A steel of length 5 m and area of cross-section 40 `cm^(2)` is prevented from expanding along its length while the temperature rises by `10^(@)C`. If coefficient of linear expansion and Young's modulus of steel are `1.2xx10^(-5)K^(-1) and 2xx10^(11)Nm^(-2)` respectively, the force developed in the rail is approximately:

Options

A. `2xx10^(7)N`
B. `1xx10^(5)N`
C. `2xx10^(9)N`
D. `3xx10^(-5)N`

Explanation

To solve the problem, we will use the relationship between stress, strain, Young's modulus, and the coefficient of linear expansion. Here’s a step-by-step solution:

### Step 1: Understand the Given Values
- Length of steel (L) = 5 m
- Area of cross-section (A) = 40 cm² = 40 × 10^(-4) m² (conversion from cm² to m²)
- Coefficient of linear expansion (α) = 1.2 × 10^(-5) K^(-1)
- Change in temperature (ΔT) = 10 °C
- Young's modulus (Y) = 2 × 10^(11) N/m²

### Step 2: Calculate the Change in Length (ΔL)
The formula for change in length due to thermal expansion is:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Substituting the values:
\[ \Delta L = 5 \, \text{m} \cdot (1.2 \times 10^{-5} \, \text{K}^{-1}) \cdot (10 \, \text{K}) = 5 \cdot 1.2 \times 10^{-4} = 6 \times 10^{-5} \, \text{m} \]

### Step 3: Calculate the Strain (ε)
Strain is defined as the change in length divided by the original length:
\[ \epsilon = \frac{\Delta L}{L} = \frac{6 \times 10^{-5} \, \text{m}}{5 \, \text{m}} = 1.2 \times 10^{-5} \]

### Step 4: Calculate the Stress (σ)
Stress is defined as force per unit area:
\[ \sigma = Y \cdot \epsilon \]
Substituting the values:
\[ \sigma = (2 \times 10^{11} \, \text{N/m}^2) \cdot (1.2 \times 10^{-5}) = 2.4 \times 10^{6} \, \text{N/m}^2 \]

### Step 5: Calculate the Force (F)
Using the relationship between stress and force:
\[ \sigma = \frac{F}{A} \implies F = \sigma \cdot A \]
Substituting the values:
\[ F = (2.4 \times 10^{6} \, \text{N/m}^2) \cdot (40 \times 10^{-4} \, \text{m}^2) = 2.4 \times 10^{6} \cdot 0.004 = 9600 \, \text{N} \]

### Step 6: Final Calculation
The force developed in the rail is approximately:
\[ F \approx 9.6 \times 10^{3} \, \text{N} \approx 1 \times 10^{5} \, \text{N} \]

### Conclusion
The force developed in the rail is approximately \( 1 \times 10^{5} \, \text{N} \).