A material has Poisson's ratio 0.20. If a uniform rod suffers a longitudinal strain `2xx10^(-3)`, then the percentage change in volume is

Options

A. `+0.12`
B. `-0.12`
C. `0.28`
D. `-0.28`

Answer

A

Explanation

Here, poisson's ratio `sigma=(-Delta D//D)/(Delta L//L=0.20`,
`(Delta L)/(L)=2xx10^(-3)`
As `V=(pi D^(2) L)/(4)`,
`Delta V=(pi)/(4)[D^(2)Delta L+L2D Delta D]`
`=(pi)/(4)D[D Delta L+2L Delta D]`
`(Delta V)/(V)=(D Delta L+2L Delta D)/(DL)=(Delta L)/(L)+(2Delta D)/(D)`
Now `(Delta V)/(V)=(Delta L)/(L)[1+(2Delta D//D)/(Delta L//L)]=(Delta L)/(L)[1-2rho]`
`=2xx10^(-3)[1-2xx0.20]=2xx10^(-3)[0.6]`
`1.2xx10^(-3)`
% change `=(Delta V)/(V)xx100=1.2xx10^(-3)xx100=0.12`