A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tenston with

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8.Mechanical Properties of Solids

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A piece of copper having a rectangular cross-section of $15.2 \;mm \times 19.1 \;mm$ is pulled in tenston with $44,500\; N$ force, productng only elastic deformation. Calculate the resulting strain?

A

$3.65 \times 10^{-3}$

B

$6.52 \times 10^{-4}$

C

$5.88 \times 10^{-4}$

D

$7.46 \times 10^{-5}$

Solution

Area of the copper piece:

$A=l \times b=19.1 \times 10^{-3} \times 15.2 \times 10^{-3}=2.9 \times 10^{-4} m ^{2}$

Tension force applied on the piece of copper, $F=44500 N$

Modulus of elasticity of copper, $\eta=42 \times 10^{9} N / m ^{2}$

Modulus of elasticity, $\eta=\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\text { Strain }}$

$\therefore$ Strain $=\frac{F}{A \eta}$

$=\frac{44500}{2.9 \times 10^{-4} \times 42 \times 10^{9}}$

$=3.65 \times 10^{-3}$

Standard 11

Physics

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