Answer
(a) 126300 Pa (b) 19.4 m
Work Step by Step
(a) Let's apply equation 11.4 $P_{2}=P_{1}+h\rho g$ to find the gauge pressure at the second-floor faucet.
$P_{2}=P_{1}+h\rho g$ ; Let's plug known values into this equation.
$1.9\times10^{5}Pa=P_{1}+6.5\space m\times 1000\space kg/m^{3}\times 9.8\space m/s^{2}$
$P_{1}=126300\space Pa$
(b) Let's apply equation 11.4 $P_{2}=P_{1}+h\rho g$ to find the height of the second faucet above the first faucet.
$P_{2}=P_{1}+h\rho g$
$P_{2}=0+h\rho g$
$h=\frac{P_{2}}{\rho g}$ ; Let's plug known values into this equation.
$h=\frac{1.9\times10^{5}Pa}{1000\space kg/m^{3}\times 9.8\space m/s^{2}}=19.4\space m$
