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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