Answer
$1.6\times 10^{-5}\space m$
Work Step by Step
Let's apply equation 10.17 $\Delta L=\frac{FL_{0}}{YA}$ for both types of forces.
$\Delta L_{T}=\frac{FL_{0}}{Y_{T}A}-(1)$ and $\Delta L_{C}=\frac{FL_{0}}{Y_{C}A}-(2)$
$(1)\div(2)=>$
$\frac{\Delta L_{T}}{\Delta L_{C}}=\frac{\frac{FL_{0}}{Y_{T}A}}{\frac{FL_{0}}{Y_{C}A}}=\frac{Y_{C}}{Y_{T}}$
${\Delta L_{T}}=(\frac{Y_{C}}{Y_{T}})\Delta L_{C}$ ; Let's plug known values into this equation.
$\Delta L_{T}=(\frac{9.4\times10^{9}\space N/m^{2}}{1.6\times10^{10}N/m^{2}})(2.7\times10^{-5}\space m)=1.6\times10^{-5}m$
