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JEE Main Mechanics: practice questions with solutions

22 questions on Circular Motion, Collisions, Elasticity, Fluid Mechanics, Friction, Gravitation, Kinematics, Laws of Motion, Oscillations (SHM), Relative Motion, Rotational Motion, Units & Errors, Work, Energy & Power. Try each one first, then open the solution.

Tip: solve on paper before opening a solution. Once you open it, that question is counted as practised and won't appear in your HeyGyan tests.
KinematicsMedium

Q1. A projectile's horizontal range is 4√3 times its maximum height. What is its angle of projection with the horizontal?

  1. 30°
  2. 45°
  3. 60°
  4. 75°
Show answer and solution

Answer: (A) 30°

SolutionR = u²sin2θ/g and H = u²sin²θ/2g, so R/H = 4/tanθ. Setting 4/tanθ = 4√3 gives tanθ = 1/√3, so θ = 30°.
Common trapMany students remember R = 4H at 45° and pick 45° without redoing the ratio.
Rotational MotionMedium

Q2. A solid sphere rolls without slipping down an incline of 30°. What is the acceleration of its centre of mass?

  1. g/2
  2. 5g/14
  3. 5g/7
  4. 2g/7
Show answer and solution

Answer: (B) 5g/14

SolutionFor rolling, a = g sinθ / (1 + k²/R²). For a solid sphere k²/R² = 2/5. So a = (g/2)/(7/5) = 5g/14.
Common trapg/2 ignores rotation. 5g/7 is the formula's factor before multiplying by sin30°.
GravitationMedium

Q3. A satellite orbiting close to Earth's surface has period T₀. What is the period of a satellite orbiting at a height 3R above the surface (R = Earth's radius)?

  1. 3T₀
  2. 4T₀
  3. 8T₀
  4. 3√3 T₀
Show answer and solution

Answer: (C) 8T₀

SolutionKepler's law: T ∝ r^(3/2). The orbit radius is R + 3R = 4R, so T = (4)^(3/2) T₀ = 8T₀.
Common trapUsing the height (3R) instead of the orbit radius (4R) gives 3√3 T₀.
Work, Energy & PowerMediumNumerical

Q4. A 2 kg block is released from rest at a height of 5 m on a smooth curved track. At the bottom it enters a rough horizontal floor with μ = 0.25. How far (in m) does it slide on the floor before stopping?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 20

SolutionAll potential energy is lost to friction on the floor: mgh = μmg·d. Mass and g cancel, so d = h/μ = 5/0.25 = 20 m.
Common trapMass is a distractor here; it cancels out.
Oscillations (SHM)Medium

Q5. A particle performs SHM with amplitude A. At what distance from the mean position is its kinetic energy three times its potential energy?

  1. A/4
  2. A/2
  3. A/√2
  4. √3A/2
Show answer and solution

Answer: (B) A/2

SolutionKE = ½k(A² − x²) and PE = ½kx². Setting A² − x² = 3x² gives x² = A²/4, so x = A/2.
Common trap√3A/2 is where PE is three times KE, the reverse condition.
Units & ErrorsEasy

Q6. A quantity is P = a²b³ / (c·√d). The percentage errors in a, b, c and d are 1%, 2%, 3% and 4%. What is the maximum percentage error in P?

  1. 7%
  2. 10%
  3. 13%
  4. 14%
Show answer and solution

Answer: (C) 13%

SolutionErrors add with powers as weights: 2(1) + 3(2) + 1(3) + ½(4) = 2 + 6 + 3 + 2 = 13%.
Common trapSubtracting the errors of quantities in the denominator. Errors always add.
Laws of MotionMedium

Q7. Blocks of 4 kg and 2 kg are in contact on a horizontal floor (μ = 0.1 for both). A horizontal force of 18 N pushes the 4 kg block, which pushes the 2 kg block. What is the contact force between the blocks? (g = 10 m/s²)

  1. 4 N
  2. 6 N
  3. 8 N
  4. 12 N
Show answer and solution

Answer: (B) 6 N

SolutionTotal friction = 0.1 × 6 × 10 = 6 N, so a = (18 − 6)/6 = 2 m/s². For the 2 kg block: N − 0.1 × 2 × 10 = 2 × 2, so N = 6 N.
Common trap4 N ignores friction on the 2 kg block. It still needs force to overcome its own friction.
Circular MotionMedium

Q8. A stone tied to a string is whirled in a vertical circle. What is the difference between the string's tension at the lowest point and at the highest point?

  1. 2mg
  2. 4mg
  3. 5mg
  4. 6mg
Show answer and solution

Answer: (D) 6mg

SolutionAt the bottom: T₁ − mg = mv₁²/l. At the top: T₂ + mg = mv₂²/l. Energy conservation gives v₁² − v₂² = 4gl. Subtracting: T₁ − T₂ = 2mg + 4mg = 6mg. This holds at any speed.
Common trap5mg comes from mixing up the minimum-speed condition v = √(5gl) with the tension difference.
CollisionsHardNumerical

Q9. A ball is dropped from a height of 20 m onto a hard floor. The coefficient of restitution is 1/√2. What total distance (in m) does the ball travel before coming to rest?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 60

SolutionEach bounce height is e² = 1/2 of the previous one. Distance = h + 2(e²h + e⁴h + …) = h(1 + e²)/(1 − e²) = 20 × 1.5/0.5 = 60 m.
Common trapForgetting that the ball travels each bounce height twice, up and down.
Fluid MechanicsMedium

Q10. Water fills a tall tank on the ground to height H. A small hole is made in the side wall. For the water jet to land farthest from the tank, the hole should be at what depth below the water surface?

  1. H/4
  2. H/2
  3. 3H/4
  4. H/3
Show answer and solution

Answer: (B) H/2

SolutionAt depth h, the jet speed is √(2gh) and it falls a height (H − h). Range R = 2√(h(H − h)), which is greatest when h = H/2. Maximum range = H.
Common trapAssuming the deepest hole always gives the longest jet. It is fastest but has little height to fall.
Rotational MotionMedium

Q11. What is the moment of inertia of a thin uniform disc (mass M, radius R) about a tangent lying in the plane of the disc?

  1. MR²/2
  2. 3MR²/4
  3. 5MR²/4
  4. 3MR²/2
Show answer and solution

Answer: (C) 5MR²/4

SolutionAbout a diameter, I = MR²/4. The parallel-axis theorem shifts the axis by R: I = MR²/4 + MR² = 5MR²/4.
Common trap3MR²/2 is for a tangent perpendicular to the disc's plane.
KinematicsMedium

Q12. A particle moves along a straight line with velocity v = 3t² − 6t (m/s). What distance does it cover in the first 3 seconds?

  1. 0 m
  2. 4 m
  3. 8 m
  4. 12 m
Show answer and solution

Answer: (C) 8 m

SolutionPosition x = t³ − 3t² (starting at 0). The particle stops and turns back at t = 2 s, where x = −4 m. At t = 3 s, x = 0. Distance = 4 + 4 = 8 m.
Common trap0 m is the displacement. Distance counts both legs of the journey.
Relative MotionEasy

Q13. A swimmer can swim at 5 km/h in still water. A river 1 km wide flows at 3 km/h. If she crosses along the shortest path (straight across), how long does it take?

  1. 12 min
  2. 15 min
  3. 20 min
  4. 25 min
Show answer and solution

Answer: (B) 15 min

SolutionTo go straight across, part of her velocity cancels the current. Her speed across = √(5² − 3²) = 4 km/h. Time = 1/4 h = 15 min.
Common trap12 min uses 5 km/h directly, which is the shortest-time route, not the shortest path.
FrictionMedium

Q14. A block just begins to slide on a rough incline when the incline makes 37° with the horizontal. If the incline is raised to 53°, what is the block's acceleration? (g = 10 m/s², sin 37° = 3/5)

  1. 2.5 m/s²
  2. 3.5 m/s²
  3. 4.5 m/s²
  4. 8 m/s²
Show answer and solution

Answer: (B) 3.5 m/s²

SolutionSliding just begins at the angle of repose, so μ = tan 37° = 3/4. At 53°: a = g(sin 53° − μ cos 53°) = 10(0.8 − 0.75 × 0.6) = 3.5 m/s².
Common trap8 m/s² is g sin 53°, which ignores friction.
Rotational MotionMedium

Q15. A skater spinning freely pulls in her arms so that her moment of inertia becomes one-fourth. Her rotational kinetic energy becomes:

  1. one-fourth
  2. unchanged
  3. 4 times
  4. 16 times
Show answer and solution

Answer: (C) 4 times

SolutionAngular momentum Iω is conserved, so ω becomes 4 times. KE = L²/2I, with L fixed and I one-fourth, so KE becomes 4 times. The extra energy comes from her muscles.
Common trapAssuming kinetic energy is conserved along with angular momentum.
GravitationHard

Q16. What is the ratio of the acceleration due to gravity at a depth R/2 below Earth's surface to that at a height R/2 above it? (R = Earth's radius)

  1. 9 : 8
  2. 8 : 9
  3. 1 : 1
  4. 3 : 2
Show answer and solution

Answer: (A) 9 : 8

SolutionAt depth d: g(1 − d/R) = g/2. At height h: g/(1 + h/R)² = g/(3/2)² = 4g/9. Ratio = (1/2)/(4/9) = 9/8.
Common trapUsing the approximation g(1 − 2h/R) for height, which only works when h is much smaller than R.
ElasticityEasy

Q17. A wire stretches by 1 mm under a load. A second wire of the same material has twice the length and twice the diameter. How much does it stretch under the same load?

  1. 0.25 mm
  2. 0.5 mm
  3. 1 mm
  4. 2 mm
Show answer and solution

Answer: (B) 0.5 mm

SolutionExtension ΔL = FL/(AY). Doubling the length doubles ΔL; doubling the diameter makes the area 4 times, dividing ΔL by 4. Net factor = 2/4 = ½, so 0.5 mm.
Common trapTreating diameter like area. Area grows with the square of the diameter.
KinematicsEasy

Q18. A ball is thrown horizontally at 20 m/s from the top of a 45 m high cliff. How far from the foot of the cliff does it land? (g = 10 m/s²)

  1. 30 m
  2. 45 m
  3. 60 m
  4. 90 m
Show answer and solution

Answer: (C) 60 m

SolutionTime to fall 45 m: t = √(2h/g) = √9 = 3 s. Horizontal speed stays 20 m/s, so distance = 20 × 3 = 60 m.
Common trap45 m confuses the cliff height with the horizontal distance.
Work, Energy & PowerEasy

Q19. A 1 kg block moving at 2 m/s on a smooth floor hits a spring of force constant 100 N/m. What is the maximum compression of the spring?

  1. 0.02 m
  2. 0.1 m
  3. 0.2 m
  4. 0.4 m
Show answer and solution

Answer: (C) 0.2 m

SolutionAll kinetic energy becomes spring energy: ½mv² = ½kx². x = v√(m/k) = 2 × √(1/100) = 0.2 m.
Common trapUsing mv = kx (a force balance) instead of energy conservation gives 0.02 m.
Laws of MotionMedium

Q20. A 10 g bullet moving at 400 m/s embeds itself in a 2 kg wooden block resting on a frictionless surface. What is the velocity of the block just after?

  1. ≈ 0.2 m/s
  2. ≈ 2 m/s
  3. ≈ 4 m/s
  4. ≈ 20 m/s
Show answer and solution

Answer: (B) ≈ 2 m/s

SolutionMomentum is conserved: 0.01 × 400 = (2 + 0.01)v, so v = 4/2.01 ≈ 1.99 m/s.
Common trapForgetting to convert 10 g to 0.01 kg.
Rotational MotionHard

Q21. A uniform rod of length L, hinged at one end, is released from rest in the horizontal position. What is its angular velocity as it passes the vertical position?

  1. √(g/L)
  2. √(2g/L)
  3. √(3g/L)
  4. √(6g/L)
Show answer and solution

Answer: (C) √(3g/L)

SolutionThe centre of mass falls L/2. Energy: mg(L/2) = ½ × (mL²/3) × ω², so ω² = 3g/L and ω = √(3g/L).
Common trap√(2g/L) treats the rod as a point mass at its end. A rod's mass is spread out, so use I = mL²/3 about the hinge.
Circular MotionMediumNumerical

Q22. A road curve of radius 90 m is banked at 45°. At what speed (in m/s) can a car take the curve without needing any friction? (g = 10 m/s²)

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 30

SolutionFor no friction, tan θ = v²/(rg). With tan 45° = 1: v² = 90 × 10 = 900, so v = 30 m/s.
Common trapForgetting the square root and answering 900.

Frequently asked questions

Which Mechanics chapters are covered here?

This page covers Circular Motion, Collisions, Elasticity, Fluid Mechanics, Friction, Gravitation, Kinematics, Laws of Motion, Oscillations (SHM), Relative Motion, Rotational Motion, Units & Errors, Work, Energy & Power. Questions follow the JEE Main pattern, with multiple-choice and numerical value types.

Are these previous year JEE questions?

No. These are original JEE Main-level practice questions written on the latest pattern, with solutions and the common trap for each.