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Steel wire (Y = 2 × 10¹¹ Pa) of length 2 m and cross-section 1 mm² stretches by 1 mm under load. Load force:
A200 N
B100 N
CForce = (Y × A × ΔL)/L = (2 × 10¹¹ × 10⁻⁶ × 10⁻³)/2 = 100 N
D500 N
Answer & Solution
Correct answer: C. Force = (Y × A × ΔL)/L = (2 × 10¹¹ × 10⁻⁶ × 10⁻³)/2 = 100 N
F = Y A (ΔL/L) = 2×10¹¹ × 10⁻⁶ × (10⁻³/2) = 100 N.
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