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2014-09-30T23:39:36+05:30

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Diameter(d) = 1mm
length(L) = 2.5 mm
Force(F) = 30N
E = 2 x 10^11 Nm^-2 ( What you have provided is wrong and you will get an unrealistic answer with that)
Area of cross section (A) =
π×d²÷4 = 0.7855mm² = 7.855 10^{-3}
Let change in length = (ΔL)

E =  \frac{stress}{strain}
⇒ 2 x 10^11 =  \frac{F/A}{ΔL/L}
⇒2 x 10^11 =  [30/(7.855 X 10^-3)]/[ΔL/2.5 X 10^-3]
⇒2 x 10^11 = (30 X 2.5 X 10^-3) / ((7.855 X 10^-3) X ΔL)
⇒ΔL = (30 X 2.5 X 10^-3) / ((7.855 X 10^-3) X (2 x 10^11))
⇒ΔL = 4.77 X 10^-11 m
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