First slide

Elastic Modulie

Question

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

Moderate

A

0.1
B

0.2
C

0.4
D

0.5

Solution

Let L be the length, r be the radius of the wire.
Volume of the wire is
$V = {πr}^{2} L$
Differentiating both sides, we get
$∆ V = π (2 r ∆ r) L + {πr}^{2} ∆ L$
As the volume of the wire remains uncharged when it gets stretched, so $∆ V = 0$ . Hence
$0 = 2 πrL ∆ r + {πr}^{2} ∆ L$
$∴ \frac{\frac{∆ r}{r}}{\frac{∆ L}{L}} = - \frac{1}{2}$
Poisson's ratio = $\frac{Lateral strain}{Longitudinal strain}$

= $- \frac{\frac{∆ r}{r}}{\frac{∆ L}{L}} = \frac{1}{2} = 0.5$

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