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(A) The volume $V$ of a wire of length $L$ and radius $r$ is given by $V = \pi r^2 L$.Taking the logarithmic derivative,we get $\frac{dV}{V} = 2\frac{

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$0.5$

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(A) The volume $V$ of a wire of length $L$ and radius $r$ is given by $V = \pi r^2 L$.Taking the logarithmic derivative,we get $\frac{dV}{V} = 2\frac{dr}{r} + \frac{dL}{L}$.Since the volume remains unchanged,$\frac{dV}{V} = 0$,so $2\frac{dr}{r} = -\frac{dL}{L}$,which implies $\frac{dr/r}{dL/L} = -\frac{1}{2}$.The Poisson's ratio $\sigma$ is defined as $\sigma = -\frac{	ext{lateral strain}}{	ext{longitudinal strain}} = -\frac{dr/r}{dL/L}$.Substituting the value,we get $\sigma = -(-0.5) = 0.5$.

If the volume of a wire remains unchanged when it is stretched,what is the Poisson's ratio for the wire?

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