Discussion
A dc motor developing 60 N·m at 1,200 rpm delivers a shaft power of nearest
Answer: D
Correct answer: 7.5 kW
Shaft power is torque times angular velocity:
\[ P = T\omega = 60 \times \frac{2\pi \times 1200}{60} = 60 \times 125.7 \approx 7540 \text{ W} \approx 7.5 \text{ kW} \]The speed must be converted from revolutions per minute to radians per second before multiplying - the 72 kW answer comes from using 1200 directly, and the 1.2 kW answer from mishandling the same conversion the other way.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development