The Poisson's ratio of a material is 0.5. If | vedclass

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(D) Given: Poisson's ratio $\sigma = 0.5$. The percentage decrease in cross-sectional area is $\frac{\Delta A}{A} 	imes 100 = 4 \%$,so $\frac{\Delta

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(D) Given: Poisson's ratio $\sigma = 0.5$. The percentage decrease in cross-sectional area is $\frac{\Delta A}{A} 	imes 100 = 4 \%$,so $\frac{\Delta A}{A} = 0.04$.Since the area $A = \pi r^2$,the fractional change in area is related to the fractional change in radius by $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.Thus,$2 \frac{\Delta r}{r} = 0.04$,which gives the lateral strain $\frac{\Delta r}{r} = 0.02$.Poisson's ratio is defined as $\sigma = - \frac{	ext{Lateral Strain}}{	ext{Longitudinal Strain}} = - \frac{\Delta r / r}{\Delta l / l}$.Taking magnitudes,$0.5 = \frac{0.02}{\Delta l / l}$.Therefore,the longitudinal strain $\frac{\Delta l}{l} = \frac{0.02}{0.5} = 0.04$.The percentage increase in length is $\frac{\Delta l}{l} 	imes 100 = 0.04 	imes 100 = 4 \%$.

The Poisson's ratio of a material is $0.5$. If a force is applied to a wire of this material,there is a decrease in the cross-sectional area by $4 \%$. The percentage increase in the length is ........ $\%$

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