Answer
$2.3\times10^{-6}m$
Work Step by Step
Let's apply equation 10.18 $\Delta X=\frac{FL_{0}}{SA}$ to find the shear deformation of the disk.
$\Delta X=\frac{FL_{0}}{SA}=\frac{FL_{0}}{S\pi r^{2}}$ ; Let's plug known values into this equation.
$\Delta X=\frac{(11\space N)(7\times10^{-3}m)}{(1.2\times10^{7}N/m^{2})\pi(3\times10^{-2}m)^{2}}\approx2.3\times10^{-6}m$
So, The shear deformation = $2.3\times10^{-6}m$
