Answer
$90 \text{ horsepower}$
Work Step by Step
The variation model described by the problem is $
H=ksd^3
,$ where $H$ is the horsepower, $s$ is the angular speed of rotation, and $d$ is the diameter.
Substituting the known values of the variables results to
\begin{array}{l}\require{cancel}
40=k(120)(2)^3
\\
40=k(120)(8)
\\
40=960k
\\
\dfrac{40}{960}=k
\\\\
k=\dfrac{\cancel{40}}{\cancel{40}\cdot24}
\\\\
k=\dfrac{1}{24}
.\end{array}
Hence, the equation of variation is
\begin{array}{l}\require{cancel}
H=\dfrac{1}{24}sd^3
.\end{array}
Using the variation equation above, then
\begin{array}{l}\require{cancel}
H=\dfrac{1}{24}(80)(3)^3
\\\\
H=\dfrac{1}{24}(80)(27)
\\\\
H=\dfrac{2160}{24}
\\\\
H=90
.\end{array}
Hence, the amount of horsepower, $H,$ that can be safely transmitted is $
90 \text{ horsepower}
.$
