Question

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

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

Answer 1

Answer: (B)

Poisson's ratio
$\sigma =\frac{lateralstrain}{longitudinalstrain}$
$=-\frac{\Delta R/R}{\Delta l/l}$
$\sigma =-\frac{\Delta R}{R}\times \frac{l}{\Delta l}or|\sigma |=\frac{\Delta R}{R}\times \frac{l}{\Delta l}$
= $\frac{\Delta R}{R}|=|\sigma |\left(\frac{\Delta }{l}\right)$
or $\Delta R=0.5\times \frac{0.1}{100}\times \left(\frac{2\times 10^{-3}}{2}\right)$
= 0.0005 mm