A wire is stretched such that its volume remains constant. The Poisson's ratio of the material of the wire is

Options

A. `0.25`
B. `0.50`
C. `-0.25`
D. `-0.50`0

Answer

D

Explanation

(d): Poisson's ratio, `sigma = (Deltar//r)/(Deltal//l)`
Given volume of wire remains constant. So,
`pir^(2)l = k rArr r^(2)l = k’` (where’ = `k/pi`)
`2r dr l + r^(2) dl = 0`
`-2 (dr)/(r) = (dl)/l rArr (dr//r)/(dl//l) = - ½ = -0.5`
`:.` Poisson’s ratio = -0.5