1.

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 :

Answer»

0.6
0.4
0.2
zero.

Solution :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


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