Question

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

A) 9.17 Pa.s Olive oil
B) 1.83 Pa.s Motor oil
C) 5.50 Pa.s Vegetable oil
D) 3.66 Pa.s Corn oil

Answer 1

The viscosity of the oil is calculated to be 1.375 Pas, which is closest to motor oil with a viscosity of 1.83 Pas, as given in option B.

Step-by-step explanation:

To find the oil's viscosity, we use the formula for viscous force in a situation similar to this:
F = ηA(v/d), where F is the force, A is the area of contact, v is the velocity, d is the distance between the slides, and η (eta) is the viscosity.

Given:
F = 5.50 x 10-4 N
A = 6.00 cm2 = 6.00 x 10-4 m2
v = 1.00 cm/s = 1.00 x 10-2 m/s
d = 1.50 mm = 1.50 x 10-3 m
Plugging these values into the formula, we get:

5.50 x 10-4 N = η(6.00 x 10-4 m2)(1.00 x 10-2 m/s) / (1.50 x 10-3 m)

After rearranging the equation to solve for η, we have:

η = (5.50 x 10-4 N)(1.50 x 10-3 m) / (6.00 x 10-4 m2 x 1.00 x 10-2 m/s)
η = 1.375 Pas

The calculated viscosity does not match any options provided directly, but it is closest to B) 1.83 Pas (Motor oil), considering that the numbers provided in the options might be rounded or approximate.