A wire has Poisson’s ratio of $0.5$ . It is stretched by an external force to produce a longitudinal strain ${{10}^{-3}}$ . If the original diameter was $2mm$ . The final diameter after stretching is:
$\begin{align}
  & (a)2.002mm \\
 & (b)1.998mm \\
 & (c)1.980mm \\
 & (d)1.999mm \\
\end{align}$

Hint: The Poisson’s ratio $(\mu )$ for any material is given by, negative of lateral strain upon longitudinal strain. Here, lateral strain is the strain in the body which acts along the direction perpendicular to the direction of applied force. And, longitudinal strain is the strain in the body which acts along the direction of the applied force. The negative sign is to counter the negative value of lateral strain which is obtained under stretching. Since, longitudinal strain is always along the direction of force applied, it is always positive.

Complete step by step answer:
We have, the formula of Poisson’s ratio as follows:
$\Rightarrow \mu =-\dfrac{\left( \dfrac{\vartriangle d}{d} \right)}{\left( \dfrac{\vartriangle l}{l} \right)}$
Where, $\left( \dfrac{\vartriangle l}{l} \right)$ is the longitudinal strain and $\left( \dfrac{\vartriangle d}{d} \right)$ is the lateral strain.
Also, the respective terms in the equation are as follows:
$(\vartriangle d)$ is the small change in the diameter of the wire.
$d$ is the initial diameter of the wire.
$(\vartriangle l)$ is the change in longitudinal length of the wire.
$l$ is the initial length of the wire, that is, before applying any external force on the wire.
In the above equation, we have the values of:
$\begin{align}
  & \Rightarrow \mu =0.5 \\
 & \Rightarrow d=2mm \\
 & \Rightarrow \left( \dfrac{\vartriangle l}{l} \right)={{10}^{-3}} \\
\end{align}$
Thus, $\vartriangle d$ can be calculated as:
$\begin{align}
  & \Rightarrow \vartriangle d=-\mu \times d\times \left( \dfrac{\vartriangle l}{l} \right) \\
 & \Rightarrow \vartriangle d=-0.5\times 2\times {{10}^{-3}}mm \\
 & \Rightarrow \vartriangle d=-0.001mm \\
\end{align}$
Therefore, the final diameter is:
$\begin{align}
  & \Rightarrow {{d}^{'}}=d+\vartriangle d \\
 & \Rightarrow {{d}^{'}}=2+(-0.001)mm \\
 & \Rightarrow {{d}^{'}}=1.999mm \\
\end{align}$
Hence, the final diameter after stretching is $1.999mm$ .

So, the correct answer is “Option d”.

Note: In the above question, we had to find the final diameter of the wire after stretching it. This is the main reason why the diameter of the wire decreased. If it had been the case of compressing the wire by applying force on it from its two opposite ends, the diameter of the wire should have increased and its length would have decreased.