Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Ideal and real fluids, properties of fluids
Question Statement:
Which of the following correctly pairs the fundamental physical dimension and SI unit of dynamic viscosity ($\mu$) and kinematic viscosity ($\nu$) in fluid mechanics?
Options:
(A) Dynamic viscosity: $[M L^{-1} T^{-1}]$ and $\text{Pa}\cdot\text{s}$; Kinematic viscosity: $[L^2 T^{-1}]$ and $\text{m}^2/\text{s}$
(B) Dynamic viscosity: $[M L^{-2} T^{-1}]$ and $\text{N}/\text{m}^2$; Kinematic viscosity: $[L T^{-2}]$ and $\text{m/s}^2$
(C) Dynamic viscosity: $[M L T^{-2}]$ and $\text{N}$; Kinematic viscosity: $[M L^{-3}]$ and $\text{kg/m}^3$
(D) Dynamic viscosity: $[L^2 T^{-1}]$ and $\text{m}^2/\text{s}$; Kinematic viscosity: $[M L^{-1} T^{-1}]$ and $\text{Pa}\cdot\text{s}$
Answer Key:
A
Step-by-Step Solution & Explanation
From Newton's law of viscosity $\tau = \mu \frac{du}{dy}$:
$\mu = \frac{\tau}{du/dy} = \frac{\text{N/m}^2}{(\text{m/s})/\text{m}} = \text{N}\cdot\text{s/m}^2 = \text{Pa}\cdot\text{s} \implies [M L^{-1} T^{-1}]$
Kinematic viscosity is defined as ratio of dynamic viscosity to density:
$\nu = \frac{\mu}{\rho} = \frac{\text{kg}/(\text{m}\cdot\text{s})}{\text{kg/m}^3} = \text{m}^2/\text{s} \implies [L^2 T^{-1}]$
Therefore, Option A is correct.