Physics — Mechanics for the JEE Main Exam

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 resolving forces into components in the chosen inertial frame
• !Confusing moment of inertia with torque — missing the parallel-axis theorem when the pivot is off-centre
• !Sign errors in work-energy theorem when friction and external forces act simultaneously
• !Forgetting that gravitational potential energy is negative (bound system)
• !Treating pseudo-forces as real forces in non-inertial frames without explicitly labelling the frame

Study tips

• 1Draw free-body diagrams for every mechanics problem before writing any equation — JEE Advanced markers look for methodical setup.
• 2Memorise the standard moments of inertia (disc, sphere, rod, hollow cylinder) and the parallel-axis theorem for off-centre pivots.
• 3Practice converting between translational and rotational variables (v = rω, a = rα) until it is automatic.
• 4For JEE Advanced, attempt integer-type mechanics problems daily; they penalise guessing and reward exact reasoning.
• 5Review SHM as a special case of mechanics — it bridges to oscillations and electromagnetism (LC circuits).

Sample JEE Main Physics — Mechanics questions

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

1. 1

   A disc of mass M and radius R rolls without slipping on a horizontal surface. Its total kinetic energy is:

   • A(1/2)Mv²
   • B(3/4)Mv²Correct
   • C(2/3)Mv²
   • D(5/4)Mv²
   Why this answer?

   Illustrative JEE-style: Total KE = translational KE + rotational KE = (1/2)Mv² + (1/2)Iω². For a disc, I = (1/2)MR², and for rolling without slipping ω = v/R. So KE = (1/2)Mv² + (1/4)Mv² = (3/4)Mv².

2. 2

   A block of mass m is placed on a frictionless incline at angle θ. The acceleration of the block is:

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

   Illustrative JEE-style: Resolving forces along the incline, the net force is mg sin θ (gravity component along incline). Since the surface is frictionless, Newton's second law gives ma = mg sin θ, so a = g sin θ.

3. 3

   A particle moves in a circle of radius r with uniform speed v. The magnitude of centripetal acceleration is:

   • Av/r
   • Bv²/rCorrect
   • Cv²r
   • Dvr²
   Why this answer?

   Illustrative JEE-style: For uniform circular motion, the centripetal acceleration is directed toward the centre and has magnitude a = v²/r. This is derived from the rate of change of the velocity vector direction.

4. 4

   Two masses m₁ and m₂ are connected by a light string over a frictionless pulley (Atwood machine). The acceleration of the system is:

   • A(m₁ − m₂)g / (m₁ + m₂)Correct
   • B(m₁ + m₂)g / (m₁ − m₂)
   • C(m₁ × m₂)g / (m₁ + m₂)
   • Dg / 2
   Why this answer?

   Illustrative JEE-style: Applying Newton's second law to each mass and assuming m₁ > m₂, the net driving force is (m₁ − m₂)g and the total inertia is (m₁ + m₂). Hence a = (m₁ − m₂)g / (m₁ + m₂).