Extension inversely proportional to cross-sectional area

Steps:

• Use Hooke's law formula: extension ΔL = (F L) / (Y A), where F is load, L length, Y Young's modulus, A area.
• Both wires share F, L, Y; only A differs.
• A ∝ d², so for second wire d=2D, A₂ = 4 A₁.
• ΔL₂ = ΔL₁ / 4 = x/4.

Why B is correct:

• Area quadruples (d² relation), so extension quarters per ΔL = F L / (Y A).

Why the others are wrong:

• A: x/3 misapplies scaling, ignoring d² factor.
• C: 3x assumes area decreases, opposite of reality.
• D: x/2 treats extension linear with diameter, not area.

Final answer: B