Determine the poisson's ratio of the material of a wire whose volume remains constant under an external normal stress.

Explanation

Volume of a wire, `V=(piD^(2))/(4)l`
As volume remains constant, the differentiation of above equation gives
`0=(pil)/(4).2D.dD+(piD^(2))/(4).dl`
`-2ldD=Ddl`
`implies (dD)/(D)=-(1)/(2)(dl)/(l)`
`because` By definition, Poisson's ratio
`sigma=(-dD//D)/(dl//l)=(-(1)/(2)(dl)/(l))/((dl)/(l))=(1)/(2)=0.5`