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**Hint:**In this question first we will use the relation between the volume, length and the side of the wire, and then we will deduce an expression for the Poisson’s ratio. Further, we will also study the basics of Poisson’s effect and Poisson’s ratio, for our better understanding.

**Formula used:**

$V = l \times {s^2}$

$\eta = - \dfrac{{\dfrac{{ds}}{s}}}{{\dfrac{{dl}}{l}}}$

**Complete answer:**

Here, let length of the material l and side s. As the material maintains constant volume during stretching, this volume V is given as:

$V = l \times {s^2}$

Now, we differentiate the above equation with respect to small length dl:

$dV = dl \times 1.2s.ds$

Further, solving this we get:

$\eqalign{& dl.s = 2l.ds \cr

& \Rightarrow \dfrac{{ds}}{{dl}} = - \dfrac{s}{{2l}} \cr} $

$\therefore \eta = - \dfrac{{\dfrac{{ds}}{s}}}{{\dfrac{{dl}}{l}}} = \dfrac{1}{2}$

**Therefore, the correct option is A) i.e., the Poisson’s ratio of the material of the wire is 0.5, which is given by the above result.**

**Additional information:**

As we know that Poisson’s ratio is measured in materials science and solid mechanics, here it is a measure of the Poisson effect. The Poisson’s ratio is named after the French mathematician and physicist Siméon Poisson.

Now, Poisson’s effect is known as the deformation i.e., expansion or contraction of a material in different directions perpendicular to the direction of loading. Mathematically it can be expressed as the negative of the ratio of transverse strain to axial strain.

In general, Poisson's ratio values of most materials range between 0.0 and 0.5. Here, incompressible materials, like rubber, have a ratio near 0.5.

Further, if a material is stretched rather than compressed, it mostly tends to contract in the directions transverse to the direction of stretching. We can understand this with a common example: when a rubber band is stretched, it becomes a bit thinner. Here, the Poisson ratio will be the ratio of relative contraction to relative expansion and will have the same value.

Now, when we see certain observed values of the Poisson's ratio of a stable, isotropic, linear elastic material that must be between −1.0 and +0.5 because of the requirement for Young's modulus, the shear modulus and bulk modulus to have positive values.

**Note:**

There are few cases where a material will actually shrink in the transverse direction when compressed or expand when stretched; this will give us a negative value of the Poisson ratio. We have to remember that most materials have Poisson's ratio values ranging from 0.0 to 0.5.

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