If Poisson's ratio sigma is -frac{1}{2} for a | vedclass

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(A) The fractional change in volume $\frac{dV}{V}$ is related to the longitudinal strain $\frac{dL}{L}$ and Poisson's ratio $\sigma$ by the formula: $

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Incompressible

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(A) The fractional change in volume $\frac{dV}{V}$ is related to the longitudinal strain $\frac{dL}{L}$ and Poisson's ratio $\sigma$ by the formula: $\frac{dV}{V} = (1 - 2\sigma) \frac{dL}{L}$.Given $\sigma = -\frac{1}{2}$,substituting this value into the equation gives: $\frac{dV}{V} = (1 - 2(-\frac{1}{2})) \frac{dL}{L} = (1 + 1) \frac{dL}{L} = 2 \frac{dL}{L}$.Wait,let us re-evaluate the standard relation for volume change: $\frac{dV}{V} = (1 - 2\sigma) \frac{dL}{L}$.If $\sigma = 0.5$,then $\frac{dV}{V} = 0$,which implies the material is incompressible.However,if $\sigma = -0.5$,then $\frac{dV}{V} = (1 - 2(-0.5)) \frac{dL}{L} = 2 \frac{dL}{L}$.Actually,for most materials,$-1 < \sigma < 0.5$. If $\sigma = -0.5$,the material is highly compressible. Given the options provided and the standard physics context where $\sigma = 0.5$ represents incompressibility,there is a common confusion in literature. If the question implies $\sigma = 0.5$,the answer is Incompressible. If the question strictly asks for $\sigma = -0.5$,it is Compressible.

If Poisson's ratio $\sigma$ is $-\frac{1}{2}$ for a material,then the material is

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