Wind Problem — JEE Main Physics MCQs with Solutions
Free JEE Main 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 heads due north with an airspeed of 150 km/h while a steady wind blows at 80 km/h towards the east. The speed of the plane relative to the ground is:
A. 150 km/h
B. 70 km/h
C. 170 km/h ✓ Correct
D. 230 km/h
Solution: Ground velocity = air velocity + wind velocity; the components are perpendicular, so v = √(150² + 80²) = √28900 = 170 km/h.
Q2 — Wind Problem · easy · numerical
An aircraft can fly at 290 km/h in still air. It flies directly into a headwind of 90 km/h. Its ground speed is:
A. 200 km/h ✓ Correct
B. 290 km/h
C. 260 km/h
D. 380 km/h
Solution: Against a headwind the speeds subtract: ground speed = 290 − 90 = 200 km/h.
Q3 — Wind Problem · easy · numerical
A plane whose airspeed is 200 km/h flies with a tailwind of 90 km/h. How long does it take to cover 725 km?
A. 6.6 h
B. 2.5 h ✓ Correct
C. 2 h
D. 3.6 h
Solution: Ground speed = 200 + 90 = 290 km/h; t = 725/290 = 2.5 h.
Q4 — Wind Problem · easy · numerical
A bird heads due north at 7 m/s (its speed relative to the air) while the wind blows at 24 m/s towards the east. The speed of the bird relative to the ground is:
A. 24 m/s
B. 25 m/s ✓ Correct
C. 17 m/s
D. 31 m/s
Solution: v = √(7² + 24²) = √(49 + 576) = √625 = 25 m/s.
Q5 — Wind Problem · easy · numerical
A plane heads due north with airspeed 350 km/h, but because of a crosswind blowing towards the east its ground speed becomes 370 km/h. The wind speed is:
A. 60 km/h
B. 150 km/h
C. 120 km/h ✓ Correct
D. 20 km/h
Solution: w = √(370² − 350²) = √(136900 − 122500) = √14400 = 120 km/h.
Q6 — Wind Problem · easy · numerical
With a tailwind an aircraft has a ground speed of 320 km/h; flying back against the same wind its ground speed is 180 km/h. The wind speed is:
A. 50 km/h
B. 70 km/h ✓ Correct
C. 140 km/h
D. 250 km/h
Solution: v + w = 320 and v − w = 180, so w = (320 − 180)/2 = 70 km/h.
Q7 — Wind Problem · medium · numerical
A plane heads north-east with airspeed 120√2 km/h. The wind blows at 40 km/h towards the east. The ground speed of the plane is:
A. 160 km/h
B. 200 km/h ✓ Correct
C. 170 km/h
D. 210 km/h
Solution: Air velocity components = (120 east, 120 north) km/h; adding the wind gives (160, 120), so v = √(160² + 120²) = √40000 = 200 km/h.
Q8 — Wind Problem · medium · numerical
A pilot wants to travel due north. Her airspeed is 200 km/h and the wind blows at 120√2 km/h towards the north-east (sin 37° = 0.6). What should she do, and what is her ground speed?
A. Head 53° west of north; ground speed 200 km/h
B. Head 37° east of north; ground speed 280 km/h
C. Head 45° west of north; ground speed 240 km/h
D. Head 37° west of north; ground speed 280 km/h ✓ Correct
Solution: The wind has components (120 east, 120 north) km/h; to cancel the eastward part, 200 sin θ = 120 ⇒ θ = 37° west of north, and ground speed = 200 cos 37° + 120 = 160 + 120 = 280 km/h.
Q9 — Wind Problem · medium · numerical
An aircraft with airspeed 250 km/h makes a round trip between two towns 720 km apart, with a steady 70 km/h wind blowing along the line joining them. The total time for the round trip is:
A. 5.5 h
B. 6.25 h ✓ Correct
C. 7.2 h
D. 5.76 h
Solution: With the wind: 720/(250 + 70) = 2.25 h; against it: 720/(250 − 70) = 4 h; total = 6.25 h (equivalently T = 2dv/(v² − w²)).
Q10 — Wind Problem · medium · numerical
A plane flies between two cities and back along the wind direction. Its airspeed is 400 km/h and the wind speed is 200 km/h. The ratio of the time taken with this wind to the time taken in still air is:
A. 4 : 3 ✓ Correct
B. 3 : 4
C. 2 : 1
D. 1 : 1
Solution: T_wind/T_still = v²/(v² − w²) = 400²/(400² − 200²) = 160000/120000 = 4/3.
Q11 — Wind Problem · medium · numerical
A plane with airspeed 370 km/h flies from town A to town B, 700 km away, and back, while a steady 120 km/h wind blows perpendicular to the line AB. The total time for the round trip is:
A. 4 h ✓ Correct
B. 3.8 h
C. 2 h
D. 5 h
Solution: Each way the plane must crab into the wind, so its effective speed along AB is √(370² − 120²) = √122500 = 350 km/h; T = 1400/350 = 4 h.
Q12 — Wind Problem · medium · numerical
A plane noses due north with airspeed 320 km/h, but its GPS shows a ground velocity of 70 km/h towards the east combined with 80 km/h towards the north. The wind speed is:
A. 240 km/h
B. 70 km/h
C. 250 km/h ✓ Correct
D. 170 km/h
Solution: Wind = ground velocity − air velocity = (70, 80 − 320) = (70, −240) km/h, so w = √(70² + 240²) = √62500 = 250 km/h.
Q13 — Wind Problem · medium · numerical
An aircraft covers the 1200 km between two airports in 3 h flying with the wind and takes 4 h on the return flight against the same wind. The airspeed of the aircraft and the wind speed are:
A. 350 km/h and 50 km/h ✓ Correct
B. 350 km/h and 100 km/h
C. 300 km/h and 100 km/h
D. 400 km/h and 100 km/h
Solution: v + w = 1200/3 = 400 km/h and v − w = 1200/4 = 300 km/h, so v = 350 km/h and w = 50 km/h.
Q14 — Wind Problem · medium · numerical
A ship steams due north at 20 km/h. The true wind blows at 20√2 km/h towards the north-west. The wind as experienced on the ship (apparent wind) is:
A. 20 km/h, towards the east
B. 40 km/h, towards the north-west
C. 20 km/h, towards the west ✓ Correct
D. 20√2 ≈ 28 km/h, towards the south-west
Solution: True wind components = (−20 east, +20 north) km/h; subtracting the ship velocity (0, 20) leaves (−20, 0), i.e. 20 km/h towards the west.
Q15 — Wind Problem · medium · numerical
A balloon rises vertically at 7 m/s while a horizontal wind carries it sideways at 24 m/s. The distance of the balloon from its starting point after 80 s is:
A. 2000 m ✓ Correct
B. 1920 m
C. 2480 m
D. 560 m
Solution: Resultant speed = √(7² + 24²) = 25 m/s; displacement = 25 × 80 = 2000 m.
Q16 — Wind Problem · medium · numerical
A bird points itself 37° west of north and flies with airspeed 20 m/s while the wind blows at 12 m/s towards the east (sin 37° = 0.6). The velocity of the bird relative to the ground is:
A. 16 m/s, 37° west of north
B. 16 m/s due north ✓ Correct
C. 20 m/s due north
D. 32 m/s due north
Solution: The westward air component 20 sin 37° = 12 m/s exactly cancels the wind, leaving the northward component 20 cos 37° = 16 m/s.
Q17 — Wind Problem · medium · numerical
A plane with airspeed 290 km/h covers 680 km with a tailwind in 2 h. How long will the return flight take against the same wind?
A. 2 h
B. 3 h 20 min
C. 2 h 20 min
D. 2 h 50 min ✓ Correct
Solution: Tailwind ground speed = 680/2 = 340 km/h ⇒ wind = 340 − 290 = 50 km/h; return ground speed = 290 − 50 = 240 km/h, so t = 680/240 = 17/6 h = 2 h 50 min.
Q18 — Wind Problem · medium · numerical
An aeroplane heads due north at 240 km/h (airspeed) while the wind blows at 70 km/h towards the east. How far is the plane from its starting point after 2 hours?
A. 500 km ✓ Correct
B. 620 km
C. 480 km
D. 340 km
Solution: Ground speed = √(240² + 70²) = √62500 = 250 km/h; distance = 250 × 2 = 500 km.
Q19 — Wind Problem · medium · numerical
A drone has airspeed 25 m/s in a wind of speed 7 m/s. The ratio of the time for a round trip parallel to the wind to the time for a round trip of the same total distance perpendicular to the wind is:
A. 24 : 25
B. 25 : 24 ✓ Correct
C. 1 : 1
D. 25 : 7
Solution: T_parallel = 2dv/(v² − w²) and T_perp = 2d/√(v² − w²), so T_parallel/T_perp = v/√(v² − w²) = 25/√(625 − 49) = 25/24.
Q20 — Wind Problem · medium · numerical
A train runs due east at 20 m/s while the true wind blows due north at 21 m/s. To a passenger on the train, the apparent wind speed is:
A. 29 m/s ✓ Correct
B. 41 m/s
C. 21 m/s
D. 20 m/s
Solution: Apparent wind = true wind − train velocity = (−20 east, 21 north) m/s, so its speed = √(20² + 21²) = √841 = 29 m/s.