Question

If a wire having initial diameter of $2mm$ produced the longitudinal strain of $0.1$%, then the final diameter of wire is $(\sigma =0.5)$

• A. $2.002mm$
• B. $1.999mm$
• C. $1.998mm$
• D. $2.001mm$

Answer 1

The correct option is B $1.999mm$

When a deforming force is applied at the free end of a suspend wire of length $l$ and radius $R$, then its length increases by $dl$, but its radius decreases by $dR$. Now two types of strains are produced by a single force
(i) Longitudinal strain $=\frac{\Delta l}{l}$
(ii) Lateral strain $=-\frac{\Delta R}{R}$
Then the Poisson's ratio
$\sigma =\frac{lateralstrain}{longitudinalstrain}$
$=-\frac{\Delta R/R}{\Delta l/l}$
$\sigma =-\frac{\Delta R}{R}\times \frac{l}{\Delta l}$
or $\left|\sigma \right|=\frac{\Delta R}{R}\times \frac{l}{\Delta l}$
$\therefore \frac{\Delta R}{R}=\left|\sigma \right|\left(\frac{\Delta l}{l}\right)$
or $\Delta R=0.5\times \frac{0.1}{100}\times \left(\frac{2\times {10}^{-3}}{2}\right)$
$=0.0005mm$
$\therefore$ Final radius $R-\Delta R=1mm-0.0005mm$
$=0.9995mm$
$\therefore$ The final diameter $=2\times 0.9995mm$
$=1.9990mm$