If the Young's modulus of steel is ` 2xx10^(11) Nm^(-2)`, calculate the work done in stretching a steel wire 100 cm in length and of cross-sectional area `0.03 cm^(3)` when a load of 20 kg is slowly applied without the elastic limit being reached.

Answer

To solve the problem of calculating the work done in stretching a steel wire, we will follow these steps:

### Step 1: Understand the Given Data
- Young's modulus of steel, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Length of the wire, \( L = 100 \, \text{cm} = 1 \, \text{m} \)
- Cross-sectional area, \( A = 0.03 \, \text{cm}^2 = 0.03 \times 10^{-4} \, \text{m}^2 \) (since \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \))
- Load applied, \( m = 20 \, \text{kg} \)

### Step 2: Calculate the Force
The force \( F \) exerted by the load can be calculated using the formula:
\[ F = m \cdot g \]
where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

Substituting the values:
\[ F = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N} \]

### Step 3: Calculate the Extension (\( \Delta L \))
Using the formula for Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Rearranging gives:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

Substituting the values:
- \( F = 196 \, \text{N} \)
- \( L = 1 \, \text{m} \)
- \( A = 0.03 \times 10^{-4} \, \text{m}^2 \)
- \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

Calculating:
\[ \Delta L = \frac{196 \, \text{N} \cdot 1 \, \text{m}}{0.03 \times 10^{-4} \, \text{m}^2 \cdot 2 \times 10^{11} \, \text{N/m}^2} \]

### Step 4: Calculate the Work Done
The work done \( W \) in stretching the wire is given by:
\[ W = \frac{1}{2} F \Delta L \]

Substituting the values we calculated:
1. Calculate \( \Delta L \) from the previous step.
2. Then substitute \( F = 196 \, \text{N} \) and \( \Delta L \) into the work done formula.

### Final Calculation
After calculating \( \Delta L \), substitute it back into the work done formula to find the final answer.

### Conclusion
The work done in stretching the steel wire is approximately \( 0.032 \, \text{J} \).

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