Question

A layer of oil 1.50 mm thick is placed between two microscope slides. researchers find that a force of 5.50Ã10â4 n is required to glide one over the other at a speed of 1.00 cm/s when their contact area is 6.00 cm2. what is the oil's viscosity? what type of oil might it be?

Answer 1

The viscosity of the oil is calculated using Newton's law of viscosity. By rearranging the formula and plugging in the provided values of force, contact area, velocity, and thickness of the layer, we can calculate the viscosity and compare it to known standards to identify the type of oil.

Step-by-step explanation:

To determine the oil's viscosity, a formula derived from Newton's law of viscosity can be used, which states the force (F) required to move a layer of fluid over another at a constant velocity (v) is proportional to the contact area (A) and the velocity gradient (Δv/Δx), and inversely proportional to the thickness of the fluid layer (d). The viscosity (η) can be given by the following equation:

F = η(AΔv/Δx)

Here,

The oil's viscosity (η) can be calculated by rearranging the formula:

η = Fd/(A×v)

Where d represents the thickness of the oil layer. By inserting the given values:

η = (5.50 x 10⁻⁴ N) x (1.50 x 10⁻⁳ m) / (6.00 x 10⁻⁴ m²× 1.00 x 10⁻⁲ m/s)

Calculating this gives us the viscosity of the oil, which can be compared to known values to speculate on the type of oil it might be.

Answer 2

Force to slide the two surface over each other is given as

`F = \eta A (dv)/(dx)`

as we know that

`F = 5.50 * 10^(-4) N`

`A = 6 * 10^(-4)`

`(dv)/(dx) = (1* 10^(-2))/(1.50 * 10^(-3))`

now plug in all data in above formula

`5.50 * 10^(-4) = \eta 6 * 10^(-4) (1 * 10^(-4))/(1.50 * 10^(-3))`

`\eta = 0.1375 Pa s`

This is approximate range of Motor Oil