Download JEE MCQs for JEE Physics: System Of Particles and Rotational Motion
Access targeted multiple-choice questions for System Of Particles and Rotational Motion designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.
Chapter-wise Objective Questions: System Of Particles and Rotational Motion
View or download the dedicated System Of Particles and Rotational Motion MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.
Multiple Choice Questions
(a) \( 6ma^2 \)
(b) \( 12ma^2 \)
(c) \( 18ma^2 \)
(d) \( 24ma^2 \)
Show Answer & Explanation
Answer: (d) \( 24ma^2 \)
Explanation:
1. Distance of each mass from the Y-axis is its x-coordinate: 0, 2a, 4a, 2a.
2. I = m(0 + 4a² + 16a² + 4a²) = 24ma².
Teacher's Note:
1. Moment of inertia about an axis uses the perpendicular distance from that axis.
2. For the Y-axis this is the x-coordinate.
(a) \( 2m\ell^2 \)
(b) \( \dfrac{5}{4}m\ell^2 \)
(c) \( \dfrac{3}{2}m\ell^2 \)
(d) \( \dfrac{3}{4}m\ell^2 \)
Show Answer & Explanation
Answer: (b) \( \dfrac{5}{4}m\ell^2 \)
Explanation:
1. Particle A lies on the axis: distance 0.
2. Particle B is at distance ℓ from the axis.
3. Particle C is at horizontal distance ℓ/2 from the axis (its projection falls on the midpoint of AB).
4. I = m(0 + ℓ² + ℓ²/4) = 5mℓ²/4.
Teacher's Note:
1. Find the perpendicular distance of every particle from the axis.
2. The height of the triangle is not needed here.
(a) \( \dfrac{MR^2}{2} \)
(b) \( MR^2 \)
(c) \( \dfrac{3}{2}MR^2 \)
(d) \( 2MR^2 \)
Show Answer & Explanation
Answer: (c) \( \dfrac{3}{2}MR^2 \)
Explanation:
1. About the centre: I = ½MR².
2. Parallel-axis theorem with distance R: I = ½MR² + MR² = (3/2)MR².
Teacher's Note:
1. Use I = I_cm + Md².
2. The tangent is a distance R from the centre.
(a) \( \dfrac{2}{5}MR^2 \)
(b) \( \dfrac{MR^2}{2} \)
(c) \( \dfrac{3}{4}MR^2 \)
(d) \( \dfrac{5}{4}MR^2 \)
Show Answer & Explanation
Answer: (a) \( \dfrac{2}{5}MR^2 \)
Explanation:
1. For a uniform solid sphere I = (2/5)MR² about any diameter.
Teacher's Note:
1. Memorise: solid sphere 2/5, hollow sphere 2/3, ring 1, disc 1/2 (times MR²).
2. The sphere is symmetric, so every diameter gives the same value.
(a) Ring
(b) Disc
(c) Hollow sphere
(d) Solid sphere
Show Answer & Explanation
Answer: (d) Solid sphere
Explanation:
1. Values of I/MR²: ring 1, disc 1/2, hollow sphere 2/3, solid sphere 2/5.
2. The solid sphere has the smallest value.
Teacher's Note:
1. More mass near the axis means smaller I.
2. Compare the coefficients.
(a) 0.5 kg m²
(b) 5 kg m²
(c) 7.5 kg m²
(d) 10 kg m²
Show Answer & Explanation
Answer: (c) 7.5 kg m²
Explanation:
1. For a hollow cylinder I = MR² about its axis.
2. I = 30 × 0.25 = 7.5 kg m².
3. The length does not matter.
Teacher's Note:
1. Length is irrelevant for the geometrical axis.
2. Treat the hollow cylinder like a ring.
(a) \( \dfrac{ML^2}{12} \)
(b) \( \dfrac{ML^2}{6} \)
(c) \( \dfrac{ML^2}{3} \)
(d) \( ML^2 \)
Show Answer & Explanation
Answer: (c) \( \dfrac{ML^2}{3} \)
Explanation:
1. About the centre I = ML²/12.
2. Parallel axes, d = L/2: I = ML²/12 + M(L/2)² = ML²/3.
Teacher's Note:
1. Apply the parallel-axis theorem from the centre.
2. End axis gives four times the centre value.
(a) \( \dfrac{5}{3}m\ell^2 \)
(b) \( \dfrac{4}{3}m\ell^2 \)
(c) \( m\ell^2 \)
(d) None of these
Show Answer & Explanation
Answer: (a) \( \dfrac{5}{3}m\ell^2 \)
Explanation:
1. The rod along the x-axis contributes 0.
2. The opposite rod, at distance ℓ, contributes mℓ².
3. Each of the two side rods turns about its end: mℓ²/3, so together 2mℓ²/3.
4. I = mℓ² + 2mℓ²/3 = (5/3)mℓ².
Teacher's Note:
1. Add the contributions of all four rods separately.
2. A rod about its end is mℓ²/3.
(a) \( \dfrac{ML^2}{3} \)
(b) \( \dfrac{2ML^2}{3} \)
(c) \( \dfrac{ML^2}{6} \)
(d) \( ML^2 \)
Show Answer & Explanation
Answer: (b) \( \dfrac{2ML^2}{3} \)
Explanation:
1. The rod along z lies on the axis: I = 0.
2. The rods along x and y are perpendicular to the z-axis and turn about their ends: ML²/3 each.
3. Total = 2ML²/3.
Teacher's Note:
1. A rod along the axis adds nothing.
2. Use ML²/3 for rods turning about an end.
(a) \( \dfrac{m\ell^2}{12} \)
(b) \( \dfrac{m\ell^2}{6} \)
(c) \( \dfrac{m\ell^2}{3} \)
(d) \( \dfrac{m\ell^2}{2} \)
Show Answer & Explanation
Answer: (b) \( \dfrac{m\ell^2}{6} \)
Explanation:
1. Each rod about this axis (through its centre, perpendicular to it): mℓ²/12.
2. Total = 2 × mℓ²/12 = mℓ²/6.
Teacher's Note:
1. Moments of inertia about the same axis add.
2. The crossing does not change the individual values.
(a) \( \dfrac{2}{3}MR^2 \)
(b) \( \dfrac{1}{2}MR^2 \)
(c) \( \dfrac{2}{5}MR^2 \)
(d) \( \dfrac{2}{7}MR^2 \)
Show Answer & Explanation
Answer: (d) \( \dfrac{2}{7}MR^2 \)
Explanation:
1. The smallest moment of inertia occurs about an axis through the centre: (2/5)MR².
2. Any other axis gives a larger value, e.g. 2/3 or 1/2 MR² at suitable distances.
3. (2/7)MR² is less than (2/5)MR², so it is impossible.
Teacher's Note:
1. The centre-of-mass axis gives the minimum I (parallel-axis theorem).
2. Compare 2/7 with 2/5.
(a) \( \dfrac{ML^2}{6} \)
(b) \( \dfrac{\sqrt{2}ML^2}{24} \)
(c) \( \dfrac{ML^2}{24} \)
(d) \( \dfrac{ML^2}{12} \)
Show Answer & Explanation
Answer: (d) \( \dfrac{ML^2}{12} \)
Explanation:
1. Each half has mass M/2 and length L/2.
2. About an end, each half has I = (M/2)(L/2)²/3 = ML²/24.
3. Total I = 2 × ML²/24 = ML²/12.
Teacher's Note:
1. Treat each half as a rod about its end.
2. The angle between the halves does not matter for this axis.
(a) 4I
(b) 8I
(c) 12I
(d) 16I
Show Answer & Explanation
Answer: (d) 16I
Explanation:
1. Each rod has mass m, I = mℓ²/12.
2. For the square of side ℓ, a rod's centre is ℓ/2 from the centre: I_rod = mℓ²/12 + m(ℓ/2)² = mℓ²/3 = 4I.
3. Four rods: 4 × 4I = 16I.
Teacher's Note:
1. Use the parallel-axis theorem for each side.
2. Multiply the single-rod value by 4 twice.
(a) \( I_1 > I_2 > I_3 \)
(b) \( I_3 > I_2 > I_1 \)
(c) \( I_2 > I_1 > I_3 \)
(d) \( I_2 > I_3 > I_1 \)
Show Answer & Explanation
Answer: (b) \( I_3 > I_2 > I_1 \)
Explanation:
1. I₁ = (2/5)MR², I₂ = (2/3)MR², I₃ = MR².
2. So I₃ > I₂ > I₁.
Teacher's Note:
1. Compare the coefficients 2/5, 2/3 and 1.
2. The more the mass is pushed outward, the larger I.
(a) \( \dfrac{mr^2}{4} \)
(b) \( 2mr^2 \)
(c) \( \dfrac{mr^2}{2} \)
(d) \( 4mr^2 \)
Show Answer & Explanation
Answer: (c) \( \dfrac{mr^2}{2} \)
Explanation:
1. The centre of mass is at the midpoint, so each mass is r/2 from the axis.
2. I = 2 × m(r/2)² = mr²/2.
Teacher's Note:
1. Locate the centre of mass first.
2. Two identical terms add.
(a) \( MR^2 + 4mR^2 \)
(b) \( \dfrac{1}{2}MR^2 + 4mR^2 \)
(c) \( \dfrac{8}{5}MR^2 + 4mR^2 \)
(d) None of these
Show Answer & Explanation
Answer: (a) \( MR^2 + 4mR^2 \)
Explanation:
1. Ring: MR².
2. Each point mass is at distance R: mR², four give 4mR².
3. Total = MR² + 4mR².
Teacher's Note:
1. Moments of inertia about the same axis add.
2. The ring is thin, so I_ring = MR².
(a) \( \left(M + \dfrac{m}{4}\right)R^2 \)
(b) \( (M + m)R^2 \)
(c) \( (M + 3m)R^2 \)
(d) \( \left(\dfrac{M+m}{2}\right)R^2 \)
Show Answer & Explanation
Answer: (b) \( (M + m)R^2 \)
Explanation:
1. Ring: MR².
2. Each spoke turns about its end: mR²/3, so three spokes give mR².
3. Total = (M + m)R².
Teacher's Note:
1. Treat each spoke as a rod about one end.
2. Add the ring and the spokes.
(a) \( m\ell^2 \)
(b) \( \dfrac{m\ell^2}{4\pi^2} \)
(c) \( \dfrac{m\ell^2}{2\pi^2} \)
(d) \( \dfrac{m\ell^2}{8\pi^2} \)
Show Answer & Explanation
Answer: (b) \( \dfrac{m\ell^2}{4\pi^2} \)
Explanation:
1. Circumference ℓ = 2πR, so R = ℓ/2π.
2. I = mR² = mℓ²/4π².
Teacher's Note:
1. Convert the length into a radius first.
2. I = mR² for a ring.
(a) 1 : 2
(b) \( \sqrt{3} : \sqrt{5} \)
(c) \( \sqrt{5} : \sqrt{3} \)
(d) 5 : 4
Show Answer & Explanation
Answer: (c) \( \sqrt{5} : \sqrt{3} \)
Explanation:
1. (2/5)MR₁² = (2/3)MR₂².
2. R₁²/R₂² = (2/3)/(2/5) = 5/3.
3. R₁ : R₂ = √5 : √3.
Teacher's Note:
1. Equate the two moments of inertia.
2. Take the square root at the end.
(a) \( \dfrac{1}{2}MR^2 \)
(b) \( \dfrac{1}{4}MR^2 \)
(c) \( \dfrac{1}{8}MR^2 \)
(d) \( \sqrt{2}MR^2 \)
Show Answer & Explanation
Answer: (a) \( \dfrac{1}{2}MR^2 \)
Explanation:
1. A full disc of mass 4M has I = ½(4M)R² = 2MR².
2. The quarter section has one quarter of it: I = ½MR².
Teacher's Note:
1. By symmetry every quarter contributes equally.
2. The mass of the quarter is M, so the disc has mass 4M.
(a) \( \dfrac{M\ell^2}{18} \)
(b) \( \dfrac{M\ell^2}{12} \)
(c) \( \dfrac{M\ell^2}{6} \)
(d) \( \dfrac{M\ell^2}{4} \)
Show Answer & Explanation
Answer: (b) \( \dfrac{M\ell^2}{12} \)
Explanation:
1. For a triangle, I about its base = Mh²/6, where h is the height on that base.
2. Taking AC as the base: AC = ℓ√2 and the height from B is ℓ/√2, so h² = ℓ²/2.
3. I = M(ℓ²/2)/6 = Mℓ²/12.
Teacher's Note:
1. The hypotenuse of a right isosceles triangle is the base here.
2. I about the base of a triangular lamina is Mh²/6.
(a) AB
(b) BC
(c) EG
(d) FH
Show Answer & Explanation
Answer: (d) FH
Explanation:
1. The lamina is twice as long as it is wide. The axis FH runs along the long direction through the centre.
2. I about FH = M(AB)²/12, about EG = M(BC)²/12 = 4M(AB)²/12.
3. Edge axes AB and BC are larger still (parallel-axis theorem).
4. So FH gives the minimum.
Teacher's Note:
1. Mass nearer the axis means smaller I.
2. Compare M(width)²/12 for the centre axes.
(a) 20 kg m²
(b) 50 kg m²
(c) 30 kg m²
(d) zero
Show Answer & Explanation
Answer: (c) 30 kg m²
Explanation:
1. Perpendicular-axes theorem: I_y = I_x + I_z.
2. I_z = 50 − 20 = 30 kg m².
Teacher's Note:
1. The axis perpendicular to the plane has the largest moment.
2. Use I_perp = I_x + I_z for a plane body.
(a) 10 kg m²
(b) 30 kg m²
(c) 40 kg m²
(d) 60 kg m²
Show Answer & Explanation
Answer: (d) 60 kg m²
Explanation:
1. Perpendicular axis: I_perp = 30 = I_diameter + I_diameter, so each in-plane axis through the centre has 15 kg m².
2. I_perp = Ma²/6 = 30 gives Ma² = 180.
3. Axis along a side: I = 15 + M(a/2)² = 15 + 180/4 = 60 kg m².
Teacher's Note:
1. Use the perpendicular-axes theorem first, then the parallel-axis theorem.
2. The side is a/2 from the centre line.
Free study material for Physics
Practice MCQs for JEE Physics System Of Particles and Rotational Motion
JEE Physics System Of Particles and Rotational Motion Objective Test Questions
Review structured objective questions for JEE Physics System Of Particles and Rotational Motion. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.
NCERT-Aligned Objective Questions and Solutions
Built using the official NCERT book for JEE, these Physics objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.
Next Steps in Your Exam Preparation
Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.
FAQs
You can get most exhaustive JEE Physics System Of Particles and Rotational Motion MCQs Set 23 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.
Yes, our JEE Physics System Of Particles and Rotational Motion MCQs Set 23 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.
By solving our JEE Physics System Of Particles and Rotational Motion MCQs Set 23, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.
Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.
Yes, you can also access online interactive tests for JEE Physics System Of Particles and Rotational Motion MCQs Set 23 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.