Correct Answer - Option 2 : Higher the frequency of shaft lower will be the torque
Concept:
Angular velocity (ω) = 2πf; where f is the frequency
The power transmitted by any solid shaft is given by,
\({\rm{P}} = {\rm{Torque\;}}\left( {\rm{{\rm T}}} \right) \times {\rm{Angular\;velocity\;}}({\rm{\omega }})\)
\({\rm{P}} = {\rm{T}} \times 2{\rm{\pi f}}\)
\(T=\frac{P}{2\pi f}\)
For a given Power (P); Torque is inversely proportional to frequency i.e. Higher the frequency of shaft, lower will be the torque.