To solve the problem of finding the tension in a steel wire when the temperature drops, we can follow these steps:
### Step 1: Identify the given values
- Diameter of the wire, \(d = 2 \, \text{mm} = 0.002 \, \text{m}\)
- Linear coefficient of expansion of steel, \(\alpha = 11 \times 10^{-6} \, \text{K}^{-1}\)
- Young's modulus of steel, \(Y = 2 \times 10^{11} \, \text{N/m}^2\)
- Initial temperature, \(T_1 = 20^\circ \text{C}\)
- Final temperature, \(T_2 = 10^\circ \text{C}\)
### Step 2: Calculate the change in temperature
\[
\Delta T = T_1 - T_2 = 20^\circ \text{C} - 10^\circ \text{C} = 10 \, \text{K}
\]
### Step 3: 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:
\[
A = \frac{\pi d^2}{4}
\]
Substituting the diameter:
\[
A = \frac{\pi (0.002)^2}{4} = \frac{\pi \times 0.000004}{4} = \frac{\pi \times 10^{-6}}{4} \approx 7.854 \times 10^{-7} \, \text{m}^2
\]
### Step 4: Calculate the thermal stress
The thermal stress \(\sigma\) in the wire due to the temperature change can be calculated using the formula:
\[
\sigma = \alpha \Delta T Y
\]
Substituting the known values:
\[
\sigma = (11 \times 10^{-6} \, \text{K}^{-1}) \times (10 \, \text{K}) \times (2 \times 10^{11} \, \text{N/m}^2)
\]
Calculating this gives:
\[
\sigma = 11 \times 10^{-6} \times 10 \times 2 \times 10^{11} = 22 \times 10^{6} \, \text{N/m}^2 = 2.2 \times 10^{7} \, \text{N/m}^2
\]
### Step 5: Calculate the tension in the wire
The tension \(T\) in the wire can be found by multiplying the stress by the cross-sectional area:
\[
T = \sigma A
\]
Substituting the values:
\[
T = (2.2 \times 10^{7} \, \text{N/m}^2) \times (7.854 \times 10^{-7} \, \text{m}^2)
\]
Calculating this gives:
\[
T \approx 17.29 \, \text{N}
\]
### Final Answer
The tension in the wire when the temperature falls to 10°C is approximately \(17.29 \, \text{N}\).
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