A metal wire having Poisson's ratio 1/4 and Y | vedclass

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(A) Given:Poisson's ratio $(\sigma)$ = $1/4 = 0.25$Young's modulus $(Y)$ = $8 	imes 10^{10} \, N/m^2$Lateral strain = $0.02 \% = 0.02 / 100 = 2 	ime

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$2.56 	imes 10^4$

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(A) Given:Poisson's ratio $(\sigma)$ = $1/4 = 0.25$Young's modulus $(Y)$ = $8 	imes 10^{10} \, N/m^2$Lateral strain = $0.02 \% = 0.02 / 100 = 2 	imes 10^{-4}$Formula for Poisson's ratio:$\sigma = \frac{	ext{Lateral strain}}{	ext{Longitudinal strain}}$Calculating longitudinal strain $(\epsilon)$:$\epsilon = \frac{	ext{Lateral strain}}{\sigma} = \frac{2 	imes 10^{-4}}{0.25} = 8 	imes 10^{-4}$Elastic potential energy per unit volume $(u)$:$u = \frac{1}{2} 	imes Y 	imes \epsilon^2$Substituting the values:$u = \frac{1}{2} 	imes (8 	imes 10^{10}) 	imes (8 	imes 10^{-4})^2$$u = 4 	imes 10^{10} 	imes 64 	imes 10^{-8}$$u = 256 	imes 10^2 = 2.56 	imes 10^4 \, J/m^3$

$A$ metal wire having Poisson's ratio $1/4$ and Young's modulus $8 \times 10^{10} \, N/m^2$ is stretched by a force,which produces a lateral strain of $0.02 \%$ in it. The elastic potential energy stored per unit volume in the wire is [in $J/m^3$]

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