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If the volume of a wire remains constant...

If the volume of a wire remains constant when subjected to tensile stress, the value of Poisson's ratio of the material of the wire is

A

0.1

B

0.2

C

0.4

D

0.5

Text Solution

Verified by Experts

The correct Answer is:
D
Let L be the length, r be the radius of the wire. Volume of the wire is `V = pir^(2)L`
Differentiating both side, we get `DeltaV = pi (2rDeltar)L + pir^(2) DeltaL`
As the volume of the wire remains unchanged when it gets stretched, so AV = 0. Hence
`0 = 2pirLDelta+ pir^(2)DeltaL)`
`therefore (Deltar//r)/(DeltaL//L) = -(1)/(2)`
`"Poisson's ratio"=("Lateral strain")/("Longitudinal strain")=-(Delta r//r)/(DeltaL//L)= (1)/(2) = 0.5`
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