Answer:
1) Dimensions of shear rate is [tex][T^{-1}][/tex] .
2)Dimensions of shear stress are [tex][ML^{-1}T^{-2}][/tex]
Explanation:
Since the dimensions of velocity 'v' are [tex][LT^{-1}][/tex] and the dimensions of distance 'y' are [tex][L][/tex] , thus the dimensions of [tex]\frac{dv}{dy}[/tex] become
[tex]\frac{[LT^{-1}]}{[L]}=[T^{-1}][/tex] and hence the units become [tex]s^{-1}[/tex].
Now we know that the dimensions of coefficient of dynamic viscosity [tex]\mu [/tex] are [tex][ML^{-1}T^{-1}][/tex] thus the dimensions of shear stress can be obtained from the given formula as
[tex][\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}][/tex]
Now we know that dimensions of momentum are [tex][MLT^{-1}][/tex]
The dimensions of [tex]Area\times time[/tex] are [tex][L^{2}T][/tex]
Thus the dimensions of [tex]\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}][/tex]
Which is same as that of shear stress. Hence proved.
