A steel rod of length `25cm` has a cross-sectional area of `0.8cm^(2)` . The force required to stretch this rod by the same amount as the expansion produced by heating it through `10^(@)C` is `(alpha_(steel)=10^(-5)//^(@)C` and `Y_(steel)=2xx10^(10)N//m^(2))`

Options

A. `40N`
B. `80N`
C. `120N`
D. `160N`

Explanation

To solve the problem, we need to find the force required to stretch a steel rod by the same amount as the expansion produced by heating it through \(10^\circ C\). We will use the formulas for linear thermal expansion and Young's modulus.

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the steel rod, \(L = 25 \, \text{cm} = 0.25 \, \text{m}\)
- Cross-sectional area, \(A = 0.8 \, \text{cm}^2 = 0.8 \times 10^{-4} \, \text{m}^2\) (since \(1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2\))
- Coefficient of linear expansion, \(\alpha = 10^{-5} \, \text{°C}^{-1}\)
- Young's modulus, \(Y = 2 \times 10^{10} \, \text{N/m}^2\)
- Change in temperature, \(\Delta T = 10 \, \text{°C}\)

2. **Calculate the change in length (\(\Delta L\)) due to thermal expansion:**
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Substituting the values:
\[ \Delta L = 0.25 \, \text{m} \cdot 10^{-5} \, \text{°C}^{-1} \cdot 10 \, \text{°C} = 0.25 \cdot 10^{-4} \, \text{m} = 2.5 \times 10^{-5} \, \text{m} \]

3. **Relate the force (\(F\)) to the change in length using Young's modulus:**
The formula relating stress, strain, and Young's modulus is:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \quad \Rightarrow \quad \text{Stress} = Y \cdot \text{Strain} \]
Where:
\[ \text{Strain} = \frac{\Delta L}{L} \]
Therefore, substituting for stress:
\[ \text{Stress} = Y \cdot \frac{\Delta L}{L} \]

4. **Calculate the stress:**
\[ \text{Stress} = 2 \times 10^{10} \, \text{N/m}^2 \cdot \frac{2.5 \times 10^{-5} \, \text{m}}{0.25 \, \text{m}} = 2 \times 10^{10} \cdot 10^{-4} = 2 \times 10^{6} \, \text{N/m}^2 \]

5. **Calculate the force (\(F\)):**
The force can be calculated using:
\[ F = \text{Stress} \cdot A \]
Substituting the values:
\[ F = 2 \times 10^{6} \, \text{N/m}^2 \cdot 0.8 \times 10^{-4} \, \text{m}^2 = 160 \, \text{N} \]

### Final Answer:
The force required to stretch the rod by the same amount as the expansion produced by heating it through \(10^\circ C\) is \(F = 160 \, \text{N}\).

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