Question

A metal plate 5 cm × 5 cm rests on a layer of castor oil 1 mm thick whose coefficient of viscosity is 1.55 Nsm^-2. Find the horizontal force required to move the plate with a speed of 2 cms^-1
1. 0.0066 N
2. 0.8745 N
3. 0.0775 N
4. 0.0055 N

Answer 1

Correct Answer - Option 3 : 0.0775 N

Option(3)

CONCEPT:

• Viscosity is the property of fluid by virtue of which an internal force of friction comes into play when a fluid is in motion and which opposes the relative motion between its different layers.
• The coefficient of viscosity of liquids is analogous to the Modulus of rigidity of solids

EXPLANATION:

• Newton's formula for the viscous force between two parallel layers is given by

 
\(F =η A \frac {dv}{dx}\)

Here A = 25 cm² = 25 × 10^-4 m² ,dx = 1 mm =10^-3 m , η =1.55 Nsm^-2 ,dv =2 × 10^-2 ms^-1

On putting the value of the above we get 

\(F =η A \frac {dv}{dx}\) = 1.55 × 25 × 10^-4 × 2 × 10^-2 / 1 × 10^-3 = 0.0775 N

The horizontal force required to move the plate with the speed of 2 cms-1  is  0.0775 N

•  The coefficient of viscosity of a fluid can be defined as the ratio of shearing stress to the strain rate.

Coefficient of Viscosity.

η =\(\frac{F/A}{v/x}\) where v/x is dx.x/t ( velocity = dx/dt)=\(\frac{ Shearing stress}{Shear Strain/t} \)

• Shearing Stress is the internal restoring force set up per unit area of cross-section of the deformed body is called stress.
  • Shearing Strain the ratio of change in length to the original length.