Wires in series: same force, total extension sums individual δ = FL/(AY) Steps:

• δ_steel = (40 N × 0.5 m) / (2×10^{-6} m² × 2.0×10^{11} Pa) = 20 / 4×10^5 = 5×10^{-5} m
• δ_brass = (40 N × 1 m) / (2×10^{-6} m² × 1.0×10^{11} Pa) = 40 / 2×10^5 = 2×10^{-4} m
• Total δ = 5×10^{-5} m + 2×10^{-4} m = 2.5×10^{-4} m ≈ 3×10^{-4} m Why B is correct:
• Hooke's law gives extension δ = FL/(AY); for series wires, same F and A, total δ sums δ_steel and δ_brass to ≈3×10^{-4} m Why the others are wrong:
• A overestimates by using lower Y_brass (e.g., 10^{10} Pa), yielding larger δ_brass ≈2×10^{-3} m
• C adds unrelated values or misapplies lengths, exceeding calculated total
• D ignores extension, violating Hooke's law for elastic wires under force

Final answer: B