Physics — Mechanics for the JEE Main Exam — Korean 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 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).
• 6한국 응시자에게 JEE Main 대비의 핵심은 독해 속도와 듣기 정확도입니다 — 한국식 시험 문화와 다른 출제 패턴에 익숙해지세요.

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).

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