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The ratio of lateral contractional strain and the longitudinal elongational strain of a stretched wire is(a) ½(b) 2[c] 3/2(d) 1

Jay , 8 Years ago
Grade 12
anser 1 Answers
Saurabh Kumar

Last Activity: 8 Years ago

Poisson's ration for most metals is close to 1/3. That means for a stretch of 1%, you get a reduction in the diameter of 0.33% Adding up the changes in all three dimensions gives you a volume increase of 0.33%. (The "rubber-band" deformation with no volume change corresponds to a Poisson's Ration of 1/2.)
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When a wire, or a rod(whose diameter is not negligible) is subjected to a tensile stress, is there any change in its volume?

If yes, does the volume increase or decrease? Take into consideration both the longitudinal elongation and the lateral contraction.

If no, then what is wrong with the following derivation?

V=πd2l/4(Vis the volume of the wire)

⟹ΔV/V=2Δd/d+Δl/l

⟹ΔV/V=(1−2σ)Δl/l(Δd/d=−σΔl/l)where σis Poisson's ratio.

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