3 \times 10 \times 10^{-3}} \]
\[ = \frac{10 \times 10^{-2}}{10^3 \times 10^{-2}} = \frac{10}{10^3} = 0.01 \, \text{m} \]
Converting meters to centimeters:
\[ Z_0 = 0.01 \, \text{m} = 1 \, \text{cm} \]

### Final Answer:
\[ Z_0 = 1 \, \text{cm} \]