To solve the problem step by step, we will calculate the tension developed in the brass wire when it is cooled from 27°C to -33°C.
### Step 1: Identify the given values
- Length of the wire, \( L = 2 \, m \)
- Initial temperature, \( T_i = 27°C \)
- Final temperature, \( T_f = -33°C \)
- Coefficient of linear expansion of brass, \( \alpha = 2.0 \times 10^{-5} \, /°C \)
- Young's modulus of brass, \( Y = 0.91 \times 10^{11} \, Pa \)
- Diameter of the wire, \( d = 2 \, mm = 0.002 \, m \)
### Step 2: Calculate the change in temperature
\[
\Delta T = T_f - T_i = -33°C - 27°C = -60°C
\]
### Step 3: Calculate the change in length due to temperature change
The change in length (\( \Delta L \)) can be calculated using the formula:
\[
\Delta L = L \cdot \alpha \cdot \Delta T
\]
Substituting the values:
\[
\Delta L = 2 \, m \cdot (2.0 \times 10^{-5} \, /°C) \cdot (-60°C)
\]
\[
\Delta L = 2 \cdot 2.0 \times 10^{-5} \cdot (-60) = -2.4 \times 10^{-3} \, m = -2.4 \, mm
\]
### Step 4: Calculate the area of the wire
The area (\( A \)) of the wire can be calculated using the formula for the area of a circle:
\[
A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.002}{2} \right)^2 = \pi \left( 0.001 \right)^2 = \pi \times 10^{-6} \, m^2
\]
### Step 5: Calculate the stress in the wire
The stress (\( S \)) in the wire can be expressed as:
\[
S = \frac{T}{A} = Y \cdot \text{strain}
\]
Where strain is given by:
\[
\text{strain} = \frac{\Delta L}{L}
\]
Substituting the values:
\[
\text{strain} = \frac{-2.4 \times 10^{-3}}{2} = -1.2 \times 10^{-3}
\]
### Step 6: Relate tension to stress
Rearranging the stress equation gives:
\[
T = S \cdot A = Y \cdot \text{strain} \cdot A
\]
Substituting the values:
\[
T = 0.91 \times 10^{11} \cdot (-1.2 \times 10^{-3}) \cdot (\pi \times 10^{-6})
\]
### Step 7: Calculate the tension
\[
T = 0.91 \times 10^{11} \cdot (-1.2 \times 10^{-3}) \cdot (3.14 \times 10^{-6})
\]
Calculating this:
\[
T \approx 0.91 \times 10^{11} \cdot (-1.2 \times 3.14) \times 10^{-9}
\]
\[
T \approx 0.91 \times 10^{11} \cdot (-3.768) \times 10^{-9}
\]
\[
T \approx -3.43 \times 10^{2} \, N \approx -343 \, N
\]
### Final Result
The tension developed in the wire is approximately \( 340 \, N \).
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