Answer
(a) $2.5\times10^{-4}$
(b) $7.5\times10^{-5}m$
Work Step by Step
(a) Here we use the equation $Strain=\frac{\Delta L}{L_{0}}=\frac{F}{YA}$ to find the strain.
$Strain=\frac{F}{YA}$
$Strain=\frac{mg}{Y\pi (r_{out}^{2}-r_{in}^{2})}$ ; Let's plug known values into this equation.
$Strain=\frac{(63\space kg)(9.8\space m/s^{2})}{(9.4\times10^{9}Pa)\pi[(1\times10^{-2})^{2}-(4\times10^{-3})^{2}]m^{2}}\approx2.5\times10^{-4}$
(b) We can write, $\Delta L=Strain\times L_{0}=(2.5\times10^{-4})(0.3\space m)=7.5\times10^{-5}m$
