Question

What is the dimensional formula of dynamic viscosity?
A. [$M^1{L}^{-1}{T}^{-2}$]
B. [$M^1{L}^1{T}^{-1}$]
C. [$M^1{L}^{-1}{T}^{-1}$]
D. [$M^1{L}^1{T}^1$]

Answer 1

Hint: You can always derive dimension by using formula. Split formula until end. Use basic algebra. Use the definition of Newton’s law of viscous force. Coefficient of viscosity will give dynamic viscosity.

Complete step by step solutions:
Take an example of a pipe through which water is coming out. If you will observe then velocity at the near surface of the pipe is less and as we move away from the surface the velocity is maximised. The velocity at centre is maximum. So we can assume that there must be a layer which is acting there forces on each other.

Newton’s law of viscous law:
According to this law, force exerted on a faster layer by an adjacent slower layer is proportional to 1) area of layer and 2) velocity gradient, perpendicular to direction of flow and it is given by,
$\begin{array}{rl}& F\propto A.{\displaystyle \frac{dv}{dy}}\\ & F=-\eta .A.{\displaystyle \frac{dv}{dy}}\end{array}$
Where, $\eta$= dynamic viscosity
     v = velocity of fluid
     A = Area of layer
$\begin{array}{rl}& \eta ={\displaystyle \frac{F}{A.{\displaystyle \frac{dv}{dy}}}}\\ & \end{array}$
Dimension of force is [$ML{T}^{-2}$]
Dimension of area (A) is [$L^2$]
Dimension of v is [$L{T}^{-1}$]
Dimension of y is [$L^{-1}$]
Therefore dimension of equation of $\eta$ is,
[$\eta$] = ${\displaystyle \frac{[ML{T}^{-2}]}{[L^2.L{T}^{-1}.L^{-1}]}}$
L gets canceled in the denominator and the inverse of T gets shifted in the numerator. Final dimension is,
[$\eta$] = $[M{L}^{-1}{T}^{-1}]$
Dimension of dynamic viscosity is $[M{L}^{-1}{T}^{-1}]$.
Option c) is correct.

Note: Use examples to keep concepts in mind. Like in this particular problem you can remember the pipe through which water is flowing as an example to understand the concept of dynamic viscosity. Just remember one thing either formula or definition, from this you can easily derive dimension.