A wire of cross-sectional area `3 mm^(2)` is first stretched between two fixed points at a temperature of `20^(@)C`. Determine the tension when the temperature falls to `10^(@)C` Coefficient of linear expansion `alpha=10^(-5).^(@)C^(-1) and Y =2 xx10^(11) N//m^(2)`

Options

A. 20 N
B. 30 N
C. 60 N
D. 120 N

Answer

To solve the problem, we need to determine the tension in a wire when the temperature decreases from \(20^\circ C\) to \(10^\circ C\). Given the coefficient of linear expansion (\(\alpha\)) and Young's modulus (Y), we can use the following steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Cross-sectional area, \(A = 3 \, \text{mm}^2 = 3 \times 10^{-6} \, \text{m}^2\)
- Coefficient of linear expansion, \(\alpha = 10^{-5} \, \text{°C}^{-1}\)
- Young's modulus, \(Y = 2 \times 10^{11} \, \text{N/m}^2\)
- Initial temperature, \(T_1 = 20^\circ C\)
- Final temperature, \(T_2 = 10^\circ C\)

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

3. **Determine the change in length (\(\Delta L\)) if the wire were free to contract:**
The change in length due to temperature change can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
However, since the wire is fixed at both ends, this contraction will create tension in the wire.

4. **Using Young's modulus to relate tension to change in length:**
The relationship between stress, strain, and Young's modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Rearranging this gives:
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]

5. **Substituting \(\Delta L\) into the equation:**
Since \(\Delta L = L \cdot \alpha \cdot \Delta T\), we can substitute this into the force equation:
\[ F = Y \cdot A \cdot \frac{\alpha \cdot \Delta T}{L} \]

6. **Cancelling \(L\):**
The length \(L\) cancels out, leading to:
\[ F = Y \cdot A \cdot \alpha \cdot \Delta T \]

7. **Substituting the known values:**
Now we can substitute the values into the equation:
\[ F = (2 \times 10^{11} \, \text{N/m}^2) \cdot (3 \times 10^{-6} \, \text{m}^2) \cdot (10^{-5} \, \text{°C}^{-1}) \cdot (-10 \, \text{°C}) \]

8. **Calculating the force:**
\[ F = 2 \times 10^{11} \cdot 3 \times 10^{-6} \cdot 10^{-5} \cdot (-10) \]
\[ F = 2 \times 3 \times (-10) \times 10^{11} \times 10^{-6} \times 10^{-5} \]
\[ F = -60 \, \text{N} \]
Since tension is a force, we take the absolute value:
\[ F = 60 \, \text{N} \]

### Final Answer:
The tension in the wire when the temperature falls to \(10^\circ C\) is \(60 \, \text{N}\).

---