Given:

$D = 20\hspace{0.05cm}mm,\hspace{5cm}L = 200\hspace{0.05cm}mm\\ \delta d = 0.0035\hspace{0.05cm}mm(Lateral),\hspace{2.5cm}\delta l = 0.1\hspace{0.05cm}mm(Linear)$

To Find

$\mu = E = ?$

Solution

$(i)\hspace{0.05cm}\mu = ?\\ \mu = \frac{Lateral strain}{Linear strain}\\ Lateral strain = \frac{\delta d}{d} = 1.75\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\\ Linear strain = \frac{\delta l}{l} = 5\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\\ \mu = \frac{1.75\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}}{5\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}}\\ \mu = 0.35$

$(ii)\hspace{0.05cm}\textit{From Hook's Law}\\ E = \frac{\sigma}{\rho}\\ \hspace{0.08cm} = \frac{\frac{P}{A}}{\frac{\delta l}{L}}\\ E = \frac{P\hspace{0.05cm}\times\hspace{0.05cm}L}{A\hspace{0.05cm}\times\hspace{0.05cm}\delta l} = \frac{50\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}200}{\frac{\pi}{4}\hspace{0.05cm}\times\hspace{0.05cm}20^2\hspace{0.05cm}\times\hspace{0.05cm}0.1}\\ \hspace{0.08cm}= 31.83\hspace{0.05cm}\times\hspace{0.05cm}10^4\hspace{0.05cm}N/mm^2$