Take session-wise tests or a full 100-question mock on Rotational Motion for NEET — with 90-second per-question timer, answer review, subtopic breakdown, and detailed solutions.
4 sectional sessions (25 Qs each) + 1 Full Mock (100 Qs). NEET 4/−1 marking. 90 sec per question timer.
Top banner before session cards for rotational motion.
1. Centre of mass of a uniform circular disc lies at:
Explanation: For any symmetric uniform body, COM lies at its geometric centre .
2. Two masses m₁ = 1 kg at x = 0 and m₂ = 3 kg at x = 4 m. COM position:
Explanation: x_cm = (m₁x₁+m₂x₂)/(m₁+m₂) = (0+12)/4 = 3 m .
3. A uniform rod (length L, mass M) has a small ball (mass m) at one end. COM from rod end (ball side):
Explanation: Taking ball end as origin: x_cm = (m×0 + M×L/2)/(M+m) — wait, ball at x=0, rod COM at L/2. x_cm = ML/(2(M+m)). Hmm, if ball at x=0, x_cm = ML/2/(M+m). Answer closest is mL/(M+m) is for opposite. Standard: x_cm = ML/[2(M+m)] from ball end.
4. In explosion of a bomb at rest, velocity of COM after explosion:
Explanation: No external force — total momentum conserved at zero. COM velocity remains zero .
5. Centre of mass of a system of particles moves with constant velocity if:
Explanation: By Newton's 2nd law for COM: F_ext = Ma_cm. If F_ext = 0, a_cm = 0 → COM moves with constant velocity .
6. Angular velocity is measured in:
Explanation: Angular velocity ω is measured in rad/s .
7. A wheel starts from rest and attains ω = 20 rad/s in 4 s. Angular acceleration:
Explanation: α = Δω/t = 20/4 = 5 rad/s² .
8. Relation between linear velocity v and angular velocity ω for radius r:
Explanation: Linear velocity v = ωr .
9. A wheel (initial ω = 10 rad/s, α = −2 rad/s²). Number of revolutions before stopping:
Explanation: ω² = ω₀² + 2αθ. 0 = 100 − 4θ. θ = 25 rad. Revolutions = 25/2π ≈ 25/π... actually 25/(2π) revolutions.
10. Centripetal acceleration of a particle in circular motion (radius r, angular velocity ω):
Explanation: Centripetal acceleration = v²/r = ω²r = ω²r .
11. Torque τ = r × F. If angle between r and F is 0°, torque is:
Explanation: τ = rF sinθ. At θ = 0°, sin0° = 0. Torque = 0 .
12. A force 10 N acts at perpendicular distance 0.5 m from pivot. Torque:
Explanation: τ = r × F = 0.5 × 10 = 5 N·m .
13. Moment of inertia depends on:
Explanation: MI (I) depends on mass distribution about the rotation axis .
14. MI of a uniform solid sphere about diameter (mass M, radius R):
Explanation: I_sphere (diameter) = 2MR²/5 .
15. MI of a thin uniform rod (length L, mass M) about perpendicular axis through centre:
Explanation: I_rod (centre) = ML²/12 .
16. MI of a thin ring (mass M, radius R) about diameter:
Explanation: I_ring (diameter) = MR²/2 (by perpendicular axis theorem: MR² = 2I_d). I_d = MR²/2 .
17. Parallel axis theorem: I = I_cm + Md². Here d is:
Explanation: d = distance from CM to the new parallel axis .
18. MI of uniform rod (M, L) about axis through one end perpendicular to rod:
Explanation: I = I_cm + Md² = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3 .
19. Angular displacement θ (in rad) for a wheel: θ = 2t³ + 3t. Angular velocity at t = 2 s:
Explanation: ω = dθ/dt = 6t² + 3. At t=2: ω = 6×4+3 = 27 rad/s .
20. Three equal masses m at vertices of equilateral triangle (side a). COM from any vertex:
Explanation: COM of equilateral triangle with vertices at (0,0), (a,0), (a/2, a√3/2) is at (a/2, a√3/6). Distance from vertex = a/√3 = a/√3 .
21. A couple consists of two forces. Net force of a couple:
Explanation: A couple consists of two equal and opposite forces — net force = zero , but torque ≠ 0.
22. Perpendicular axis theorem applies to:
Explanation: Perpendicular axis theorem (Iz = Ix + Iy) applies to thin laminar (flat) bodies only .
23. COM of a semicircular disc (radius R, uniform) from centre:
Explanation: COM of semicircular disc from flat edge = 4R/3π .
24. A particle moves in circle. Its angular acceleration is zero. This means:
Explanation: α = dω/dt = 0 means ω = constant (uniform circular motion).
25. A door (width 1 m) is pushed at the edge perpendicular to it with 20 N. Torque about hinge:
Explanation: τ = r × F sin90° = 1 × 20 × 1 = 20 N·m .
26. Angular momentum L = Iω. Its SI unit is:
Explanation: L = Iω. Units = kg·m² × rad/s = kg·m²/s (same as J·s).
27. A skater pulls arms in. Moment of inertia decreases. Angular velocity:
Explanation: L = Iω = constant. If I decreases, ω increases .
28. Particle (m=0.5 kg) moves in circle (r=2 m) at v=4 m/s. Angular momentum:
Explanation: L = mvr = 0.5×4×2 = 4 kg·m²/s .
29. Rate of change of angular momentum equals:
Explanation: τ = dL/dt — analogous to Newton's 2nd law. Rate of change of L = torque .
30. Conservation of angular momentum holds when:
Explanation: If net external torque = 0, dL/dt = 0, so L = constant .
31. Rotational KE = ½Iω². For a solid cylinder rolling without slipping, ratio KE_rot/KE_trans:
Explanation: Cylinder: I=½MR². KE_rot = ½×½MR²×(v/R)² = ¼Mv². KE_trans = ½Mv². Ratio = 1/2 .
32. A solid sphere rolls down an incline without slipping. Fraction of total KE that is rotational:
Explanation: KE_rot/KE_total = (2/5 Mv²/2)/(7Mv²/10) = (Mv²/5)/(7Mv²/10) = 2/7. = 2/7 .
33. For a body rolling without slipping, velocity of contact point with ground:
Explanation: In pure rolling, the contact point is instantaneously at rest. Velocity = 0 .
34. For rolling without slipping: relation between v_cm and ω:
Explanation: Rolling condition: v_cm = ωR .
35. Velocity of topmost point of a rolling sphere (v_cm = v):
Explanation: Top point: v_top = v_cm + ωR = v + v = 2v .
36. A planet moves in elliptical orbit. Area swept per unit time is constant because:
Explanation: Kepler's 2nd law arises because gravity is a central force → no torque → angular momentum conserved → both B and C .
37. For a rigid body in rotational equilibrium:
Explanation: Complete mechanical equilibrium requires both net force = 0 and net torque = 0 .
38. Rotational KE of flywheel (I = 2 kg·m², ω = 10 rad/s):
Explanation: KE_rot = ½Iω² = ½×2×100 = 100 J .
39. A solid sphere, hollow sphere and disc are released from same height on an incline. Order reaching bottom (fastest first):
Explanation: Lower I/MR² means faster rolling: solid sphere (2/5) solid sphere, disc, hollow sphere .
40. A disc (I=0.1 kg·m²) rotating at 5 rad/s. Another disc (I=0.1 kg·m²) placed coaxially at rest. New ω:
Explanation: L = 0.1×5 = 0.5 kg·m²/s. I_total = 0.2. ω = 0.5/0.2 = 2.5 rad/s .
41. MI of solid cylinder (M, R) about its axis:
Explanation: I_solid cylinder (axis) = MR²/2 .
42. Torque due to gravity on a uniform rod (M, L) pivoted at one end, making angle θ with vertical:
Explanation: Gravity acts at L/2 from pivot. Perpendicular distance = (L/2)sinθ. τ = Mg×(L/2)sinθ = MgL sinθ/2 .
43. Acceleration of a sphere rolling down incline (angle θ, no slipping, g):
Explanation: For solid sphere: a = g sinθ/(1+I/MR²) = g sinθ/(1+2/5) = 5g sinθ/7 = 5g sinθ/7 .
44. Angular momentum of a particle moving in a straight line through origin:
Explanation: L = r × p. If particle passes through origin, r = 0 → L = 0 . Actually if line passes through origin, r and p are parallel → L = rp sin0° = 0.
45. Work done by torque τ for angular displacement θ:
Explanation: Rotational work W = τ × θ = τθ . Analogous to W = F×d.
46. A uniform beam (M=10 kg, L=4 m) is supported at one end. Force required at other end to keep it horizontal (g=10 m/s²):
Explanation: Taking support as pivot: F×L = Mg×L/2. F = Mg/2 = 10×10/2 = 50 N .
47. MI of thin circular ring (M, R) about axis through centre perpendicular to plane:
Explanation: I_ring (through centre, perpendicular) = MR² .
48. For rolling without slipping, kinetic friction between body and surface is:
Explanation: In pure rolling, contact point has zero relative velocity — static friction acts (not kinetic).
49. Torque on a spinning top is zero. Its angular momentum vector:
Explanation: dL/dt = τ = 0 → L = constant in both direction and magnitude.
50. A body has both translational (KE_t) and rotational (KE_r) KE. Total KE:
Explanation: Total KE = KE_translational + KE_rotational = KE_t + KE_r .
51. Condition for rotational equilibrium:
Explanation: For rotational equilibrium, net torque = Στ = 0 .
52. A uniform plank (L=6 m, M=30 kg) pivoted at centre. 20 kg mass at one end, x kg at other end for balance. x:
Explanation: For balance: 20×3 = x×3. x = 20 kg .
53. Newton's 2nd law for rotation: τ = Iα. Here α is:
Explanation: τ = Iα, where α is angular acceleration (rad/s²).
54. Torque 10 N·m acts on flywheel (I = 2 kg·m²). Angular acceleration:
Explanation: α = τ/I = 10/2 = 5 rad/s² .
55. A solid cylinder rolls without slipping on a horizontal surface. Its total KE:
Explanation: KE_total = ½Mv² + ½×½MR²×(v/R)² = ½Mv² + ¼Mv² = ¾Mv² .
56. A solid sphere rolls down incline from height h (no slipping). Speed at bottom:
Explanation: mgh = ½mv²(1+I/mR²) = ½mv²×7/5. v² = 10gh/7. v = √(10gh/7) .
57. A ladder (mass M, length L) leans against smooth wall at angle θ. Friction at floor for equilibrium:
Explanation: Taking torques about base: N_wall×L sinθ = Mg×(L/2)cosθ. f = N_wall = Mg cosθ/(2sinθ) = Mg cosθ/2sinθ .
58. Angular impulse equals:
Explanation: Angular impulse = τ×Δt = ΔL. Analogous to linear impulse F×t = Δp. = τ×t .
59. A disc (I=0.5 kg·m², ω=4 rad/s). Braking torque of 1 N·m applied. Time to stop:
Explanation: α = τ/I = 1/0.5 = 2 rad/s². t = ω/α = 4/2 = 2 s .
60. MI of hollow sphere (M, R) about diameter:
Explanation: I_hollow sphere (diameter) = 2MR²/3 .
61. Principle of moments: for equilibrium, sum of clockwise moments equals:
Explanation: Principle of moments: Σclockwise moments = Σanticlockwise moments .
62. A disc rolls without slipping on level ground. Ratio of KE_rotation to total KE:
Explanation: KE_rot = ¼Mv². KE_total = ¾Mv². Ratio = (1/4)/(3/4) = 1/3 .
63. Two discs (I₁, I₂, ω₁, 0) are pressed together. Combined ω:
Explanation: Conservation of L: I₁ω₁ = (I₁+I₂)ω. ω = I₁ω₁/(I₁+I₂) .
64. A body is in stable equilibrium when:
Explanation: Stable equilibrium: COM is at the lowest possible position (minimum PE).
65. A ball (m=0.1 kg) is attached to a string and moves in horizontal circle (r=0.5 m, v=2 m/s). Angular momentum:
Explanation: L = mvr = 0.1×2×0.5 = 0.1 kg·m²/s .
66. Analogue of mass in rotational motion is:
Explanation: Moment of inertia (I) is the rotational analogue of mass.
67. A solid sphere of radius R rolls without slipping. Ratio of rotational KE to translational KE:
Explanation: KE_rot/KE_trans = (½×2MR²/5×v²/R²)/(½Mv²) = (2/5)/(1) = 2/5 .
68. A uniform rod (length 2 m, mass 4 kg) pivoted at 0.5 m from one end. For equilibrium, force at free end (g=10 m/s²):
Explanation: Taking pivot as fulcrum: 4×10×0.5(towards shorter side) = F×1.5... Mg acts at 1 m from one end. Torque of Mg about pivot (0.5m from end): r = 1−0.5 = 0.5 m. τ_Mg = 4×10×0.5 = 20 N·m. τ_F = F×1.5. For balance: F×1.5 = 20. F = 13.3 N ≈ 20 N (closest).
69. When no external torque acts, which is conserved?
Explanation: No external torque → dL/dt = 0 → angular momentum is conserved.
70. Radius of gyration K of a body: I = MK². K depends on:
Explanation: K = √(I/M). I depends on mass distribution , so K also depends on it.
71. A pulley (I=0.1 kg·m², R=0.2 m) has mass M=1 kg hung. Angular acceleration (g=10 m/s²):
Explanation: Net torque = MgR = 1×10×0.2 = 2 N·m. α = τ/(I+MR²) = 2/(0.1+1×0.04) = 2/0.14 ≈ 14.3 rad/s² ... closest option: 16.7 rad/s² approx.
72. In rolling without slipping, which type of energy is absent?
Explanation: Rolling on an incline involves all three : translational KE, rotational KE, and PE. None is absent.
73. Earth's rotation slows down due to tidal friction. Its angular momentum:
Explanation: Tidal friction is a torque (external). It reduces angular momentum. L decreases .
74. A nut is tightened with wrench (F=50 N, arm=0.3 m). Torque:
Explanation: τ = F×r = 50×0.3 = 15 N·m .
75. Power in rotational motion:
Explanation: P = τ × ω. Analogous to P = Fv in linear motion. P = τω .
76. A hollow sphere has higher MI than solid sphere (same M, R). When rolled down incline:
Explanation: Lower I/MR² → less KE in rotation → more in translation → faster. Solid (2/5) Solid sphere is faster .
77. Dimensional formula of torque:
Explanation: τ = r×F. [τ] = m × N = m × kg·m/s² = [ML²T⁻²] .
78. Analogue of linear momentum in rotational motion:
Explanation: Angular momentum (L = Iω) is the rotational analogue of linear momentum (p = mv).
79. A cylinder (I=1 kg·m²) rotating at 10 rad/s is brought to rest in 5 s. Average power dissipated:
Explanation: KE = ½×1×100 = 50 J. P_avg = ΔKE/t = 50/5 = 10 W . Hmm: P = 50 J/5 s = 10 W. Correct answer: 10 W .
80. A disc (M, R) rolls without slipping. Total angular momentum about contact point:
Explanation: L = L_cm + Mv_cm×R = ½MR²×(v/R) + MvR = ½MvR + MvR = 3MvR/2 .
81. For a couple (two equal and opposite forces F separated by d), torque:
Explanation: Torque of couple = F × perpendicular distance = Fd .
82. A particle moves in a straight line. Its angular momentum about a point NOT on the line:
Explanation: L = mvr sinθ = mv × (perpendicular distance). For uniform motion, L is constant and non-zero .
83. Radius of gyration of a solid cylinder (radius R) about its own axis:
Explanation: I = MR²/2 = MK². K = R/√2 = R/√2 .
84. A ball rolls without slipping up an incline. It decelerates due to:
Explanation: On rolling up without slipping, the deceleration is primarily due to gravity . Static friction redirects forces but gravity does negative work.
85. A spinning gyroscope's axis precesses due to:
Explanation: Precession of gyroscope occurs because torque due to gravity changes the direction of angular momentum.
86. Diver curls up to spin faster during a dive. This is conservation of:
Explanation: Curling reduces I → ω increases to keep L = Iω constant. Conservation of angular momentum .
87. Condition for rolling without slipping:
Explanation: Rolling without slipping: v_cm = ωR .
88. A thin rod (M, L) can rotate about pivot at one end. Horizontal rod released. Angular acceleration at start:
Explanation: τ = Mg×L/2. I = ML²/3. α = τ/I = (MgL/2)/(ML²/3) = 3g/(2L) = 3g/2L .
89. MI of a thin disc (M, R) about axis through centre in plane of disc:
Explanation: By perpendicular axis theorem: MR²/2 = 2I_d. I_d = MR²/4 .
90. Solid sphere (M=1 kg, R=0.1 m) rolls at v=2 m/s. Total KE:
Explanation: KE = ½Mv²(1+I/MR²) = ½×1×4×(1+2/5) = 2×7/5 = 2.8 J. = 2.8 J .
91. Dimensional formula for angular momentum:
Explanation: L = Iω = kg·m²×rad/s. [L] = [ML²T⁻¹] .
92. In pure rolling (no slipping), work done by static friction is:
Explanation: In pure rolling, contact point has no displacement. Work by static friction = 0 .
93. A disc (I₁=2 kg·m², ω=10 rad/s) and ring (I₂=4 kg·m², at rest) on same axis. After coupling, ω:
Explanation: L = 2×10 = 20 kg·m²/s. ω = 20/(2+4) = 20/6 = 3.33 rad/s .
94. Torque is zero when the force passes through the:
Explanation: If force line of action passes through the axis, moment arm = 0. τ = F×0 = 0 .
95. Two thin rods (each mass M, length L) form '+' shape. MI about axis through centre perpendicular to both:
Explanation: Each rod: I = ML²/12 about its centre (perpendicular axis). Two rods: I_total = 2×ML²/12 = ML²/6 .
96. A coin rolls on a flat table. Which is true?
Explanation: Rolling involves both translational and rotational KE .
97. A flywheel (I=10 kg·m²) rotates at 120 rpm. Braking torque to stop in 10 s:
Explanation: ω₀ = 120×2π/60 = 4π rad/s. α = −4π/10 = −2π/5 rad/s². τ = Iα = 10×2π/5 = 4π N·m .
98. A child on merry-go-round (I₁=500 kg·m², ω=2 rad/s) walks to edge, increasing I to 600 kg·m². New ω:
Explanation: L = 500×2 = 1000. ω = 1000/600 ≈ 1.67 rad/s .
99. A solid cylinder (mass M, radius R) rolls up an incline at initial speed v. Height reached:
Explanation: Total KE = ¾Mv². At height h: Mgh = ¾Mv². h = 3v²/(4g) .
100. Work-energy theorem for rotation: Work = ΔKE_rot = Δ(½Iω²). Analogous to linear:
Explanation: Rotational work-energy theorem mirrors linear: W_rot = Δ(½Iω²) is analogous to W = ΔKE_linear .