Answer
$1.29mm$
Work Step by Step
We know that
$\Delta L=\frac{FL}{AY}$
$\implies \Delta L=\frac{\frac{1}{2}mgL}{\frac{\pi D^2}{4Y}}$
$\implies \Delta L=\frac{2mgL}{\pi D^2Y}$
We plug in the known values to obtain:
$\Delta L=\frac{2(66)(9.8)(0.005)}{\pi(0.04)^2(1)(10^6)}$
This simplifies to:
$\Delta L=1.29mm$
