Question

A wire has Poisson’s ratio $0.3$. If the change in length of the wire on stretching it, is $1\%$, then the percentage change in volume of the wire will be given as,
A. 0.4%
B. 0
C. 1.6%
D. 0.6%

Answer 1

Hint: The Poisson’s ratio is given by taking the ratio of the lateral strain to longitudinal strain. Using this relation, find the value for the change in diameter divided by the original diameter value. Now for finding out the volume, the area has been multiplied with the length of the wire. Change in volume can be calculated by differentiating this equation. These all may help you to solve this question.

Complete step-by-step answer:
First of all let us look at the Poisson’s ratio of a material. It is given as the ratio of the lateral strain to longitudinal strain. This can be written as,
$\text{poisson }\!\!'\!\!\text{ s ratio=}\sigma \text{=}\dfrac{\text{lateral strain}}{\text{longitudinal strain}}$
This can be written as,
$\sigma \text{=}\left( \dfrac{\dfrac{\Delta D}{D}}{\dfrac{\Delta L}{L}} \right)$
The values mentioned in the question is,
The Poisson’s ratio is given as
$\sigma =0.3$
The change in length to the original length is given as,
$\dfrac{\Delta L}{L}=0.01$
Therefore we can write that, the change in diameter to the original diameter is given as,
$\dfrac{\Delta D}{D}=\sigma \times \dfrac{\Delta L}{L}=0.01\times 0.3=3\times {{10}^{-3}}$
As we all know, the area of cross section of the wire is given as,
\[A=\pi {{\left( \dfrac{D}{2} \right)}^{2}}\]
The volume is the product of area and length. It can be written as,
\[V=\pi {{\left( \dfrac{D}{2} \right)}^{2}}\times L\]
Therefore the change in volume will be calculated by differentiating the equation,
\[\begin{align}
  & dV=d\left( \pi \left( \dfrac{{{D}^{2}}}{4} \right)\times L \right) \\
 & dV=\dfrac{\pi \times L\times 2D\times dD}{4}+\dfrac{\pi \times {{D}^{2}}\times dL}{4} \\
\end{align}\]
Therefore the change in volume to the original volume, can be written as,
\[\dfrac{dV}{V}=\dfrac{\dfrac{\pi }{4}\times \left( L\times 2D\times dD+\pi \times {{D}^{2}}\times dL \right)}{\pi \times L\times \dfrac{{{D}^{2}}}{4}}\]
Let us cancel the common terms in both the equations. Therefore we can write that,
\[\dfrac{dV}{V}=\dfrac{\Delta D}{D}+\dfrac{\Delta L}{L}\]
Substituting the values in it will give,
$ \dfrac{dV}{V}=3\times {{10}^{-3}}+0.01 $
$ \dfrac{dV}{V}=0.016=1.6\% $
Therefore the correct answer is obtained.

So, the correct answer is “Option C”.

Note: If a substance is stretched in one direction, it will tend to compress in the direction perpendicular to the of force application. This is true in the opposite sense also. This phenomenon is known as Poisson’s effect. This measure is given as Poisson’s ratio.