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**Hint:**Calculate the volume of the wire using the formula, differentiate it with the radius and the length and from it frame the Poisson’s ratio as the formula given below. Poisson’ ratio is the ratio of the change in the radius and the change in the length of the wire.

**Formula used:**

(1) The formula of the volume of the cylinder is given by

$V = \pi {r^2}l$

Where $V$ is the volume of the cylinder, $r$ is the radius of the cylinder and $l$ is the length of the cylinder.

(2) The formula of the Poisson’s ratio is given by

$\sigma = \dfrac{{\Delta r}}{{\Delta l}}$

Where $\sigma $ is the Poisson’s ratio, $\Delta r$ is the change in the radius after stretching and $\Delta l$ is the change in the length after stretching.

**Complete step by step solution:**

It is given that there is no change in the volume of the wire when stretching.

Since the wire is in the form of the cylinder, the formula of the cylinder is used.

$V = \pi {r^2}l$

Differentiating the above formula with respect to the radius and the length of the wire, we get

$\Rightarrow dV = 2\pi rldr + \pi {r^2}dl$

Since there is no change in the volume, $dV = 0$ .

$\Rightarrow 2\pi rldr + \pi {r^2}dl = 0$

By solving the above equation, we get

$\Rightarrow \dfrac{{dr}}{r} = - \dfrac{1}{2}\dfrac{{dl}}{l}$

$\Rightarrow \dfrac{{\dfrac{{dr}}{r}}}{{\dfrac{{dl}}{l}}} = - \dfrac{1}{2}$

From the formula of the Poisson’s ratio, we get

$\Rightarrow \sigma = - \dfrac{1}{2}$

The negative value of the Poisson’s ratio indicates that when the length of the wire is increased by stretching, the radius of the wire decreases. From the above explanation, we get

$\Rightarrow \sigma = 0.5$

**Thus the option (A) is correct.**

**Note:**If the Poisson’s ratio is greater than $0.5$ , it indicates that the materials are isotropic in nature. The isotropic materials have the same physical property when measured in all directions. For the hexagonal cells, the Poisson’s ratio is always one.

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