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 find the longitudinal strain developed in the steel rod when it is heated, we can follow these steps:

### Step 1: Understand the Concept of Strain
Strain is defined as the change in length per unit length of the material. The formula for longitudinal strain (ε) is given by:
\[ \text{Strain} (\varepsilon) = \frac{\Delta L}{L} \]
where \(\Delta L\) is the change in length and \(L\) is the original length.

### Step 2: Identify the Parameters
Given:
- Original length of the 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, \(\alpha = 12 \times 10^{-6} \, /^\circ C\)

### Step 3: Calculate the Change in Temperature
The change in temperature (\(\Delta \theta\)) can be calculated as:
\[ \Delta \theta = T_2 - T_1 = 75^\circ C - 25^\circ C = 50^\circ C \]

### Step 4: Calculate the Change in Length
The change in length (\(\Delta L\)) due to thermal expansion can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta \theta \]
Substituting the values:
\[ \Delta L = 1 \, \text{m} \cdot (12 \times 10^{-6} \, /^\circ C) \cdot (50^\circ C) \]

### Step 5: Perform the Calculation
Calculating \(\Delta L\):
\[ \Delta L = 1 \cdot 12 \times 10^{-6} \cdot 50 = 600 \times 10^{-6} \, \text{m} = 6 \times 10^{-4} \, \text{m} \]

### Step 6: Substitute into the Strain Formula
Now, substituting \(\Delta L\) back into the strain formula:
\[ \varepsilon = \frac{\Delta L}{L} = \frac{6 \times 10^{-4}}{1} = 6 \times 10^{-4} \]

### Step 7: Consider the Direction of Strain
Since the rod is heated and its length is kept constant, the longitudinal strain will be negative (indicating compression):
\[ \varepsilon = -6 \times 10^{-4} \]

### Final Answer
The longitudinal strain developed in the rod is:
\[ \varepsilon = -6 \times 10^{-4} \]