A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of 2 x10-3, then the percentage change in volume is

σ =lateral strain/longitudinal strain∴ σ=(Δr/r)/(Δl/l)or 0⋅5=(Δr/r)/2×10−3  or  (Δr/r)=10−3Let initial and final volumes be V and V + dV, thenHere, V=πr2land V+dV=π(r−Δr)2(l+Δl)or ΔV=πr2Δl−2πrlΔrΔVV=Δll−2Δrr=2×10−3−2×10−3= zero