Question

A rod $100cm$ long and of $2cm\times 2cm$ cross-section is subjected to a pull of $1000kg$ force. Modulus of elasticity of the material is $2.0\times {10}^{6}kg/c{m}^2$. If the elongation of the rod is $x$ mm, find the value of $40x$.

Answer 1

Youmng's modulus $=\frac{longitudinalstress}{longitudinalstrain}$
i.e, $Y=\frac{F/A}{\Delta l/l}$
$\Delta l=\frac{Fl}{AY}=\frac{(1000\times {10}^3)\left(100\right)}{(2\times 2)(2\times {10}^6\times {10}^3)}=0.0125$ cm