Respuesta :
The total gauge pressure at the bottom of the cylinder would simply be the sum of the pressure exerted by water and pressure exerted by the oil.
The formula for calculating pressure in a column is:
P = ρ g h
Where,
P = gauge pressure
ρ = density of the liquid
g = gravitational acceleration
h = height of liquid
Adding the two pressures will give the total:
P total = (ρ g h)_water + (ρ g h)_oil
P total = (1000 kg / m^3) (9.8 m / s^2) (0.30 m) + (900 kg / m^3) (9.8 m / s^2) (0.4 - 0.30 m)
P total = 2940 Pa + 882 Pa
P total = 3,822 Pa
Answer:
The total gauge pressure at the bottom is 3,822 Pa.
The total gauge pressure at the bottom of cylinder due to the oil and the water is [tex]\boxed{3822\,{\text{Pa}}}[/tex].
Further Explanation:
Given:
The water in the cylinder is up to the height of 30 cm .
The total height of the liquid column in the cylinder is 40 cm .
The density of oil is [tex]900\,{{{\text{kg}}}\mathord{\left/{\vphantom {{{\text{kg}}} {{{\text{m}}^{\text{3}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{m}}^{\text{3}}}}}[/tex].
Concept:
The gauge pressure is the amount of pressure exerted by the liquid column on the surface below it.
The gauge pressure due to the height of the liquid pressure is given by:
[tex]P =\rho gh[/tex]
Here, [tex]P[/tex] is the gauge pressure, [tex]\rho[/tex] is the density of the liquid, g is the acceleration due to gravity and [tex]h[/tex] is the height of the liquid column.
The height of the oil present in the cylinder is:
[tex]\begin{aligned}{h_{oil}}&={h_{total}} - {h_{water}}\\&= 40 - 30\,{\text{cm}}\\&=10\,{\text{cm}}\\&\approx {\text{0}}{\text{.1}}\,{\text{m}}\\\end{aligned}[/tex]
The total gauge pressure at the bottom of the cylinder will be:
[tex]{P_{total}} = {\left({\rho gh}\right)_{water}} + {\left( {\rho gh}\right)_{oil}}[/tex]
Substitute the values in the above expression.
[tex]\begin{aligned}{P_{total}}&=\left({1000 \times 9.8 \times 0.30} \right) + \left( {900 \times 9.\times 0.1}\right)\\&=2940 + 882\\&=3822\,{\text{Pa}}\\\end{aligned}[/tex]
Thus, the total gauge pressure at the bottom of cylinder due to the oil and the water is [tex]\boxed{3822\,{\text{Pa}}}[/tex]
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Answer details:
Grade: High School
Subject: Physics
Chapter: Pressure
Keywords: Gauge pressure, water, oil, bottom of cylinder, 30cm of water, density of oil, 900kg/m^3, poured, floats on top, total liquid depth.
