Question

Dimensions of coefficient of viscosity is:
$\text{A}\text{.}\left[ M{{T}^{2}} \right]$
$\text{B}\text{.}\left[ M{{L}^{-3}}{{T}^{-4}} \right]$
$\text{C}\text{.}\left[ M{{L}^{-1}}{{T}^{-2}} \right]$
$\text{D}\text{.}\left[ M{{L}^{-1}}{{T}^{-1}} \right]$

Answer 1

Hint: Convert the derived physical quantities into fundamental physical quantities. Viscosity depends on the force area and the velocity gradient. So, we can find the dimensional formula of viscosity from the dimension of these three quantities.

Complete Step-by-Step solution:
All the derived physical quantities can be expressed in terms of the fundamental quantities. The derived units are dependent on the 7 fundamental quantities. Fundamental units are mutually independent of each other.
Dimension of a physical quantity is the power to which the fundamental quantities are raised to express that physical quantity.
Now, Viscosity can be defined as the resistive force with which the fluid resists the flow under applied pressure.
$f=\eta A\dfrac{dv}{dx}$
Where f is the viscous force, $\eta $ is the coefficient of viscosity, A is the area and $\dfrac{dv}{dx}$is the velocity gradient.
Coefficient of viscosity can be defined as the force of friction that will be requires to maintain a difference of velocity of 1cm/s between parallel layers of fluid.
$\eta =\dfrac{f}{A\dfrac{dv}{dx}}$
The dimensional formula of area is$\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right]$
Now, force is a derived quantity. We have to express in terms of the fundamental quantities.
Now, dimension of force is given as,
$\text{Force}=\text{mass}\times \text{acceleration}$
$\text{Acceleration}=\text{displacement}\times \text{tim}{{\text{e}}^{(-2)}}$
Now. dimension of acceleration= $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]$
Dimension of force=$[{{M}^{1}}]\times [{{M}^{0}}{{L}^{1}}{{T}^{-2}}]=[{{M}^{1}}{{L}^{1}}{{T}^{-2}}]$
Hence, dimension of velocity gradient $\dfrac{dv}{dx}$ = \[\dfrac{\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]}{\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]}=\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]\]
Dimension of coefficient of viscosity, $\eta $ = $\begin{align}
  & \text{force }\times \text{ are}{{\text{a}}^{-1}}\text{ }\times \text{ (velocity gradient}{{\text{)}}^{-1}} \\
 & [{{M}^{1}}{{L}^{1}}{{T}^{-2}}]\times {{[{{M}^{0}}{{L}^{2}}{{T}^{0}}]}^{-1}}\times {{\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]}^{-1}} \\
 & =[{{M}^{1}}{{L}^{-1}}{{T}^{-1}}] \\
\end{align}$
So, the correct answer is option (D).

Note: Don’t try to remember the dimensional formula. You may get confused. Always express the derived quantities in terms of the fundamental quantities and you will get the dimension of quantities.