Question

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

• A. 0.5
• B. - 0.50
• C. 0.25
• D. - 0.25

Answer 1

The correct option is A 0.5

Volume of a wire of radius $r$ and length $l$ is,

$V=\pi r^{2}l$

$\therefore dV=2\pi rldr+\pi r^{2}dl$

$0=2\pi rldr+\pi r^{2}dl$

$(dv=0,$ as volume is unchanged$)$

$\therefore 2rldr=-r^{2}dl$

$\frac{dr}{r}=-\frac{1}{2}\frac{dl}{l}$

$\therefore \frac{\frac{dr}{r}}{\frac{dl}{l}}=-\frac{1}{2}$

$\therefore \sigma =-\frac{1}{2}$

As, here $-ve$ sign implies that if, length increased, radius decreased, so, we can write

$\sigma =\frac{1}{2}=0.5$