Answer
$\mu =0.6[N.s/{m^2}$]=1.3*$10^{-2}$ [$\frac{lb.s}{ft^{2}}$]
Work Step by Step
$T_{C}={\frac{5}{9}}(T_{F}-32)={\frac{5}{9}}(85^{\circ}{\bf F}-32)=29.4$$^{\circ}\mathbf{C}$
from Fig B.1 in appendix B
${\mu}$ for glycerin at $85^{\circ}\mathbf{F}$ =0.6 $\frac{N.s}{m^{2}}$ (SI unit)
${\mu}$=(0.6)(2.089*$10^{-2}$)=1.3*$10^{-2}$ [$\frac{lb.s}{ft^{2}}$] (british unit)
