Answer with Explanation:
By the equation or Torque we have
[tex]\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}[/tex]
where
T is the torque applied on the shaft
[tex]I_{p}[/tex] is the polar moment of inertia of the shaft
[tex]\tau [/tex] is the shear stress developed at a distance 'r' from the center of the shaft
[tex]\theta [/tex] is the angle of twist of the shaft
'G' is the modulus of rigidity of the shaft
We know that for solid shaft [tex]I_{p}=\frac{\pi R^4}{2}[/tex]
For a hollow shaft [tex]I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}[/tex]
Since the two shafts are subjected to same torque from the relation of Torque we have
1) For solid shaft
[tex]\frac{2T}{\pi R^4}\times r=\tau _{solid} [/tex]
2) For hollow shaft we have
[tex]\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4} [/tex]
Comparing the above 2 relations we see
[tex]\frac{\tau _{solid}}{\tau _{hollow}}=0.76[/tex]
Similarly for angle of twist we can see
[tex]\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316[/tex]
Part b)
Strength of solid shaft = [tex]\tau _{max}=\frac{T\times R}{I_{solid}}[/tex]
Weight of solid shaft =[tex]\rho \times \pi R^2\times L[/tex]
Strength per unit weight of solid shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}[/tex]
Strength of hollow shaft = [tex]\tau '_{max}=\frac{T\times R}{I_{hollow}}[/tex]
Weight of hollow shaft =[tex]\rho \times \pi (R^2-0.7R^2)\times L[/tex]
Strength per unit weight of hollow shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}[/tex]
Thus [tex]\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16[/tex]
