Question

# A material has Poisson's ratio $$0.50$$. If a uniform rod of it suffers a longitudinal strain of $$2\times {10}^{-3}$$, then the percentage change in volume is

A
0.6
B
0.4
C
0.2
D
zero

Solution

## The correct option is C zeroGiven: $$\dfrac{dL}{L} = 2 \times 10^{-3}$$Poisson's ratio, $$\sigma = -\dfrac{\frac{dr}{r}}{\frac{dL}{L}}$$              where, $$r$$ and $$L$$ are the radius and length of the rod respectively$$\Rightarrow$$   $$0.5 = \dfrac{- \dfrac{dr}{r}}{2 \times 10^{-3}}$$              $$\Rightarrow \dfrac{dr}{r} = -10^{-3}$$Volume of the rod, $$V = \pi r^2 L$$Differentiating, we get, $$dV = \pi (r^2 dL + 2Lrdr)$$$$\Rightarrow \dfrac{dV}{V} \times 100 = \dfrac{\pi(r^2 dL + 2rL dr)}{\pi r^2 L} \times 100 = \bigg( \dfrac{dL}{L} + 2\dfrac{dr}{r}\bigg)\times 100$$$$\Rightarrow \dfrac{dV}{V} \times 100 = [2 \times 10^{-3 } + 2 (-10^{-3} ) ] \times 100 = 0$$Physics

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