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A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 is
  • a)
    200 GPa
  • b)
    150 GPa
  • c)
    125 GPa
  • d)
    100 GPa
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A bar produces a lateral strain of magnitude 60 x 10−5 when subjected...
Lateral strain = Axial strain x poisson′ s ratio ∈y=∈x x μ
= 150 GPa
Community Answer
A bar produces a lateral strain of magnitude 60 x 10−5 when subjected...
Given:
- Lateral strain (ε) = 60 x 10^(-5)
- Tensile stress (σ) = 300 MPa
- Poisson's ratio (ν) = 0.3

To find:
Elastic modulus (E)

Formula:
Poisson's ratio is defined as the ratio of lateral strain to axial strain.
ν = -ε_lat / ε_axial

Elastic modulus is defined as the ratio of stress to strain.
E = σ / ε_axial

Solution:
Step 1: Calculate the axial strain (ε_axial) using Poisson's ratio and lateral strain.
ε_axial = -ε_lat / ν
ε_axial = -60 x 10^(-5) / 0.3
ε_axial = -2 x 10^(-4)

Step 2: Calculate the elastic modulus (E) using stress and axial strain.
E = σ / ε_axial
E = 300 x 10^6 / (-2 x 10^(-4))
E = -1500 x 10^9
E = 150 GPa

Answer:
The elastic modulus of the material, when the Poisson's ratio is 0.3, is 150 GPa. Therefore, the correct answer is option B.
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