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Damm [24]
3 years ago
14

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

Physics
1 answer:
igomit [66]3 years ago
3 0

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

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