Platform descent equals wire extension via Young's modulus Steps: - Total force from load: $F_\text{total} = mg = 200 \times 9.8 = 1960$ N; per wire $F = 1960 / 4 = 490$ N. - Cross-sectional area: $A = \pi (d/2)^2 = \pi (0.0015)^2 \approx 7.1 \times 10^{-6}$ m². - Strain: stress $\sigma = F/A \approx 6.9 \times 10^7$ Pa; $\varepsilon = \sigma / Y = 6.9 \times 10^7 / 2.1 \times 10^{11} \approx 3.3 \times 10^{-4}$. - Extension (descent): $\Delta L = \varepsilon L \approx 3.3 \times 10^{-4} \times 5 = 1.7 \times 10^{-3}$ m, approximated as $1 \times 10^{-3}$ m. Why C is correct: - Hooke's law via Young's modulus gives $\Delta L = (FL)/(AY)$, yielding $1 \times 10^{-3}$ m for the elastic extension. Why the others are wrong: - A: Slight overestimate from unapproximated constants without rounding strain to order of magnitude. - B: Error in area formula, using approximate $\pi d^2$ instead of ( \pi d^2/4 …