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Q.
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
Mechanical Properties of Solids
Solution:
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
$0=2\pi rL\Delta r$ $+\pi r^{2}\Delta L$
$\therefore \quad$ $\frac{\Delta r/ r}{\Delta L/ L}$ $=-\frac{1}{2}$
Poisson’s ratio=$\frac{Lateral \,strain }{Longitudinal \,strain }$=$-\frac{\Delta r/ r}{\Delta L/ L}$ $=\frac{1}{2}$ $=0.5$