Question

According to Newton, the viscous force acting between liquid layers of area $A$ and velocity gradient $\dfrac{\Delta v}{\Delta z}$ is given by $F=-\eta A\dfrac{dv}{dz}$, then the dimensions of $\eta $ are
$[M{{L}^{-2}}{{T}^{-2}}]$
\[[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]
$[M{{L}^{2}}{{T}^{-2}}]$
$[M{{L}^{-1}}{{T}^{-1}}]$

Answer 1

Hint: For the formula $F=-\eta A\dfrac{dv}{dz}$, perform the dimensional analysis by bringing terms other than $\eta $ to the one side of the equation. As the dimensions of other measures are known.

Formulae used: The dimensional formula of force $F=[ML{{T}^{-2}}]$, Area \[A=[{{M}^{0}}{{L}^{2}}{{T}^{0}}]\] and the terms in velocity gradient $\dfrac{\Delta v}{\Delta z}=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]$ are used.


Complete step by step answer:
Do not get overwhelmed by looking at the complex alien formula in the question. Another way to solve this question, if you are unfamiliar with the concepts of viscosity, is to identify the familiar quantities which you can dimensionally remember.
Every quantity can be expressed in the terms of the following seven dimensions
Dimension Symbol
Length L
Mass M
Time T
Electric charge Q
Luminous intensity C
Temperature K
Angle None

For the question, we are concerned with only three dimensions : mass, length and time. If we rearrange the equation
$F=-\eta A\dfrac{dv}{dz}$
to obtain $\eta $on the left side. Then, the equation becomes
$\eta =-\dfrac{F}{A\dfrac{dv}{dz}}=\dfrac{Fdz}{Adv}$
Where, $F$ is force which is mass times acceleration, $A$ is area, $v$ is velocity which and $z$ is depth at which viscosity is being calculated.
The dimensional formulae are:
\[\begin{align}
  & F=[ML{{T}^{-2}}] \\
 & A=[{{M}^{0}}{{L}^{2}}{{T}^{0}}] \\
 & v=[{{M}^{0}}L{{T}^{-1}}] \\
 & z=[{{M}^{0}}L{{T}^{0}}] \\
\end{align}\]
The equation for dimensional formula of $\eta $ becomes
\[\begin{align}
  & \Rightarrow \dfrac{[ML{{T}^{-2}}][{{M}^{0}}L{{T}^{0}}]}{[{{M}^{0}}{{L}^{2}}{{T}^{0}}][{{M}^{0}}L{{T}^{-1}}]} \\
 & \Rightarrow \dfrac{[M{{L}^{2}}{{T}^{-2}}]}{[{{M}^{0}}{{L}^{3}}{{T}^{-1}}]} \\
 & \Rightarrow [M{{L}^{2-3}}{{T}^{-2-(-1)}}] \\
 & \Rightarrow [M{{L}^{-1}}{{T}^{-1}}] \\
\end{align}\]
Therefore, the correct answer to this question is option D, $[M{{L}^{-1}}{{T}^{-1}}]$.
Note: Taking derivative or change ($\Delta $) of a quantity does not change its dimension because it is essentially the difference of the quantity. It can be explained by the fact that the dimensions of two quantities added or subtracted has to be the same to perform the operation, and the resulting quantity also has the same dimension.