Correct Answer - Option 1 : 2
Explanation:
In the case of uniaxial loading:
Stress developed in bar(σ) = \(\frac FA\)
where F = Force/load applied, A = Cross-section Area
For same material bars
\(Safety ∝\frac {1}{stress ~developed}\)
In case of circular bar A
\(\sigma_A =\frac FA\)
In case of circular bar B
Area = 4A as diameter doubled and Area ∝ d2
load = 2F as force doubled
\(\sigma_B =\frac {2F}{4A} = \frac 12\frac FA= \frac {\sigma _A}{2}\)
∴ Ratio of safety of bar B to bar A= \(\frac {\sigma _A}{\sigma _B}=\frac {\sigma _A}{\frac {\sigma _A}2}=2\)