Correct Answer - Option 1 : 80/81

__Concept:__

The equation for Shaft Subjected to Torsion T

\(\frac{τ }{R}\) = \(\frac{T}{J}\) = \(\frac{{Gθ }}{L}\)

Solid shaft **J** = \(\frac{{\pi D_s^4}}{{32}}\)

Hollow Shaft **J** = \(\frac{{\pi \left( {D_0^4 - \;D_i^4} \right)}}{{32}}\)

τ = Shear stress induced due to torsion T, G = Modulus of Rigidity, θ = Angular deflection of the shaft, R, L = Shaft radius and length respectively, Ds = Diameter of the solid shaft, D0 = Outer Diameter of Hollow shaft, Di = Inner Diameter of Hollow shaft

__Calculation;__

__Given__

D_{0} = 3 × D_{i}

D_{s} = D_{0} (Solid shaft diameter is equal to an outer diameter of hollow shaft)

Material is the same for the hollow and solid shaft (i.e. G is the same for both because G depends upon the material)

\(\frac{T}{\tau }\) = \(\frac{{G\theta }}{L}\)

∴ T \( \propto \;\) J (θ, G, L same for both or constant)

\(\frac{{{T_H}}}{{{T_s}}}\) = \(\frac{{{J_H}}}{{{J_s}}}\) = \(\frac{{\frac{\pi }{{32}}\left( {D_0^4 - D_i^4} \right)}}{{\frac{\pi }{{32}}{{\left( {{D_s}} \right)}^4}}}\)

Put, D_{0} = 3 × D_{i}

D_{s} = D_{0}

\(J_H\over J_s\) = \(80\over 81\)

**Polar moment of inertia**

**J **= \(\mathop \smallint \limits_0^R 2\pi \)r^{3}dr

For Solid Shaft, **J** = \(\pi D^4 \over 32\)

For Hollow Shaft, **J** = \(\mathop \smallint \limits_r^R 2\pi \)r^{3}dr = \(\frac{{\pi \left( {D_0^4 - \;D_i^4} \right)}}{{32}}\)

**The assumption for Torsion equation:**

\(\tau\over R\) = \(T\over J\) = \(G\theta\over L\)

- The bar is acted upon by pure torque.
- The section under consideration is remote from the point of application of the load and from a change in diameter.
- The material must obey Hooke's Law
- Cross-sections rotate as if rigid, i.e. every diameter rotates through the same angle.