To solve the problem of determining the ratio of torsional strengths between a hollow shaft and a solid shaft, we need to apply the torsion formula for each type of shaft.

The torsional strength of a shaft can be expressed in terms of its polar moment of inertia (J). The formula for torsional strength T of a shaft is:

T = \frac{\tau J}{r},

where \tau is the shear stress and r is the outer radius.

Since these shafts are made from the same material and have the same outer diameter, let's denote:

• For the hollow shaft: outer diameter = D, inner diameter = \frac{D}{2}.
• For the solid shaft: diameter = D.

The polar moment of inertia for a solid shaft is given by:

J_s = \frac{\pi}{32} D^4

For the hollow shaft, the polar moment of inertia is:

J_h = \frac{\pi}{32} (D^4 - (\frac{D}{2})^4)

Simplifying this, we get:

J_h = \frac{\pi}{32} (D^4 - \frac{D^4}{16}) = \frac{\pi}{32} \cdot \frac{15D^4}{16} = \frac{15\pi D^4}{512}

Now, the ratio of the torsional strengths is given by the ratio of their polar moments of inertia, assuming the same shear stress conditions:

\frac{T_h}{T_s} = \frac{J_h}{J_s} = \frac{\frac{15\pi D^4}{512}}{\frac{\pi D^4}{32}}

This simplifies to:

\frac{T_h}{T_s} = \frac{15}{16}

Thus, the ratio of their torsional strength is \frac{15}{16}, confirming that the correct answer is \frac{15}{16}.