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A 50 cm long x 20 cm diameter cylinder of brass was subjected to a tensile load of 0.1 MN. The resulting increase in length and decrease in diameter were noted 1 mm and 0.1 mm respectively. Then the brass has a Poisson’s ratio equal to
  • a)
    0.25
  • b)
    0.20
  • c)
    0.26
  • d)
    0.22
Correct answer is option 'A'. Can you explain this answer?
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Calculation of Poisson's ratio:
- Given data:
- Load (F) = 0.1 MN = 100 kN
- Original length (L) = 50 cm = 0.5 m
- Original diameter (D) = 20 cm = 0.2 m
- Increase in length (ΔL) = 1 mm = 0.001 m
- Decrease in diameter (ΔD) = 0.1 mm = 0.0001 m

Calculations:
- Strain in axial direction (ε) = ΔL / L = 0.001 / 0.5 = 0.002
- Strain in radial direction (ε') = -ΔD / D = -0.0001 / 0.2 = -0.0005
- Poisson's ratio (ν) = ε' / ε = -0.0005 / 0.002 = -0.25

Explanation:
- Poisson's ratio is a measure of the Poisson effect, which describes the behavior of a material when subjected to a load.
- In this case, the negative sign in the calculation indicates that the material contracted in the radial direction when stretched in the axial direction.
- The Poisson's ratio of brass is typically around 0.25, which is consistent with the calculated value.
- Therefore, the correct answer is option 'A' (0.25).
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A 50 cm long x 20 cm diameter cylinder of brass was subjected to a tensile load of 0.1 MN. The resulting increase in length and decrease in diameter were noted 1 mm and 0.1 mm respectively. Then the brass has a Poisson’s ratio equal toa)0.25b)0.20c)0.26d)0.22Correct answer is option 'A'. Can you explain this answer?
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