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

Explanation

Let l, D be the length and diameter of the given wire respectively. As volume of wire remains constant. So`(piD^(2)l)/(4)=a"constant"" "...(1)`
Differentiating it, we have,
`(pi)/(4)(2DDeltaDxxl+D^(2)Deltal)=0`
Dividing it by `D^(2)l` we get
`(2DeltaD)/(D)+(Deltal)/(l)=0`
(or) `(Deltal)/(l)=-(2DeltaD)/(D)`
`sigma=(-DeltaD//D)/(Deltal//l)=(-DeltaD//D)/(-(2DeltaD//D))=(1)/(2)`
`=0.5`