Question

Find the elongation produced in a copper wire of length 2m and radius 5mm, when suspended by a block

Answer 1

Answer:

0.104 m

Explanation:

Stress, $\sigma=\frac {F}{A}$

Where F is force and A is area. Also, F=mg where m is mass and g is acceleration due to gravity

$Area= \pi r^{2}$

$Strain=\frac {\triangle l}{l}$ where $\triangle l$ is the elongation and l is the original length

$E=\frac {stress}{strain}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}$

Making $\triangle l$ the subject then

$\triangle l=\frac {Fl}{AE}=\frac{mg l}{\pi r^{2} E}$

By substituting the given values and taking g as 9.81 then

$\triangle l=\frac {Fl}{AE}=\frac{500\times 9.81\times 2 m}{\pi \times 0.005^{2}\times 1.2\times 10^{9}}=0.104087333 m\approx 0.104 m$