If the longitudinal strain of a stretched wir | vedclass

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(B) The longitudinal strain is given by $\epsilon_L = 0.2 \% = 0.002$.Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d$

Vedclass · Accepted Answer

$0.08$

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(B) The longitudinal strain is given by $\epsilon_L = 0.2 \% = 0.002$.Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d$ to longitudinal strain $\epsilon_L$,so $\epsilon_d = -\sigma \cdot \epsilon_L$.Here,$\sigma = 0.3$,so the lateral strain is $\epsilon_d = -0.3 	imes 0.002 = -0.0006$.The volume strain $\frac{\Delta V}{V}$ for a wire is given by the sum of longitudinal strain and two lateral strains: $\frac{\Delta V}{V} = \epsilon_L + 2\epsilon_d$.Substituting the values: $\frac{\Delta V}{V} = 0.002 + 2(-0.0006) = 0.002 - 0.0012 = 0.0008$.Converting to percentage: $0.0008 	imes 100 \% = 0.08 \%$.

If the longitudinal strain of a stretched wire is $0.2 \%$ and the Poisson's ratio of the material of the wire is $0.3$,then the volume strain of the wire is (in $\%$)

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