To develop this problem it is necessary to apply the concepts related to the definition of absolute, manometric and atmospheric pressure.
The pressure would be defined as
[tex]P = P_0 + P_w +P_m[/tex]
Where,
[tex]P_0 =[/tex] Standard atmosphere pressure
[tex]P_W =[/tex]Pressure of Water
[tex]P_m =[/tex]Pressure of mercury
Therefore we have that
[tex]P = P_0 \rho_w g \frac{h}{2} + \rho_m g\frac{h}{2}[/tex]
Here g is the gravity, \rho the density at normal conditions and h the height of the cylinder.
Replacing with our values we have that,
[tex]P = 1.013*10^5+10^3*9.81*\frac{0.25}{2}+13.6*10^3*9.81*\frac{0.25}{2}[/tex]
[tex]P = 1.21*10^5Pa \approx 0.121Mpa[/tex]
Therefore the pressure at the Bottom of a graduated cylinder is 0.121Mpa
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
A graduated cylinder is half full of mercury and half full of water. If the height of the cylinder is 0.25 m, the height of each column of liquid is:
[tex]\frac{0.25m}{2} = 0.13 m[/tex]
We can calculate the pressure at the bottom of the cylinder (P) using the following expression.
[tex]P = P_{atm} + P_{Hg} + P_w[/tex]
where,
We can calculate the pressure exerted by each liquid (P) as:
[tex]P = \rho \times g \times h[/tex]
where,
The pressure exerted by the column of Hg is:
[tex]P_{Hg} = \frac{13.6 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 17 \times 10^{3} Pa[/tex]
The pressure exerted by the column of water is:
[tex]P_{w} = \frac{1.00 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 1.3 \times 10^{3} Pa[/tex]
The pressure at the bottom of the cylinder is:
[tex]P = P_{atm} + P_{Hg} + P_w = 101325 Pa + 17 \times 10^{3} Pa + 1.3 \times 10^{3} Pa = 1.2 \times 10^{5} Pa[/tex]
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
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