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) $0.5$
Concept. Poisson's ratio is $\sigma = -\dfrac{\Delta r/r}{\Delta L/L}$. Use the constant-volume condition to relate the two strains.
Step 1. Constant volume. $V = \pi r^2L$, so
$$\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta L}{L} = 0 \;\Rightarrow\; \frac{\Delta r}{r} = -\frac12\frac{\Delta L}{L}$$
Step 2. Poisson's ratio. $\sigma = \dfrac12 = 0.5$.
Answer. $0.5$ (Option B).
Solution by Sreeraj P, M.Sc Physics