If there is no change in the volume of a wire | vedclass

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(C) The volume $V$ of a wire of length $\ell$ and diameter $D$ is given by $V = \ell \cdot \frac{\pi D^2}{4}$.Taking the differential, we get $dV = \f

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$0.5$

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(C) The volume $V$ of a wire of length $\ell$ and diameter $D$ is given by $V = \ell \cdot \frac{\pi D^2}{4}$.Taking the differential, we get $dV = \frac{\pi D^2}{4} d\ell + \ell \cdot \frac{\pi}{2} D dD$.Since the volume remains constant, $dV = 0$.Therefore, $\frac{\pi D^2}{4} d\ell = -\ell \cdot \frac{\pi}{2} D dD$.Rearranging the terms, we get $\frac{d\ell}{\ell} = -2 \frac{dD}{D}$.Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{dD/D}{d\ell/\ell}$.Substituting the relation $\frac{dD}{D} = -\frac{1}{2} \frac{d\ell}{\ell}$, we get $\sigma = -\frac{-\frac{1}{2} \frac{d\ell}{\ell}}{\frac{d\ell}{\ell}} = 0.5$.

If there is no change in the volume of a wire upon stretching, then the Poisson's ratio for the material of the wire is:

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