A steel wire 2mm in diameter is stretched between two clamps, when its temperature is `40^@C` Calculate the tension in the wire, when its temperature falls to `30^@C` Given, coefficient Y for steel ` = 21 xx 10^(11) dyn e//cm^2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Diameter of the steel wire, \( d = 2 \, \text{mm} = 0.2 \, \text{cm} \)
- Radius of the wire, \( r = \frac{d}{2} = 1 \, \text{mm} = 0.1 \, \text{cm} \)
- Initial temperature, \( T_1 = 40^\circ C \)
- Final temperature, \( T_2 = 30^\circ C \)
- Change in temperature, \( \Delta \theta = T_1 - T_2 = 40 - 30 = 10^\circ C \)
- Young's modulus for steel, \( Y = 21 \times 10^{11} \, \text{dyn/cm}^2 \)
- Coefficient of linear expansion for steel, \( \alpha = 11 \times 10^{-6} \, \text{°C}^{-1} \)

### Step 2: Calculate the change in length (\( \Delta L \))
The formula for change in length due to temperature change is:
\[ \Delta L = \alpha \cdot L \cdot \Delta \theta \]
Since we do not have the original length \( L \) of the wire, we will keep it as \( L \) for now.

### Step 3: Calculate the area of cross-section (\( A \))
The area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (0.1 \, \text{cm})^2 = \pi \times 0.01 \, \text{cm}^2 = 0.0314 \, \text{cm}^2 \]

### Step 4: Relate stress and strain
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 us:
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]

### Step 5: Substitute \( \Delta L \) into the force equation
Substituting \( \Delta L \) into the force equation:
\[ F = Y \cdot A \cdot \frac{\alpha \cdot L \cdot \Delta \theta}{L} = Y \cdot A \cdot \alpha \cdot \Delta \theta \]

### Step 6: Calculate the tension in the wire
Substituting the known values into the equation:
\[ F = (21 \times 10^{11} \, \text{dyn/cm}^2) \cdot (0.0314 \, \text{cm}^2) \cdot (11 \times 10^{-6} \, \text{°C}^{-1}) \cdot (10 \, \text{°C}) \]
Calculating this gives:
\[ F = 21 \times 10^{11} \cdot 0.0314 \cdot 11 \times 10^{-6} \cdot 10 \]
\[ F = 21 \times 10^{11} \cdot 0.0314 \cdot 1.1 \times 10^{-5} \]
\[ F \approx 7.26 \times 10^6 \, \text{dyn} \]

### Final Answer
The tension in the wire when its temperature falls to \( 30^\circ C \) is approximately \( 7.26 \times 10^6 \, \text{dyn} \).

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