A cylindrical metal bar of 12 mm diameter is located by an axial force of 20 kN results in a change in diameter by 0.003 mm. Poisson’s ratio is given by: (Assume modulus of rigidity = 80 GPa)

Explanation:We know that, Poisson ratio is the ratio of Lateral Strain to Longitudinal Strain, i.e. The lateral strain is given as, change in dimension to the original dimension, i.e. \(\epsilon_{late}=\frac{δ d}{d}\)And the longitudinal strain is given as, \(\epsilon _{long}=\frac{\sigma}{E}\)We know the relation between Modulus of ELASTICITY, Modulus of Rigidity and Poisson ratio is, E = 2G (1 + μ)Calculation:Given:δd = 0.003 mm, d = 12 mm, F = 20 kN, G = 80 GPaLongitudinal strain:\(\epsilon _{long}=\frac{\sigma}{E}=\frac{F/A}{2G (1 + μ)}\)\(​​​​\epsilon_{long}=\frac{20000/\pi6^2}{2× 80000 (1+μ)}\)\(\epsilon_{long}=\frac{176.8388}{160000(1+μ)}\).........(1)Lateral strain:\(\epsilon_{late}=\frac{δ d}{d}=\frac{0.003}{12}=2.5×10^{-4}\)..............(2)Poisson ratio:From EQUATION (1) and (2)\(μ = \frac{\epsilon_{lateral} }{\epsilon_{Long}}=\frac{2.5×10^{-4}}{\frac{176.8388}{160000(1+μ)}}\)176.8388μ = 2.5 × 10-4 × 160000(1 + μ)μ = 0.2923