A uniform wire of length 10 m and diameter 0.6 mm is stretched by 6 mm with certain force. If the Poisson's ratio of the material of the wire is 0.3, then the change in diameter of the wire is

Options

A. `108 xx 10^(-8)m`
B. `108 xx 10^(-6)m`
C. `10.8 xx 10^(-8)m`
D. `1.08 xx 10^(-8)m`

Answer

C

Explanation

Given, length of the wire, L = 10 m, diameter, `D = 0.6 xx 10^(-3) m`, poisson.s ratio, `sigma = 0.3` and change in wire length, `Delta L = 6 xx 10^(-3) m`
as poisson.s ratio, `sigma = ("lateral strain")/("longitudinal strain")`
`sigma = ((Delta D)/(D))/((Delta L)/(L))`
`:. sigma = (Delta D L)/(D Delta L)`
`rArr Delta D = (sigma D Delta L)/(L)`
`= (0.3 xx 0.6 xx 10^(-3) xx 6 xx 10^(-3))/(10)`
`rArr Delta D = 10.8 xx 10^(8) m`
Hence, the change in diameter of wire is `10.8 xx 10^(-8)m`. So, the correct option is (C ).