The poisson's ratio for a materia is 0.1. If longitudinal strain of a rod of this material is `1xx10^(-3)`, then calculate the percentage change in the volume of the rod.

Explanation

For a rod, `V=pir^(2)L`
`therefore(DeltaV)/(V)=2(Deltar)/(r)+(DeltaL)/(L)`
`sigma=0.1,and(DeltaL)/(L)=10^(-3)` (given)
Thus from the definition of Poisson's ratio,
`sigma=((-Deltar//r)/(DeltaL//L))`
(or) `(Deltar)/(r)=-0.1xx10^(-3)`
Thus `(DeltaV)/(V)=-0.2xx10^(-3)+1xx10^(-3)`
`=0.8xx10^(-3)`
(or) percentage change in volume
`=0.8xx100=0.08%`