Viscosity from torque on rotating cylinder

Data: inner radius $R=0.075$ m, outer radius $0.0755$ m, so the gap $t=0.5$ mm $=0.0005$ m. Height $h=0.25$ m, $T=12$ N·m, $N=100$ rpm, so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s.

Surface velocity of the inner cylinder: $V=\omega R=10.472\times0.075=0.7854$ m/s.

For a thin gap the profile is linear: $\tau=\mu\dfrac{V}{t}$. The torque on the inner cylinder (bottom neglected):

$$T=\tau\,(2\pi Rh)\,R=\mu\frac{\omega R}{t}\,2\pi R^{2}h$$ $$\mu=\frac{T\,t}{2\pi\omega R^{3}h}=\frac{12\times0.0005}{2\pi(10.472)(0.075)^{3}(0.25)}$$ $$\mu=0.8646\ \text{Pa·s}$$

Answer: $\mu\approx0.865\ \text{N·s/m}^2$ (8.65 poise).