Answer:
![\\u =0.000108\ (m^2)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/bh6eprvg5eg7gn5uf554puejm062irzfi5.png)
![\\u =0.001161\ (ft^2)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/q5xtn4bui5vnj7emhhvh0dbcjj02rinek4.png)
Step-by-step explanation:
Given that
![\mu =0.1\ Pa-s](https://img.qammunity.org/2020/formulas/engineering/college/xdjy73y9nhrgwwzlyuf0bq9plkpeflcykg.png)
Specific gravity of fluid S=0.92
So that density of fluid
![\rho=920\ (kg)/(m^3)](https://img.qammunity.org/2020/formulas/engineering/college/bheiglo2tpjzz9n256ycbjx46gxxn9tw7s.png)
As we know that ,kinematic viscosity of fluid is given as
![\\u =(\mu )/(\rho)](https://img.qammunity.org/2020/formulas/engineering/college/l3yndgq4bso6rz5nlk1dand4mlrbs5wg04.png)
Now by putting the values
![\\u =(0.1 )/(920)](https://img.qammunity.org/2020/formulas/engineering/college/j4t4lrn0zu89oxe0xvbeotcqqcrcqoreq3.png)
![\\u =0.000108\ (m^2)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/bh6eprvg5eg7gn5uf554puejm062irzfi5.png)
So the kinematic viscosity of fluid in SI
![\\u =0.000108\ (m^2)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/bh6eprvg5eg7gn5uf554puejm062irzfi5.png)
The kinematic viscosity of fluid in SI
![\\u =0.001161\ (ft^2)/(sec)](https://img.qammunity.org/2020/formulas/engineering/college/q5xtn4bui5vnj7emhhvh0dbcjj02rinek4.png)