NEET (UG) 2026 Physics - Unit 2: Kinematics Practice Test 60 MCQs

NEET (UG) 2026 Physics - Unit 2: Kinematics Practice Test

NEET (UG) 2026 Physics - Unit 2: Kinematics Practice Test

Targeted 60 MCQ Practice Module focusing on: Unit Vector, Resolution of Vectors, Relative Velocity, Motion in a Plane, Projectile Motion, and Uniform Circular Motion.

Select your answers and click Submit Assessment Answers at the bottom to calculate your total score and review explanations.

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SECTION I: Unit Vector [Questions 1 to 10]
1. The unit vector along A = 3i + 4j is:
Explanation: Unit vector = A/|A|. Here |A| = √(3² + 4²) = 5, so the unit vector is (3i + 4j)/5 = 0.6i + 0.8j.
2. A unit vector has:
Explanation: A unit vector is the ratio of a vector to its own magnitude, so it is dimensionless with magnitude exactly 1. It only specifies direction.
3. The unit vector along A = 2i − 2j + k is:
Explanation: |A| = √(4 + 4 + 1) = 3, so the unit vector is (2i − 2j + k)/3.
4. The value of i·i + j·j + k·k is:
Explanation: The dot product of a unit vector with itself is 1 (cos 0° = 1). Hence 1 + 1 + 1 = 3.
5. The cross product k × j equals:
Explanation: In the right-handed system, j × k = i, so reversing the order gives k × j = −i.
6. A unit vector perpendicular to (i + j) and lying in the xy-plane is:
Explanation: Dot product of (i + j) and (i − j) is 1 − 1 = 0, so they are perpendicular. Dividing by the magnitude √2 gives a unit vector.
7. If A = a i + 0.5 j is a unit vector, the positive value of a is:
Explanation: Unit vector means a² + (0.5)² = 1, so a² = 0.75 and a = √3/2.
8. The unit vector along the velocity v = 6i − 8j m/s is:
Explanation: |v| = √(36 + 64) = 10 m/s. Unit vector = v/10 = 0.6i − 0.8j.
9. A unit vector making an angle of 60° with the x-axis in the xy-plane is:
Explanation: A unit vector at angle θ with the x-axis is cosθ i + sinθ j. For 60°: cos 60° = 1/2 and sin 60° = √3/2.
10. The value of (i × j)·(j × k) is:
Explanation: i × j = k and j × k = i. Then k·i = 0 because the unit vectors are mutually perpendicular.
SECTION II: Resolution of Vectors [Questions 11 to 20]
11. A vector of magnitude 10 units makes an angle of 30° with the x-axis. Its x and y components are:
Explanation: Ax = A cos30° = 10 × (√3/2) = 5√3 and Ay = A sin30° = 10 × 1/2 = 5.
12. A vector of magnitude 20 has an x-component of 12. Its (positive) y-component is:
Explanation: Ay = √(A² − Ax²) = √(400 − 144) = √256 = 16.
13. A vector has equal x and y components of 5 units each. Its magnitude and angle with the x-axis are:
Explanation: Magnitude = √(5² + 5²) = 5√2. tanθ = Ay/Ax = 1, so θ = 45°.
14. A force of 100 N acts at 60° to the horizontal. Its vertical component is approximately:
Explanation: Fy = F sin60° = 100 × 0.866 = 86.6 N (equal to 50√3 N).
15. A block of mass m rests on a smooth incline of angle 30°. The component of its weight along the incline is:
Explanation: Component of weight along the incline = mg sinθ = mg sin30° = mg/2.
16. A vector A makes an angle θ with the y-axis. Its x-component is:
Explanation: When the angle is measured from the y-axis, the y-component is A cosθ and the x-component is A sinθ.
17. The magnitude of the resultant of two perpendicular vectors of magnitudes 6 and 8 is:
Explanation: R = √(6² + 8²) = √100 = 10.
18. Two vectors, each of magnitude A, act at an angle of 120°. The magnitude of the resultant is:
Explanation: R = √(A² + A² + 2A² cos120°) = √(2A² − A²) = A.
19. The maximum and minimum possible magnitudes of the resultant of two forces 7 N and 4 N are:
Explanation: Maximum occurs at 0° (7 + 4 = 11 N) and minimum at 180° (7 − 4 = 3 N).
20. The angle made by the vector i + √3 j with the x-axis is:
Explanation: tanθ = Ay/Ax = √3/1 = √3, so θ = 60°.
SECTION III: Relative Velocity [Questions 21 to 30]
21. Two trains move in opposite directions on parallel tracks at 60 km/h and 40 km/h. The velocity of one relative to the other is:
Explanation: For opposite directions, relative speed is the sum of the speeds: 60 + 40 = 100 km/h.
22. Car A moves at 80 km/h and car B at 60 km/h in the same direction. The velocity of A relative to B is:
Explanation: vAB = vA − vB = 80 − 60 = 20 km/h in the direction of motion.
23. Rain falls vertically at 10 m/s. A man walks horizontally at 10 m/s. The rain appears to him to fall at:
Explanation: Relative velocity of rain = √(10² + 10²) = 10√2 m/s. tanθ = 10/10 = 1, so θ = 45° with the vertical.
24. A boat with speed 4 m/s in still water heads perpendicular to the bank of a river flowing at 3 m/s. The resultant speed of the boat relative to the ground is:
Explanation: The two velocities are perpendicular: v = √(4² + 3²) = 5 m/s.
25. A boat can move at 10 m/s in still water. The river flows at 5 m/s. To reach the point directly opposite, the boat must head at what angle to the perpendicular to the bank (towards upstream)?
Explanation: Upstream component must cancel river flow: vb sinθ = vr, so sinθ = 5/10 = 1/2 and θ = 30°.
26. A river is 100 m wide. A boat with speed 5 m/s in still water crosses in the minimum possible time. That time is:
Explanation: Minimum time occurs when the boat heads perpendicular to the bank: t = width/vb = 100/5 = 20 s.
27. A river 100 m wide flows at 3 m/s. A boat of speed 4 m/s heads perpendicular to the bank. The drift downstream is:
Explanation: Time to cross = 100/4 = 25 s. Drift = vr × t = 3 × 25 = 75 m.
28. A man walks at 3 m/s in rain falling vertically at 4 m/s. To protect himself the umbrella should be tilted from the vertical by (take tan 37° = 3/4):
Explanation: Rain relative to the man has a horizontal component 3 m/s (backwards) and vertical 4 m/s. tanθ = 3/4, so θ = 37° from the vertical, tilted forward.
29. Particle A moves with 10 m/s due east and particle B with 10 m/s due north. The magnitude of the velocity of A relative to B is:
Explanation: vAB = vA − vB = 10i − 10j, whose magnitude is √(100 + 100) = 10√2 m/s.
30. An aeroplane has an air speed of 120 km/h due east and a wind blows at 50 km/h due north. The ground speed of the aeroplane is:
Explanation: The velocities are perpendicular, so ground speed = √(120² + 50²) = 130 km/h.
SECTION IV: Motion in a Plane [Questions 31 to 40]
31. The position of a particle is r = 2t i + t² j (SI units). Its speed at t = 2 s is:
Explanation: v = dr/dt = 2i + 2t j. At t = 2 s, v = 2i + 4j, so speed = √(4 + 16) = 2√5 m/s.
32. The position of a particle is r = 3t² i + 4t j (SI units). The magnitude of its acceleration is:
Explanation: v = 6t i + 4 j, a = dv/dt = 6i. Magnitude = 6 m/s².
33. A particle moves so that x = t and y = t². The shape of its path is:
Explanation: Eliminating t gives y = x², which is the equation of a parabola.
34. A particle has x = a cosωt and y = a sinωt. Its path is:
Explanation: x² + y² = a²(cos²ωt + sin²ωt) = a², which is a circle of radius a.
35. A particle has velocity v = 4i + 3j m/s. Its speed is:
Explanation: Speed = |v| = √(4² + 3²) = 5 m/s.
36. A particle starts with velocity 3i m/s and has constant acceleration 2j m/s². Its speed after 3 s is:
Explanation: v = u + at = 3i + 6j. Speed = √(9 + 36) = √45 = 3√5 m/s.
37. A particle moves from the origin to the point (3 m, 4 m). The magnitude of its displacement is:
Explanation: Displacement = √(3² + 4²) = 5 m.
38. A particle is at r1 = i + 2j at t = 0 and at r2 = 4i + 6j at t = 2 s. Its average velocity is:
Explanation: Average velocity = Δr/Δt = (3i + 4j)/2 = 1.5i + 2j m/s.
39. A particle's velocity changes from 2i m/s to 2j m/s in 4 s. The magnitude of its average acceleration is:
Explanation: Δv = 2j − 2i, so |Δv| = 2√2 m/s. Average acceleration = 2√2/4 = 1/√2 ≈ 0.71 m/s².
40. In two-dimensional motion, the direction of the instantaneous velocity at any point is:
Explanation: Instantaneous velocity is dr/dt, which is always tangential to the trajectory.
SECTION V: Projectile Motion [Questions 41 to 50]
41. For a projectile launched with a fixed speed on level ground, the horizontal range is maximum when the angle of projection is:
Explanation: R = u² sin2θ / g is maximum when sin2θ = 1, i.e. 2θ = 90° and θ = 45°.
42. A projectile fired at 30° has a certain range. Another projectile with the same speed gives the same range at an angle of:
Explanation: Ranges are equal for angles θ and (90° − θ), because sin2θ = sin(180° − 2θ). So 30° pairs with 60°.
43. A projectile is thrown at 20 m/s at 30° to the horizontal (g = 10 m/s²). Its time of flight is:
Explanation: T = 2u sinθ / g = 2 × 20 × 0.5 / 10 = 2 s.
44. For the same projectile (u = 20 m/s, θ = 30°, g = 10 m/s²), the maximum height reached is:
Explanation: H = u² sin²θ / 2g = 400 × (1/4) / 20 = 5 m.
45. For the same projectile (u = 20 m/s, θ = 30°, g = 10 m/s²), the horizontal range is approximately:
Explanation: R = u² sin2θ / g = 400 × sin60° / 10 = 40 × 0.866 = 20√3 ≈ 34.6 m.
46. At the highest point of its trajectory, the acceleration of a projectile (ignoring air resistance) is:
Explanation: Gravity acts throughout the flight. Only the vertical velocity is zero at the top; acceleration remains g downward.
47. For a projectile launched at 45°, the ratio of horizontal range to maximum height (R/H) is:
Explanation: R/H = 4 cotθ. For θ = 45°, cot45° = 1, so R/H = 4.
48. A stone is thrown horizontally from the top of a 80 m high cliff (g = 10 m/s²). The time taken to reach the ground is:
Explanation: For horizontal projection, h = ½gt², so t = √(2h/g) = √(160/10) = 4 s.
49. A ball is thrown horizontally at 20 m/s from the top of a 45 m high tower (g = 10 m/s²). The horizontal distance it travels before hitting the ground is:
Explanation: t = √(2 × 45/10) = 3 s. Range = u × t = 20 × 3 = 60 m.
50. For what angle of projection is the horizontal range of a projectile equal to its maximum height?
Explanation: R = H means 2u² sinθ cosθ / g = u² sin²θ / 2g, which simplifies to 4 cosθ = sinθ. Hence tanθ = 4 and θ = tan⁻¹(4) ≈ 76°.
SECTION VI: Uniform Circular Motion [Questions 51 to 60]
51. A particle completes 5 revolutions in 10 s while moving uniformly in a circle. Its angular velocity is:
Explanation: Frequency = 5/10 = 0.5 Hz. ω = 2πf = 2π × 0.5 = π rad/s.
52. A particle moves in a circle of radius 5 m with a uniform speed of 10 m/s. Its centripetal acceleration is:
Explanation: a = v²/r = 100/5 = 20 m/s².
53. A wheel rotates at 50 revolutions per minute. Its angular velocity in rad/s is:
Explanation: 50 rpm = 50/60 rev/s. ω = 2π × (50/60) = 5π/3 rad/s.
54. In uniform circular motion, the direction of the acceleration is:
Explanation: Speed is constant, so there is no tangential acceleration. The only acceleration is centripetal, directed towards the centre.
55. A particle moves in a circle of radius 2 m with angular velocity 3 rad/s. Its linear speed is:
Explanation: v = rω = 2 × 3 = 6 m/s.
56. The ratio of the angular speed of the hour hand to that of the minute hand of a clock is:
Explanation: The hour hand completes one rotation in 12 h and the minute hand in 1 h, so the angular speeds are in the ratio 1 : 12.
57. A particle moves in a circle of radius 0.5 m with a constant angular velocity of 2 rad/s. The magnitude of its centripetal acceleration is:
Explanation: a = ω²r = 4 × 0.5 = 2 m/s².
58. A particle moves in a circle with uniform speed v. The magnitude of the change in its velocity after half a revolution is:
Explanation: After half a turn the velocity reverses direction: Δv = (−v) − (v) = −2v, so the magnitude is 2v.
59. Which of the following quantities remains constant in uniform circular motion?
Explanation: Speed (and angular speed) is constant. Velocity, acceleration and momentum are vectors whose directions keep changing.
60. A particle in uniform circular motion makes 20 revolutions per minute. Its time period is:
Explanation: Frequency = 20/60 = 1/3 Hz. T = 1/f = 3 s.

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