The poisson's ratio of a material is 0.1. If the longitudinal strain of a rod of this material is `10^(-3)`, then the percentage change in the volume of the rod will be

Options

A. 0.008%
B. 0.08%
C. 0.8%
D. 8%

Answer

b

Explanation

Longitudinal strain, `alpha = (I_(2) - I_(1))/(I_(1)) = 10^(-3)` and `(I_(2))/(I_(1)) = 1.001`
Poisson's ratio `sigma = ("Lateral strain")/("Longitudinal strain") = (beta)/(alpha)`
or `beta = sigma alpha = 0.1 xx 10^(-3) = 10^(-4) = (r_(1) - r_(2))/(r_(1))`
or `(r_(2))/(r_(1)) = 1 - 10^(-4) = 0.9999`
% increase in volume `= ((V_(2) - V_(1))/(V_(1))) xx 100`
`= ((pi r_(2)^(2) I_(2) - pi r_(1)^(2) I_(1))/(pi r_(1)^(2) I_(1))) xx 100 = ((r_(2)^(2) I_(2))/(r_(1)^(2) I_(1)) - 1) xx 100`
`= [(0.9999)^(2) xx 1.001 - 1] xx 100 = 0.08%`