Question

There is no change in the volume of a wire due to change in its length on stretching. The poisson's ratio of the material of the wire is

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-\frac{1}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

Let $V$ be the volume of wire and $A$ and $L$ be the area of cross-section and length of the wire respectively.
$V=A\times L$
$V=\pi r^{2}L$ ($r=$ radius)
Taking log on both sides,
$\log V=\log \pi +2\log r+\log L$
Differentiating the above equation, we get
$\frac{dV}{V}=0+2\frac{dr}{r}+\frac{dL}{L}$
$\Rightarrow \frac{dV}{V}=2\frac{dr}{r}+\frac{dL}{L}$
But given that $dV=0$
$\therefore 0=2\frac{dr}{r}+\frac{dL}{L}\Rightarrow \frac{dL}{L}=-2\frac{dr}{r}...\left(1\right)$
Now we know that,$\text{Poisson Ratio}\left(\mu \right)=\frac{\text{Lateral strain}}{\text{Longitudinal strain}}=-\frac{dr/r}{dL/L}...\left(2\right)$
Using $\left(1\right)$ and $\left(2\right)$, we get
$\mu =\frac{-dr/r}{2\left(\frac{-dr}{r}\right)}=\frac{1}{2}$
$\Rightarrow \mu =\frac{1}{2}$