Question

The Poisson's ratio of a material is $0.8$. If a force is applied to a wire of this material decreases its cross-sectional area by $4\%$, then the percentage increase in its length will be.

• A. $1\%$
• B. $2\%$
• C. $2.5\%$
• D. $4\%$

Answer 1

Answer: (C)

$2.5\%$

The Poisson's ratio $\sigma$ is given by
$\sigma =\frac{lateralstrain}{longitudinalstrain}=\frac{\Delta d/d}{\Delta l/l}$ ........(i)
where, l$=$length of wire
d$=$diameter of wire
The cross-sectional area of wire is given by
$A=\pi r^2=\frac{\pi d^2}{4}$
$\therefore \frac{\Delta A}{A}=\frac{2\Delta d}{d}$
$\Rightarrow \frac{\Delta d}{d}=\frac{\Delta A}{2A}$ ........(ii)
From Eqs. (i) and (ii), we get
$\sigma =\frac{\frac{\Delta A}{2A}}{\Delta l/l}$
$\Rightarrow \frac{\Delta l}{l}=\frac{\Delta A}{2\sigma \cdot A}=\frac{1}{2\times \sigma }\left(\frac{\Delta A}{A}\right)$
$=\frac{1}{2\times 0.8}\times 4\%$
$\frac{\Delta l}{l}=2.5\%$.