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

Options

A. `+(1)/(2)`
B. `-(1)/(2)`
C. `+(1)/(4)`
D. `-(1)/(4)`

Answer

B

Explanation

volume of cylindrical weire,
`V=(pix^2L)/(4)`, where x is diameter of wire
Differentiating both sides
`(dV)/(dx)=(pi)/(4)[2xL+x^2.(dL)/(dx)]
Also volume remains constant `therefore (dV)/(dx)=0`
`therefore 2xL+x^(2)(dL)/(dx)=0rArr2xL=-x^(2)(dL)/(dx)`
`rArr((dx)/(x))/((dL)/(L))=-(1)/(2)`
`rArr"Poisson's ratio"=-(1)/(2)`
So the correct choice is `(b)`.