A steel rod of length `1 m` is heated from `25^@ "to" 75^@ C` keeping its length constant. The longitudinal strain developed in the rod is ( Given, coefficient of linear expansion of steel = `12 xx 10^-6//^@ C`).

Options

A. `6 xx 10^-4`
B. `-6 xx 10^-5`
C. `-6 xx 10^-4`
D. zero

Explanation

To solve the problem of determining the longitudinal strain developed in a steel rod when it is heated, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of the steel rod, \( L = 1 \, \text{m} \)
- Initial temperature, \( T_1 = 25^\circ C \)
- Final temperature, \( T_2 = 75^\circ C \)
- Coefficient of linear expansion of steel, \( \alpha = 12 \times 10^{-6} \, \text{°C}^{-1} \)

2. **Calculate the Change in Temperature (\( \Delta T \)):**
\[ \Delta T = T_2 - T_1 = 75^\circ C - 25^\circ C = 50^\circ C \]

3. **Understand the Concept of Longitudinal Strain:**
- Longitudinal strain (\( \epsilon \)) is defined as the change in length (\( \Delta L \)) divided by the original length (\( L \)):
\[ \epsilon = \frac{\Delta L}{L} \]

4. **Relate Change in Length to Temperature Change:**
- The change in length due to thermal expansion can be expressed as:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
- Since the length is kept constant, we consider the strain caused by the thermal expansion:
\[ \epsilon = -\alpha \cdot \Delta T \]

5. **Substitute the Values into the Strain Formula:**
\[ \epsilon = -\alpha \cdot \Delta T = - (12 \times 10^{-6} \, \text{°C}^{-1}) \cdot (50 \, \text{°C}) \]

6. **Calculate the Longitudinal Strain:**
\[ \epsilon = - (12 \times 10^{-6}) \cdot 50 = -6 \times 10^{-4} \]

7. **Final Result:**
- The longitudinal strain developed in the steel rod is:
\[ \epsilon = -6 \times 10^{-4} \]