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A layer of oil \(1.50 \mathrm{~mm}\) thick is placed between two microscope slides. Researchers find that a force of \(5.50 \times 10^{-4} \mathrm{~N}\) is required to glide one over the other at a speed of \(1.00 \mathrm{~cm} / \mathrm{s}\) when their contact area is \(6.00 \mathrm{~cm}^{2}\). What is the oil's viscosity? What type of oil might it be?

Short Answer

Expert verified
The oil's viscosity is approximately \(1.375 \,\mathrm{Pa \cdot s}\). Based on this value, it might be a light lubricating oil.

Step by step solution

01

Understand the Concept of Viscosity

Viscosity is a measure of a fluid's resistance to flow. It is described by the equation \( F = \eta \frac{A v}{d} \) where \(F\) is the force required to move the layers, \(A\) is the area of contact, \(v\) is the relative velocity, \(d\) is the separation distance between layers, and \(\eta\) is the fluid's viscosity.
02

Convert Units

Since the standard unit for viscosity is \( \frac{N \cdot s}{m^2} \), convert the thickness of the oil layer to meters and the speed to meters per second. \(1.50 \,\mathrm{mm} = 0.0015 \,\mathrm{m}\) and \(1.00 \,\mathrm{cm/s} = 0.01 \,\mathrm{m/s}\).
03

Calculate the Viscosity

Rearrange the viscosity equation and solve for \(\eta\). \[ \eta = \frac{F d}{A v} \] Substitute the given values \( F = 5.50 \times 10^{-4} \,\mathrm{N} \), \( d = 0.0015 \,\mathrm{m} \), \( A = 6.00 \,\mathrm{cm}^2 = 6.00 \times 10^{-4} \,\mathrm{m}^2 \), and \( v = 0.01 \,\mathrm{m/s} \) into the equation.
04

Perform the Calculation

Plug the values into the viscosity equation. \[ \eta = \frac{(5.50 \times 10^{-4} \,\mathrm{N}) \times 0.0015 \,\mathrm{m}}{(6.00 \times 10^{-4} \,\mathrm{m}^2)\times 0.01 \,\mathrm{m/s}} \] This yields \( \eta \approx 1.375 \,\mathrm{Pa \cdot s}\).
05

Identify the Oil Type

Compare the calculated viscosity with viscous properties of known oils to speculate the type of oil. A viscosity of \(1.375 \,\mathrm{Pa \cdot s}\) might suggest a relatively thin oil, possibly a light lubricating oil.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fluid Resistance to Flow
Understanding the concept of fluid resistance to flow paints a picture of how substances like water, oil, and honey behave when subjected to force. This resistance stems from the internal friction within the fluid and between the fluid and surfaces it interacts with. Imagine trying to push your hand through water as opposed to through honey; you'll notice that the honey offers more resistance, indicating it has a higher viscosity.

To improve comprehension in this area, students should consider real-world examples, such as the difference in resistance felt when stirring different liquids, or why certain oils make better lubricants because they provide a thin, low resistance layer, whereas others, due to their higher resistance, are better suited for use as dressings and toppings in culinary applications.
Viscosity Equation
The viscosity equation is foundational in calculating the viscosity of a fluid. It mathematically relates the force needed to move one layer of fluid over another to the area of contact, the velocity, and the separation distance. The equation is denoted as:
\[ F = \frac{et \times A \times v}{d} \]
Where F is the force, et is viscosity, A is the area, v is velocity, and d is the separation distance. The students should focus on how each factor influences the viscosity. For instance, a larger force or smaller area will increase the calculated viscosity, as well as how a greater velocity or thicker layer between slides reduces the fluid's measured resistance.
SI Units Conversion
The importance of converting to SI units, the International System of Units, cannot be overstated in scientific calculations. It ensures consistency and correct interpretation of results. For viscosity calculation, the force is in newtons (N), area in square meters (m²), velocity in meters per second (m/s), and separation distance in meters (m).

Providing additional example conversions midway in the lesson can reinforce learning. For example, convert a given velocity from kilometers per hour to meters per second or explain how to switch from inches to meters, as these are common conversions students might encounter in various exercises.
Viscous Properties of Oils
Oils can have a wide range of viscous properties, affecting their use in everyday life and industry. Light oils, like those used in machinery, have low viscosity and flow easily, providing a smooth, thin film between moving parts. In contrast, heavy oils, like motor oil, have high viscosity and are better at maintaining a thick protective layer.

Students can benefit from a tactile exercise like feeling the resistance by dipping their fingers into oils of different viscosities, which helps build an intuitive understanding of the concept. Additionally, introducing a chart of common oils with their respective viscosities can provide a reference point for identifying unknown oils in real-world applications.

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Most popular questions from this chapter

Based on Bernoulli's equation, what are three forms of energy in a fluid? (Note that these forms are conservative, unlike heat transfer and other dissipative forms not included in Bernoulli's equation.)

A rugby player passes the ball \(7.00 \mathrm{m}\) across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was \(12.0 \mathrm{m} /\) s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

Water emerges straight down from a faucet with a \(1.80-\mathrm{cm}\) diameter at a speed of \(0.500 \mathrm{~m} / \mathrm{s}\). (Because of the construction of the faucet, there is no variation in speed across the stream.) (a) What is the flow rate in \(\mathrm{cm}^{3} / \mathrm{s} ?\) (b) What is the diameter of the stream \(0.200 \mathrm{~m}\) below the faucet? Neglect any effects due to surface tension.

Water that has emerged from a hose into the atmosphere has a gauge pressure of zero. Why? When you put your hand in front of the emerging stream you feel a force, yet the water's gauge pressure is zero. Explain where the force comes from in terms of energy.

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