One end of a steel wire is fixed to the ceiling of an elevator moving up with an acceleration \( 2 \, \text{m/s}^2 \) and a load of 10 kg hangs from the other end. If the cross-section of the wire is \( 2 \, \text{cm}^2 \), then the longitudinal strain in the wire will be (Take \( g = 10 \, \text{m/s}^2 \) and \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)).
Show Hint
When calculating strain, the force is related to both the gravitational force and any additional forces due to acceleration. The strain is the ratio of stress to Young's modulus.
Step 1: Problem Definition
Determine the longitudinal strain in the wire when the elevator accelerates upwards with \( a = 2 \, \text{m/s}^2 \). Strain relates to stress, which relates to the force on the wire.
Step 2: Force Calculation
The force has two components:
1. Weight of the 10 kg load.
2. Force from the elevator's acceleration.
Total force \( F \) is:
\[
F = m(g + a)
\]
Where:
- \( m = 10 \, \text{kg} \),
- \( g = 10 \, \text{m/s}^2 \),
- \( a = 2 \, \text{m/s}^2 \).
Therefore:
\[
F = 10 \times (10 + 2) = 10 \times 12 = 120 \, \text{N}
\]
Step 3: Stress Calculation
Stress \( \sigma \) is force per unit area:
\[
\sigma = \frac{F}{A}
\]
Where:
- \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \).
Thus:
\[
\sigma = \frac{120}{2 \times 10^{-4}} = 6 \times 10^5 \, \text{N/m}^2
\]
Step 4: Strain Calculation
Longitudinal strain \( \epsilon \) relates to stress and Young's modulus \( Y \):
\[
\epsilon = \frac{\sigma}{Y}
\]
Where:
- \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \).
Substituting values:
\[
\epsilon = \frac{6 \times 10^5}{2.0 \times 10^{11}} = 3 \times 10^{-6}
\]
Step 5: Conclusion
The longitudinal strain is \( 3 \times 10^{-6} \).
Therefore, the correct answer is:
\[
\boxed{(D)} \, 2 \times 10^{-6}
\]