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 ∝ d^{2}

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= **\(\frac {\sigma _A}{\sigma _B}=\frac {\sigma _A}{\frac {\sigma _A}2}=2\)**