Find the maximum energy per unit volume that can be stored in a metal wire when stretched, if its breaking stress is `1.3 xx 10^(8) Nm^(-2)` and Young's modulus of the material is `2.2 xx 10^(11) Nm^(-2)` .

Explanation

To find the maximum energy per unit volume that can be stored in a metal wire when stretched, we will use the relationship between stress, strain, and Young's modulus.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Breaking Stress (σ) = \(1.3 \times 10^8 \, \text{N/m}^2\)
- Young's Modulus (Y) = \(2.2 \times 10^{11} \, \text{N/m}^2\)

2. **Formula for Energy per Unit Volume:**
The energy per unit volume (U) stored in a stretched wire can be expressed as:
\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]

3. **Relate Strain to Stress and Young's Modulus:**
From the definition of Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \implies \text{Strain} = \frac{\text{Stress}}{Y} \]
Substituting the breaking stress (σ) into this equation gives:
\[ \text{Strain} = \frac{σ}{Y} \]

4. **Substitute Strain in the Energy Formula:**
Now, substituting the expression for strain back into the energy formula:
\[ U = \frac{1}{2} \times σ \times \left(\frac{σ}{Y}\right) = \frac{σ^2}{2Y} \]

5. **Plug in the Values:**
Now, substituting the values of σ and Y into the equation:
\[ U = \frac{(1.3 \times 10^8)^2}{2 \times (2.2 \times 10^{11})} \]

6. **Calculate σ²:**
\[ (1.3 \times 10^8)^2 = 1.69 \times 10^{16} \]

7. **Calculate the Denominator:**
\[ 2 \times (2.2 \times 10^{11}) = 4.4 \times 10^{11} \]

8. **Final Calculation of Energy per Unit Volume:**
\[ U = \frac{1.69 \times 10^{16}}{4.4 \times 10^{11}} \approx 0.384 \times 10^5 \, \text{J/m}^3 \]

9. **Final Result:**
\[ U \approx 3.84 \times 10^4 \, \text{J/m}^3 \]

### Conclusion:
The maximum energy per unit volume that can be stored in the metal wire when stretched is approximately \(3.84 \times 10^4 \, \text{J/m}^3\).