Answer:
Approximately [tex]10.3\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex] and that the density of water is [tex]10^{3}\; {\rm kg \cdot m^{-3}}[/tex].)
Explanation:
The standard atmospheric pressure is [tex]101.325 \; {\rm kPa}[/tex].
Convert the unit of standard atmospheric pressure to standard units:
[tex]\begin{aligned}& 101.325 \; {\rm kPa} \\ =\; & 101.325 \; {\rm kPa} \times \frac{10^{3}\; {\rm Pa}}{1\; {\rm kPa}} \\ =\; & 1.01325 \times 10^{5}\; {\rm Pa}\end{aligned}[/tex].
Since [tex]1\; {\rm Pa} = 1\; {\rm N \cdot m^{-2}}[/tex] ([tex]1\; {\rm N}[/tex] of normal force exerted over an area of [tex]1\; {\rm m^{2}}[/tex],) [tex]1.01325 \times 10^{5}\; {\rm Pa}[/tex] would be equivalent to [tex]1.01325 \times 10^{5}\; {\rm N \cdot m^{-2}}[/tex].
Let [tex]g[/tex] denote the gravitational field strength. The pressure of a column of liquid of density [tex]\rho[/tex] and height [tex]h[/tex] would be [tex]P = \rho\, g \, h[/tex].
In a liquid barometer, the density at the base of the liquid column should be equal to the pressure outside the barometer.
Assume that the barometer in this question is set up properly. The pressure at the base of the water column would be equal to the atmospheric pressure, [tex]1.01325 \times 10^{5}\; {\rm N \cdot m^{-2}}[/tex]. Thus, [tex]P = 1.01325 \times 10^{5}\; {\rm N \cdot m^{-2}}[/tex].
The density of water is [tex]\rho = 10^{3}\; {\rm kg \cdot m^{-3}}[/tex]. The gravitational field strength [tex]g[/tex] near the surface of the earth is approximately [tex]9.81\; {\rm N \cdot kg^{-1}}[/tex]. Rearrange the equation [tex]P = \rho\, g \, h[/tex] to find the value of height [tex]h[/tex]:
[tex]\begin{aligned}h &= \frac{P}{\rho\, g} \\ &= \frac{1.01325 \times 10^{5}\; {\rm N \cdot m^{-2}}}{10^{3}\; {\rm kg \cdot m^{-3}} \times 9.81\; {\rm N \cdot kg^{-1}}} \\ &\approx 10.3\; {\rm m}\end{aligned}[/tex].
