Motion in a Plane NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Motion in a Plane question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Motion in a Plane previous year questions
32 questions. Tap an option to check your answer.
Q1NEET 2021
A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s ? (Take g = 10 m/s²)
Not quite — the answer is D.
At t = 4 s the car's speed is 5 × 4 = 20 m/s, which the ball keeps horizontally. After 2 s of free fall its vertical speed is 10 × 2 = 20 m/s, so v = √(20² + 20²) = 20√2 m/s. Once released, only gravity acts, so its acceleration is g = 10 m/s², not the car's 5 m/s².
Q2NEET 2021
A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle 'θ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by :
Not quite — the answer is D.
Speed v = 2πR/T. Maximum height H = v²sin²θ/(2g) = 4R gives sin²θ = 8gR/v² = 8gR·T²/(4π²R²) = 2gT²/(π²R). Hence θ = sin⁻¹(2gT²/π²R)^(1/2).
Q3NEET 2019 · Q32
The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by:
Not quite — the answer is A.
For the shortest path his resultant velocity must point due north, so the westward component of his swimming velocity must cancel the current: 20 sinθ = 10, θ = 30° west of north. Swimming straight north (0°) gives minimum time, not the shortest path.
Q4NEET 2019 · Q34
Two particles A and B are moving in uniform circular motion in concentric circles of radii r_A and r_B with speed v_A and v_B respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be:
Not quite — the answer is D.
Angular speed ω = 2π/T; equal time periods mean equal angular speeds, ratio 1 : 1. The ratio v_A : v_B equals r_A : r_B here, which is the ratio of linear, not angular, speeds.
Q5NEET 2017 · Q169
The x and y coordinates of the particle at any time are x = 5t − 2t² and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is:
Not quite — the answer is C.
aₓ = d²x/dt² = −4 m/s² (constant) and a_y = d²y/dt² = 0 since y grows uniformly. The acceleration is therefore −4 m/s² along x at every instant, including t = 2 s. −8 m/s² comes from substituting t = 2 into −4t, which is not the acceleration.
Q6NEET 2017 (Gujarati paper) · Q147
A ball of mass 1 kg is thrown vertically upwards and returns to the ground after 3 seconds. Another ball, thrown at 60° with vertical also stays in air for the same time before it touches the ground. The ratio of the two heights are:
Not quite — the answer is B.
Time of flight T = 2u_y/g depends only on the vertical component of velocity, and maximum height H = u_y²/2g = gT²/8. Equal flight times mean equal vertical components, so both balls rise to the same height, 1 : 1 (each H = 10 × 9/8 ≈ 11.25 m with g = 10 m s⁻²). The 60° angle changes only the horizontal motion.
Q7NEET 2017 (Gujarati paper) · Q169
The angle between A⃗ − B⃗ and A⃗ × B⃗ is (A⃗ ≠ B⃗):
Not quite — the answer is A.
A⃗ × B⃗ is perpendicular to the plane containing A⃗ and B⃗, and A⃗ − B⃗ lies in that plane. So (A⃗ − B⃗)·(A⃗ × B⃗) = 0 and the angle is 90° for any non-parallel A⃗ and B⃗.
Q8NEET 2016 Phase 1 · Q24
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10⁻⁴ J by the end of the second revolution after the beginning of the motion?
Not quite — the answer is A.
½mv² = 8 × 10⁻⁴ J with m = 0.01 kg gives v² = 0.16 m²/s². The distance covered in two revolutions is s = 2 × 2πr = 4π × 0.064 ≈ 0.804 m. From v² = 2a_t s, a_t = 0.16/(2 × 0.804) ≈ 0.0995 ≈ 0.1 m/s²; 0.15 m/s² does not satisfy this.
Q9NEET 2016 Phase 1 · Q27
A particle moves so that its position vector is given by r⃗ = cos ωt x̂ + sin ωt ŷ, where ω is a constant. Which of the following is true?
Not quite — the answer is C.
This is uniform circular motion of unit radius. v⃗ = ω(−sin ωt x̂ + cos ωt ŷ), and v⃗·r⃗ = 0, so velocity is perpendicular to r⃗. a⃗ = −ω²r⃗, so acceleration points towards the origin, not away from it.
Q10NEET 2016 Phase 1 · Q37
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is:
Not quite — the answer is B.
|A + B|² = |A − B|² gives A² + B² + 2AB cosθ = A² + B² − 2AB cosθ, so cosθ = 0 and θ = 90°. At 0° the sum would exceed the difference, and at 180° the difference would exceed the sum.
Q11AIPMT 2015
A ship A is moving Westwards with a speed of 10 km h⁻¹ and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h⁻¹. The time after which the distance between them becomes shortest, is:
Not quite — the answer is C.
Take A as origin: B starts at (0, −100) km and its velocity relative to A is (10, 10) km/h, i.e. 10√2 km/h at 45°. Time of closest approach t = −(r·v)/v² = (100 × 10)/200 = 5 h. The minimum separation is then 50√2 km, which is the trap behind 5√2 h.
Q12AIPMT 2014
A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. Average velocity vector (V⃗av) from t = 0 to t = 5 s is:
Not quite — the answer is C.
Average velocity = total displacement/total time. Displacement from (2, 3) to (13, 14) is 11î + 11ĵ in 5 s, giving (11/5)(î + ĵ); dividing the final position by time ignores the starting point.
Q13AIPMT 2014
A projectile is fired from the surface of the earth with a velocity of 5 ms⁻¹ and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms⁻¹ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ms⁻²) is: (given g = 9.8 ms⁻²)
Not quite — the answer is D.
The trajectory y = x tan θ − gx²/(2u²cos²θ) is identical when g/u² is the same. g′ = 9.8 × 3²/5² = 3.53 ≈ 3.5 m s⁻²; scaling with u instead of u² gives the wrong value.
Q14NEET 2013
The velocity of a projectile at the initial point A is (2î + 3ĵ) m/s. Its velocity (in m/s) at point B (the point where it returns to the same horizontal level as A) is
Not quite — the answer is C.
With no horizontal force the horizontal component stays 2î, while at the same level the vertical component has the same size but reverses, becoming −3ĵ. So the velocity at B is 2î − 3ĵ; reversing both components would wrongly reverse the horizontal motion too.
Q15AIPMT 2012 Prelims
The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
Not quite — the answer is C.
Range R = u² sin 2θ/g = 2u² sin θ cos θ/g and maximum height H = u² sin²θ/(2g). Setting R = H gives 2 cos θ = sin θ/2, i.e. tan θ = 4, so θ = tan⁻¹ 4. An angle of 45° gives maximum range, not R = H.
Q16AIPMT 2012 Prelims
A particle has initial velocity (2î + 3ĵ) and acceleration (0.3î + 0.2ĵ). The magnitude of velocity after 10 seconds will be
Not quite — the answer is C.
v = u + at = (2 + 0.3 × 10)î + (3 + 0.2 × 10)ĵ = 5î + 5ĵ. Magnitude = √(25 + 25) = 5√2 units. Adding the components as scalars (5 + 5 or 9) is the trap.
Q17AIPMT 2011 Mains
A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
Not quite — the answer is D.
At the top the projectile is at height H and horizontal distance R/2. tan α = H/(R/2) = (u² sin²θ/2g)/(u² sinθ cosθ/g) = tanθ/2 = 1/2 for θ = 45°. So α = tan⁻¹(1/2); 45° would be the launch angle, not the elevation of the highest point.
Q18AIPMT 2011 Prelims
A particle moves in a circle of radius 5 cm with constant speed and time period 0.2π s. The acceleration of the particle is
Not quite — the answer is A.
ω = 2π/T = 2π/(0.2π) = 10 rad/s, and the centripetal acceleration is a = ω²r = 100 × 0.05 = 5 m/s². Forgetting to convert 5 cm to metres gives a value 100 times too large.
Q19AIPMT 2011 Prelims
A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is
Not quite — the answer is A.
The velocity changes from 30 m/s east to 40 m/s north; the change Δv has magnitude √(30² + 40²) = 50 m/s. Average acceleration = 50/10 = 5 m/s². Taking the difference of speeds (1 m/s²) ignores the change in direction.
Q20AIPMT 2011 Prelims
A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s², the range of the missile is
Not quite — the answer is B.
Maximum range occurs at 45°, R = u²/g = 20²/10 = 40 m. The value 20 m is the range for u²/2g, which is the maximum height formula at 90°, not the range.
Q21AIPMT 2010 Mains · Q72
The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
Not quite — the answer is A.
At the highest point only the horizontal component u cos θ remains. Setting u cos θ = u/2 gives cos θ = 1/2, so θ = 60°. An angle of 30° would leave a speed of (√3/2)u at the top.
Q22AIPMT 2010 Mains · Q74
A particle moves in x-y plane according to rule x = a sin ωt and y = a cosωt. The particle follows
Not quite — the answer is B.
Squaring and adding x = a sin ωt and y = a cos ωt gives x² + y² = a²(sin²ωt + cos²ωt) = a², the equation of a circle of radius a. An ellipse would need unequal amplitudes, and a straight line would need the two motions to be in phase.
Q23AIPMT 2010 Prelims · Q4
A particle has initial velocity (3î + 4ĵ) and has acceleration (0.4î + 0.3ĵ). Its speed after 10 s is
Not quite — the answer is C.
v = u + at = (3 + 0.4×10)î + (4 + 0.3×10)ĵ = 7î + 7ĵ. Speed = √(49 + 49) = 7√2 ≈ 9.9 units. Adding the initial speed 5 and the gain 5 to get 10 units wrongly adds magnitudes of non-parallel vectors.
Q24AIPMT 2007 Prelims · Q4
A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are (√3, 3). The path of the particle makes with the x-axis an angle of
Not quite — the answer is B.
For a straight line from the origin, tan θ = y/x = 3/√3 = √3, so θ = 60°. 30° would be the angle with the y-axis instead.
Q25AIPMT 2007 Prelims · Q32
A⃗ and B⃗ are two vectors and θ is the angle between them, if |A⃗ × B⃗| = √3 (A⃗ · B⃗), the value of θ is
Not quite — the answer is D.
AB sin θ = √3 AB cos θ gives tan θ = √3, so θ = 60°. 30° would come from tan θ = 1/√3.
Q26AIPMT 2006 Prelims · Q14
For angles of projection of a projectile at angles (45° − θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of :
Not quite — the answer is A.
The range is R = u² sin 2θ/g. For (45° − θ), sin(90° − 2θ) = cos 2θ; for (45° + θ), sin(90° + 2θ) = cos 2θ. The two ranges are therefore equal, a ratio of 1 : 1, for any θ.
Q27AIPMT 2006 Prelims · Q45
A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is :
Not quite — the answer is B.
After a complete lap the car is back at its start, so displacement and hence average velocity are zero. Average speed = distance/time = 2π × 100/62.8 = 10 m/s. Average speed cannot be zero for a moving car.
Q28AIPMT 2006 Prelims · Q49
The vectors A⃗ and B⃗ are such that |A⃗ + B⃗| = |A⃗ − B⃗|. The angle between the two vectors is :
Not quite — the answer is A.
Squaring both sides, A² + B² + 2A·B = A² + B² − 2A·B, so A·B = 0. The vectors are therefore perpendicular, at 90°. At 0° the sum would be larger than the difference.
Q29AIPMT 2005 Prelims · Q29
If a vector 2î + 3ĵ + 8k̂ is perpendicular to the vector 4ĵ − 4î + αk̂, then the value of α is :
Not quite — the answer is C.
Perpendicular vectors have zero dot product: 2(−4) + 3(4) + 8α = 0, so 4 + 8α = 0 and α = −1/2. Ignoring the sign of the î component gives the wrong −2 or 2.
Q30AIPMT 2005 Prelims · Q32
A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone ?
Not quite — the answer is C.
f = 22/44 = 0.5 rev/s, so ω = 2πf = π rad/s. a = ω²r = π² × 1 = π² m s⁻², directed along the radius towards the centre (centripetal). Acceleration of uniform circular motion is never tangential or outward.
Q31AIPMT 2005 Prelims · Q40
Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v₁. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is :
Not quite — the answer is B.
The catcher runs along the hypotenuse of a right triangle with legs a and v₁t: (vt)² = a² + (v₁t)², so t = √(a²/(v² − v₁²)). a/(v − v₁) would apply only if both ran along the same line.
Q32AIPMT 2005 Prelims · Q45
If the angle between the vectors A⃗ and B⃗ is θ, the value of the product (B⃗ × A⃗) · A⃗ is equal to :
Not quite — the answer is D.
B⃗ × A⃗ is perpendicular to A⃗, so its dot product with A⃗ is zero for any θ. The sin θ and cos θ forms wrongly treat the product as non-zero.
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