A vertical test tube has 2.0 cm of oil (ρ=0.80 g / cm^3)...

A vertical test tube has $2.0 \mathrm{~cm}$ of oil $\left(\rho=0.80 \mathrm{~g} / \mathrm{cm}^{3}\right)$ floating on $8.0 \mathrm{~cm}$ of water. What is the pressure at the bottom of the tube due to the liquid in it? $\begin{aligned} P &=\rho_{1} g h_{1}+\rho_{2} g h_{2}=\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.020 \mathrm{~m})+\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.080 \mathrm{~m}) \\ &=0.94 \mathrm{kPa} \end{aligned}$

A vertical test tube has $2.0 \mathrm{~cm}$ of oil $\left(\rho=0.80 \mathrm{~g} / \mathrm{cm}^{3}\right)$ floating on $8.0 \mathrm{~cm}$ of water. What is the pressure at the bottom of the tube due to the liquid in it?
$\begin{aligned} P &=\rho_{1} g h_{1}+\rho_{2} g h_{2}=\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.020 \mathrm{~m})+\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.080 \mathrm{~m}) \\ &=0.94 \mathrm{kPa} \end{aligned}$

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Key Concepts

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the gravitational force acting on it. This pressure is directly proportional to the density of the fluid, the gravitational acceleration, and the height of the fluid column, making it a fundamental concept in fluid mechanics.

Density

Density is the mass per unit volume of a substance and plays a key role in determining the pressure within fluids. In problems involving fluids, different layers may have different densities, which directly affect the pressure calculated at various depths.

Multi-Layer Fluids

When dealing with immiscible fluids layered one on top of the other, the overall pressure at a given depth is the sum of the pressures exerted by each of the fluid layers above that point. Each individual layer contributes pressure proportional to its own density and thickness, illustrating the additive nature of pressures in a stratified fluid system.

Unit Conversion and Consistency

Ensuring that all units are consistent is essential in physics problems involving pressure calculation. This involves converting measurements into a common system, such as meters for height when using SI units, ensuring that the computed pressure values are accurate and meaningful.

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