Question

A flat square plate of side 20cm moves over other similar plate with a thin layer of 0.4cm of a liquid between them with force 1 Kgw moves of the plates uniformly with a velocity 1m/s. Calculate co-efficient of viscosity of the liquid.

Answer 1

`\mu=0.98\ Pa.s`

Step-by-step explanation:

Given:

• dimension of square plate,
  `l=0.2\ m`

• thickness of fluid layer,
  `dy=0.004\ m`

• force on the fluid due to plate,
  `F=1\ kgw=1* 9.8=9.8\ N`

• velocity of plate,
  `du=1\ m.s^(-1)`

Using Newton's law of viscosity:

`\tau=\mu.(du)/(dy)` ..........................................(1)

where:

`\tau=` shear force on the surface on the fluid

`\mu=` coefficient of (dynamic) viscosity

Now, shear force:

`\tau=(shear\ force)/(area)`

`\tau=(9.8)/(0.2*0.2) \ Pa`

Putting respective values in eq.(1)

`(9.8)/(0.2*0.2)=\mu*(1)/(0.004)`

`\mu=0.98\ Pa.s`