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 expansion, `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 \(27^\circ C\) to \(-39^\circ C\), we can follow these steps:

### Step 1: Determine the change in temperature
The change in temperature (\(\Delta T\)) can be calculated as:
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = -39^\circ C - 27^\circ C = -66^\circ C \]

### Step 2: Calculate the change in length due to temperature change
The change in length (\(\Delta L\)) of the wire can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Where:
- \(L = 1.8 \, m\) (initial length of the wire)
- \(\alpha = 2 \times 10^{-5} \, ^\circ C^{-1}\) (coefficient of linear expansion)
- \(\Delta T = -66 \, ^\circ C\)

Substituting the values:
\[ \Delta L = 1.8 \, m \cdot (2 \times 10^{-5} \, ^\circ C^{-1}) \cdot (-66 \, ^\circ C) \]
\[ \Delta L = 1.8 \cdot 2 \cdot 10^{-5} \cdot -66 = -0.0002376 \, m \]

### Step 3: Calculate the stress in the wire
The stress (\(\sigma\)) in the wire can be calculated using Young's modulus (\(Y\)):
\[ \sigma = Y \cdot \frac{\Delta L}{L} \]
Where:
- \(Y = 9 \times 10^{10} \, N/m^2\)

Substituting the values:
\[ \sigma = 9 \times 10^{10} \cdot \frac{-0.0002376}{1.8} \]
\[ \sigma = 9 \times 10^{10} \cdot -0.000131333 = -1.189 \times 10^{7} \, N/m^2 \]

### Step 4: Calculate the area of the wire
The area (\(A\)) of the wire can be calculated using the diameter:
\[ A = \pi r^2 \]
Where:
- Diameter \(d = 2 \, mm = 0.002 \, m\)
- Radius \(r = \frac{d}{2} = 0.001 \, m\)

Substituting the values:
\[ A = \pi (0.001)^2 = \pi \times 10^{-6} \, m^2 \approx 3.14 \times 10^{-6} \, m^2 \]

### Step 5: Calculate the force (tension) in the wire
The force (tension \(F\)) can be calculated using the formula:
\[ F = \sigma \cdot A \]
Substituting the values:
\[ F = -1.189 \times 10^{7} \cdot 3.14 \times 10^{-6} \]
\[ F \approx -37.34 \, N \]

Since tension cannot be negative, we take the absolute value:
\[ F \approx 37.34 \, N \]

### Final Answer
The tension developed in the wire is approximately \(37.34 \, N\).

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