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

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0.50

-0.50

+0.25

-0.25

answer is B.

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Detailed Solution

Let D be the diameter and Y, the volume. Then
$V = \frac{{πD}^{2} L}{4}$
There is no change in volume
$∴ 0 = π 2 DdDL + {πD}^{2} dL$
or $\frac{(dD / D)}{(dL / L)} = - 0 \cdot 5$

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