Question
A test tube standing vertically in a test-tube rack contains $2.5 \mathrm{cm}$ of oil $\left(\rho=0.81 \mathrm{g} / \mathrm{cm}^{3}\right)$ and $6.5 \mathrm{cm}$of water. What is the pressure exerted by the two liquids on the bottom of the test tube?
Step 1
The pressure exerted by a liquid column is given by the formula $P=\rho g h$, where $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the height of the liquid column. Show more…
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A vertical tube open at the top contains $5.0 \mathrm{cm}$ of oil with density $0.82 \mathrm{g} / \mathrm{cm}^{3},$ floating on $5.0 \mathrm{cm}$ of water. Find the gauge pressure at the bottom of the tube.
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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