Q: When a wire of length 10 m subjected to a force of 100 N along its length, the lateral stain produced is 0.01 × 10^{-3}. The Poisson’s ratio was found to be 0.4. If area of cress section of wire is 0.025 m^{2}, its Young’s modulus is

Sol: Poisson’s ratio $\large \sigma = \frac{Lateral \; Strain}{Longitudinal\; Strain}$

$\large \sigma = \frac{-\frac{\Delta r}{r}}{\frac{\Delta l}{l}}$

$\large \frac{\Delta l}{l} = \frac{-\frac{\Delta r}{r}}{\sigma}$

$\large \frac{\Delta l}{l} = \frac{10^{-5}}{0.4} = 25 \times 10^{-6}$

$\large Y = \frac{F/A}{\frac{\Delta l}{l}} $

$\large Y = \frac{100}{25 \times 10^{-3}\times 25 \times 10^{-6}} $

= 1.6 × 10^{8} N/m^{2}