Correct Answer - Option 3 : 0.25
Concept:
Relation between modulus of rigidity (G), young’s modulus (E) & Poisson’s ratio (μ) is
E=2G(1+μ)
Calculation:
Given:
\(\frac {{G}}{{E}} \) = 0.4
E = 2G (1 + μ)
\(\frac {{E}}{{G}} \) = 2 (1 + μ)
\(\frac {{1}}{{0.4}} \) = 2 (1 + μ)
1.25 = 1 + μ
μ = 0.25
Other relations between elastic constants
E = \( \frac {{9KG}}{{3K+G}}\)
E = 3K (1 - 2μ)