Chapter 9 - Solids and Fluids - Learning Path Questions and Exercises - Exercises - Page 349: 10

$3.8\times10^{-6}m$

$L=2m$ $A=15\times20\times10^{-4}m^{2}$ $Y=2.1\times10^{11}Pa$ $F=12\times10^{3}N$ So, from $\frac{F}{A}=Y\frac{\Delta L}{L}$ $\Delta L=\frac{FL}{AY}=\frac{12\times10^{3}\times2}{3\times10^{-2}\times2.1\times10^{11}}=3.8\times10^{-6}m$