A steel wire is 4.0m long and 2mm in diameter. How much is it elongated by a suspended body of mass 20 kg? Young’s modulus for steel is 1,96,000 Mpa.

Let $\Delta L$  be the elongation. Then by Hooke’s law,

$\frac{F}{A}=Y\frac{\Delta L}{L}$  where Y is Young’s modulus.

The elongation is $\Delta L=\frac{1}{Y}\frac{F}{A}L=\frac{mgL}{YA}$

$=\frac{\left(20\right)\left(9.8\right)\left(4.0\right)}{\left(196\times {10}^9\right)\pi {\left(0.001\right)}^2}$

$=1.273\times {10}^{-3}m=1.273mm$