Answer:
Approximately [tex]2.0\times 10^{3}\; {\rm Pa}[/tex], assuming that [tex]\rho = 0.8 \; {\rm g \cdot cm^{-3}}[/tex] for kerosene and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
The pressure at the bottom of a liquid column is:
[tex]P = \rho\, g\, h[/tex],
Where:
Ensure that all quantities are measured in standard units:
Hence, the pressure at the bottom of this kerosene column would be:
[tex]\begin{aligned} P &= \rho\, g\, h \\ &\approx (0.8\times 10^{3}\; {\rm kg\cdot m^{-3}}) \times (9.81\; {\rm m\cdot s^{-2}}) \times (0.25\; {\rm m}) \\ &\approx 2.0 \times 10^{3}\; {\rm kg \cdot (m \cdot s^{-2}) \cdot m^{-2}} \\ &= 2.0 \times 10^{3}\; {\rm Pa}\end{aligned}[/tex].