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(D) The fractional change in volume $\frac{\Delta V}{V}$ is given by the formula: $\frac{\Delta V}{V} = \epsilon_l (1 - 2\sigma)$,where $\epsilon_l$ i

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(D) The fractional change in volume $\frac{\Delta V}{V}$ is given by the formula: $\frac{\Delta V}{V} = \epsilon_l (1 - 2\sigma)$,where $\epsilon_l$ is the longitudinal strain and $\sigma$ is the Poisson's ratio.Given,$\epsilon_l = 2 	imes 10^{-3}$ and $\sigma = 0.5$.Substituting these values into the formula:$\frac{\Delta V}{V} = (2 	imes 10^{-3}) 	imes (1 - 2 	imes 0.5)$$\frac{\Delta V}{V} = (2 	imes 10^{-3}) 	imes (1 - 1)$$\frac{\Delta V}{V} = (2 	imes 10^{-3}) 	imes 0 = 0$.Therefore,the percentage change in volume is $0 	imes 100 = 0\%$.

$A$ material has a Poisson's ratio $0.5$. If a uniform rod of this material suffers a longitudinal strain of $2 \times 10^{-3}$,then the percentage change in its volume is

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