Important Questions Class 11 Physics Chapter 6: Systems of Particles and Rotational Motion

A system of particles can be studied through its centre of mass and total external force.
Rotational motion depends on torque, angular momentum and moment of inertia about a chosen axis.

Systems of particles and rotational motion extend Newton’s laws from point particles to rigid bodies with size, shape and rotation. Important Questions Class 11 Physics Chapter 6 help students practise centre of mass, linear momentum of a system, vector product, torque, angular momentum, equilibrium, centre of gravity, moment of inertia, rotational kinematics and rotational dynamics. The CBSE 2026 chapter also builds the base for rolling motion, levers, flywheels, conservation of angular momentum and rigid body equilibrium.

Key Takeaways

• Centre of Mass: A system moves as if its total mass acts at its centre of mass under external force.
• Torque: Moment of force is τ = r × F and depends on force arm.
• Moment of Inertia: Rotational inertia is I = Σmiri² about a chosen axis.
• Angular Momentum: For fixed-axis rotation of symmetric bodies, L = Iω.

Important Questions Class 11 Physics Chapter 6 Structure 2026

Important Questions Class 11 Physics Chapter 6 with Answers

Rigid body motion often combines translation of centre of mass and rotation about an axis.
Students should identify external force, torque, axis and mass distribution before solving.
These systems of particles and rotational motion class 11 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 11 Physics Chapter 6 mainly test?

Important Questions Class 11 Physics Chapter 6 mainly test centre of mass, torque, angular momentum, rigid body equilibrium, moment of inertia and rotational dynamics. The chapter connects translational and rotational laws.

1. Centre of Mass Skill: Use R = Σmiri/M.
2. Torque Skill: Use τ = r × F.
3. Inertia Skill: Use I = Σmiri².
4. Rotation Skill: Use τ = Iα.
5. Final Result: The chapter tests extended body mechanics.

2. Why is the particle model inadequate for rigid bodies?

The particle model is inadequate because real bodies have finite size and rotation. Different points of a rotating body can have different velocities.

1. Particle Model: Ignores size.
2. Rigid Body: Has fixed shape and size.
3. Rolling Cylinder: Combines translation and rotation.
4. Final Result: Rigid bodies need centre of mass and rotational analysis.

3. What types of motion can a rigid body have?

A rigid body can have pure translation, pure rotation, or a combination of both. Rolling motion is translation plus rotation.

1. Pure Translation: All particles have same velocity.
2. Pure Rotation: Particles move in circles about an axis.
3. Rolling Motion: Centre moves and body rotates.
4. Final Result: Rigid body motion can combine translation and rotation.

Centre of Mass Class 11 Questions

The centre of mass represents the mass-weighted position of a system.
It helps treat extended bodies as equivalent point masses for translational motion.
These centre of mass class 11 questions cover particles, rigid bodies and symmetry.

4. What is centre of mass?

Centre of mass is the point whose position is the mass-weighted average position of all particles. It represents the system for translational motion.

1. Masses: m1, m2, m3 and so on.
2. Positions: r1, r2, r3 and so on.
3. Formula Used: R = Σmiri/M.
4. Final Result: Centre of mass is the mass-weighted mean position.

5. What is centre of mass of two particles on a line?

For two particles, centre of mass is X = (m1x1 + m2x2)/(m1 + m2). It lies nearer the heavier mass.

1. Particle 1: Mass m1 at x1.
2. Particle 2: Mass m2 at x2.
3. Formula Used: X = (m1x1 + m2x2)/(m1 + m2).
4. Final Result: Centre of mass divides distance in inverse mass ratio.

6. Where is centre of mass of two equal masses?

The centre of mass lies exactly midway between two equal masses. Both masses contribute equally.

1. Mass Condition: m1 = m2.
2. Formula: X = (x1 + x2)/2.
3. Position: Midpoint.
4. Final Result: Equal masses have centre of mass at midpoint.

7. Where is centre of mass of a uniform sphere, cylinder, ring and cube?

The centre of mass lies at the geometric centre for each uniform body. Symmetry places equal mass around the centre.

1. Sphere: Centre.
2. Cylinder: Midpoint of axis.
3. Ring: Centre of ring.
4. Cube: Intersection of body diagonals.
5. Final Result: Uniform symmetric bodies have centre of mass at geometric centre.

8. Does centre of mass always lie inside the body?

No, centre of mass need not always lie inside the body. A ring has its centre of mass at its empty centre.

1. Solid Sphere: Centre lies inside.
2. Ring: Centre lies in empty space.
3. L-shaped Lamina: Centre may lie outside material region.
4. Final Result: Centre of mass can lie outside a body.

9. Find centre of mass of masses 2 kg and 3 kg placed at x = 0 m and x = 10 m.

The centre of mass is 6 m from the 2 kg mass. Use X = (m1x1 + m2x2)/(m1 + m2).

1. Given Data:
   m1 = 2 kg, x1 = 0 m
   m2 = 3 kg, x2 = 10 m
2. Formula Used: X = (m1x1 + m2x2)/(m1 + m2).
3. Calculation:
   X = (2 × 0 + 3 × 10)/(2 + 3)
   X = 30/5 = 6 m
4. Final Result: Centre of mass is at x = 6 m.

10. Find centre of mass of three equal masses at the vertices of an equilateral triangle.

The centre of mass is at the centroid of the triangle. Equal masses make the result match the geometric centroid.

1. Masses: Equal at all three vertices.
2. Symmetry: Each vertex contributes equally.
3. Centroid: Intersection of medians.
4. Final Result: Centre of mass lies at the centroid.

Motion of Centre of Mass Class 11 Questions

The motion of centre of mass depends only on external forces.
Internal forces cancel in pairs because of Newton’s third law.
These motion of centre of mass class 11 questions explain MA = Fext.

11. State the equation of motion of centre of mass.

The equation is MA = Fext. The centre of mass accelerates due to total external force only.

1. Total Mass: M.
2. Centre of Mass Acceleration: A.
3. External Force: Fext.
4. Final Result: MA = Fext.

12. Why do internal forces not affect centre of mass motion?

Internal forces do not affect centre of mass motion because they occur in equal and opposite pairs. Their vector sum is zero.

1. Newton’s Third Law: Internal forces pair up.
2. Equal Magnitude: Same strength.
3. Opposite Direction: Net internal force is zero.
4. Final Result: Only external force changes centre of mass motion.

13. What happens to centre of mass when a projectile explodes mid-air?

The centre of mass continues along the same parabolic path. Explosion forces are internal and cannot change centre of mass motion.

1. Before Explosion: Projectile follows parabola.
2. During Explosion: Internal forces act.
3. External Force: Gravity remains same.
4. Final Result: Centre of mass follows the original projectile path.

14. What happens to centre of mass if total external force is zero?

The centre of mass moves with constant velocity when total external force is zero. It may also remain at rest.

1. External Force: Fext = 0.
2. Equation: MA = 0.
3. Result: A = 0.
4. Final Result: Centre of mass has uniform motion.

15. Why can a system have internal motion while centre of mass moves uniformly?

Particles can move relative to each other due to internal forces. These forces cannot change total centre of mass motion.

1. Internal Motion: Particles interact inside the system.
2. Internal Force Sum: Zero.
3. Centre of Mass: Depends on external force.
4. Final Result: Internal motion and centre of mass motion are separable.

Linear Momentum of System of Particles Class 11 Questions

Total linear momentum of a system equals total mass times centre of mass velocity.
Conservation of linear momentum follows when total external force is zero.
These linear momentum of system of particles class 11 questions cover P = MV.

16. What is total linear momentum of a system of particles?

Total linear momentum is the vector sum of momenta of all particles. It is written as P = Σmivi.

1. Particle Momentum: pi = mivi.
2. Total Momentum: P = p1 + p2 + ...
3. Vector Nature: Direction matters.
4. Final Result: Total momentum is vector sum of individual momenta.

17. How is total momentum related to centre of mass velocity?

Total momentum is P = MV, where M is total mass and V is centre of mass velocity. This relation holds for any particle system.

1. Total Mass: M.
2. Centre of Mass Velocity: V.
3. Formula Used: P = MV.
4. Final Result: System momentum equals mass times centre of mass velocity.

18. State conservation of linear momentum for a system.

If total external force on a system is zero, total linear momentum remains constant. Internal forces do not change total momentum.

1. Condition: Fext = 0.
2. Equation: dP/dt = 0.
3. Result: P = constant.
4. Final Result: Total linear momentum is conserved.

19. Why do decay products move back-to-back in the centre of mass frame?

Decay products move back-to-back because initial total momentum is zero in the centre of mass frame. Final momenta must add to zero.

1. Initial Momentum: Zero.
2. Momentum Conservation: Pfinal = 0.
3. Two Products: Equal and opposite momenta.
4. Final Result: The products move in opposite directions.

20. What is the centre of mass motion in a binary star system without external force?

The centre of mass moves uniformly in a straight line. In the centre of mass frame, both stars orbit around the centre of mass.

1. External Force: Negligible.
2. Centre of Mass: Uniform motion.
3. Relative Motion: Stars orbit around centre of mass.
4. Final Result: Binary stars share a common centre of mass.

Vector Product Class 11 Physics Questions

Vector product creates a vector perpendicular to the plane of two vectors.
It is required for torque and angular momentum.
These vector product class 11 physics questions cover cross product rules.

21. What is vector product of two vectors?

Vector product of a and b is a × b = ab sin θ n̂. Its direction is perpendicular to both vectors.

1. Magnitude: ab sin θ.
2. Direction: Right-hand rule.
3. Symbol: Cross product.
4. Final Result: Vector product gives a perpendicular vector.

22. Is vector product commutative?

No, vector product is not commutative. It follows a × b = −b × a.

1. Magnitude: Same for both products.
2. Direction: Opposite.
3. Reason: Order decides right-hand rule direction.
4. Final Result: Cross product changes sign when order reverses.

23. When is vector product zero?

Vector product is zero when the vectors are parallel, anti-parallel, or one vector is zero. Then sin θ equals zero.

1. Parallel Vectors: θ = 0°.
2. Anti-parallel Vectors: θ = 180°.
3. Zero Vector: Magnitude is zero.
4. Final Result: a × b = 0 for collinear vectors.

24. What are the cross product relations among i, j and k?

The cyclic relations are i × j = k, j × k = i, and k × i = j. Reversing order gives negative signs.

1. Positive Cyclic Order: i, j, k.
2. Reverse Order: j × i = −k.
3. Use: Component cross products.
4. Final Result: Cyclic order gives positive cross products.

25. Find scalar product and vector product of a = 3i − 4j + 5k and b = −2i + j − 3k.

The scalar product is −25, and vector product is 7i − k − 5j after simplification order. Use component multiplication and determinant.

1. Scalar Product:
   a · b = 3(−2) + (−4)(1) + 5(−3)
   a · b = −6 − 4 − 15 = −25
2. Vector Product:
   a × b = |i j k; 3 −4 5; −2 1 −3|
3. Calculation:
   i(12 − 5) − j(−9 + 10) + k(3 − 8)
4. Final Result: a · b = −25 and a × b = 7i − j − 5k.

Angular Velocity and Rotational Motion Class 11 Questions

Angular velocity describes how fast a rigid body rotates about an axis.
For fixed-axis rotation, all particles share the same angular velocity.
These rotational motion questions connect angular velocity with linear velocity.

26. What is angular velocity?

Angular velocity is the time rate of change of angular displacement. It is written as ω = dθ/dt.

1. Angular Displacement: θ.
2. Time: t.
3. Formula Used: ω = dθ/dt.
4. Final Result: Angular velocity measures rotation rate.

27. What is the direction of angular velocity?

Angular velocity points along the axis of rotation. Its direction follows the right-hand screw rule.

1. Curl Fingers: In direction of rotation.
2. Thumb Direction: Gives angular velocity.
3. Fixed Axis: Direction remains fixed.
4. Final Result: Angular velocity is an axial vector.

28. What is the relation between linear velocity and angular velocity?

The vector relation is v = ω × r. Its magnitude is v = ωr⊥.

1. Angular Velocity: ω.
2. Position Vector: r.
3. Perpendicular Distance: r⊥.
4. Final Result: Linear velocity equals angular velocity cross position vector.

29. Why do particles on the axis of rotation remain stationary?

Particles on the axis remain stationary because their perpendicular distance from the axis is zero. Hence v = ωr⊥ = 0.

1. On Axis: r⊥ = 0.
2. Formula: v = ωr⊥.
3. Result: v = 0.
4. Final Result: Axis particles have zero linear speed.

30. What is angular acceleration?

Angular acceleration is the time rate of change of angular velocity. It is written as α = dω/dt.

1. Angular Velocity: ω.
2. Time: t.
3. Formula Used: α = dω/dt.
4. Final Result: Angular acceleration measures change in rotation rate.

Torque Class 11 Physics Questions

Torque measures the turning effect of a force about a point or axis.
It depends on force magnitude, point of application and angle.
These torque class 11 physics questions cover moment of force and couple.

31. What is torque or moment of force?

Torque is the rotational effect of a force about a point. It is defined as τ = r × F.

1. Position Vector: r.
2. Force: F.
3. Magnitude: τ = rF sin θ.
4. Final Result: Torque is moment of force.

32. What is SI unit of torque?

The SI unit of torque is newton metre. Its symbol is N m.

1. Force Unit: Newton.
2. Distance Unit: Metre.
3. Torque Unit: N m.
4. Final Result: Torque is measured in N m.

33. When is torque zero?

Torque is zero when force is zero, position vector is zero, or force line passes through the origin. Then rF sin θ becomes zero.

1. Case 1: F = 0.
2. Case 2: r = 0.
3. Case 3: θ = 0° or 180°.
4. Final Result: Torque vanishes when force has no turning arm.

34. Why is a door easier to open by pushing near the handle?

A door is easier to open near the handle because the perpendicular distance from hinge is larger. Larger force arm gives larger torque.

1. Hinge: Axis of rotation.
2. Handle: Far from hinge.
3. Torque: τ = rF sin θ.
4. Final Result: Greater distance gives greater torque.

35. A force of 20 N acts perpendicular at 0.5 m from a hinge. Find torque.

The torque is 10 N m. Use τ = rF sin 90°.

1. Given Data:
   F = 20 N
   r = 0.5 m
   θ = 90°
2. Formula Used: τ = rF sin θ.
3. Calculation:
   τ = 0.5 × 20 × 1
   τ = 10 N m
4. Final Result: Torque = 10 N m.

36. What is a couple?

A couple is a pair of equal and opposite forces with different lines of action. It produces rotation without translation.

1. Force Sum: Zero.
2. Torque: Non-zero.
3. Example: Turning a bottle cap.
4. Final Result: A couple gives pure turning effect.

Angular Momentum Class 11 Questions

Angular momentum is the rotational analogue of linear momentum.
It depends on position vector and linear momentum.
These angular momentum class 11 questions cover l = r × p and conservation.

37. What is angular momentum of a particle?

Angular momentum of a particle is l = r × p. It is the moment of linear momentum about a point.

1. Position Vector: r.
2. Linear Momentum: p.
3. Magnitude: l = rp sin θ.
4. Final Result: Angular momentum is a vector product.

38. What is relation between torque and angular momentum?

Torque equals the time rate of change of angular momentum. The relation is τ = dl/dt for a particle.

1. Angular Momentum: l.
2. Torque: τ.
3. Relation: τ = dl/dt.
4. Final Result: Torque changes angular momentum.

39. State conservation of angular momentum.

If total external torque on a system is zero, total angular momentum remains constant. This is conservation of angular momentum.

1. Condition: τext = 0.
2. Equation: dL/dt = 0.
3. Result: L = constant.
4. Final Result: Angular momentum is conserved without external torque.

40. Why does a skater spin faster when arms are pulled inward?

A skater spins faster because moment of inertia decreases when arms move inward. Angular momentum Iω remains constant.

1. External Torque: Nearly zero.
2. Arms Inward: I decreases.
3. Conservation: Iω = constant.
4. Final Result: Angular speed increases when moment of inertia decreases.

41. Why does angular momentum of a particle moving with constant velocity remain constant about any point?

Angular momentum remains constant because perpendicular distance from the line of motion stays constant. No external torque acts on the particle.

1. Velocity: Constant.
2. Line of Motion: Fixed.
3. Perpendicular Distance: Constant.
4. Final Result: Angular momentum remains unchanged.

Equilibrium of Rigid Body Class 11 Questions

A rigid body in mechanical equilibrium has no linear or angular acceleration.
Both total external force and total external torque must be zero.
These equilibrium of rigid body class 11 questions cover translational and rotational equilibrium.

42. What are conditions for equilibrium of a rigid body?

A rigid body is in equilibrium when ΣF = 0 and Στ = 0. These conditions ensure no translational or rotational acceleration.

1. Translational Equilibrium: ΣF = 0.
2. Rotational Equilibrium: Στ = 0.
3. Mechanical Equilibrium: Both conditions hold.
4. Final Result: A rigid body needs force and torque balance.

43. What is translational equilibrium?

Translational equilibrium means total external force on the body is zero. Linear momentum remains constant.

1. Condition: ΣF = 0.
2. Acceleration: Centre of mass acceleration is zero.
3. Motion: Rest or uniform motion.
4. Final Result: Translational equilibrium requires zero net force.

44. What is rotational equilibrium?

Rotational equilibrium means total external torque on the body is zero. Angular momentum remains constant.

1. Condition: Στ = 0.
2. Angular Acceleration: Zero.
3. Rotation: Rest or uniform rotation.
4. Final Result: Rotational equilibrium requires zero net torque.

45. What is principle of moments?

Principle of moments states that clockwise moments equal anticlockwise moments in rotational equilibrium. It applies to levers.

1. Load Moment: Load × load arm.
2. Effort Moment: Effort × effort arm.
3. Equilibrium: Load arm × load = effort arm × effort.
4. Final Result: Balanced lever has equal opposite moments.

46. What is mechanical advantage of a lever?

Mechanical advantage is the ratio of load to effort. For a lever, it equals effort arm divided by load arm.

1. Definition: M.A. = load/effort.
2. Lever Relation: M.A. = effort arm/load arm.
3. Use: Larger effort arm reduces effort.
4. Final Result: Mechanical advantage measures force multiplication.

47. A 2 m bar has weight W and is supported by two strings making 36.9° and 53.1° with vertical. What is the equilibrium idea?

The bar stays in equilibrium because vertical force balance and torque balance hold together. Its centre of gravity location follows from these two conditions.

1. Vertical Force Balance: Upward components support weight.
2. Horizontal Force Balance: Horizontal components cancel.
3. Torque Balance: Net moment about any point is zero.
4. Final Result: Rigid body equilibrium needs both ΣF = 0 and Στ = 0.

Centre of Gravity Class 11 Questions

Centre of gravity is the point where total gravitational torque becomes zero.
It coincides with centre of mass in uniform gravitational field.
These questions clarify the difference between centre of mass and centre of gravity.

48. What is centre of gravity?

Centre of gravity is the point where total gravitational torque on a body is zero. Weight acts effectively through this point.

1. Gravity Forces: Act on all particles.
2. Net Torque: Zero about centre of gravity.
3. Effective Weight: Acts through CG.
4. Final Result: Centre of gravity is the balance point under gravity.

49. When do centre of mass and centre of gravity coincide?

They coincide when gravitational field is uniform across the body. This holds for small bodies near Earth’s surface.

1. Uniform Gravity: Same g for all particles.
2. Small Body: Gravity variation is negligible.
3. Result: CM and CG match.
4. Final Result: Centre of gravity equals centre of mass in uniform gravity.

50. How can centre of gravity of an irregular lamina be found?

It can be found by suspending the lamina from different points and drawing vertical lines. Their intersection gives centre of gravity.

1. First Suspension: Draw vertical line.
2. Second Suspension: Draw another vertical line.
3. Intersection: Gives CG.
4. Final Result: The plumb-line method locates centre of gravity.

51. Why does a balanced cardboard stay horizontal on a pencil tip?

It stays balanced because the pencil tip passes through the centre of gravity. Net torque due to gravity becomes zero.

1. Support Force: Acts upward at tip.
2. Weight: Acts downward through CG.
3. Torque: Zero when lines of action match.
4. Final Result: Balance occurs when support lies below centre of gravity.

52. Can centre of gravity lie outside a body?

Yes, centre of gravity can lie outside a body. A ring’s centre of gravity lies at its empty centre.

1. Ring: No material at centre.
2. Symmetry: Weight balances around centre.
3. Result: CG lies outside material.
4. Final Result: Centre of gravity need not lie within the body.

Moment of Inertia Class 11 Questions

Moment of inertia measures resistance to angular acceleration.
It depends on mass distribution and chosen axis of rotation.
These moment of inertia class 11 questions cover formulas and physical meaning.

53. What is moment of inertia?

Moment of inertia is rotational inertia of a body about an axis. It is defined as I = Σmiri².

1. Mass Element: mi.
2. Distance from Axis: ri.
3. Formula Used: I = Σmiri².
4. Final Result: Moment of inertia depends on axis and mass distribution.

54. What is SI unit of moment of inertia?

The SI unit of moment of inertia is kg m². Its dimensions are ML².

1. Mass Unit: kg.
2. Distance Squared Unit: m².
3. Unit: kg m².
4. Final Result: Moment of inertia is measured in kg m².

55. Why does moment of inertia depend on axis?

Moment of inertia depends on axis because particle distances from the axis change. Larger distances increase I.

1. Formula: I = Σmiri².
2. Distance Factor: Squared distance.
3. Axis Shift: Changes ri values.
4. Final Result: Same body has different I about different axes.

56. What is radius of gyration?

Radius of gyration is the distance from the axis where total mass can be assumed concentrated to give same moment of inertia. It is k in I = Mk².

1. Moment of Inertia: I.
2. Total Mass: M.
3. Formula Used: k = √(I/M).
4. Final Result: Radius of gyration represents mass distribution.

57. What is moment of inertia of a thin ring about its central axis?

The moment of inertia is I = MR². All mass elements lie at distance R from the axis.

1. Mass: M.
2. Radius: R.
3. Formula Used: I = MR².
4. Final Result: Thin ring has I = MR².

58. What is moment of inertia of a solid cylinder about its axis?

The moment of inertia is I = MR²/2. It applies to a uniform solid cylinder about its symmetry axis.

1. Mass: M.
2. Radius: R.
3. Axis: Cylinder axis.
4. Final Result: Solid cylinder has I = MR²/2.

59. What is moment of inertia of a solid sphere about its diameter?

The moment of inertia is I = 2MR²/5. It applies to a uniform solid sphere.

1. Mass: M.
2. Radius: R.
3. Axis: Diameter.
4. Final Result: Solid sphere has I = 2MR²/5.

60. A solid cylinder of mass 20 kg and radius 0.25 m rotates at 100 rad/s. Find rotational kinetic energy.

The rotational kinetic energy is 3125 J. Use K = 1/2 Iω² and I = MR²/2.

1. Given Data:
   M = 20 kg
   R = 0.25 m
   ω = 100 rad/s
2. Moment of Inertia:
   I = MR²/2 = 20 × 0.25²/2
   I = 0.625 kg m²
3. Kinetic Energy:
   K = 1/2 × 0.625 × 100²
   K = 3125 J
4. Final Result: Rotational kinetic energy = 3125 J.

61. Find angular momentum of the same solid cylinder.

The angular momentum is 62.5 kg m²/s. Use L = Iω.

1. Moment of Inertia: I = 0.625 kg m².
2. Angular Speed: ω = 100 rad/s.
3. Formula Used: L = Iω.
4. Calculation: L = 0.625 × 100 = 62.5.
5. Final Result: Angular momentum = 62.5 kg m²/s.

Rotational Kinematics Class 11 Questions

Rotational kinematics uses angular displacement, angular velocity and angular acceleration.
The equations match linear kinematics when angular acceleration is constant.
These rotational kinematics class 11 questions cover ω, α and θ relations.

62. What are rotational kinematic equations for constant angular acceleration?

The equations are ω = ω0 + αt, θ = θ0 + ω0t + 1/2αt², and ω² = ω0² + 2α(θ − θ0).

1. Angular Velocity: ω.
2. Angular Acceleration: α.
3. Angular Displacement: θ.
4. Final Result: Rotational equations mirror linear kinematics.

63. What is the rotational analogue of displacement, velocity and acceleration?

The analogues are angular displacement, angular velocity and angular acceleration. They describe rotation instead of straight-line motion.

1. Displacement Analogue: θ.
2. Velocity Analogue: ω.
3. Acceleration Analogue: α.
4. Final Result: Rotation uses angular kinematic quantities.

64. A wheel speeds up from 1200 rpm to 3120 rpm in 16 s. Find angular acceleration.

The angular acceleration is 4π rad/s². Convert rpm to rad/s first.

1. Initial Angular Speed:
   ω0 = 2π × 1200/60 = 40π rad/s
2. Final Angular Speed:
   ω = 2π × 3120/60 = 104π rad/s
3. Formula Used: α = (ω − ω0)/t.
4. Calculation:
   α = (104π − 40π)/16
   α = 4π rad/s²
5. Final Result: α = 4π rad/s².

65. How many revolutions does the wheel make during this time?

The wheel makes 576 revolutions. Use θ = ω0t + 1/2αt².

1. Given Data:
   ω0 = 40π rad/s
   α = 4π rad/s²
   t = 16 s
2. Angular Displacement:
   θ = 40π × 16 + 1/2 × 4π × 16²
   θ = 1152π rad
3. Revolutions:
   N = θ/(2π) = 576
4. Final Result: Number of revolutions = 576.

Rotational Dynamics Class 11 Questions

Rotational dynamics relates torque, moment of inertia and angular acceleration.
It gives the rotational form of Newton’s second law.
These rotational dynamics class 11 questions cover τ = Iα, power and work.

66. What is rotational form of Newton’s second law?

The rotational form is τ = Iα for rotation about a fixed axis. Torque produces angular acceleration.

1. Torque: τ.
2. Moment of Inertia: I.
3. Angular Acceleration: α.
4. Final Result: Torque equals moment of inertia times angular acceleration.

67. What is work done by torque?

Work done by torque is dW = τdθ. For constant torque, W = τθ.

1. Torque: τ.
2. Angular Displacement: dθ.
3. Formula: dW = τdθ.
4. Final Result: Rotational work depends on angular displacement.

68. What is power in rotational motion?

Power in rotational motion is P = τω. It is the rate of doing rotational work.

1. Torque: τ.
2. Angular Velocity: ω.
3. Formula Used: P = τω.
4. Final Result: Rotational power equals torque times angular speed.

69. A flywheel has mass 20 kg and radius 0.20 m. A 25 N pull acts on its rim. Find angular acceleration.

The angular acceleration is 12.5 rad/s². Use τ = Iα.

1. Given Data:
   M = 20 kg
   R = 0.20 m
   F = 25 N
2. Torque:
   τ = FR = 25 × 0.20 = 5 N m
3. Moment of Inertia:
   I = MR²/2 = 20 × 0.20²/2 = 0.4 kg m²
4. Angular Acceleration:
   α = τ/I = 5/0.4 = 12.5 rad/s²
5. Final Result: α = 12.5 rad/s².

70. If 2 m of cord unwinds from the flywheel, find work done by pull.

The work done is 50 J. Work equals force times length of cord pulled.

1. Given Data:
   F = 25 N
   s = 2 m
2. Formula Used: W = Fs.
3. Calculation:
   W = 25 × 2
   W = 50 J
4. Final Result: Work done = 50 J.

71. A rotor runs at 200 rad/s and needs torque 180 N m. Find power.

The power required is 36 kW. Use P = τω.

1. Given Data:
   τ = 180 N m
   ω = 200 rad/s
2. Formula Used: P = τω.
3. Calculation:
   P = 180 × 200
   P = 36000 W
4. Final Result: Power = 36 kW.

Conservation of Angular Momentum Class 11 Questions

Angular momentum remains constant when external torque is zero.
This explains spinning skaters, divers and rotating chairs.
These conservation of angular momentum class 11 questions cover Iω = constant.

72. What is conservation of angular momentum for fixed-axis rotation?

For fixed-axis rotation, if external torque is zero, Iω = constant. A change in I changes ω inversely.

1. Condition: τext = 0.
2. Angular Momentum: L = Iω.
3. Conservation: I1ω1 = I2ω2.
4. Final Result: Angular momentum remains constant without external torque.

73. A child on a turntable rotates at 40 rev/min. If moment of inertia becomes 2/5 of initial value, find new angular speed.

The new angular speed is 100 rev/min. Use I1ω1 = I2ω2.

1. Given Data:
   ω1 = 40 rev/min
   I2 = 2I1/5
2. Formula Used: I1ω1 = I2ω2.
3. Calculation:
   ω2 = I1ω1/I2
   ω2 = 40 ÷ (2/5)
   ω2 = 100 rev/min
4. Final Result: New angular speed = 100 rev/min.

74. Why does kinetic energy increase when the child folds hands inward?

Kinetic energy increases because the child does internal work while pulling hands inward. Angular momentum remains conserved.

1. Moment of Inertia: Decreases.
2. Angular Speed: Increases.
3. Energy Source: Muscular work.
4. Final Result: Extra kinetic energy comes from internal muscular work.

75. Why does a diver tuck in during a somersault?

A diver tucks in to reduce moment of inertia and increase angular speed. This helps complete rotation before entering water.

1. Tucked Position: Smaller I.
2. Angular Momentum: Conserved.
3. Angular Speed: Increases.
4. Final Result: Tucking increases spin rate.

NCERT Class 11 Physics Chapter 6 Questions

NCERT questions combine centre of mass, torque, equilibrium, moment of inertia and angular momentum.
Students should choose the axis carefully before writing torque equations.
These NCERT Class 11 Physics Chapter 6 questions follow the 2026 exercise pattern.

76. In HCl, nuclei are separated by 1.27 Å. Chlorine is 35.5 times hydrogen mass. Find centre of mass from hydrogen.

The centre of mass is about 1.235 Å from hydrogen. It lies closer to chlorine.

1. Given Data:
   mH = m
   mCl = 35.5m
   xH = 0
   xCl = 1.27 Å
2. Formula Used: X = (mH xH + mCl xCl)/(mH + mCl).
3. Calculation:
   X = (35.5m × 1.27)/(36.5m)
   X ≈ 1.235 Å
4. Final Result: Centre of mass lies 1.235 Å from hydrogen.

77. A child runs on a uniformly moving trolley. What is speed of centre of mass of trolley-child system?

The centre of mass speed remains V if no external horizontal force acts. Internal motion cannot change centre of mass velocity.

1. System: Trolley plus child.
2. External Horizontal Force: Zero on smooth floor.
3. Momentum: Conserved.
4. Final Result: Centre of mass speed remains V.

78. Which gains greater angular speed under equal torque, hollow cylinder or solid sphere?

The solid sphere gains greater angular speed. It has smaller moment of inertia than hollow cylinder for same mass and radius.

1. Hollow Cylinder: I = MR².
2. Solid Sphere: I = 2MR²/5.
3. Angular Acceleration: α = τ/I.
4. Final Result: Solid sphere gets larger angular acceleration.

79. A hollow cylinder of mass 3 kg and radius 0.40 m is pulled by 30 N rope. Find angular acceleration.

The angular acceleration is 25 rad/s². Use I = MR² and τ = FR.

1. Given Data:
   M = 3 kg
   R = 0.40 m
   F = 30 N
2. Torque:
   τ = FR = 30 × 0.40 = 12 N m
3. Moment of Inertia:
   I = MR² = 3 × 0.40² = 0.48 kg m²
4. Angular Acceleration:
   α = τ/I = 12/0.48 = 25 rad/s²
5. Final Result: α = 25 rad/s².

80. What is the linear acceleration of the rope in the same case?

The linear acceleration is 10 m/s². Use a = Rα.

1. Given Data:
   R = 0.40 m
   α = 25 rad/s²
2. Formula Used: a = Rα.
3. Calculation:
   a = 0.40 × 25
   a = 10 m/s²
4. Final Result: Linear acceleration = 10 m/s².

Class 11 Physics Chapter 6 Numericals

Numericals in rotational motion require consistent SI units and correct axis selection.
Use radians for angular displacement and kg m² for moment of inertia.
These Class 11 Physics Chapter 6 numericals support stepwise CBSE 2026 practice.

81. A metre stick balances at 45 cm when two 5 g coins are placed at 12 cm. Find mass of stick.

The mass of the metre stick is 66 g. Take moments about the new balance point.

1. Coin Mass: 10 g at 12 cm.
2. Stick Centre: 50 cm.
3. Balance Point: 45 cm.
4. Torque Balance:
   10(45 − 12) = M(50 − 45)
5. Calculation:
   10 × 33 = 5M
   M = 66 g
6. Final Result: Mass of stick = 66 g.

82. A car of mass 1800 kg has axle distance 1.8 m. Centre of gravity is 1.05 m behind front axle. Find total reaction on back axle.

The total reaction on back axle is 10290 N. Take torque about the front axle.

1. Given Data:
   M = 1800 kg
   g = 9.8 m/s²
   Wheelbase = 1.8 m
   CG distance from front axle = 1.05 m
2. Weight: W = 1800 × 9.8 = 17640 N.
3. Torque Balance: Rback × 1.8 = W × 1.05.
4. Calculation:
   Rback = 17640 × 1.05/1.8
   Rback = 10290 N
5. Final Result: Back axle reaction = 10290 N.

83. Find force on each back wheel in the same car.

The force on each back wheel is 5145 N. The back axle load divides equally between two wheels.

1. Total Back Reaction: 10290 N.
2. Number of Back Wheels: 2.
3. Calculation:
   Force per wheel = 10290/2
   Force per wheel = 5145 N
4. Final Result: Each back wheel gets 5145 N.

84. Find total reaction on front axle and force on each front wheel.

The total front reaction is 7350 N, and each front wheel gets 3675 N.

1. Total Weight: 17640 N.
2. Back Reaction: 10290 N.
3. Front Reaction:
   Rfront = 17640 − 10290
   Rfront = 7350 N
4. Each Front Wheel:
   7350/2 = 3675 N
5. Final Result: Front axle reaction = 7350 N and each front wheel = 3675 N.

85. A solid cylinder and hollow cylinder have same mass and radius. Which has greater rotational kinetic energy at same angular speed?

The hollow cylinder has greater rotational kinetic energy. Its moment of inertia is larger.

1. Solid Cylinder: I = MR²/2.
2. Hollow Cylinder: I = MR².
3. Energy: K = 1/2 Iω².
4. Final Result: Larger I gives larger rotational kinetic energy at same ω.

CBSE Class 11 Physics Chapter-Wise Important Questions