Question

A uniform cylindrical wire of length $4m$ and diameter $0.6mm$ is stretched by a certain force such that its length is increased by $4mm$. If the Poisson's ratio of the material is $0.3$ then, calculate the change in diameter of the wire.

Answer 1

Given : $L=4m$ $D=0.6mm$ $\sigma =0.3$ $\Delta L=4mm$

Using $\sigma =\frac{-\Delta D/D}{\Delta L/L}$

Or $0.3=\frac{-\Delta D/(0.6\times {10}^{-3})}{(4\times {10}^{-3})/4}$

$⟹\Delta D=-5\times {10}^{-7}m$