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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 \(892.84 \mathrm{Pa}\).

Step by step solution

01

Calculate the pressure due to the oil.

Start by finding the pressure at the bottom of the oil layer, using the formula: Pressure = Density * Gravity * Height. Here, the density of the oil is \(0.82 \mathrm{g} / \mathrm{cm}^3\), which is actually \(820 \mathrm{kg}/\mathrm{m}^3\) if we convert it into an SI unit. We are assuming the value of gravity to be \(9.81 \mathrm{m/s}^2\). The depth (height) of oil is given as \(5.0 \mathrm{cm}\), which is \(0.05 \mathrm{m}\). Therefore, Pressure (due to oil) = \(820 \times 9.81 \times 0.05 = 402.34 \mathrm{Pa}\)
02

Calculate the pressure due to the water.

Next, calculate the pressure at the bottom of the water layer. Use the same formula but note that the density of water is \(1 \mathrm{g}/\mathrm{cm}^3\) or \(1000 \mathrm{kg}/\mathrm{m}^3\) in SI units. The depth (height) of water is also \(0.05 \mathrm{m}\). Therefore, Pressure (due to water) = \(1000 x 9.81 x 0.05 = 490.5 \mathrm{Pa}\)
03

Calculate the total pressure.

The overall gauge pressure at the bottom of the tube is the sum of the pressures due to the oil and the water. Therefore, the total pressure = Pressure (due to oil) + Pressure (due to water) = \(402.34 \mathrm{Pa} + 490.5 \mathrm{Pa} = 892.84 \mathrm{Pa}\)

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Most popular questions from this chapter

Archimedes purportedly used his principle to verify that the king's crown was pure gold by weighing the crown submerged in water. Suppose the crown's actual weight was \(25.0 \mathrm{N}\). What would be its apparent weight if it were made of (a) pure gold and (b) \(75 \%\) gold and \(25 \%\) silver, by volume? The densities of gold, silver, and water are \(19.3 \mathrm{g} / \mathrm{cm}^{3}, 10.5 \mathrm{g} / \mathrm{cm}^{3},\) and \(1.00 \mathrm{g} / \mathrm{cm}^{3},\) respectively.

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