A hollow shaft of same cross-section area as solid shaft transmits:

Explanation:

As the cross-section area is the same,

\(\frac{\pi }{4}\left( {d_1^2 - d_2^2} \right) = \frac{\pi }{4}d_s^2\)

\(\therefore d_1^2 - d_2^2 = d_s^2\)

\(\frac{τ }{r} = \frac{T}{J} = \frac{{G\theta }}{L}\)

\(T=\frac{τJ }{r}\)

Now torque ∝ polar moment

\(\therefore \frac{{{T_h}}}{{{T_s}}} = \frac{{\frac{\pi }{{32}}\left( {d_1^4 - d_2^4} \right)}}{{\frac{\pi }{{32}}\left( {d_s^4} \right)}} = \frac{{d_1^4 - d_2^4}}{{d_s^4}}= \frac{{(d_1^2 - d_2^2)}{(d_1^2 + d_2^2)}}{{d_s^4}}\)

\(\frac{{{T_h}}}{{{T_s}}} = \frac{{d_1^2 + d_2^2}}{{d_s^2}}\)