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

Answer: 0.5

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: $V_1=V_2$
$\therefore \pi r^{2}l=\pi (r-\Delta r{)}^2(l+\Delta l)$
$=\pi \left[r^2-2r(\Delta r)+(\Delta r{)}^2\right](l+\Delta l)$
$=\pi r^2(l+\Delta l)-2\pi r\Delta r(l+\Delta l)+\pi (\Delta r{)}^2(l+\Delta l)$
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.
$\therefore \pi r^{2}l=\pi r^{2}l+\pi r^2\Delta l-2\pi rl\Delta r$
$\therefore r\Delta l=2l\Delta r$
$\Rightarrow \frac{\Delta l}{l}=2\frac{\Delta r}{r}$
$\therefore \sigma =\frac{\Delta r/r}{\Delta l/l}=\frac{1}{2}=0.5$
Alternate method
We know that,
$\frac{dV}{V}=(1-2\sigma )\frac{dL}{L}$
$\therefore$ when $\frac{dV}{V}=0,$ then $\sigma =+\frac{1}{2}=0.5$