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If there is no change in the volume of wire on stretching, then poisson's ratio for material of wire is :

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0.25

$- 1$

0.5

answer is C.

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Detailed Solution

$V = l \times π \frac{D^{2}}{4}, (r = \frac{D}{2})$

$dV = dl \times \frac{{πD}^{2}}{4} + l \times \frac{π}{4} \times 2 DdD, if$

dV = 0

$then dl \times \frac{{πD}^{2}}{4} + l \frac{π}{4} \times 2 DdD = 0$

$\Rightarrow \frac{dl}{l} = - 2 \frac{dD}{D}, \Rightarrow σ = \frac{- \frac{dD}{D}}{\frac{d}{l}} = + \frac{1}{2}$

= 0.5

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