# What extension is produced in a steel wire of 1mm in diameter and 2.5mm long by hanging a force of 30N from its end ? Take young mOdulus of the steel wire to be 2 x 10^-2 Nm^-2 and pi =3.142

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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

Let change in length = (ΔL)

E =

⇒ 2 x 10^11 =

⇒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

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

Let change in length = (ΔL)

E =

⇒ 2 x 10^11 =

⇒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