# A Material Has Poisson's Ratio 0.50. If A Uniform Rod Of It Suffers A Longitudinal Strain Of 2×10 −3, Then The Percentage Change In Volume Is

Given:

$$\frac{dL}{L} = 2 * 10^{-3}$$

Poisson’s Ratio, $$\sigma = -\frac{\frac{dr}{r}}{\frac{dL}{L}}$$

Here r and L are the radius and length of the rod, respectively.

= $$0.5 = -\frac{\frac{dr}{r}}{2 * 10^{-3}}$$

=$$\frac{dr}{r} = -10^{-3}$$

The volume of the rod V =$$\pi r^{^{2}}L$$

By differentiating we get V =$$\pi(r^{^{2}}dL + 2Lrdr)$$

=$$\frac{dV}{V} * 100$$

= $$\frac{\pi (r^{2}dL + 2rLdr)}{\pi r^{2}L} * 100 = (\frac{dL}{L} + 2\frac{dr}{r})*100$$

=$$\frac{dV}{V} * 100 = [2 * 10^{-3} + 2 (-10^{-3})] * 100 = 0$$

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