Question

A square metal plate of 10 cm side moves parallel to another plate with a velocity of 10 cms^-1, both plates immersed in water. If the viscous force is 200 dyne and the viscosity of water is 0.01 poise, what is their distance apart?
1. 0.067 cm
2. 0.04cm
3. 0.05 cm
4. 0.06 cm

Answer 1

Correct Answer - Option 3 : 0.05 cm

Concept:

Newton law of viscosity:

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

\(\rm{F =η A \frac {dv}{dy}}\)

Calculation:

Given:

Here A = 100 cm2 , η = 0.01 poise ,dv = 10 cms^-1, dy =?

On putting the value of the above we get 

\(\rm{F =η A \frac {dv}{dy}}\) ⇒  \(\rm{dy=\frac{\eta Adv}{F}}=\rm{\frac {0.01 \;\times \;100 \;\times\; 10}{200}=0.05~ cm}\)

Hence, the distance will be 0.05 cm.

•  The coefficient of viscosity(η ) 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.dt ( 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.