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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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A 50 cm long x 20 cm diameter cylinder of brass was subjected to a ten...

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A 50 cm long x 20 cm diameter cylinder of brass was subjected to a ten...
The Poisson's ratio, denoted by the Greek letter ν (nu), is a measure of the ratio of transverse strain to axial strain for a material under axial loading.

To find the Poisson's ratio of the brass cylinder, we can use the formula:

ν = (ΔD / D) / (ΔL / L)

Where:
ν = Poisson's ratio
ΔD = Change in diameter
D = Original diameter
ΔL = Change in length
L = Original length

Given:
Original length (L) = 50 cm = 0.5 m
Original diameter (D) = 20 cm = 0.2 m
Change in length (ΔL) = 1 mm = 0.001 m
Change in diameter (ΔD) = 0.1 mm = 0.0001 m

Plugging the values into the formula:

ν = (0.0001 / 0.2) / (0.001 / 0.5)
ν = 0.0005 / 0.002
ν = 0.25

Therefore, the Poisson's ratio of the brass is 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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