There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of lhe material of the wire is :

Options

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

Answer

B

Explanation

Volume of cylindrical wire, V = `(pix^(2)L)/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`
`therefore2xL+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).