Correct Answer - Option 2 : [ML
-1T
-1]
Explanation:
From Newton’s law of viscosity
\(\tau = \mu \cdot \frac{{du}}{{dy}}\)
μ = Proportionality constant/ coefficient of viscosity or viscosity
μ = dynamic viscosity (since it involves force)
\(\therefore \mu = \frac{\tau }{{du/dy}}\)
Dimensional formula for μ
∴ μ = M.L-1.T-1
Units:
Dynamic Viscosity (μ):
SI system:
μ = N.s/m2 or Pa.s = kg/m.s
CGS system:
μ = Dyne.sec/cm2 = 1 poise
1 N = 105 dynes
1 N.s/m2 = 105 dynes.sec/104 cm2
1 N.s/m2 = 10 dyne.sec/cm2
1 N.s/m2 = 10 poise = 1 Pa.s
Now,
Kinematic Viscosity (v): It is the ratio of dynamic viscosity to mass density.
\(v = \frac{{Dynamic\;viscosity}}{{Mass\;density}} = \frac{\mu }{\rho }\)
Dimensional formula for Kinematic viscosity (ν)
∴ ν = L2T-1
The dimensions of the kinematic viscosity show that they involves the magnitudes of length and time only.
The name kinematic viscosity has been given to the ratio (μ/ρ) because its unit (m2/s) is similar to the unit of kinematic quantities like velocity (m/s) and acceleration (m/s2).
Units
SI system:
m2/sec
CGS system:
cm2/sec or stoke
1 stoke = 1 cm
2/sec = 10
-4 m
2/sec