A material has Poisson's ratio 0.50 . If a uniform rod made of this material suffers a longitudinal strain…

A material has Poisson's ratio 0.50 . If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is
  1. 0.6
  2. 0.4
  3. 0.2
  4. 0

Solution

Given, Poisson's ratio, $ \sigma=0.5 $ Longitudinal strain, $ \frac{\Delta l}{l}=2 \times 10^{-3} $ Volumetric strain $\left(\frac{\Delta V}{V}\right)$ and Iongitudinal strain $\left(\frac{\Delta l}{l}\right)$ are related as $ \begin{array}{rlrl} \frac{\Delta V}{V} & =(1-2 \sigma) \frac{\Delta l}{l} \\ \Rightarrow & & =(1-2 \times 0.5) \times 2 \times 10^{-3} \\ & =(1-1) \times 2 \times 10^{-3}=0 \times 2 \times 10^{-3}=0 \\ \therefore \quad & \frac{\Delta V}{V} \times 100 & =0 \times 100 \%=0 \end{array} $

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

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