A rod is fixed between two points at `20^(@)C`. The coefficient of linear expansion of material of rod is `1.1 xx 10^(-5)//.^(C` and Young's modulus is `1.2 xx 10^(11) N//m`. Find the stress develop in the rod if temperature of rod becomes `10^(@)C`

Options

A. `1.32 xx10^(7)N//m^(2)`
B. `1.10xx10^(15)N//m^(2)`
C. `1.32xx10^(8) N//m^(2)`
D. `1.10xx10^(6) N//m^(2)`

Answer

To solve the problem, we need to find the stress developed in a rod when the temperature decreases from 20°C to 10°C. The stress can be calculated using the formula derived from Young's modulus and the linear expansion of the rod.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Coefficient of linear expansion, \( \alpha = 1.1 \times 10^{-5} \, \text{°C}^{-1} \)
- Young's modulus, \( Y = 1.2 \times 10^{11} \, \text{N/m}^2 \)
- Initial temperature, \( T_1 = 20 \, \text{°C} \)
- Final temperature, \( T_2 = 10 \, \text{°C} \)

2. **Calculate the Change in Temperature:**
\[ \Delta T = T_2 - T_1 = 10 \, \text{°C} - 20 \, \text{°C} = -10 \, \text{°C} \]

3. **Calculate the Change in Length Due to Temperature Change:**
The change in length \( \Delta L \) due to temperature change can be expressed as:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Since \( \Delta T = -10 \, \text{°C} \):
\[ \Delta L = L \cdot (1.1 \times 10^{-5}) \cdot (-10) = -1.1 \times 10^{-4} L \]
This indicates that the rod would want to contract by \( 1.1 \times 10^{-4} L \).

4. **Since the Rod is Fixed:**
The rod cannot contract because it is fixed between two points. Therefore, the stress developed in the rod is due to this inability to change length.

5. **Using Young's Modulus to Find Stress:**
The stress \( \sigma \) developed in the rod can be calculated using the relationship:
\[ \sigma = Y \cdot \frac{\Delta L}{L} \]
Substituting \( \Delta L \):
\[ \sigma = Y \cdot \frac{-1.1 \times 10^{-4} L}{L} = Y \cdot (-1.1 \times 10^{-4}) \]
\[ \sigma = 1.2 \times 10^{11} \cdot (-1.1 \times 10^{-4}) \]

6. **Calculate the Stress:**
\[ \sigma = 1.2 \times 10^{11} \cdot -1.1 \times 10^{-4} = -1.32 \times 10^{7} \, \text{N/m}^2 \]
The negative sign indicates that the stress is compressive.

7. **Final Result:**
The stress developed in the rod is:
\[ \sigma = 1.32 \times 10^{7} \, \text{N/m}^2 \]