Answer
$1mm$
Work Step by Step
We know that
$\Delta L=\frac{FL}{AY}$
$\Delta L=\frac{\rho_w gVL}{4a^2 Y}$
We plug in the known values to obtain:
$\Delta L=\frac{1000(9.8)(10)(0.8)}{4(0.04)^2(10)^{10}}$
This simplifies to:
$\Delta L=1.225mm\approx 1mm$
