A uniform wire of length 6m and a cross-sectional area 1.2 `cm^(2)` is stretched by a force of 600N . If the Young's modulus of the material of the wire `20xx 10^(10) Nm^(-2)` , calculate (i) stress (ii) strain and (iii) increase in length of the wire.

Explanation

To solve the problem step by step, we will calculate the stress, strain, and increase in length of the wire based on the given parameters.

### Given Data:
- Length of the wire, \( L = 6 \, \text{m} \)
- Cross-sectional area, \( A = 1.2 \, \text{cm}^2 = 1.2 \times 10^{-4} \, \text{m}^2 \) (conversion from cm² to m²)
- Force applied, \( F = 600 \, \text{N} \)
- Young's modulus, \( Y = 20 \times 10^{10} \, \text{N/m}^2 \)

### (i) Calculate Stress
Stress (\( \sigma \)) is defined as the force applied per unit area.
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{600 \, \text{N}}{1.2 \times 10^{-4} \, \text{m}^2} \]
Calculating:
\[ \sigma = \frac{600}{1.2 \times 10^{-4}} = 5 \times 10^6 \, \text{N/m}^2 \]

### (ii) Calculate Strain
Strain (\( \epsilon \)) is defined as the ratio of stress to Young's modulus.
\[ \epsilon = \frac{\sigma}{Y} \]
Substituting the values:
\[ \epsilon = \frac{5 \times 10^6 \, \text{N/m}^2}{20 \times 10^{10} \, \text{N/m}^2} \]
Calculating:
\[ \epsilon = \frac{5}{20 \times 10^4} = 0.25 \times 10^{-4} \]

### (iii) Calculate Increase in Length
The increase in length (\( \Delta L \)) can be calculated using the formula:
\[ \Delta L = \epsilon \times L \]
Substituting the values:
\[ \Delta L = (0.25 \times 10^{-4}) \times 6 \, \text{m} \]
Calculating:
\[ \Delta L = 1.5 \times 10^{-4} \, \text{m} \]

### Final Results:
- **Stress**: \( 5 \times 10^6 \, \text{N/m}^2 \)
- **Strain**: \( 0.25 \times 10^{-4} \)
- **Increase in Length**: \( 1.5 \times 10^{-4} \, \text{m} \)

---