Wind Problem — NEET Physics MCQs with Solutions
Free NEET Physics Wind Problem MCQs with step-by-step solutions (20 questions). Part of Motion in 2D. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Wind Problem · easy · numerical
An aeroplane has an airspeed of 240 km/h and flies with a tailwind of 60 km/h. Its ground speed is:
A. 180 km/h
B. 360 km/h
C. 240 km/h
D. 300 km/h ✓ Correct
Solution: A tailwind adds directly: 240 + 60 = 300 km/h.
Q2 — Wind Problem · easy · numerical
An aeroplane with an airspeed of 320 km/h flies directly into a headwind of 80 km/h. Its ground speed is:
A. 240 km/h ✓ Correct
B. 160 km/h
C. 400 km/h
D. 320 km/h
Solution: A headwind subtracts directly: 320 − 80 = 240 km/h.
Q3 — Wind Problem · easy · numerical
An aeroplane points due north with an airspeed of 120 km/h while a wind blows at 50 km/h towards the east. The magnitude of its velocity with respect to the ground is:
A. 120 km/h
B. 70 km/h
C. 130 km/h ✓ Correct
D. 170 km/h
Solution: v = √(120² + 50²) = √16900 = 130 km/h.
Q4 — Wind Problem · easy · numerical
A plane points due north with an airspeed of 60 m/s while a crosswind blows at 60 m/s towards the east. Its ground track deviates from north by an angle of:
A. 30°
B. 60°
C. 90°
D. 45° ✓ Correct
Solution: tanθ = wind/airspeed = 60/60 = 1 ⇒ θ = 45°.
Q5 — Wind Problem · easy · numerical
A pilot wants to fly due north while a wind of 20 m/s blows towards the west. If the plane's airspeed is 40 m/s, the pilot must point the plane east of north by an angle of:
A. 15°
B. 60°
C. 30° ✓ Correct
D. 45°
Solution: sinθ = w/v = 20/40 = 1/2 ⇒ θ = 30°.
Q6 — Wind Problem · easy · numerical
A plane with an airspeed of 500 km/h heads into a crosswind of 300 km/h at the correct angle so that it tracks along a straight line. Its ground speed is:
A. 500 km/h
B. 400 km/h ✓ Correct
C. 300 km/h
D. 200 km/h
Solution: Ground speed while crabbing = √(500² − 300²) = √160000 = 400 km/h.
Q7 — Wind Problem · easy · numerical
An aeroplane with an airspeed of 150 km/h has a 30 km/h tailwind. The time it takes to cover 540 km is:
A. 2 h
B. 3.6 h
C. 3 h ✓ Correct
D. 4.5 h
Solution: Ground speed = 150 + 30 = 180 km/h; t = 540/180 = 3 h.
Q8 — Wind Problem · easy · numerical
A plane with an airspeed of 160 km/h flies 480 km against a 40 km/h headwind. The flight takes:
A. 2.4 h
B. 4 h ✓ Correct
C. 3 h
D. 6 h
Solution: Ground speed = 160 − 40 = 120 km/h; t = 480/120 = 4 h.
Q9 — Wind Problem · easy · numerical
A plane's airspeed is 220 km/h, but its ground speed while flying straight into the wind is only 190 km/h. The wind speed is:
A. 410 km/h
B. 30 km/h ✓ Correct
C. 15 km/h
D. 60 km/h
Solution: w = airspeed − ground speed = 220 − 190 = 30 km/h.
Q10 — Wind Problem · easy · numerical
A balloon rises vertically at 8 m/s while a horizontal wind blows at 6 m/s. The speed of the balloon with respect to the ground is:
A. 10 m/s ✓ Correct
B. 7 m/s
C. 2 m/s
D. 14 m/s
Solution: v = √(8² + 6²) = √100 = 10 m/s.
Q11 — Wind Problem · easy · numerical
A balloon rises vertically at 4 m/s in a horizontal wind of 3 m/s. When it reaches a height of 80 m, its horizontal drift from the starting point is:
A. 80 m
B. 100 m
C. 40 m
D. 60 m ✓ Correct
Solution: Time to rise: t = 80/4 = 20 s; drift = 3 × 20 = 60 m.
Q12 — Wind Problem · easy · numerical
A cyclist rides at 3 m/s while a true crosswind blows at 4 m/s perpendicular to his motion. The speed of the wind as felt by the cyclist (apparent wind) is:
A. 3.5 m/s
B. 5 m/s ✓ Correct
C. 7 m/s
D. 1 m/s
Solution: Apparent wind is the vector difference of wind and cyclist velocities: √(3² + 4²) = 5 m/s.
Q13 — Wind Problem · medium · numerical
A plane with an airspeed of 100 km/h flies 240 km with the wind and returns 240 km against the same 20 km/h wind. The total time for the round trip is:
A. 5.5 h
B. 4.8 h
C. 4 h
D. 5 h ✓ Correct
Solution: Outbound: 240/(100 + 20) = 2 h; return: 240/(100 − 20) = 3 h; total = 5 h.
Q14 — Wind Problem · medium · numerical
A plane with an airspeed of 250 km/h must track due north while a wind of 125 km/h blows towards the east. After heading into the wind at the correct angle, its ground speed is:
A. 125√3 ≈ 216.5 km/h ✓ Correct
B. 375 km/h
C. 125 km/h
D. 250 km/h
Solution: Ground speed = √(250² − 125²) = 125√3 ≈ 216.5 km/h.
Q15 — Wind Problem · medium · numerical
To keep moving straight along its intended track, a plane with an airspeed of 60 m/s must point 30° away from the track, into the crosswind. The wind speed is:
A. 15 m/s
B. 30√3 ≈ 52 m/s
C. 30 m/s ✓ Correct
D. 60 m/s
Solution: w = v sinθ = 60 sin30° = 30 m/s.
Q16 — Wind Problem · medium · numerical
A balloon rises vertically at 5 m/s in a horizontal wind of 12 m/s. The magnitude of its displacement from the starting point when it reaches a height of 200 m is:
A. 200 m
B. 520 m ✓ Correct
C. 680 m
D. 480 m
Solution: t = 200/5 = 40 s gives a drift of 12 × 40 = 480 m; displacement = √(200² + 480²) = 520 m.
Q17 — Wind Problem · medium · numerical
A pilot wishes to have a velocity of 240 km/h due north with respect to the ground while a wind of 100 km/h blows towards the east. The required airspeed of the plane is:
A. 140 km/h
B. 240 km/h
C. 260 km/h ✓ Correct
D. 340 km/h
Solution: v_air = v_ground − v_wind, so airspeed = √(240² + 100²) = √67600 = 260 km/h.
Q18 — Wind Problem · medium · numerical
A plane with an airspeed of 50 m/s must fly straight to a station 12 km away while a crosswind of 30 m/s blows perpendicular to the line joining them. The time of flight is:
A. 5 min ✓ Correct
B. 6 min
C. 4 min
D. 10 min
Solution: Ground speed along the line = √(50² − 30²) = 40 m/s; t = 12000/40 = 300 s = 5 min.
Q19 — Wind Problem · medium · numerical
A plane points at the proper angle into a crosswind so that it travels in a straight line. Its airspeed is 20 m/s and its resulting ground speed is 16 m/s. The wind speed is:
A. 4 m/s
B. 8 m/s
C. 12 m/s ✓ Correct
D. 16 m/s
Solution: w = √(20² − 16²) = √144 = 12 m/s.
Q20 — Wind Problem · medium · numerical
A cyclist rides due north at 12 km/h while the true wind blows at 16 km/h towards the east. The apparent wind speed felt by the cyclist is:
A. 14 km/h
B. 28 km/h
C. 4 km/h
D. 20 km/h ✓ Correct
Solution: Apparent wind = v_wind − v_cyclist, with perpendicular components 16 and 12: √(16² + 12²) = 20 km/h.