Stress inversely proportional to cross-sectional area

Steps:

• Let narrow diameter be $d$, so wide diameter is $2d$.
• Cross-sectional area of narrow portion: $A_n = \pi (d/2)^2 = \pi d^2 / 4$.
• Cross-sectional area of wide portion: $A_w = \pi (2d/2)^2 = \pi d^2$.
• Tensile force $T$ is the same in both portions for equilibrium.
• Stress in narrow portion: $\sigma_n = T / A_n = 4T / (\pi d^2)$, so .

Why D is correct:

• Stress ratio $\sigma_n / \sigma_w = 4$ follows from $\sigma = F/A$, with areas differing by factor of 4.

Why the others are wrong:

• A: 0.25 is inverse ratio $\sigma_w / \sigma_n$.
• B: 0.50 is half the wide area ratio.
• C: 2.0 is diameter ratio, not area-based stress ratio.

Final answer: D