A steel wire of length 20 cm and uniform cross-section `1mm^(2)` is tied rigidly at both the ends. If the temperature of the wire is altered from `40^(@)C` to `20^(@)C`, the change in tension. [Given coefficient of linear expansion of steel is `1.1xx10^(5) .^(@)C^(-1)` and Young's modulus for steel is `2.0xx10^(11) Nm^(-2)`]

Options

A. 22 N
B. 44 N
C. 66 N
D. 88 N

Explanation

To solve the problem of finding the change in tension in a steel wire when its temperature is altered, we can follow these steps:

### Step 1: Identify the given values
- Length of the wire, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Cross-sectional area, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Coefficient of linear expansion of steel, \( \alpha = 1.1 \times 10^{-5} \, \text{°C}^{-1} \)
- Young's modulus for steel, \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)
- Initial temperature, \( T_1 = 40 \, \text{°C} \)
- Final temperature, \( T_2 = 20 \, \text{°C} \)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_2 - T_1 = 20 \, \text{°C} - 40 \, \text{°C} = -20 \, \text{°C} \]

### Step 3: Calculate the change in length (\( \Delta L \))
The formula for the change in length due to temperature change is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Substituting the known values:
\[ \Delta L = 0.2 \, \text{m} \cdot (1.1 \times 10^{-5} \, \text{°C}^{-1}) \cdot (-20 \, \text{°C}) \]
\[ \Delta L = 0.2 \cdot 1.1 \times 10^{-5} \cdot (-20) = -4.4 \times 10^{-5} \, \text{m} \]

### Step 4: Calculate the change in tension (\( \Delta T \))
The change in tension can be calculated using the formula:
\[ \Delta T = Y \cdot A \cdot \frac{\Delta L}{L} \]
Substituting the values:
\[ \Delta T = (2.0 \times 10^{11} \, \text{N/m}^2) \cdot (1 \times 10^{-6} \, \text{m}^2) \cdot \frac{-4.4 \times 10^{-5} \, \text{m}}{0.2 \, \text{m}} \]
Calculating the fraction:
\[ \frac{-4.4 \times 10^{-5}}{0.2} = -2.2 \times 10^{-4} \]
Now substituting this back into the equation for \( \Delta T \):
\[ \Delta T = (2.0 \times 10^{11}) \cdot (1 \times 10^{-6}) \cdot (-2.2 \times 10^{-4}) \]
\[ \Delta T = 2.0 \times 10^{11} \cdot (-2.2 \times 10^{-10}) = -44 \, \text{N} \]

### Final Answer
The change in tension in the wire is \( \Delta T = -44 \, \text{N} \).

---