Solved Problem on Fluid Mechanics

A glass cylinder contains, in the bottom mercury up to 10 cm in height and water to the height of 80 cm, measured from the base. The density of mercury and water are, respectively, 13.6 g/cm³ and 1 g/cm³. Determine the pressure at the bottom of the container.

Problem data:

Density of mercury: ρ_m = 13.6 g/cm³;
Heigh of mercury column: h_m = 10 cm;
Density of water: ρ_w = 1 g/cm³;
Height of water column from the base: h = 80 cm;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

Figure 1

Solution

First, we must convert the densities of water and mercury, given in grams per cubic centimeter (g/cm³) to kilograms per cubic meter (kg/m³), and the heights given in centimeters (cm) to meters (m) used in the International System of Units (S.I.)

\[ \begin{align} \rho_m=13.6\;\mathrm{\frac{\cancel g}{cm^3}}\times\mathrm{\frac{1\;kg}{1000\;\cancel{g}}}\times\mathrm{\frac{(100\;cm)^3}{(1\;m)^3}}=13.6\;\mathrm{\frac{1}{\cancel{cm^3}}}\times\mathrm{\frac{1\;kg}{1\cancel{000}}}\times\mathrm{\frac{1 000 \cancel{000}\;\cancel{cm^3}}{1\;m^3}} =13600\;\mathrm{\frac{kg}{m^3}} \end{align} \]

\[ \begin{gather} \rho_a=1\;\mathrm{\frac{\cancel g}{cm^3}}\times\mathrm{\frac{1\;kg}{1000\;\cancel{g}}}\times\mathrm{\frac{(100\;cm)^3}{(1\;m)^3}}=1\;\mathrm{\frac{1}{\cancel{cm^3}}}\times\mathrm{\frac{1\;kg}{1\cancel{000}}}\times\mathrm{\frac{1 000 \cancel{000}\;\cancel{cm^3}}{1\;m^3}} =1000\;\mathrm{\frac{kg}{m^3}} \end{gather} \]

\[ \begin{gather} h_m=10\;\mathrm{\cancel{cm}}\times \mathrm{\frac{1\;m}{100\;\cancel{cm}}}=0,1\;\mathrm m \end{gather} \]

\[ \begin{gather} h=80\;\mathrm{\cancel{cm}}\times \mathrm{\frac{1\;m}{100\;\cancel{cm}}}=0,8\;\mathrm m \end{gather} \]

The pressure of a liquid column, p_c, is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {p_c=\rho gh} \tag{I} \end{gather} \]

For mercury p_m

\[ \begin{gather} p_m=\rho_m g h_m\\[5pt] p_m=\left(13600\;\mathrm{\frac{kg}{m^\cancelto{2}{3}}}\right)\times\left(9.8\;\mathrm{\frac{\cancel m}{s^2}}\right)\times\left(0.1\;\mathrm{\cancel m}\right)\\[5pt] p_m=13328\;\mathrm{Pa} \tag{II} \end{gather} \]

The height of the water,h_w, column will be the difference between the measured height from the base and the height of the mercury column

\[ \begin{gather} h_w=h-h_m \tag{III} \end{gather} \]

Applying the equation (I), we have the pressure of the column of water, p_w

\[ \begin{gather} p_w=\rho_w g h_w \tag{IV} \end{gather} \]

substituting equation (III) into equation (IV)

\[ \begin{gather} p_w=\rho_w g\;(h-h_m)\\[5pt] p_w=\left(1000\;\mathrm{\frac{kg}{m^\cancelto{2}{3}}}\right)\times\left(9.8\;\mathrm{\frac{\cancel m}{s^2}}\right)\times(0.8\;\mathrm m-0.1\;\mathrm m)\\[5pt] p_w=\left(9800\;\mathrm{\frac{kg}{m^{\cancel 2}.s^2}}\right)\times(0.7\;\mathrm{\cancel m})\\[5pt] p_w=6860\;\mathrm{Pa} \tag{V} \end{gather} \]

The total pressure on the bottom of the container is given by the sum of the equations (II) and (V)

\[ \begin{gather} p=p_m+p_w\\[5pt] p=13328\;\mathrm{Pa}+6860\;\mathrm{Pa} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {p=20188\;\mathrm{Pa}} \end{gather} \]

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