A) \[\left[ M{{L}^{-1}}{{T}^{-2}} \right]\]
B) \[\left[ ML{{T}^{-1}} \right]\]
C) \[\left[ M{{L}^{-1}}{{T}^{-1}} \right]\]
D) \[\left[ M{{L}^{-2}}{{T}^{-2}} \right]\]
E) None of these
Correct Answer: C
Solution :
By Newtons formula, \[\eta =\frac{F}{A\,(\Delta {{V}_{x}}/\Delta z)}\] \[\therefore \]Dimensions of\[\eta \] \[=\frac{\text{dimensions}\,\text{of}\,\text{force}}{\begin{align} & \text{dimensions of area}\,\text{ }\!\!\times\!\!\text{ }\,\text{dimensions of}\, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{velocity gradient} \\ \end{align}}\] \[=\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}]\,[{{T}^{-1}}]}=[M{{L}^{-1}}{{T}^{-1}}]\]
You need to login to perform this action.
You will be redirected in
3 sec
