Question

Consider an iron rod of length 1 metre and cross-section 1 $c{m}^2$ with a Young's modulus of ${10}^{12}dynec{m}^{-2}$. We wish to calculate the force with which the two ends must be pulled to produce an elongation of 1 mm. It is equal to:

Answer 1

Given : $L==1m=100cm$ $A=1c{m}^2$ $Y={10}^{12}dyne/c{m}^2$ $\Delta L=1mm=0.1cm$

Using $Y=\frac{FL}{A\Delta L}$

Or ${10}^{12}=\frac{F\times 100}{1\times 0.1}$

$⟹F={10}^{9}dyne={10}^{4}N$