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(A) Let the initial length be $L$ and the cross-sectional area be $A$. The volume $V = A 	imes L$.Since the volume remains constant during stretching

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(A) Let the initial length be $L$ and the cross-sectional area be $A$. The volume $V = A 	imes L$.Since the volume remains constant during stretching,$dV = 0$.Taking the derivative,$d(AL) = A \cdot dL + L \cdot dA = 0$.This implies $\frac{dA}{A} = -\frac{dL}{L}$.The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain:$\sigma = -\frac{dA/A}{dL/L}$.Substituting the relation $\frac{dA}{A} = -\frac{dL}{L}$ into the formula:$\sigma = -\frac{-dL/L}{dL/L} = 1$.Wait,re-evaluating: For an incompressible material,the Poisson's ratio is exactly $0.5$.Let longitudinal strain be $\epsilon_l = \frac{dL}{L}$ and lateral strain be $\epsilon_d = \frac{dr}{r}$.Volume $V = \pi r^2 L$. Taking log and differentiating: $\frac{dV}{V} = 2\frac{dr}{r} + \frac{dL}{L} = 0$.Thus,$\frac{dr}{r} = -\frac{1}{2} \frac{dL}{L}$.Poisson's ratio $\sigma = -\frac{\epsilon_d}{\epsilon_l} = -\frac{-0.5 \epsilon_l}{\epsilon_l} = 0.5$.

There is no change in the volume of a wire due to a change in its length on stretching. The Poisson's ratio of the material of the wire is:

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