Two exactly similar wire of steel and copper are stretched by equal force. if the total elongation is 10cm. find how much each wire is elongated given young's modules for steel =2×10^10 n/m^2 and young's modules for copper 2×10^10n/m^2.

The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.

Explanation:

Young's modulus is defined as:

E = stress / strain

E = (F / A) / (dL / L)

E = (F L) / (A dL)

Solving for dL:

dL = (F L) / (A E)

The wires have the same force, length, and cross-sectional area.  So:

dL₁ + dL₂ = (FL/A) (1/E₁ + 1/E₂)

Given that dL₁ + dL₂ = 0.10 m, E₁ = 20×10¹⁰ N/m², and E₂ = 12×10¹⁰ N/m²:

0.10 = (FL/A) (1/(20×10¹⁰) + 1/(12×10¹⁰))

FL/A = 0.75×10¹⁰ N/m

Solving for dL₁ and dL₂:

dL₁ = (FL/A) / E₁

dL₁ = (0.75×10¹⁰ N/m) / (20×10¹⁰ N/m²)

dL₁ = 0.0375 m

dL₂ = (FL/A) / E₂

dL₂ = (0.75×10¹⁰ N/m) / (12×10¹⁰ N/m²)

dL₂ = 0.0625 m

The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.