A steel wire of cross sectional area `5xx10^(-7)m^(2)` is tied between two rigid clamps. It is given that at `30^(@)C` wire is just taut. If the temperature of wire is decreased to `10^(@)C`, find the tension in the wire. Given that the coefficient of linear expansion of stees is `1.1xx10^(-5).^(@)C^(-1)` and its Young's modulus is `2xx10^(11)N//m^(2)`

Answer

To find the tension in the steel wire when the temperature decreases from \(30^\circ C\) to \(10^\circ C\), we can follow these steps:

### Step 1: Identify the given data
- Cross-sectional area of the wire, \(A = 5 \times 10^{-7} \, m^2\)
- Coefficient of linear expansion of steel, \(\alpha = 1.1 \times 10^{-5} \, ^\circ C^{-1}\)
- Young's modulus of steel, \(Y = 2 \times 10^{11} \, N/m^2\)
- Initial temperature, \(T_1 = 30^\circ C\)
- Final temperature, \(T_2 = 10^\circ C\)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_1 - T_2 = 30^\circ C - 10^\circ C = 20^\circ C \]

### Step 3: Calculate the contraction of the wire
The change in length (\(\Delta L\)) due to thermal contraction can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
However, we do not have the original length \(L\) of the wire. We will express the tension in terms of \(L\) and then eliminate \(L\) later.

### Step 4: Relate tension to stress and strain
The stress (\(\sigma\)) in the wire is given by:
\[ \sigma = \frac{T}{A} \]
where \(T\) is the tension and \(A\) is the cross-sectional area.

The strain (\(\epsilon\)) in the wire is given by:
\[ \epsilon = \frac{\Delta L}{L} \]
Substituting \(\Delta L\) from Step 3:
\[ \epsilon = \frac{L \cdot \alpha \cdot \Delta T}{L} = \alpha \cdot \Delta T \]

### Step 5: Use Young's modulus
From the definition of Young's modulus:
\[ Y = \frac{\sigma}{\epsilon} \]
Substituting the expressions for stress and strain:
\[ Y = \frac{T/A}{\alpha \cdot \Delta T} \]
Rearranging gives us:
\[ T = Y \cdot A \cdot \alpha \cdot \Delta T \]

### Step 6: Substitute the known values
Now we can substitute the values into the equation:
\[ T = (2 \times 10^{11} \, N/m^2) \cdot (5 \times 10^{-7} \, m^2) \cdot (1.1 \times 10^{-5} \, ^\circ C^{-1}) \cdot (20 \, ^\circ C) \]

### Step 7: Calculate the tension
Calculating the right-hand side:
\[ T = 2 \times 10^{11} \cdot 5 \times 10^{-7} \cdot 1.1 \times 10^{-5} \cdot 20 \]
\[ = 2 \times 5 \times 1.1 \times 20 \times 10^{11 - 7 - 5} \]
\[ = 220 \times 10^{(-1)} \, N \]
\[ = 22 \, N \]

### Final Answer
The tension in the wire when the temperature decreases to \(10^\circ C\) is \(T = 22 \, N\).

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