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(D) The longitudinal strain is given by $\epsilon_L = 2 	imes 10^{-3}$.Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d

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(D) The longitudinal strain is given by $\epsilon_L = 2 	imes 10^{-3}$.Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d$ to longitudinal strain $\epsilon_L$, i.e.,$\sigma = -\frac{\epsilon_d}{\epsilon_L}$.Therefore, the lateral strain is $\epsilon_d = -\sigma \epsilon_L = -0.50 	imes (2 	imes 10^{-3}) = -1 	imes 10^{-3}$.The volumetric strain $\frac{\Delta V}{V}$ is given by the sum of the longitudinal strain and two lateral strains: $\frac{\Delta V}{V} = \epsilon_L + 2\epsilon_d$.Substituting the values: $\frac{\Delta V}{V} = (2 	imes 10^{-3}) + 2(-1 	imes 10^{-3}) = 2 	imes 10^{-3} - 2 	imes 10^{-3} = 0$.Since the volumetric strain is $0$, the percentage change in volume is $0$.

$A$ 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

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