In a multi-story building, the pressure of water on the ground floor is $6\times {10}^4$ Pa, and the pressure of water on the second floor is $1\times {10}^4$ Pa. Given that the density of water is $1000 {\text{kg/m}}^3$ and the acceleration due to gravity g = 10 m/s², determine the height (in meters) of the second floor above the ground floor.

Given:
  - Pressure on the ground floor, ${\mathrm{P}}_1=6\times {10}^4 \text{Pa}$
  - Pressure on the second floor, ${\mathrm{P}}_2=1\times {10}^4 \text{Pa}$
  - Density of water, $\mathrm{\rho }=1000 {\text{kg/m}}^3$
  - Acceleration due to gravity, $\mathrm{g}=10 {\text{m/s}}^2$

Pressure difference between the ground floor and the second floor:
$∆\mathrm{P}={\mathrm{P}}_1-{\mathrm{P}}_2=(6\times {10}^4 \text{Pa})-(1\times {10}^4 \text{Pa})=5\times {10}^4 \text{Pa}$

Relationship between pressure difference and height:
  $\mathrm{\Delta P}=\mathrm{\rho gh}$, where h is the height between the ground floor and the second floor.

$$\begin{gathered} \mathrm{h}=\frac{\mathrm{\Delta P}}{\mathrm{\rho g}}=\frac{5\times {10}^4 \text{Pa}}{1000 {\text{kg/m}}^3\times 10 {\text{m/s}}^2} \\ \mathrm{h}=\frac{5\times {10}^4}{{10}^4}=5 \text{m} \end{gathered}$$

The height of the second floor is 5 meters.