A material has a Poisson's ratio 0.3. If a uniform of it suffers longitudinal strain `4.5 xx 10^(-3)`, calculate the percentage change in its volume.

Explanation

Here `sigma = 0.3 , Deltal//l =4.5 xx 10^(-3)`
`sigma =-(DeltaR//R)/(Deltal//l)`
or ` (DeltaR)/( R) = - sigma (l)/(l) = -0.3 xx(4.5 xx 10^(-3))`
`-1.35 xx 10^(-3)`
Volume , `V=pi R^(2)l`
`(DeltaV)/(V) = (2 DeltaR)/( R) +(Deltal)/(l)`
% change in volume `= (DeltaV)/(V) xx 100`
`=(2DeltaR)/(R ) + (Deltal)/(l) xx 100`
`=[2 x (-1.35 xx 10^(-3))+4.5xx10^(-3)] xx 100`
`=(1.8 xx 10^(-3)) xx 100 = 0.18%`.