Question

Determine the elongation of a copper (E=11×10¹⁰Pa) rod having a length of 6m and a cross sectional area of 0.7cm² under
the action of a force 10kN

Answer 1

Answ

the action of a force 10kNDetermine the elongation of a copper (E=11×10¹⁰Pa) rod having a length of 6m and a cross sectional area of 0.7cm² under

the action of a force 10kNDetermine the elongation of a copper (E=11×10¹⁰Pa) rod having a length of 6m and a cross sectional area of 0.7cm² under

the action of a force 10kN

Answer 2

The elongation in the copper rod is 6.007 m

Explanation:

We know that

$E=\frac{\text{Stress}}{\text{Strain}}$

Where,

$\text{Stress}=\frac{F}{A}$

Given, F = 10 kN

Cross sectional area A = 0.7 cm² = 0.7 × 10⁻⁴ m²

∴ $\text{Stress}=\frac{10\times 1000}{0.7\times 10^{-4}}$

$\implies \text{Stress}=\frac{1}{7}\times 10^9$ Pa

Also,

$\text{Strain}=\frac{\Delta l}{l}$

Given,

l = 6m

Δl = ?

$\text{Strain}=\frac{\Delta l}{6}$

Thus,

$11\times 10^{10}=\frac{1/7\times 10^9}{\Delta l/6}$

$\implies 11\times 10^{10}\times\frac{\Delta l}{6}=\frac{1}{7}\times 10^9$

$\implies 110\times\Delta l=\frac{6}{7}$

$\implies \Delta l =\frac{6}{770}=0.0077$ m

or, $\Delta l=0.77$ cm

Elongation

$=\Delta l+l=0.0077+6=6.0077$ m

Hope this answer is helpful.

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