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Chapter 15: Problem 26

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.

Short Answer

Expert verified
The gauge pressure at the bottom of the tube is \(891.8 \mathrm{Pa}\).

Step by step solution

01

Find the pressure of the oil layer.

Starting with the oil layer, which is at the top, we will use the formula for pressure in a fluid column, which is \(P = \rho g h\), where \(P\) is the pressure, \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(h\) is the height of the fluid column. Here, \(\rho = 0.82 \mathrm{g/cm^3} = 820 \mathrm{kg/m^3}\) (we convert to kg/m鲁 because the unit for acceleration due to gravity is in m/s虏), \(g = 9.8 \mathrm{m/s^2}\), and \(h = 5.0 cm = 0.05 m\). So the pressure due to the oil is \(P_o = \rho g h = 820 \mathrm{kg/m^3} * 9.8 \mathrm{m/s^2} * 0.05 m = 401.8 \mathrm{Pa}\).
02

Find the pressure of the water layer.

Next is the water layer below the oil. Using the same pressure formula, with \(\rho = 1.00 \mathrm{g/cm^3} = 1000 \mathrm{kg/m^3}\), \(g = 9.8 \mathrm{m/s^2}\), and \(h = 5.0 cm = 0.05 m\), we find the pressure due to the water is \(P_w = \rho g h = 1000 \mathrm{kg/m^3} * 9.8 \mathrm{m/s^2} * 0.05 m = 490 \mathrm{Pa}\).
03

Find the total gauge pressure.

The total gauge pressure is the sum of the pressures due to the oil and water. So the total pressure is \(P = P_o + P_w = 401.8 \mathrm{Pa} + 490 \mathrm{Pa} = 891.8 \mathrm{Pa}\).

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