Question

In CGS system, the unit of kinematic viscosity is stoke, where 1 stoke =_____.
1. 10⁶ cm³/s
2. 10^-2 m²
3. 10⁴ cm³/s
4. 10^-4 m²/s

Answer 1

Correct Answer - Option 4 : 10^-4 m²/s

Explanation:

Dynamic viscosity (μ) : 

• It is the property of a fluid which offers resistance to the movement of one layer of fluid over adjacent layer of the fluid.

• Shear stress between two layers of fluid is directly proportional to rate of change of velocity with respect to perpendicular distance from fixed point.
  \({\rm{\tau \;\alpha \;}}\frac{{{\rm{du}}}}{{{\rm{dy}}}}\)
  \({\rm{\tau }} = {\rm{\mu \;}}\frac{{{\rm{du}}}}{{{\rm{dy}}}}\)
  \({\rm{\mu }} = \frac{{\rm{\tau }}}{{\frac{{{\rm{du}}}}{{{\rm{dy}}}}}}\)
  Where,  τ – Shear stress
    \(\frac{{{\rm{du}}}}{{{\rm{dy}}}}\)– Velocity gradient
• Unit – kgm-1s-1
• 1 poise = 0.1 Pa.s = 0.1 N.s/m2
• Dimensions - [ M1 L-1 T-1 ]

Kinematic Viscosity (ϒ):

• It is the ratio of dynamic viscosity to the density of fluid flowing
• \({\rm{Kinematic\;Viscosity\;}}\left( {\rm{\gamma }} \right) = {\rm{\;}}\frac{{{\rm{Dynamic\;viscosity\;}}\left( {\rm{\mu }} \right)}}{{{\rm{density\;}}\left( {\rm{\rho }} \right)}}\)
• Unit is – m2s-1
• 1 stoke = 0.0001 m2/s = 10^-4 m2/s
• Dimensions – [ M0 L2 T-1 ]