# 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? - Physics

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 1

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/(Aeta)

= 44500/(2.9xx10^(-4)xx42xx10^(9))

= 3.65 xx 10^(-3)

#### Solution 2

A = 15.2 x 19.2 x 10-6 m2; F  = 44500 N; η = 42 x 109 Nm-2

Strain =  "Stress"/"modulus of elasticity" = "F/A"/eta

=F/(Aeta) = 44500/((15.2xx19.2xx10^(-6))xx42xx10^9) = 3.65 xx 10^(-3)

Concept: Hooke’s Law
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#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 9 Mechanical Properties of Solids
Q 8 | Page 244