Answer
$0.284$
Work Step by Step
We know that
$M=\frac{2}{3}\mu P(\frac{R_2^3-R_1^3}{R_2^2-R_1^2})$
This can be rearranged as:
$\mu=\frac{M}{\frac{2}{3}P(\frac{R_2^3-R_1^3}{R_2^2-R_1^2})}$
We plug in the known values to obtain:
$\mu=\frac{200}{\frac{2}{3}(8000)[\frac{(0.100)^3-(0.075)^3}{(0.100)^2-(0.075)^2}]}$
This simplifies to:
$\mu=0.284$
