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. 1.998 mm
C. 1.999 mm
D. 2.001 mm

Answer

C

Explanation

`sigma=(Delta d)/(D)xx(L)/(l)`
`(Delta d)/(D)=-D xx 5 xx 10^(-4) =-2 xx 10^(-3) xx 5 xx10^(-4)`
`=-1xx10^(-6)`
`d_(2)-d_(1)=-0.001 mm`
`d_(2)=d_(1)-0.001 =2-0.001=2.0000-0.001`
`d_(2)=1.999 mm`.