The Poisson's ratio of a material is 0.4 . If a force is applied to a wire of this material, there is a…

The Poisson's ratio of a material is 0.4 . If a force is applied to a wire of this material, there is a decrease of cross-sectional area by $2 \%$. The percentage increase in its length is
  1. $3 \%$
  2. $2.5 \%$
  3. $1 \%$
  4. $0.5 \%$

Solution

$\begin{aligned} & \text { Poission's ratio } \sigma=\frac{\Delta R / R}{\Delta l / l} \\ & \frac{\Delta l}{l}=\frac{\Delta R}{R} \times \frac{1}{\sigma} \\ & =\frac{\Delta R}{R} \times \frac{1}{0.4} \\ & \therefore \text { Increase in length }=\left(\frac{\Delta R}{R} \times \frac{1}{0.4}\right) \times 100 \\ & =\frac{2}{2} \times \frac{1}{0.4}=2.5 \% \\ & \end{aligned}$

Asked in: AP EAMCET 2002

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