Question

If the volume of a wire remains constant when subjected to tensile stress, the value of Poisson’s ratio of the material of the wire is

• A. 0.1
• B. 0.2
• C. 0.4
• D. 0.5

Answer 1

Answer: (D)

0.5

Answer 2

: Let L be the length r be the radius of the wire,

Volume of the wire is

$v = \pi r^{2}L$

Differentiating both sides, we get

$\Delta V = \pi(2r\Delta r)L + \pi r^{2}\Delta L$

As the volume of the wire remains unchanged when it gets stretched, so$\Delta V = 0.$

Hence

$0 = 2\pi rL\Delta r + \pi r^{2}\Delta L.$

$\therefore\frac{\Delta r/r}{\Delta L/L} = - \frac{1}{2}$

Poisson’s ratio

$= \frac{LateralStrain}{Longitudinalstrain} = - \frac{\Delta r/r}{\Delta L/L}$

$= \frac{1}{2} = 0.5$