Answer:
1) Dimensions of shear rate is
.
2)Dimensions of shear stress are
Step-by-step explanation:
Since the dimensions of velocity 'v' are
and the dimensions of distance 'y' are
, thus the dimensions of
become
and hence the units become
.
Now we know that the dimensions of coefficient of dynamic viscosity
are
thus the dimensions of shear stress can be obtained from the given formula as
![[\tau ]=[ML^(-1)T^(-1)]* [T^(-1)]\\\\[\tau ]=[ML^(-1)T^(-2)]](https://img.qammunity.org/2020/formulas/engineering/college/ukgxy7igfvjknwdajdpf26o674mh759lgh.png)
Now we know that dimensions of momentum are
![[MLT^(-1)]](https://img.qammunity.org/2020/formulas/engineering/college/b124qho5zq4premxg2qc7wrf5uwz5mlgkd.png)
The dimensions of
are
![[L^(2)T]](https://img.qammunity.org/2020/formulas/engineering/college/cwt45iinl3fv11wazj2xdf9u2ah2zliczi.png)
Thus the dimensions of
![(Moumentum)/(Area* time)=(MLT^(-1))/(L^(2)T)=[MLT^(-2)]](https://img.qammunity.org/2020/formulas/engineering/college/jv1raohuvr4dyrpc5q8oy48az7tc9n6vbe.png)
Which is same as that of shear stress. Hence proved.