Question

The wire subjected to tension $T$ does not undergo volume changes. The Poisson's ratio of the material of the wire is______.

Answer 1

Let $r$ be radius of wire, $l$ be its length, $\Delta r$ be change in $r$ and $\Delta l$ be the change in $l$ when the wire is subjected to tension.
$V _{1}=\pi r ^{2} l$
As length of the wire increases, its radius decreases,
$\therefore $ Volume of wire after elongation is,
$V _{2}=\pi( r -\Delta r )^{2}(l+\Delta l)$
Given: $V _{1}= V _{2}$
$ \therefore \pi r ^{2} l =\pi( r -\Delta r )^{2}(l+\Delta l)$
$=\pi\left[ r ^{2}-2 r (\Delta r )+(\Delta r )^{2}\right](l+\Delta l) $
$=\pi r ^{2}(l+\Delta l)-2 \pi r \Delta r (l+\Delta l)+\pi(\Delta r )^{2} (l + \Delta l)$
Since, $\Delta r$ and $\Delta l$ are very small, terms of order $(\Delta r \times \Delta l)$ and $(\Delta r )^{2}$ and higher can be ignored.
$\therefore \pi r ^{2} l=\pi r ^{2} l+\pi r ^{2} \Delta l-2 \pi r l \Delta r $
$\therefore r \Delta l=2 l \Delta r $
$\Rightarrow \frac{\Delta l}{l}=2 \frac{\Delta r }{ r } $
$\therefore \sigma=\frac{\Delta r / r }{\Delta l / l}=\frac{1}{2}=0.5$
Alternate method
We know that,
$ \frac{d V}{V}=(1-2 \sigma) \frac{d L}{L} $
$\therefore $ when $\frac{d V}{V}=0, $ then $\sigma=+\frac{1}{2}=0.5$