Which of the following is correct?

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NHPC JE Electrical 5 April 2022 (Shift 1) Official Paper
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  1. Dimension of ‘dynamic viscosity’ is given by MLT-2
  2. 1.0 Poise = 10.0 N-s/m2
  3. Dimension of ‘dynamic viscosity’ is given by ML-1T-1
  4. 1.0 stoke = 0.001 m2/s

Answer (Detailed Solution Below)

Option 3 : Dimension of ‘dynamic viscosity’ is given by ML-1T-1
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Detailed Solution

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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.

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:

∴ 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.

Units of kinematic viscosity:

∴ 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.

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