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

Longitudinal strain, `(Delta l)/(l)= 2 xx 10^(-3)`
Poisson.s ratio, `sigma= 0.5`
As `sigma= ("Lateral strain")/("longitudinal strain")= (-DeltaR//R)/(Delta l//l)`
`therefore (Delta R)/(R)= -sigma (Delta l)/(l)= -0.5 xx 2 xx 10^(-3) = -1 xx 10^(-3)`
Volume of rod, `V= pi R^(2)l`
Percentage increase in volume is
`(Delta V)/(V) xx 100= (2 (Delta R)/(R)+ (Delta l)/(l)) xx 100`
`= [2 xx (-1) xx 10^(-3) +2 xx 10^(-3) ] xx 100= 0`