Question

A steel wire has a cross sectional area 5×10^-5m^2.calculate the amount of tensile force to be applied to double it's length.Ysteel=2×10^11Nm^-2.

Answer 1

Elongation (e) = L

Y = FL / Ae

F = YAe / L
= (2 * 10^11 * 5 * 10^-5 * L) / L
= 10^7 N
= 10 MN [where MN is mega Newton]

Tensile Force to be applied is 10 meganewton.