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

Answer:

$\mu=0.98\ Pa.s$

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, $F=1\ kgw=1\times 9.8=9.8\ N$

• velocity of plate, $du=1\ m.s^{-1}$

Using Newton's law of viscosity:

$\tau=\mu.\frac{du}{dy}$ ..........................................(1)

where:

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

$\mu=$ coefficient of (dynamic) viscosity

Now, shear force:

$\tau=\frac{shear\ force}{area}$

$\tau=\frac{9.8}{0.2\times0.2} \ Pa$

Putting respective values in eq.(1)

$\frac{9.8}{0.2\times0.2}=\mu\times\frac{1}{0.004}$

$\mu=0.98\ Pa.s$