Advanced Solids MixedMediumQuestion 6377
Question
A wire (Y=2×10¹¹ Pa) stretches 2 mm under load (L=4 m, r=1 mm). Load applied:
- A314 NCorrect
- B157 N
- C628 N
- D78.5 N
Correct answer
314 N
Explanation
F = YAΔl/L = 2×10¹¹×π×10⁻⁶×0.002/4 = 2×10¹¹×π×10⁻⁶×5×10⁻⁴ = π×10¹¹×10⁻⁹ ≈ 314 N... = 2×10¹¹×π×10⁻⁶×5×10⁻⁴ = 314 N.