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Question

(B) The Poisson's ratio $\sigma$ is defined as the negative ratio of transverse strain to longitudinal strain: $\sigma = -\frac{\Delta D/D}{\Delta L/L

Vedclass · Accepted Answer

$0.50$

Vedclass · Answer

(B) The Poisson's ratio $\sigma$ is defined as the negative ratio of transverse strain to longitudinal strain: $\sigma = -\frac{\Delta D/D}{\Delta L/L}$.For a wire of length $L$ and diameter $D$,the volume $V = \frac{\pi D^2 L}{4}$.Since the volume $V$ remains constant,$\frac{\Delta V}{V} = 2\frac{\Delta D}{D} + \frac{\Delta L}{L} = 0$.This implies $2\frac{\Delta D}{D} = -\frac{\Delta L}{L}$,or $\frac{\Delta D/D}{\Delta L/L} = -0.5$.Substituting this into the definition of Poisson's ratio: $\sigma = -(-0.5) = 0.5$.Therefore,for a material with constant volume under stretching,the Poisson's ratio is $0.5$.Thus,option $(B)$ is correct.

$A$ wire is stretched such that its volume remains constant. The Poisson's ratio of the material of the wire is

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