Torsion capacity comparison:\nTwo shafts of the same material are compared. Shaft A is solid with diameter 100 mm. Shaft B is hollow with outer diameter 100 mm and inner diameter 50 mm. The torque that shaft B can transmit (at the same allowable shear stress) is what fraction of shaft A's torque?

Introduction / Context:
Designers often prefer hollow shafts to save weight while maintaining torsional strength. Torque capacity at a given allowable shear stress is proportional to the polar moment of inertia J of the section. Comparing J for solid and hollow circular sections quantifies the efficiency of material distribution.

Given Data / Assumptions:

• Same material and allowable shear stress for both shafts.
• Solid shaft A: diameter D = 100 mm.
• Hollow shaft B: outer diameter D_o = 100 mm, inner diameter D_i = 50 mm.
• Length and boundary conditions are comparable so that torque capacity scales with J.

Concept / Approach:
For a circular shaft, maximum shear stress tau_max = T * R / J. At a fixed tau_allow, torque capacity T_allow ∝ J / R. For shafts with the same outer radius R, the ratio of allowable torques equals the ratio of J values. Here, both have the same outer radius, so T ratio simplifies to J ratio.

Step-by-Step Solution:

Verification / Alternative check:
Numeric check confirms 0.9375 fraction. The modest reduction in capacity versus large weight saving highlights the advantage of hollow shafts.

Why Other Options Are Wrong:

• 1/6, 1/8, 1/4: greatly underestimate hollow shaft capacity; they do not reflect the fourth-power diameter dependence.

Common Pitfalls:
Comparing torque per unit weight without accounting for equal outer diameter; or forgetting to use the fourth power in J calculations.

Final Answer:
15/16