Ncert Exercises & Solutions

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?

Area of the copper piece:

$A = l \times b {=19.1\times10}^{- 3} {\times15.2\times10}^{- 3} {=2.9\times10}^{- 4} m$

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

Modulus of elasticity of copper, $η = 24 \times 10^{9} N / m^{2}$

Modulus of elasticity is given by, $η = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{Strain}$

$Strain = \frac{F}{aη} = \frac{44500}{2.9 \times 10^{- 4} \times 42 \times 10^{9}}$
$= 3.65 \times 10^{- 3}$

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