Physics — Mechanics for the NEET Exam — Indian candidates

Common failure modes

These are the patterns that cause most candidates to lose marks on this topic. Recognising them in advance is half the work.

• !Applying Newton's second law without correctly resolving all force components — especially on inclined planes
• !Confusing kinetic and static friction coefficients, or forgetting that friction ≤ μN (not always equal)
• !Misidentifying which moment of inertia formula applies (disc, sphere, cylinder, ring)
• !Orbital velocity and escape velocity formula confusion: v_orbital = √(GM/r); v_escape = √(2GM/r)
• !Elastic vs inelastic collision errors — forgetting that only momentum is conserved in all collisions; KE only conserved in elastic

Study tips

• 1Memorise all NEET-relevant kinematics equations (v = u + at, s = ut + ½at², v² = u² + 2as) and their application conditions (constant acceleration only).
• 2Drill the five standard moments of inertia for NEET: solid disc (MR²/2), ring/hoop (MR²), solid sphere (2MR²/5), hollow sphere (2MR²/3), thin rod through centre (ML²/12).
• 3For gravitation, practise problems involving orbital period (T² ∝ r³, Kepler's third law), orbital velocity, and binding energy.
• 4Solve projectile motion problems with both horizontal range and time-of-flight formulas — NEET tests both.
• 5Build unit-checking habits: every answer should pass dimensional analysis before selection.
• 6For candidates in India, NEET test windows are typically denser in the spring; book test centres in metro cities (Delhi, Mumbai, Bengaluru, Chennai, Kolkata) early to secure preferred dates.

Sample NEET Physics — Mechanics questions

These sample items mirror the format and difficulty of real NEET questions. Practice with thousands more on the free Koydo question bank.

1. 1

   A body of mass 5 kg is acted upon by a net force of 20 N. The acceleration produced is:

   • A2 m/s²
   • B4 m/s²Correct
   • C10 m/s²
   • D100 m/s²
   Why this answer?

   Newton's second law: F = ma. a = F/m = 20/5 = 4 m/s².

2. 2

   The escape velocity from the surface of a planet of mass M and radius R is:

   • A√(GM/R)
   • B√(2GM/R)Correct
   • C2√(GM/R)
   • DGM/R²
   Why this answer?

   Escape velocity is derived by setting kinetic energy equal to gravitational potential energy: ½mv² = GMm/R, so v = √(2GM/R). This is √2 times the orbital velocity (which is √(GM/R)).

3. 3

   A block slides down a frictionless incline at angle θ to the horizontal. Its acceleration is:

   • Ag
   • Bg cos θ
   • Cg sin θCorrect
   • Dg tan θ
   Why this answer?

   Resolving gravity along the frictionless incline: component along incline = mg sin θ. Newton's second law: ma = mg sin θ, so a = g sin θ.

Frequently asked questions

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