Prove that the volume remains constant when the poisson ratio of a material is 0.5
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Answer:
Explanation:
Solution :
Let L be the length, r be the radius of the wire.
Volume of the wire is
V=πr2L
Differentiating both sides, we get
△V=π(2r△r)L+πr2△L
As the volume of the wire remains uncharged when it gets stretched, so △V=0. Hence
0=2πrL△r+πr2△L
(△rr/(△LL)=−12
Poisson's ration
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