Discussion
Dispersion coefficients at \( 1 \) km are \( \sigma_y = 150 \) m and \( \sigma_z = 90 \) m. The product \( \pi \sigma_y \sigma_z \) is about
Answer: C
Correct answer: 42400 m\(^2\)
Substituting the two dispersion coefficients:
\[ \pi \sigma_y \sigma_z = \pi \times 150 \times 90 = \pi \times 13500 \approx 42412\ \text{m}^2 \]
which rounds to about \( 42400 \). The distractor \( 13500 \) is the product without \( \pi \), and \( 84800 \) doubles the result - the factor of two belongs in the full expression \( 2\pi u \sigma_y \sigma_z \), not in this product. The quantity has the character of an effective cross-sectional area over which the emission is spread.
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