A steel wire , of uniform area `2mm^2` , is heated up to `50^@C` and is stretched by tying its ends rigidly . The change in tension , when the temperature falls from `50^@C " to " 30^@C` is (Take `Y=2xx10^(11)Nm^(-2) , alpha=1.1xx10^(-5^@C-1))`

Options

A. `1.5xx10^(10)N`
B. `5 N`
C. `88 N`
D. `2.510^(10)N`

Answer

To solve the problem step by step, we need to calculate the change in tension in the steel wire when its temperature decreases from 50°C to 30°C.

### Step 1: Identify the given values
- Area of cross-section, \( A = 2 \, \text{mm}^2 = 2 \times 10^{-6} \, \text{m}^2 \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Coefficient of linear expansion, \( \alpha = 1.1 \times 10^{-5} \, \text{°C}^{-1} \)
- Initial temperature, \( T_1 = 50 \, \text{°C} \)
- Final temperature, \( T_2 = 30 \, \text{°C} \)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_1 - T_2 = 50 \, \text{°C} - 30 \, \text{°C} = 20 \, \text{°C} \]

### Step 3: Calculate the change in length due to temperature change
The change in length \( \Delta L \) due to the 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, the actual change in length cannot occur, leading to a tensile stress in the wire.

### Step 4: Relate tension to Young's modulus
The relationship between stress, strain, and Young's modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress \( = \frac{T}{A} \)
- Strain \( = \frac{\Delta L}{L} \)

Since the wire does not change its length, the strain can be expressed in terms of the change in length due to thermal expansion:
\[ \text{Strain} = \frac{\Delta L}{L} = \alpha \Delta T \]

### Step 5: Substitute into the Young's modulus equation
From the Young's modulus equation:
\[ Y = \frac{T/A}{\alpha \Delta T} \]
Rearranging gives:
\[ T = Y \cdot A \cdot \alpha \cdot \Delta T \]

### Step 6: Substitute the known values
Now we can substitute the known values into the equation:
\[ T = (2 \times 10^{11} \, \text{N/m}^2) \cdot (2 \times 10^{-6} \, \text{m}^2) \cdot (1.1 \times 10^{-5} \, \text{°C}^{-1}) \cdot (20 \, \text{°C}) \]

### Step 7: Calculate the tension
Calculating the above expression:
\[ T = 2 \times 10^{11} \cdot 2 \times 10^{-6} \cdot 1.1 \times 10^{-5} \cdot 20 \]
\[ = 2 \cdot 2 \cdot 1.1 \cdot 20 \cdot 10^{11 - 6 - 5} = 88 \, \text{N} \]

### Final Answer
The change in tension when the temperature falls from 50°C to 30°C is **88 N**.

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