Exercises
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10⁻⁵ m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Given:
Steel wire: Lₛ = 4.7 m, Aₛ = 3.0 × 10⁻⁵ m²
Copper wire: Lc = 3.5 m, Ac = 4.0 × 10⁻⁵ m²
Both wires stretch by the same amount ΔL under the same load F.
Young's modulus: Y = (F × L) / (A × ΔL)
For steel: Yₛ = (F × Lₛ) / (Aₛ × ΔL)
For copper: Yc = (F × Lc) / (Ac × ΔL)
Dividing: Yₛ / Yc = (Lₛ × Ac) / (Lc × Aₛ)
= (4.7 × 4.0 × 10⁻⁵) / (3.5 × 3.0 × 10⁻⁵)
= (4.7 × 4.0) / (3.5 × 3.0)
= 18.8 / 10.5
= 1.79
Answer: Yₛ / Yc ≈ 1.79
