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A measuring cylinder is filled with Kerosene to a height of 25cm. Calculate the pressure exerted by Kerosene at the bottom of the measuring cylinder
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Answer :

jacob193

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:

  • [tex]\rho[/tex] is the density of the liquid,
  • [tex]g[/tex] is the gravitational field strength, and
  • [tex]h[/tex] is the height of the liquid column.

Ensure that all quantities are measured in standard units:

  • Density should be in kilograms per cubic meter ([tex]{\rm kg \cdot m^{-3}}[/tex]): [tex]\begin{aligned} \rho &\approx 0.8 \; {\rm g \cdot cm^{-3}} \\ &= 0.8 \; {\rm g \cdot cm^{-3}} \, \left(\frac{(10^{2}\; {\rm cm})^{3}}{(1\; {\rm m})^{3}} \times \frac{1\; {\rm kg}}{10^{3}\; {\rm g}}\right) \\ &= 0.8 \times 10^{3}\; {\rm kg\cdot m^{-3}}\end{aligned}[/tex].
  • Height should be in meters:
    [tex]\begin{aligned} h &= 25\; {\rm cm} \\ &= 25\; {\rm cm}\, \left(\frac{1\; {\rm m}}{10^{2}\; {\rm cm}}\right) \\ &= 0.25\; {\rm m}\end{aligned}[/tex].

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].

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