A steel wire of length 20 cm and cross - section `1 mm^(2)` is tied rigidly at both ends at room temperature `30^(@)C`. If the temperature falls `10^(@)C`, what will be the change in tension ? Given that `Y = 2xx10^(11)N//m^(2)` and `alpha = 1.1xx10^(-5)//^(@)C`

Options

A. 2.2 N
B. 4.4 N
C. 22 N
D. 44 N

Answer

To solve the problem step by step, we will follow the principles of elasticity and thermal expansion.

### Step 1: Identify the given values
- Length of the steel wire, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Cross-sectional area, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Coefficient of linear expansion, \( \alpha = 1.1 \times 10^{-5} \, \text{°C}^{-1} \)
- Change in temperature, \( \Delta T = -10 \, \text{°C} \) (since the temperature falls)

### Step 2: Calculate the change in length (\( \Delta L \))
The formula for change in length due to temperature change is:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]

Substituting the values:
\[ \Delta L = 0.2 \, \text{m} \cdot (1.1 \times 10^{-5} \, \text{°C}^{-1}) \cdot (-10 \, \text{°C}) \]
\[ \Delta L = 0.2 \cdot 1.1 \times 10^{-5} \cdot (-10) = -2.2 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate the change in tension (\( \Delta T \))
The change in tension can be calculated using the formula:
\[ \Delta T = \frac{Y \cdot \Delta L}{A} \]

Substituting the values:
\[ \Delta T = \frac{(2 \times 10^{11} \, \text{N/m}^2) \cdot (-2.2 \times 10^{-6} \, \text{m})}{1 \times 10^{-6} \, \text{m}^2} \]
\[ \Delta T = \frac{-4.4 \times 10^{5} \, \text{N}}{1} = -4.4 \times 10^{5} \, \text{N} \]

### Step 4: Interpret the result
The negative sign indicates that the tension in the wire decreases due to the fall in temperature. The magnitude of the change in tension is:
\[ |\Delta T| = 4.4 \, \text{N} \]

### Final Answer
The change in tension is \( 4.4 \, \text{N} \).

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