A metal rod of length 5 m placed on a smooth tabel. Given that Young's modulus of the material of the rod is `1.6xx10^(11)N//m^(2)` and the its coefficient of linear expansion is `1.2xx10^(-5).^(@)C^(-1)`. If temperature of rod is changed from `20^(@)C` to `80^(@)C` find the elastic stress developed in the rod.

Answer

To solve the problem of finding the elastic stress developed in a metal rod when its temperature is changed, we can follow these steps:

### Step 1: Understand the Given Data
We have the following information:
- Length of the rod, \( L = 5 \, \text{m} \)
- Young's modulus, \( Y = 1.6 \times 10^{11} \, \text{N/m}^2 \)
- Coefficient of linear expansion, \( \alpha = 1.2 \times 10^{-5} \, \text{°C}^{-1} \)
- Initial temperature, \( T_1 = 20 \, \text{°C} \)
- Final temperature, \( T_2 = 80 \, \text{°C} \)

### Step 2: Calculate the Change in Temperature
The change in temperature, \( \Delta T \), can be calculated as:
\[ \Delta T = T_2 - T_1 = 80 \, \text{°C} - 20 \, \text{°C} = 60 \, \text{°C} \]

### Step 3: Calculate the Linear Expansion
The linear expansion \( \Delta L \) of the rod can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Substituting the values:
\[ \Delta L = 5 \, \text{m} \cdot (1.2 \times 10^{-5} \, \text{°C}^{-1}) \cdot (60 \, \text{°C}) = 5 \cdot 1.2 \times 10^{-5} \cdot 60 \]
Calculating this gives:
\[ \Delta L = 5 \cdot 1.2 \times 10^{-5} \cdot 60 = 3.6 \times 10^{-3} \, \text{m} = 0.0036 \, \text{m} \]

### Step 4: Analyze the Stress Development
Since the rod is placed on a smooth table, it is free to expand without any constraints. Therefore, there is no external force acting on the rod to resist this expansion.

### Step 5: Conclusion on Stress
Stress is defined as the resistive force per unit area. In this case, since there is no net external force acting on the rod (the rod is free to expand), the stress developed in the rod is:
\[ \text{Stress} = 0 \]

### Final Answer
The elastic stress developed in the rod is **0 N/m²**.

---