Question

The Poisson's ratio of the material of cylindrical wire when density of the material of wire is constant is​

Answer 1

Answer:

zero

Explanation:

density same meaning material incompressible so, volume remain same means ratio of lateral strain over longitudinal strain zero , to satisfy the condition

Answer 2

Answer:

The answer is 1.

Explanation:

i) Poisson's ratio (σ)  is defined as the ratio of transverse strain to axial strain.

σ = $\frac{dr/r}{dl/l}$

ii) It is a measure of effect of stress on the dimensions of the material in the perpendicular direction of the stress.

iii) In a cylindrical wire with uniform density:

Volume = Area x Length

Also, Volume = mass x density

Since mass and density are constant, volume will remain same after the application of stress. There will be an increase in axial length dl, and radius of the wire dr.

iv) Volume = Area x Length

πr²l = (πr² - 2πr.dr) (l + dl)

v) Dividing by πr²l,

1 = $[ 1 - \frac{dr}{r} ][ 1 + \frac{dl}{l} ]$

1 = $1 - \frac{dr}{r} \frac{dl}{l} - \frac{dr}{r} + \frac{dl}{l}$

vi) The term $\frac{dr}{r} \frac{dl}{l}$ is very small and hence can be neglected.

Hence,

1 = $1 - \frac{dr}{r} + \frac{dl}{l}$

$\frac{dr}{r} = \frac{dl}{l}$

$\frac{\frac{dr}{r}}{\frac{dl}{l}\\}$ = 1

vii) Hence, by definition of Poisson's ratio:

σ = $\frac{dr/r}{dl/l}$ = 1

viii) Hence, Poisson's ratio of the material of cylindrical wire with uniform density is 1.

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