Answered

Find the elongation produced in a copper wire of length 2m and radius 5mm, when suspended by a block
of mass 500Kg. (Take Young's modulus of elasticity of copper Y = 1,2 ×109 Pa).

Answer :

opudodennis

Answer:

0.104 m

Explanation:

Stress, [tex]\sigma=\frac {F}{A}[/tex]

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

[tex]Area= \pi r^{2}[/tex]

[tex]Strain=\frac {\triangle l}{l}[/tex] where [tex]\triangle l[/tex] is the elongation and l is the original length

[tex]E=\frac {stress}{strain}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}[/tex]

Making [tex]\triangle l[/tex] the subject then

[tex]\triangle l=\frac {Fl}{AE}=\frac{mg l}{\pi r^{2} E}[/tex]

By substituting the given values and taking g as 9.81 then

[tex]\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[/tex]

Other Questions