A material has a Poisson's ratio of 0.50. If | vedclass

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(D) The fractional change in volume $\frac{dV}{V}$ for a rod under longitudinal strain $\epsilon_l = \frac{dL}{L}$ is given by the formula: $\frac{dV}

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$0$

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(D) The fractional change in volume $\frac{dV}{V}$ for a rod under longitudinal strain $\epsilon_l = \frac{dL}{L}$ is given by the formula: $\frac{dV}{V} = (1 - 2\sigma) \frac{dL}{L}$,where $\sigma$ is the Poisson's ratio.Given $\sigma = 0.50$ and $\frac{dL}{L} = 2 	imes 10^{-3}$.Substituting these values: $\frac{dV}{V} = (1 - 2 	imes 0.50) 	imes (2 	imes 10^{-3})$.$\frac{dV}{V} = (1 - 1) 	imes (2 	imes 10^{-3}) = 0$.Therefore,the percentage change in volume is $0\%$.

$A$ material has a Poisson's ratio of $0.50$. If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$,what is the percentage change in its volume (in $\%$)?

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