Answer
$4.6\times10^{-4}$
Work Step by Step
From equation 10.17, we can get,
$\frac{\Delta L}{L_{0}}=\frac{F}{YA}-(1)$
From equation 5.3, we can get,
$F=\frac{mv^{2}}{R}-(2)$
(2)=>(1),
$\frac{\Delta L}{L_{0}}=\frac{mv^{2}}{RY(\pi r^{2})}$ ; Let's plug known values into this equation.
$\frac{\Delta L}{L_{0}}=\frac{(8\space kg)(12\space m/s)^{2}}{(4\space m)(2\times10^{11}Pa)\pi(0.001\space m)^{2}}\approx4.6\times10^{-4}$
The value of Y for steel is taken from table 10.1
