A wire having a diameter of 3mm is stretched by an external force, to produce a longitudinal strain of `3xx10^(-3)`. If the Poisson's ratio of the wire is 0.4, the change in the diameter is

Options

A. `1.8xx10^(-3) mm`
B. `3.6xx10^(-3) mm`
C. `5xx10^(-3) mm`
D. `8xx10^(-3) mm`

Answer

B

Explanation

`sigma = ((delta D)/(D))/((delta L)/(L)) " " :. (delta D)/(D) = sigma ((delta L)/(L))`
`= 3 xx 10^(-3) xx 0.4 = 12 xx 10^(-4)`
`:. delta D = (12 xx 10^(-4))D = (12 xx 10^(-4)) xx 3 xx 10^(-3)`
`= 3.6 xx 10^(-6) m`
`:. delta D = 3.6 xx 10^(-3) mm`