If the volume of a wire remains constant when subjected to tensile stress, the value of poisson's ratio of material of the wire is

Explanation

Let l, D be the length and diameter of the given wire respectively, as volume of wire remains constant so,
`(pi D^(2)l)/(4)` = a constant ..(i)
Differentiating it , we have :
`(pi)/(4)[2 D Delta D xxl +D^(2) Delta l]=0`
Dividing it by `D^(2)l` we get ltbr? `(2 Delta D)/(D) +(Delta l)/(l) = 0 (Delta l)/(l)= (2Delta D)/(D)`
`sigma =(-Delta D//D)/(Delta l//l) (-DeltaD//D)/((-2 Delta D//D)) = 1/2 = 0.5`.