Which of the following is correct?

Explanation:

According to Newton's law of viscosity, shear stress is directly proportional to the rate of angular deformation (shear strain) or velocity gradient across the flow.

\(τ\propto\frac{du}{dy}\)

\(τ=μ\frac{du}{dy}\)

where, τ = shear stress, μ = absolute or dynamic viscosity, du/dy = velocity gradient ⇒ dα/dt = rate of angular deformation (shear-strain).

Units of Dynamic viscosity:

\(\mu=\frac{\tau}{\frac{du}{dy}}\Rightarrow\frac{Shear\;stress}{\frac{Change\;in\;velocity}{Change\;in\;distance}}\)

\(\mu=\frac{\frac{Force}{Area}}{\frac{Length}{Time}\times\frac{1}{Length}}\Rightarrow\frac{Force\;\times\;Time}{(Length)^2}\)

\(\mu=\frac{Ns}{m^2}\)

∴ the unit of dynamic viscosity in the SI unit is Ns/m2 or Pa-s.

The dimension of dynamic viscosity is ML-1T-1.

1 Pa-s = 10 Poise

∴ 1 Poise = 0.1 Pa-s or Ns/m2

Kinematic viscosity: 

It is defined as the ratio between the dynamic viscosity and density of the fluid.

\(\nu=\frac{\mu}{\rho}\)

Units of kinematic viscosity:

\(\nu=\frac{\mu}{\rho}\Rightarrow\frac{\frac{Force\;\times\;Time}{(Length)^2}}{\frac{Mass}{(Length)^3}}=\frac{Mass\times\frac{Length}{(Time)^2}\times{Time}}{\frac{Mass}{Length}}\)

\(∴\nu=\frac{(Length)^2}{Time}\)

∴ the SI unit of kinematic viscosity is m2/s and the CGS unit of kinematic viscosity is cm2/s or 'Stoke'.

1 Stoke = 10-4 m2/s = 0.0001 m2/s.

The dimensional formula for Kinematic viscosity (ν) is ν = L2T-1.