A tall cylinder contains 30 cm of water. oil is carefully poured into the cylinder, where it floats on top of the water, until t

Question

2 years ago

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A tall cylinder contains 30 cm of water. oil is carefully poured into the cylinder, where it floats on top of the water, until t

he total liquid depth is 40 cm. part a what is the gauge pressure at the bottom of the cylinder? suppose that the density of oil is 900 kg/m3.

Physics

2 answers:

Art [367] · Answer 1 · 2022-08-15T03:44:12+03:00

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.

Georgia [21] · Answer 2 · 2022-08-15T06:08:07+03:00

The total gauge pressure at the bottom of cylinder due to the oil and the water is $\boxed{3822\,{\text{Pa}}}$ .

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 $900\,{{{\text{kg}}}\mathord{\left/{\vphantom {{{\text{kg}}} {{{\text{m}}^{\text{3}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{m}}^{\text{3}}}}}$ .

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:

$P =\rho gh$

Here, $P$ is the gauge pressure, $\rho$ is the density of the liquid, g is the acceleration due to gravity and $h$ is the height of the liquid column.

The height of the oil present in the cylinder is:

$\begin{aligned}{h_{oil}}&={h_{total}} - {h_{water}}\\&= 40 - 30\,{\text{cm}}\\&=10\,{\text{cm}}\\&\approx {\text{0}}{\text{.1}}\,{\text{m}}\\\end{aligned}$

The total gauge pressure at the bottom of the cylinder will be:

${P_{total}} = {\left({\rho gh}\right)_{water}} + {\left( {\rho gh}\right)_{oil}}$

Substitute the values in the above expression.

$\begin{aligned}{P_{total}}&=\left({1000 \times 9.8 \times 0.30} \right) + \left( {900 \times 9.\times 0.1}\right)\\&=2940 + 882\\&=3822\,{\text{Pa}}\\\end{aligned}$