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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
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B
0.4
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C
0.2
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D
zero
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Solution

The correct option is C zero
Given: $$\dfrac{dL}{L}  = 2 \times 10^{-3}$$
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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