Answer:
The outer diameter of the shaft to 2s.f is 12.65cm
Step-by-step explanation:
We know that the expression for the area of a circular cross section is
A= [tex]\frac{\pi d^2}{4}[/tex]
given data
area of cross section= [tex]125.6cm^2[/tex]
bore diameter d= 8cm
from the area of cross section we can solve for the outer diameter D
[tex]Area of cross section=\frac{\pi D^2}{4} \\\\125.6= \frac{3.142*D^2}{4}[/tex]
solving for D we have
[tex]3.142*D^2= 125.6*4\\\3.142*D^2= 502.4\\D^2= \frac{502.4}{3.142} \\D^2= 159.8981\\D=\sqrt{159.8981} \\D= 12.6450\\D= 12.65cm[/tex]
