A material has Poisson's ratio 0.5. If a unif | vedclass

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(D) The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{\Delta r/r}{\Delta l/l} = 0.5$.Giv

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(D) The Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{\Delta r/r}{\Delta l/l} = 0.5$.Given longitudinal strain $\frac{\Delta l}{l} = 3 	imes 10^{-3}$.Therefore, lateral strain $\frac{\Delta r}{r} = -\sigma 	imes \frac{\Delta l}{l} = -0.5 	imes 3 	imes 10^{-3} = -1.5 	imes 10^{-3}$.The volume of a cylindrical rod is $V = \pi r^2 l$.Taking the logarithmic derivative, the fractional change in volume is $\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$.Substituting the values: $\frac{\Delta V}{V} = 2(-1.5 	imes 10^{-3}) + (3 	imes 10^{-3}) = -3 	imes 10^{-3} + 3 	imes 10^{-3} = 0$.Thus, the percentage increase in volume is $0\%$.

$A$ material has Poisson's ratio $0.5$. If a uniform rod of it suffers a longitudinal strain of $3 \times 10^{-3}$, what will be the percentage increase in volume? .......... $\%$

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