Answer
(a) $8.567Km$
(b) $0.995Kg/m^3;82.96\%$
Work Step by Step
(a) We can determine the required value of $z_{\circ}$ as follows:
$z_{\circ}=\frac{RT}{Mg}$
We plug in the known values to obtain:
$z_{\circ}=\frac{8.31\times 293}{29\times 10^{-3}\times 9.8}$
$\implies z_{\circ}=8567m=8.567Km$
(b) We know that
$\rho=\rho_{\circ}exp(\frac{-z}{z_{\circ}})$
We plug in the known values to obtain:
$\rho=1.2exp(-\frac{1.6}{8.567})$
$\implies \rho=0.995Kg/m^3$
Now $N=\frac{\rho}{\rho_{\circ}}$
We plug in the known values to obtain:
$N=\frac{0.995}{1.2}\times 100\%$
$\implies N=82.96\%$
