A steel string guitar is strung so that there is negligible tension in the strings at a temperature of `24.9^@C` . The guitar is taken to an outdoor winter concert where the temperature of the string decrease to `- 15.1 ^@C` . The cross-sectional area of a particular string is `5.5 xx10^(-6)m^2` . The distance between the points where the string is attached does not change . For steel , Young's modulus is `2.0 xx10^(11) N//m^2` , and the coefficient of linear expansion is `1.2 xx10^(-5)//""^@C` . Use your knowledge of liner thermal expansion and stress to calculate the tension in the string at the concert.

Options

A. 530 N
B. 240 N
C. 120 N
D. 60 N

Answer

To solve the problem of calculating the tension in the steel string of a guitar when the temperature changes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Temperatures**:
- Initial temperature \( T_1 = 24.9^\circ C \)
- Final temperature \( T_2 = -15.1^\circ C \)

2. **Calculate the Change in Temperature**:
\[ \Delta T = T_2 - T_1 = -15.1 - 24.9 = -40.0^\circ C \]

3. **Identify the Given Values**:
- Cross-sectional area \( A = 5.5 \times 10^{-6} \, m^2 \)
- Young's modulus \( E = 2.0 \times 10^{11} \, N/m^2 \)
- Coefficient of linear expansion \( \alpha = 1.2 \times 10^{-5} \, /^\circ C \)

4. **Calculate the Strain**:
The strain \( \epsilon \) due to the temperature change can be calculated using the formula:
\[ \epsilon = \alpha \Delta T \]
Substituting the values:
\[ \epsilon = (1.2 \times 10^{-5}) \times (-40) = -4.8 \times 10^{-4} \]

5. **Calculate the Stress**:
Stress \( \sigma \) is related to strain and Young's modulus by:
\[ \sigma = E \cdot \epsilon \]
Substituting the values:
\[ \sigma = (2.0 \times 10^{11}) \times (-4.8 \times 10^{-4}) = -9.6 \times 10^{7} \, N/m^2 \]

6. **Calculate the Tension**:
The tension \( T \) in the string can be calculated using the formula:
\[ T = \sigma \cdot A \]
Substituting the values:
\[ T = (-9.6 \times 10^{7}) \times (5.5 \times 10^{-6}) = -528 \, N \]
Since tension is a force, we take the absolute value:
\[ T \approx 530 \, N \]

### Final Answer:
The tension in the string at the concert is approximately **530 N**.