Question

A copper wire of length 3m is subjected to a force of 5N. Calculate the elongation produced if the young’s modulus of copper is 12X1010 N/m2, if the area of cross-section of the wire is 0.785X10-6m

Answer 1

$\sigma=E\epsilon \to\epsilon=\sigma/E=F/(AE)=$

$=5/(0.785\cdot10^{-6}\cdot 12\cdot10^{10})=5.3\cdot10^{-5}(m)$