Q 11-08-010NEETJEE MainMedium
A wire is stretched in such a way that its volume does not change. The Poisson's ratio of its material is
Answer: (B) $0.5$
$V = \pi r^2 L$. Constant volume: $2\dfrac{\Delta r}{r} + \dfrac{\Delta L}{L} = 0$.
Poisson's ratio $= -\dfrac{\Delta r/r}{\Delta L/L} = \dfrac{1}{2}$.
Solution by Sreeraj P, M.Sc Physics