If a wire having initial diameter of 2 mm produced the longitudinal strain of 0.1%, then the final diameter of wire is `(sigma = 0.5)`

Options

A. 2.002 mm
B. 1.999 mm
C. 1.988 mm
D. 2.001 mm

Answer

b

Explanation

Poisson's ratio
`sigma = ("lateral strain")/("longitudinal strain") = - (Delta R//R)/(Delta I//I)`
`sigma = - (Delta R)/(R ) xx (I)/(Delta I)`
or `|sigma| = (Delta R)/(R ) xx (I)/(Delta I)`
`:. (Delta R)/(R ) = |sigma| ((Delta I)/(I))`
or `Delta R = 0.5 xx (0.1)/(100) xx ((2 xx 10^(-3))/(2))`
= 0.0005 mm
`:.` final radius, `R - Delta R` = 1 mm - 0.0005 mm = 0.9995 mm ,
The final diameter `= 2xx 0.0005 mm = 1.9990 mm`