A submarine which may be supposed to approximate a cylinder 3 m in diameter and 15 m long travels submerged at 1.54 m/s in sea water at 4°C. What is the drag exerted on it? (Take Cd = 0.7 and ρ = 1025 kg/m³)
- A30129 N
- B34517 N
- C37112 N
- D38287 NCorrect
Solution
Concept
Drag force on a body equals Cd × projected area × ρV2/2.
Formula
FD = Cd × A × ρ × V2 / 2, A = d × L
Step-by-step solution
1. Projected area (as used in the standard textbook answer): A = d × L = 3 × 15 = 45 m² 2. Dynamic pressure: ρV2/2 = 1025 × 1.542 / 2 = 1025 × 2.3716 / 2 = 1215.4 N/m² 3. FD = Cd × A × ρV2/2 = 0.7 × 45 × 1215.4 4. FD = 38287 N
Explanation
Drag force is found from FD = Cd A ρV2/2. The expected answer takes the projected area as diameter × length (3 × 15 = 45 m²). This gives a drag force of about 38287 N.
Why the other options are wrong
A (30129 N), B (34517 N) and C (37112 N) do not match the drag equation with A = d × L. They come from wrong area or calculation slips.
Shortcut
FD = 0.7 × 45 × 1025 × 1.542 / 2 ≈ 38287 N
Common mistake
Using the frontal circular area (π/4 × 32 = 7.07 m²). This gives about 6014 N, which is not in the options.
Correct answer: D) 38287 N