A brass wire `1.8 ` m long at `27^(@)C` is held taut with negligible tension between two rigid supports. Diameter of the wire is `2` mm, its coefficient of linear expasion, `alpha_("Brass") = 2 xx 10^(-5) .^(@)C^(-1)` and its young's modulus, `Y_("Brass") = 9xx 10^(10) N m^(-2)`. If the wire is cooled to a temperature `-39^(@)C`, tension developed in the wire is

Options

A. `2.7 xx 10^(2) N`
B. `3.7 xx 10^(2) N`
C. `4.7 xx 10^(2) N`
D. `5.7 xx 10^(2) N`

Explanation

To find the tension developed in the brass wire when it is cooled from 27 °C to -39 °C, we can follow these steps:

### Step 1: Gather the Given Data
- Length of the wire, \( L = 1.8 \, \text{m} \)
- Initial temperature, \( T_1 = 27 \, \text{°C} \)
- Final temperature, \( T_2 = -39 \, \text{°C} \)
- Diameter of the wire, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Coefficient of linear expansion, \( \alpha = 2 \times 10^{-5} \, \text{°C}^{-1} \)
- Young's modulus, \( Y = 9 \times 10^{10} \, \text{N/m}^2 \)

### Step 2: Calculate the Change in Temperature
The change in temperature, \( \Delta T \), is given by:
\[ \Delta T = T_1 - T_2 = 27 - (-39) = 27 + 39 = 66 \, \text{°C} \]

### Step 3: Calculate the Radius of the Wire
The radius \( r \) of the wire is half of the diameter:
\[ r = \frac{d}{2} = \frac{2 \times 10^{-3}}{2} = 1 \times 10^{-3} \, \text{m} \]

### Step 4: Calculate the Cross-Sectional Area of the Wire
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 5: Calculate the Thermal Strain
The thermal strain \( \epsilon \) can be calculated using the coefficient of linear expansion:
\[ \epsilon = \alpha \Delta T = 2 \times 10^{-5} \times 66 = 1.32 \times 10^{-3} \]

### Step 6: Calculate the Stress in the Wire
Stress \( \sigma \) is defined as force per unit area:
\[ \sigma = Y \cdot \epsilon \]

### Step 7: Calculate the Force (Tension) Developed in the Wire
The force \( F \) (which is the tension \( T \) in the wire) can be calculated as:
\[ F = \sigma \cdot A = Y \cdot \epsilon \cdot A \]

Substituting the values:
\[ F = (9 \times 10^{10}) \cdot (1.32 \times 10^{-3}) \cdot (\pi \times 10^{-6}) \]

### Step 8: Calculate the Final Value
Calculating the above expression:
\[ F = 9 \times 10^{10} \cdot 1.32 \times 10^{-3} \cdot 3.14 \times 10^{-6} \]
\[ F \approx 3.7 \times 10^{2} \, \text{N} \]

### Conclusion
The tension developed in the wire when it is cooled to -39 °C is approximately:
\[ T \approx 370 \, \text{N} \]