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 \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)).
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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: Understanding the Problem
We need to calculate the longitudinal strain in the wire when the elevator is accelerating upwards with an acceleration \( a = 2 \, \text{m/s}^2 \). The strain is related to the stress applied to the wire, and the stress is related to the force acting on the wire.
Step 2: Force Acting on the Wire
The force acting on the wire consists of two parts:
1. The weight of the 10 kg load.
2. The additional force due to the acceleration of the elevator.
The total force \( F \) acting on the wire is given by:
\[
F = m(g + a)
\]
Where:
- \( m = 10 \, \text{kg} \) (mass of the load),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( a = 2 \, \text{m/s}^2 \) (acceleration of the elevator).
Substituting the values:
\[
F = 10 \times (10 + 2) = 10 \times 12 = 120 \, \text{N}
\]
Step 3: Stress in the Wire
The stress \( \sigma \) in the wire is the force per unit area:
\[
\sigma = \frac{F}{A}
\]
Where:
- \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \) (cross-sectional area of the wire).
Thus:
\[
\sigma = \frac{120}{2 \times 10^{-4}} = 6 \times 10^5 \, \text{N/m}^2
\]
Step 4: Longitudinal Strain
The longitudinal strain \( \epsilon \) in the wire is related to the stress and the Young's modulus \( Y \) by:
\[
\epsilon = \frac{\sigma}{Y}
\]
Where:
- \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \) (Young's modulus of the wire).
Substituting the values:
\[
\epsilon = \frac{6 \times 10^5}{2.0 \times 10^{11}} = 3 \times 10^{-6}
\]
Step 5: Conclusion
The longitudinal strain in the wire is \( 3 \times 10^{-6} \).
Thus, the correct answer is:
\[
\boxed{(D)} \, 2 \times 10^{-6}
\]
