System of Particles and Rotational Motion NEET CBT Mock Test

Physics · Chapter test · 25 questions
System of Particles and Rotational Motion NEET CBT Mock Test

25 questions in 25 minutes on the NTA computer-based test interface, all from System of Particles and Rotational Motion. +4 / -1 marking, instant score, full solutions. Free, no login.

A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from System of Particles and Rotational Motion, of which 19 are real NEET previous-year questions and 9 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.

Revising first? Read the System of Particles and Rotational Motion chapter notes, then come back and take this test to check it stuck.

NEET goes computer-based from 2027. Until NTA releases the 2027 bulletin, this mock follows the 2025-26 pattern: 180 questions, +4 / -1.

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Mock CBT Test · NEET 2027NEET 2027 CBT Mock Test
GENERAL INSTRUCTIONS

Duration: 180 minutes · 180 questions · 720 marks maximum

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Mock CBT Test · NEET 2027NEET 2027 CBT Mock
InstructionsQuestion Paper Time Left : 180:00
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NEET CBT mock test: common questions

How many questions are in this System of Particles and Rotational Motion mock test?

25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 19 of them are NEET previous-year questions.

Is this the same interface as the real NEET CBT?

Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.

Should I take this before or after revising System of Particles and Rotational Motion?

Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.

All 25 questions with answers and solutions

The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.

Show all 25 questions with answers and solutions

Physics

Questions 1 to 25 · 25 questions

  1. Q1
    The ratio of the radii of gyration of a circular disc to that of a circular ring, each of the same mass and radius, about their respective central perpendicular axes is:
    1. √(3) : √(2)
    2. 1 : √(2)
    3. √(2) : 1
    4. √(2) : √(3)

    Answer: (B) 1 : √(2)

    k = √(I/M). Disc: I = (1)/(2)MR² ⇒ k_d = R/√(2). Ring: I = MR² ⇒ kᵣ = R. So k_d : kᵣ = (R)/(√(2)) : R = 1 : √(2).

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  2. Q2
    The angular momentum of a body changes from L to 4L while its moment of inertia stays constant. By what factor does its rotational kinetic energy change?
    1. 16
    2. 8
    3. 2
    4. 4

    Answer: (A) 16

    With I constant, KE = L²/2I, so KE ∝ L². Increasing L by a factor of 4 increases KE by 4² = 16.

    Chapter: System of Particles and Rotational Motion

  3. Q3
    A wheel has angular acceleration of 3 rad/s² and an initial angular speed of 2 rad/s. In a time of 2 s it has rotated through an angle (in radian) of:
    1. 4
    2. 6
    3. 12
    4. 10

    Answer: (D) 10

    θ = ω₀ t + (1)/(2)α t² = 2(2) + (1)/(2)(3)(2)² = 4 + 6 = 10 rad.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  4. Q4
    A solid sphere rolls without slipping down an incline of height h. Its speed at the bottom is:
    1. √(6gh/5)
    2. √(10gh/7)
    3. √(4gh/3)
    4. √(2gh)

    Answer: (B) √(10gh/7)

    v = √[2gh/(1 + k²/R²)]. For a solid sphere k²/R² = 2/5, so v = √[2gh/(1 + 2/5)] = √[2gh/(7/5)] = √(10gh/7).

    Chapter: System of Particles and Rotational Motion

  5. Q5
    An automobile moves on a road with speed 54 km/h. The radius of its wheels is 0.45 m and the moment of inertia of each wheel about its axis is 3 kg·m². If the vehicle is brought to rest in 15 s, the magnitude of the average torque transmitted by its brakes to a wheel is:
    1. 6.66 kg·m² s⁻²
    2. 2.86 kg·m² s⁻²
    3. 8.58 kg·m² s⁻²
    4. 10.86 kg·m² s⁻²

    Answer: (A) 6.66 kg·m² s⁻²

    v = 54 km/h = 15 m/s, so ω₀ = v/R = 15/0.45 = 33.33 rad/s. α = (0 – ω₀)/(15) = -2.22 rad/s². τ = I|α| = 3 × 2.22 = 6.66 kg·m² s⁻².

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  6. Q6
    A torque of 5 N·m maintains a body rotating at a constant angular speed of 12 rad/s. The power delivered is:
    1. 17 W
    2. 60 W
    3. 7 W
    4. 2.4 W

    Answer: (B) 60 W

    Rotational power P = τω = 5 × 12 = 60 W.

    Chapter: System of Particles and Rotational Motion

  7. Q7
    A uniform rod of length l and mass m is free to rotate in a vertical plane about one end A. Released from a horizontal position, its initial angular acceleration is (moment of inertia of rod about A is (ml²)/(3)):
    1. (3g)/(2l)
    2. (2l)/(3g)
    3. (2g)/(3l)
    4. (3g)/(2l²)

    Answer: (A) (3g)/(2l)

    Torque about A from the weight at the centre: τ = mg(l)/(2). With I = (ml²)/(3), α = (mgl/2)/(ml²/3) = (3g)/(2l).

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  8. Q8
    A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about a horizontal axis. A massless string wound round the cylinder carries a hanging mass. The tension in the string required to produce an angular acceleration of 2 rev/s² is:
    1. 50 N
    2. 78.5 N
    3. 25 N
    4. 157 N

    Answer: (D) 157 N

    α = 2 rev/s² = 4π rad/s². For a solid cylinder I = (1)/(2)MR² = (1)/(2)(50)(0.5)² = 6.25 kg·m². Torque TR = Iα, so T = (Iα)/(R) = (6.25 × 4π)/(0.5) = 50π ≈ 157 N.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  9. Q9
    Assertion (A): If the net external torque on a system is zero, its angular momentum is conserved. Reason (R): Torque is the rate of change of angular momentum. Choose the correct option.
    1. A is true but R is false
    2. Both A and R are true and R is the correct explanation of A
    3. A is false but R is true
    4. Both A and R are true but R is not the correct explanation of A

    Answer: (B) Both A and R are true and R is the correct explanation of A

    Since τ = dL/dt, zero net torque means dL/dt = 0, so L is constant. The reason correctly explains the assertion.

    Chapter: System of Particles and Rotational Motion

  10. Q10
    The moment of inertia of a thin uniform rod of mass M and length L about an axis through its mid-point and perpendicular to its length is I₀. Its moment of inertia about an axis through one end and perpendicular to its length is:
    1. I₀ + ML²
    2. I₀ + 2ML²
    3. (I₀ + ML²)/(2)
    4. I₀ + (ML²)/(4)

    Answer: (D) I₀ + (ML²)/(4)

    By the parallel-axis theorem, shifting from the centre to the end (distance L/2): I = I₀ + M((L)/(2))² = I₀ + (ML²)/(4).

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  11. Q11
    The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing through its edge and normal (perpendicular) to the disc is:
    1. (3)/(2)MR²
    2. MR²
    3. (7)/(2)MR²
    4. (1)/(2)MR²

    Answer: (A) (3)/(2)MR²

    About the central perpendicular axis I_(cm) = (1)/(2)MR². By the parallel-axis theorem (shift R to the edge): I = (1)/(2)MR² + MR² = (3)/(2)MR².

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  12. Q12
    A rod of length 3 m has a mass per unit length directly proportional to the distance x from one end. The centre of gravity of the rod from that end is at:
    1. 1.5 m
    2. 2.5 m
    3. 3 m
    4. 2 m

    Answer: (D) 2 m

    With λ = kx, x_(cg) = (∫₀^L x(kx) dx)/(∫₀^L (kx) dx) = (∫₀^L x² dx)/(∫₀^L x dx) = (L³/3)/(L²/2) = (2L)/(3). For L = 3 m, x_(cg) = 2 m. The COM is pulled toward the denser (far) end.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  13. Q13
    In a carbon monoxide molecule the carbon and oxygen atoms are separated by 1.12 × 10⁻¹⁰ m. The distance of the centre of mass from the carbon atom is (take C = 12 amu, O = 16 amu):
    1. 0.64 × 10⁻¹⁰ m
    2. 0.51 × 10⁻¹⁰ m
    3. 0.56 × 10⁻¹⁰ m
    4. 0.48 × 10⁻¹⁰ m

    Answer: (A) 0.64 × 10⁻¹⁰ m

    Take carbon at the origin and oxygen at r = 1.12 × 10⁻¹⁰ m. x_(cm) = (12(0) + 16(r))/(12 + 16) = (16)/(28)r = (4)/(7)(1.12 × 10⁻¹⁰) = 0.64 × 10⁻¹⁰ m from the carbon atom.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  14. Q14
    From a disc of radius R and mass M, a circular hole of diameter R whose rim passes through the centre is cut out. The moment of inertia of the remaining part of the disc about a perpendicular axis through the centre is:
    1. (11MR²)/(32)
    2. (13MR²)/(32)
    3. (15MR²)/(32)
    4. (9MR²)/(32)

    Answer: (B) (13MR²)/(32)

    Hole radius = R/2, its mass m’ = M/4 (area ratio), centred at R/2 from the centre. I_(full) = (1)/(2)MR². Iₕₒₗₑ = (1)/(2)m'(R/2)² + m'(R/2)² = m'((R²)/(8) + (R²)/(4)) = (M)/(4)·(3R²)/(8) = (3MR²)/(32). Remaining = (16MR²)/(32) – (3MR²)/(32) = (13MR²)/(32).

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  15. Q15
    In rolling without slipping, the velocity of the point of contact with the ground is:
    1. Zero
    2. Equal to Rω directed forward
    3. Equal to v_cm
    4. Equal to 2v_cm

    Answer: (A) Zero

    In pure rolling the contact point is the instantaneous axis of rotation, so it is momentarily at rest: its translational velocity v_cm and rotational velocity Rω are equal and opposite, giving zero. (The top point moves at 2v_cm.)

    Chapter: System of Particles and Rotational Motion

  16. Q16
    Find the torque of the force F = -3i + j + 5k N acting at the point r = 7i + 3j + k m.
    1. -14i + 3j – 16k
    2. 4i + 4j + 6k
    3. -21i + 3j + 5k
    4. 14i – 38j + 16k

    Answer: (D) 14i – 38j + 16k

    τ = r × F = (7,3,1) × (-3,1,5). τₓ = 3(5) – 1(1) = 14; τ_y = 1(-3) – 7(5) = -38; τ_z = 7(1) – 3(-3) = 16. So τ = 14i – 38j + 16k.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  17. Q17
    Three particles, each of mass m grams, are situated at the vertices of an equilateral triangle ABC of side l cm. The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC is (in g·cm²):
    1. (3)/(4)ml²
    2. 2ml²
    3. (3)/(2)ml²
    4. (5)/(4)ml²

    Answer: (D) (5)/(4)ml²

    Take A at origin, AB along the x-axis. AX is the y-axis, so the perpendicular distance of each mass is its x-coordinate. A: x = 0; B: x = l; C: x = lcos 60° = l/2. I = m(0)² + m(l)² + m(l/2)² = ml² + (ml²)/(4) = (5)/(4)ml².

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  18. Q18
    Two discs of the same moment of inertia I rotate about their regular axis (through centre, perpendicular to plane) with angular velocities ω₁ and ω₂. They are brought into contact face to face, coinciding their axes of rotation. The loss of energy in this process is:
    1. (1)/(2)I(ω₁ + ω₂)²
    2. I(ω₁ – ω₂)²
    3. (1)/(4)I(ω₁ – ω₂)²
    4. (1)/(8)I(ω₁ – ω₂)²

    Answer: (C) (1)/(4)I(ω₁ – ω₂)²

    Common final speed ω = (Iω₁ + Iω₂)/(2I) = (ω₁ + ω₂)/(2). Loss = (1)/(2)Iω₁² + (1)/(2)Iω₂² – (1)/(2)(2I)ω² = (1)/(4)I(ω₁ – ω₂)².

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  19. Q19
    A particle of mass m moves in the XY-plane with a velocity v along the straight line AB. If its angular momentum about the origin O is L_A when it is at A and L_B when it is at B, then:
    1. L_A = L_B
    2. the relation depends on the slope of line AB
    3. L_A > L_B
    4. L_A < L_B

    Answer: (A) L_A = L_B

    L = r × mv, and L = mv × (perpendicular distance from O to line AB). That perpendicular distance is the same for every point on the line, so L_A = L_B. Angular momentum about O is constant for uniform straight-line motion.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  20. Q20
    Find the torque about the origin when a force of 3j N acts on a particle whose position vector is 2k m.
    1. -6i N·m
    2. 6i N·m
    3. 6j N·m
    4. 6k N·m

    Answer: (A) -6i N·m

    τ = r × F = 2k × 3j = 6(k × j) = 6(-i) = -6i N·m, using k × j = -i.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  21. Q21
    A thin uniform circular ring rolls down an inclined plane of inclination 30° without slipping. Its linear acceleration along the incline is:
    1. (g)/(3)
    2. (g)/(2)
    3. (g)/(4)
    4. (2g)/(3)

    Answer: (C) (g)/(4)

    For a ring (k²)/(R²) = 1. a = (gsinθ)/(1 + k²/R²) = (gsin 30°)/(1 + 1) = (g/2)/(2) = (g)/(4).

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  22. Q22
    A torque of 20 N·m is applied to a wheel of moment of inertia 4 kg·m². The angular acceleration produced is:
    1. 8 rad/s²
    2. 2 rad/s²
    3. 5 rad/s²
    4. 80 rad/s²

    Answer: (C) 5 rad/s²

    Using τ = Iα → α = τ/I = 20/4 = 5 rad/s².

    Chapter: System of Particles and Rotational Motion

  23. Q23
    A cart of mass M is tied to one end of a massless rope of length 10 m. The other end of the rope is held by a man of mass M on a smooth horizontal surface. The man is at x = 0 and the cart at x = 10 m. If the man pulls the cart by the rope, the man and the cart will meet at the point:
    1. x = 5 m
    2. x = 0
    3. they will never meet
    4. x = 10 m

    Answer: (A) x = 5 m

    Internal (rope) forces cannot move the centre of mass, and the surface is smooth. With equal masses at x = 0 and x = 10 m, the COM is at x_(cm) = (M(0) + M(10))/(2M) = 5 m. Both bodies meet at the stationary COM, i.e. at x = 5 m.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  24. Q24
    A disc of radius 2 m and mass 100 kg rolls without slipping on a horizontal floor. Its centre of mass has a speed of 20 cm/s. How much work is needed to stop it?
    1. 30 kJ
    2. 2 J
    3. 1 J
    4. 3 J

    Answer: (D) 3 J

    Total KE of a rolling disc = (1)/(2)Mv²(1 + (1)/(2)) = (3)/(4)Mv². With v = 0.2 m/s: = (3)/(4)(100)(0.2)² = (3)/(4)(100)(0.04) = 3 J. Work to stop equals this KE.

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

  25. Q25
    The moment of inertia of a disc of mass M and radius R about a tangent to its rim and in its plane is:
    1. (2)/(3)MR²
    2. (3)/(2)MR²
    3. (4)/(5)MR²
    4. (5)/(4)MR²

    Answer: (D) (5)/(4)MR²

    MOI of a disc about a diameter is (1)/(4)MR². A tangent in the plane is parallel to a diameter at distance R: I = (1)/(4)MR² + MR² = (5)/(4)MR².

    Chapter: System of Particles and Rotational Motion · NEET previous-year question

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