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

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Question

(a)

$\frac{	ext { Lateral strain }}{	ext { Longitudinal strain }}=$ Poisson's ratio

$\frac{0.02 / 100}{\Delta l / l}=\frac{1}{4}$ &nb

Vedclass · Accepted Answer

$2.56 	imes 10^4$

Vedclass · Answer

(a)

$\frac{	ext { Lateral strain }}{	ext { Longitudinal strain }}=$ Poisson's ratio

$\frac{0.02 / 100}{\Delta l / l}=\frac{1}{4}$  $\left\{\begin{array}{l}Y=	ext { (Young's modulus) } \ \quad=8 	imes 10^{10} 	ext { (given) } \ 	ext { Poission's ratio }=\frac{1}{4} 	ext { (given) } \ 	ext { Lateral strain }=0.02 \% 	ext { (given) }\end{array}ight.$

$\frac{\Delta l}{l}=\frac{0.08}{100}$

$\Delta U$ (Elastic potential energy per unit volume $=\frac{1}{2} 	imes Y 	imes($ Longitudinal strain $\left.)ight)$

Substituting values

$\Delta U=\frac{1}{2} 	imes 8 	imes 10^{10} 	imes\left(\frac{0.08}{100}ight)^2$

$\Delta U=2.56 	imes 10^4 \,J / m ^3$

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 wire is [in $\left.J / m ^3\right]$

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