Question

A steel rod 2 cm in diameter is found to stretch 0.01 cm when subjected to a tensile force of 5000 N. What is its original length?

Answer 1

$\frac{F}{A}=E\frac{\Delta l}{l}$

$A=\pi d^2/4=3.14\cdot0.02^2/4=0.000314\ (m^2)$

$l=\frac{\Delta l\cdot E\cdot A}{F}=\frac{0.0001\cdot210\cdot10^9\cdot0.000314}{5000}=1.3188\ (m)$ . Answer