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)

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
$\Delta V$ $=\pi \left(2r\Delta r\right)$ $L+\pi r^2\Delta L$
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=$\frac{Lateralstrain}{Longitudinalstrain}$=$-\frac{\Delta r/r}{\Delta L/L}$ $=\frac{1}{2}$ $=0.5$