Answer
$132.14N.m$
Work Step by Step
We know that
$M=\frac{2}{3}\mu P(\frac{R_2^3-R_1^3}{R_2^2-R_1^2})$
We plug in the known values to obtain:
$M=\frac{2}{3}(0.30)(5000)[\frac{(0.100)^3-(0.075)^3}{(0.100)^2-(0.075)^2}]$
This simplifies to:
$M=132.14N.m$
