Answer
$2.7\times10^{-4}m$
Work Step by Step
Here both cylinders experience equal force. Let's apply equation 10.17 $F=(\frac{\Delta L}{L_{0}})YA$ to find the change in length of each cylinder.
The length of the brass cylinder decreases by,
$F=(\frac{\Delta L}{L_{0}})YA=>\Delta L_{B}=\frac{FL_{0}}{YA}$
$\Delta L_{B}=\frac{(6500\space N)(5\times10^{-2}m)}{(9\times10^{10}N/m^{2})\pi(0.25\times10^{-2}m)}\approx1.8\times10^{-4}m$
Similarly, the length of the copper cylinder decreases by,
$\Delta L_{C}=\frac{(6500\space N)(3\times10^{-2}m)}{(1.1\times10^{11}N/m^{2})\pi(0.25\times10^{-2}m)}\approx9\times10^{-5}m$
So, $=\Delta L_{B}+\Delta L_{C}\approx2.7\times10^{-4}m$
