A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-9 Mechanical properties of Solids



Q.8:- A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

 

 

Answer:-



Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m
Area of the copper piece:
A = l × b
= 19.1 × 10–3 × 15.2 × 10–3
= 2.9 × 10–4 m2
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η = 42 × 109 N/m2
Modulus of elasticity, η = Stress / Strain
= (F/A) / Strain
∴ Strain = F / Aη
= 44500 / (2.9 × 10-4 × 42 × 109)
= 3.65 × 10–3.