Mechanical Properties of solids NCERT Solutions Class 11 Physics - Solved Exercise Question 9.8

Question 9.8:
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?
Solution:
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 X 10-4 X 42 X 109)
= 3.65 × 10–3

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