A steel wire of length 2 m and Young's modul | vedclass

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(B) Given: Length $L = 2 \ m$,Young's modulus $Y = 2.0 	imes 10^{11} \ N/m^2$,Poisson's ratio $\mu = 0.2$,and transverse strain $\epsilon_t = \frac{\

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$25$

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(B) Given: Length $L = 2 \ m$,Young's modulus $Y = 2.0 	imes 10^{11} \ N/m^2$,Poisson's ratio $\mu = 0.2$,and transverse strain $\epsilon_t = \frac{\Delta r}{r} = 10^{-3}$.Poisson's ratio is defined as $\mu = -\frac{	ext{transverse strain}}{	ext{longitudinal strain}} = -\frac{\epsilon_t}{\epsilon_l}$.Taking the magnitude,$\epsilon_l = \frac{\epsilon_t}{\mu} = \frac{10^{-3}}{0.2} = 5 	imes 10^{-3}$.The elastic potential energy density $u$ is given by $u = \frac{1}{2} Y \epsilon_l^2$.Substituting the values: $u = \frac{1}{2} 	imes (2.0 	imes 10^{11}) 	imes (5 	imes 10^{-3})^2$.$u = 1.0 	imes 10^{11} 	imes 25 	imes 10^{-6} = 25 	imes 10^5 \ J/m^3$.Thus,the value is $25$.

$A$ steel wire of length $2 \ m$ and Young's modulus $2.0 \times 10^{11} \ N/m^2$ is stretched by a force. If Poisson's ratio and transverse strain for the wire are $0.2$ and $10^{-3}$ respectively,then the elastic potential energy density of the wire is . . . . . . $\times 10^5 \ J/m^3$.

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