Discussion
At a wing loading of \( 5.0 \) kN/m\(^2\) with \( C_{L,max} = 2.6 \) in the landing configuration and \( \rho = 1.225 \) kg/m\(^3\), the stalling speed is
Answer: D
Correct answer: 56.0 m/s
The stalling speed follows from equating lift to weight at maximum lift coefficient:
\[ V_s = \sqrt{\frac{2(W/S)}{\rho C_{L,max}}} = \sqrt{\frac{2 \times 5000}{1.225 \times 2.6}} = \sqrt{3140} = 56.0\ \mathrm{m/s} \]
Note the wing loading must be in newtons per square metre, not kilonewtons. The distractor \( 79.2 \) omits the \( C_{L,max} \), and \( 28.0 \) halves the result. This relation is what turns the landing requirement into a vertical line on the diagram: fix the approach speed and \( C_{L,max} \), and the wing loading is bounded.
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