A material has Poisson's ratio `0.5`, If a uniform rod of it suffers a longtiudinal strain of `2xx10^(-3)` then the percentage increases in its volume is

Answer

0

Explanation

Here, `sigma = 0.5, (Delta l)/(l) = 2xx 10^(-3)`
`sigma = (-Delta D//D)/(Delta l//l)`
`or (Delta D)/(D) = -sigma(Delta l)/(l) = - 0.5 xx (2xx10^(-3)) = - 10^(-3)`
As volume, `V = (pi D^2l)/(4)`
`Delta V = (pi)/(4) [ 2D Delta Dxxl + D^2 Delta l]`
` or (Delta V)/(V) = (pi)/(4)[2DDelta Dxxl + D^2Delta l]xx(4)/(pi D^2l)`
`= (2d D)/(D) + (Delta l)/(l) = 2(-10^(-3)) + 2 xx 10^(-3) =0`
% change in volume `= (Delta V)/(V)xx100`
`= 0 xx 100 = 0%`