The increase in length on stretching a wire i | vedclass

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(A) Poisson's ratio $(\sigma)$ is defined as the ratio of lateral strain to longitudinal strain.$\sigma = \frac{	ext{Lateral strain}}{	ext{Longitudi

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Reduce by $0.02\%$

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(A) Poisson's ratio $(\sigma)$ is defined as the ratio of lateral strain to longitudinal strain.$\sigma = \frac{	ext{Lateral strain}}{	ext{Longitudinal strain}}$Given:Longitudinal strain $(\frac{\Delta L}{L})$ = $0.05\% = 0.0005$Poisson's ratio $(\sigma)$ = $0.4$Lateral strain = $\sigma 	imes 	ext{Longitudinal strain}$Lateral strain = $0.4 	imes 0.05\% = 0.02\%$Since the wire is stretched,the length increases,which causes the diameter to decrease.Therefore,the diameter reduces by $0.02\%$.

The increase in length on stretching a wire is $0.05\%$. If its Poisson's ratio is $0.4$,then its diameter

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