A rod of length `1000 mm` and co-efficient of linear expansion `alpha = 10^(-4)` per degree Celsius is placed in horizontal smooth surface symmetrically between fixed walls separated by `1001 mm `. The young's modulus of rod is `10^(11)N//m^(2)`. If the temperature is increased by `20^@C`, then the stress developed in the rod is (in ` N//m^(2)`)

Options

A. `(a)10^(5)`
B. `(b)10^(8)`
C. `(c)10^(7)`
D. `(d)10^(6)`

Explanation

To solve the problem, we need to calculate the stress developed in the rod when the temperature is increased. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- Length of the rod, \( L = 1000 \, \text{mm} = 1 \, \text{m} \)
- Coefficient of linear expansion, \( \alpha = 10^{-4} \, \text{per degree Celsius} \)
- Distance between the walls, \( D = 1001 \, \text{mm} = 1.001 \, \text{m} \)
- Young's modulus of the rod, \( Y = 10^{11} \, \text{N/m}^2 \)
- Change in temperature, \( \Delta T = 20 \, \text{°C} \)

### Step 2: Calculate the maximum thermal expansion of the rod
The formula for linear thermal expansion is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Substituting the values:
\[ \Delta L = 1 \, \text{m} \cdot 10^{-4} \, \text{per °C} \cdot 20 \, \text{°C} = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]

### Step 3: Determine the maximum permissible elongation
The distance between the walls is \( 1.001 \, \text{m} \) and the length of the rod is \( 1 \, \text{m} \). Therefore, the maximum permissible elongation of the rod is:
\[ \text{Maximum permissible elongation} = D - L = 1.001 \, \text{m} - 1 \, \text{m} = 0.001 \, \text{m} = 1 \, \text{mm} \]

### Step 4: Determine the actual elongation of the rod
Since the rod can expand by \( 2 \, \text{mm} \) due to thermal expansion, but it is constrained by the walls, the actual elongation will be limited to the maximum permissible elongation:
\[ \Delta L_{\text{actual}} = 1 \, \text{mm} = 0.001 \, \text{m} \]

### Step 5: Calculate the strain in the rod
Strain (\( \epsilon \)) is defined as the change in length divided by the original length:
\[ \epsilon = \frac{\Delta L_{\text{actual}}}{L} = \frac{0.001 \, \text{m}}{1 \, \text{m}} = 0.001 \]

### Step 6: Calculate the stress in the rod
Stress (\( \sigma \)) is related to Young's modulus and strain by the formula:
\[ \sigma = Y \cdot \epsilon \]
Substituting the values:
\[ \sigma = 10^{11} \, \text{N/m}^2 \cdot 0.001 = 10^{8} \, \text{N/m}^2 \]

### Final Answer
The stress developed in the rod is:
\[ \sigma = 10^{8} \, \text{N/m}^2 \]

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