The maximum stress that can be applied to the material of a wire used to suspend an elevator is `(3)/(pi)xx10^(8)N//m^(2)` if the mass of elevator is 900 kg and it move up with an acceleration `2.2m//s^(2)` than calculate the minimum radius of the wire.

Answer

To solve the problem, we need to calculate the minimum radius of the wire used to suspend an elevator given the maximum stress and the conditions under which the elevator is moving.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Maximum stress, \( S_{max} = \frac{3}{\pi} \times 10^8 \, \text{N/m}^2 \)
- Mass of the elevator, \( m = 900 \, \text{kg} \)
- Acceleration of the elevator, \( a = 2.2 \, \text{m/s}^2 \)
- Acceleration due to gravity, \( g \approx 9.8 \, \text{m/s}^2 \)

2. **Calculate the Total Acceleration:**
- The total effective acceleration when the elevator is moving upwards is given by:
\[ a_{total} = a + g = 2.2 + 9.8 = 12 \, \text{m/s}^2 \]

3. **Calculate the Tension in the Wire:**
- The tension \( T \) in the wire can be calculated using Newton's second law:
\[ T = m \cdot a_{total} = 900 \cdot 12 = 10800 \, \text{N} \]

4. **Relate Tension to Stress:**
- Stress \( \sigma \) is defined as the force per unit area. For a circular wire, the area \( A \) is given by:
\[ A = \pi r^2 \]
- Therefore, the stress can be expressed as:
\[ \sigma = \frac{T}{A} = \frac{T}{\pi r^2} \]

5. **Set Maximum Stress Equal to Calculated Stress:**
- Since we are looking for the minimum radius when the stress is at its maximum:
\[ S_{max} = \frac{T}{\pi r^2} \]
- Rearranging gives:
\[ r^2 = \frac{T}{\pi S_{max}} \]

6. **Substitute Values:**
- Substitute \( T = 10800 \, \text{N} \) and \( S_{max} = \frac{3}{\pi} \times 10^8 \, \text{N/m}^2 \):
\[ r^2 = \frac{10800}{\pi \left(\frac{3}{\pi} \times 10^8\right)} = \frac{10800 \cdot \pi}{3 \cdot 10^8} \]
- Simplifying gives:
\[ r^2 = \frac{10800}{3 \cdot 10^8} = \frac{3600}{10^8} = 3.6 \times 10^{-5} \]

7. **Calculate the Radius:**
- Taking the square root to find \( r \):
\[ r = \sqrt{3.6 \times 10^{-5}} \approx 0.006 \, \text{m} = 6 \, \text{mm} \]

### Final Answer:
The minimum radius of the wire is approximately \( 6 \, \text{mm} \).