Answer:
The viscosity of the fluid is 6.05 Pa-s
Explanation:
The viscosity obtained from a rotating type viscometer is given by the equation
[tex]\mu =\frac{3Tt}{2\pi \omega R_o(R_o^3-R_i^3)}[/tex]
where
T is the torque applied in the viscometer
't' is the thickness of gap
[tex]\omega [/tex] is the angular speed of rotation
[tex]R_o,R_i[/tex] are the outer and the inner raddi respectively
Since it is given that
[tex]\omega =100rpm=\frac{100\times 2\pi }{60}=10.47rad/sec[/tex]
Applying the values in the above relation we get
[tex]\mu =\frac{3\times 0.021\times 0.02\times 10^{-3}}{2\pi \times 10.47\times 37.5\times 10^{-3}(37.52^3-37.5^3)\times 10^{-9}}=6.04Pa-s[/tex]