Answer
$2.1\times10^{-5}m$
Work Step by Step
Please see the attached image first.
Let's apply Newton's second law to the two masses.
$3\space kg=>$
$\uparrow F=ma$
$T-(3\space kg)g=(3\space kg)a-(1)$
$5\space kg=>$
$\uparrow F=ma$
$T-(5\space kg)g=-(5\space kg)a-(2)$
$(1)\times5+(2)\times3,$
$5T+3T-(15\space kg)g-(15\space kg)g=0$
$8T=30\space kg\times 9.8\space m/s^{2}$
$T=36.75\space N$
Now we apply the equation $F=\frac{YA\Delta L}{L_{0}}$ to find the change in the length of the wire.
$F=\frac{YA\Delta L}{L_{0}}=>\Delta L=\frac{FL_{0}\Delta L}{YA}$ ; Let's plug known values into this equation.
$\Delta L=\frac{(36.75\space N)(1.5\space m)}{(2\times10^{11}N/m^{2})(1.3\times10^{-5})}\approx2.1\times10^{-5}m$
