A material has poisson's ratio 0.5. If a uniform rod of its suffers a longitudinal strain of `2xx10^(-3)`, what is the percentage increase in volume?

Explanation

Longitudinal strain,
`(Deltal)/(l)=2xx10^(-3)`
Poisson's ratio, `sigma=0.5`
As `sigma=("Lateral strain")/("Longitudinal strain")`
`=(-DeltaR//R)/(Deltal//l)`
`therefore(DeltaR)/(R)=-sigma(Deltal)/(l)=-0.5xx2xx10^(-3)`
volume of Rod,
`V=piR^(2)l`
Percentage increase in volume is,
`(DeltaV)/(V)xx100=(2(DeltaR)/(R)+(Deltal)/(l))xx100`
`=[2xx(-1)xx10^(-3)+2xx10^(-3)]xx100`
`=0`