A material has Poisson ratio 0.5. If a rod of the material has a longitudinal strain `2 xx 10^(-3)`, the percentage change in volume is :

Options

A. 0.6
B. 0.4
C. 0.2
D. zero.

Answer

D

Explanation

Here Poisson.s ratio =`((Deltar)/r)/((Deltal)/l)`
In this case `(Deltal)/l=2xx10^(-3)` and `sigma=0.5`
`therefore=(Deltar)/r=sigmaxx(Deltal)/l`
`=2xx10^(-3)xx0.5=10^(-3)`
Let V be the initial volume and `(V + Delta V)` be the final volume, then
`V=pir^(2)l`
`V+DeltaV=pi(r-Deltar)^(2)(l+Deltal)`
`thereforeDeltaV=pir^(2)Deltal-2pirlDeltar` (Subtracting and simplifying)
`(DeltaV)/V=(Deltal0)/l-(2Deltar)/r`
`=2xx10^(-3)-2xx10^(-3)`=Zero