Answer
a) $(4.506\times 10^{-6})$%
b) $9.012\times 10^{-18}m$
Work Step by Step
a) The fractional decrease in the rod's length is the strain. Use Eq. 9-4. The force applied is the weight of the man.
$$\frac{\Delta l}{l_0}=\frac{F}{AE}=\frac{mg}{\pi r^2E}=\frac{(65kg)(9.80m/s^2)}{\pi(0.15)^2(200\times 10^9)N/m^2}=4.506\times 10^{-8}$$
$=(4.506\times 10^{-6})$%
b) The fractional change is the same for the atoms as for the macroscopic material. Let $d$ represent the interatomic spacing.
$$\frac{\Delta d}{d_0}=\frac{\Delta l}{l_0}=4.506\times10^{-8}$$
$$\Delta d=(4.506\times10^{-8})d_0=(4.506\times10^{-8})(2.0\times10^{-10}m)$$
$$=9.012\times 10^{-18}m$$