Question

the shear stress developed in alubricating oil, of viscosity 9.81 poise, filled between two parallel plates 1 cm apart and moving with relative velocity of 2 m/s is​

Answer 1

Answer:

Explanation:

Showing results for the shear stress developed in lubricating oil, of viscosity 9.81 poise, filled between two parallel plates 1 cm apart and moving with relative velocity of 2 m/s is​]

Search instead for the shear stress developed in alubricating oil, of viscosity 9.81 poise, filled between two parallel plates 1 cm apart and moving with relative velocity of 2 m/s is​]

Answer 2

Given: A lubricating oil of viscosity 9.81 poise is filled between two parallel plates 1 cm apart and moving with relative velocity of 2 m/ss

To find: Shear stress developed

Explanation: Shear stress is the stress that acts co-planar with cross section of the material.

Let the viscosity be n, distance between the plates be dz and relative velocity be du.

du= 2m/s

dz= 1 cm

= 0.01 m ( 1 m = 100 cm)

n= 9.81 poise

= 0.981 Pa s ( 1 poise= 0.1 Pa s)

The formula for the shear stress developed in the lubricating oil is given by the formula:

Shear stress

= n × du / dz

= 0.981 × 2 / 0.01

= 196.2 Pa

Therefore, the shear stress developed in the lubricating oil is 196.2 Pascal (Pa).