NCERT Solutions for Class 11 Physics Chapter 6: System of Particles and Rotational Motion

These Class 11 Physics Chapter 6 solutions cover System of Particles and Rotational Motion with every NCERT exercise (6.1–6.17) reproduced verbatim and solved step by step, with units shown and each numerical answer cross-checked. The chapter moves you from the motion of a single particle to the motion of extended, rigid bodies — introducing the centre of mass, torque, angular momentum and moment of inertia, updated for session 2026–27.

Class: 11 Subject: Physics Chapter: 6 Name: System of Particles and Rotational Motion Exercises: 6.1 – 6.17 Session: 2026–27
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Class 11 Physics Chapter 6 – Overview

In earlier chapters a body was treated as a single particle (a point mass). Real bodies have a finite size, so Chapter 6 builds the physics of extended bodies, treating each as a system of particles and, where shape is fixed, as a rigid body. You first learn that the motion of any body can be split into the translation of its centre of mass plus rotation about the centre of mass. The centre of mass moves as if all the mass were concentrated there and all external forces acted there. The chapter then develops the rotational analogues of linear quantities: the vector (cross) product, angular velocity (v = ωr), torque (τ = r×F), angular momentum (l = r×p), and the all-important moment of inertia (I), leading to τ = Iα, L = Iω and the conservation of angular momentum. These ideas explain everyday motion from a spinning top and a ceiling fan to a skater pulling in her arms.

Key Concepts & Definitions

Rigid body: a body of perfectly definite, unchanging shape; the distance between any two of its particles stays constant.

Centre of mass (CM): the mass-weighted mean position of all particles of a system; the whole external force can be taken to act there.

Vector (cross) product: for vectors a and b, c = a×b has magnitude ab sinθ and direction perpendicular to the plane of a and b (right-hand screw rule); it is not commutative (a×b = −b×a).

Angular velocity (ω): rate of change of angular displacement; for rotation about a fixed axis every particle shares the same ω, and v = ωr.

Torque (moment of force): τ = r×F; it is the rotational analogue of force and changes a body’s state of rotation.

Angular momentum: l = r×p; the rotational analogue of linear momentum, with dL/dt = τext.

Moment of inertia (I): the rotational analogue of mass; I = Σmiri2, measuring a body’s resistance to angular acceleration about a chosen axis.

Conservation of angular momentum: if the net external torque is zero, L = Iω remains constant.

Important Formulas

Centre of mass: X = Σmixi / M, and similarly for Y, Z (M = total mass).

Motion of CM: MA = Fext; total momentum P = MV.

Linear ↔ angular: v = ωr ; at = αr ; α = dω/dt.

Torque: τ = r×F, magnitude τ = rF sinθ = r⊥F. Angular momentum: l = r×p, l = rp sinθ.

Rotational dynamics: τ = Iα ; L = Iω ; rotational KE = ½Iω2.

Power in rotation: P = τω. Conservation: I1ω1 = I2ω2 when τext = 0.

Rotational kinematics (constant α): ω = ω0 + αt ; θ = ω0t + ½αt2 ; ω2 = ω02 + 2αθ.

Common moments of inertia: solid cylinder/disc about axis = ½MR2; hollow cylinder about axis = MR2; solid sphere about diameter = (2/5)MR2.

NCERT Exercise Solutions (6.1–6.17)

Questions are reproduced verbatim from the NCERT textbook (Reprint 2026–27). Take g = 9.8 m s−2 where required. Answers are original, step by step, with units, and cross-checked against the NCERT answer key.

6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

ANSWER For each of these bodies of uniform density and regular shape, the centre of mass lies at the geometric centre, because the origin chosen at the geometric centre is a point of reflection symmetry — for every mass element at (x, y, z) there is an equal element at (−x, −y, −z), so the integrals ∫x dm = ∫y dm = ∫z dm vanish. No, the CM need not lie inside the material of the body. For a ring (and similarly a hollow sphere, hollow cylinder or a hollow cube), the geometric centre lies in the empty space, so the centre of mass is located outside the matter of the body.

6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10−10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

ANSWER Take the H nucleus at the origin (x = 0) and the Cl nucleus at x = 1.27 Å. Let the mass of hydrogen = m, so mass of chlorine = 35.5 m. XCM = (m·0 + 35.5m·1.27) / (m + 35.5m) = (35.5 × 1.27) / 36.5 Å = 45.085 / 36.5 = 1.235 ≈ 1.24 Å The centre of mass lies on the line joining the two nuclei, at about 1.24 Å from the hydrogen nucleus (i.e. very close to the much heavier chlorine atom). ✓ matches the NCERT key.

6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

ANSWER The floor is smooth, so there is no external horizontal force on the (trolley + child) system. The forces the child exerts while getting up and running are internal to the system and cannot change its total momentum. Since MA = Fext = 0, the acceleration of the centre of mass is zero. Hence the speed of the CM of the system remains unchanged, equal to V, whatever the child does.

6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

ANSWER Let the two adjacent sides of a triangle be the vectors a and b, with angle θ between them. Take a as the base, so base = |a| = a. The height of the triangle = the perpendicular distance from the tip of b to the line of a = b sinθ. Area of triangle = ½ × base × height = ½ × a × b sinθ = ½ ab sinθ. But |a × b| = ab sinθ. Therefore Area of triangle = ½|a × b|. Hence proved.

6.5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.

ANSWER Take a parallelepiped with the three coterminous edges a, b and c. Its base is the parallelogram formed by b and c, whose area is |b × c|, and (b × c) points perpendicular to that base. The height of the parallelepiped is the component of a along this perpendicular direction = |a| cosφ, where φ is the angle between a and (b × c). Volume = base area × height = |b × c| × |a| cosφ = |a| |b × c| cosφ = a.(b × c). Thus the scalar triple product a.(b × c) equals (in magnitude) the volume of the parallelepiped. Hence proved.

6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

ANSWER l = r × p. Expanding the determinant with rows (î, ĵ, k̂), (x, y, z), (px, py, pz): lx = y pz − z py ly = z px − x pz lz = x py − y px If the particle moves only in the x-y plane, then z = 0 and pz = 0. Substituting, lx = y(0) − 0(py) = 0 and ly = 0(px) − x(0) = 0, while lz = x py − y px remains. Hence the angular momentum has only a z-component — it points perpendicular to the plane of motion.

6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

ANSWER Let particle 1 move along one line and particle 2 along a parallel line a distance d away, in opposite directions, each with momentum of magnitude p = mv. Take any reference point O. Let the perpendicular distance of line 1 from O be y, so the perpendicular distance of line 2 from O is (d − y). The two momenta are anti-parallel, but because the particles lie on opposite sides their angular momenta about O point in the same direction and add up: L = mv·y + mv·(d − y) = mv·y + mvd − mv·y = mvd. The result mvd is independent of y, i.e. independent of the location of O. Hence the angular momentum of this system is the same about every point. (This is true here because the net linear momentum mv − mv = 0; a couple has the same moment about any point.)

6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

ANSWER Let the left string tension be T1 (36.9° to vertical) and right string tension T2 (53.1° to vertical). Use sin 36.9° ≈ 0.6, cos 36.9° ≈ 0.8, sin 53.1° ≈ 0.8, cos 53.1° ≈ 0.6. Horizontal balance: T1 sin 36.9° = T2 sin 53.1° ⇒ 0.6 T1 = 0.8 T2 ⇒ T1 = (4/3)T2. Vertical balance: T1 cos 36.9° + T2 cos 53.1° = W ⇒ 0.8 T1 + 0.6 T2 = W. Substituting T1: 0.8(4/3)T2 + 0.6 T2 = (1.0667 + 0.6)T2 = 1.6667 T2 = W ⇒ T2 = 0.6 W. Torque about the left end (length 2 m). The vertical component of T2 acts at the right end: T2 cos 53.1° × 2 = W × d. d = (0.6 W × 0.6 × 2) / W = 0.72 m = 72 cm from the left end. ✓ matches the NCERT key.

6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

ANSWER Total weight W = mg = 1800 × 9.8 = 17640 N. Let Rf and Rb be the total reactions at the front and back axles (each shared by two wheels). Torque about the front axle: Rb × 1.8 = W × 1.05 ⇒ Rb = (17640 × 1.05) / 1.8 = 18522 / 1.8 = 10290 N. Vertical balance: Rf + Rb = W ⇒ Rf = 17640 − 10290 = 7350 N. Force on each front wheel = Rf / 2 = 7350 / 2 = 3675 N. Force on each back wheel = Rb / 2 = 10290 / 2 = 5145 N. ✓ matches the NCERT key.

6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

ANSWER Moment of inertia of hollow cylinder about its axis, Ic = MR2. Moment of inertia of solid sphere about a diameter, Is = (2/5)MR2 = 0.4 MR2. For equal torque τ, angular acceleration α = τ/I, so α ∝ 1/I. Since Is < Ic, the sphere has the larger angular acceleration. After the same time t, ω = αt. Hence the solid sphere acquires the greater angular speed (in fact ωsphere / ωcylinder = Ic/Is = 1/0.4 = 2.5). ✓ matches the NCERT key (Sphere).

6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s−1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

ANSWER Moment of inertia of a solid cylinder about its axis: I = ½MR2 = ½ × 20 × (0.25)2 = ½ × 20 × 0.0625 = 0.625 kg m2. Rotational kinetic energy: KE = ½Iω2 = ½ × 0.625 × (100)2 = ½ × 0.625 × 10000 = 3125 J. Angular momentum: L = Iω = 0.625 × 100 = 62.5 J s (kg m2 s−1). ✓ matches the NCERT key.

6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

ANSWER (a) The turntable is frictionless, so there is no external torque and angular momentum is conserved: I1ω1 = I2ω2. Here I2 = (2/5)I1 and ω1 = 40 rev/min. So ω2 = (I1/I2)ω1 = (5/2) × 40 = 100 rev/min. ✓ matches the NCERT key. (b) Initial KE1 = ½I1ω12. New KE2 = ½I2ω22 = ½(2/5)I1·(5/2·ω1)2 = ½(2/5)I1(25/4)ω12 = (5/2)·½I1ω12. So KE2 = 2.5 × KE1 — the new rotational kinetic energy is 2.5 times the initial value, i.e. larger. The increase is not a violation of energy conservation: the child does work using his own internal (muscular) energy to pull his arms inward, and this internal work appears as the extra rotational kinetic energy.

6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

ANSWER Given M = 3 kg, R = 40 cm = 0.40 m, F = 30 N. Hollow cylinder about its axis: I = MR2 = 3 × (0.40)2 = 3 × 0.16 = 0.48 kg m2. Torque about the axis: τ = F × R = 30 × 0.40 = 12 N m. Angular acceleration: α = τ/I = 12 / 0.48 = 25 rad s−2. Linear acceleration of the rope (equals tangential acceleration of the rim, no slipping): a = αR = 25 × 0.40 = 10 m s−2. ✓ matches the NCERT key.

6.14 To maintain a rotor at a uniform angular speed of 200 rad s−1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

ANSWER Power delivered in rotation: P = τω. P = 180 N m × 200 rad s−1 = 36000 W = 36 kW. Since the engine is 100% efficient, the power it must supply equals the power transmitted to the rotor = 36 kW. ✓ matches the NCERT key.

6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

ANSWER Let the surface mass density be σ. Mass of the full disc M = σπR2, with CM at the centre O (x = 0). Mass removed (the hole of radius R/2): m = σπ(R/2)2 = σπR2/4 = M/4, with its CM at x = R/2. Treat the hole as a negative mass. The CM of the remaining body: X = (M·0 − (M/4)(R/2)) / (M − M/4). X = (−MR/8) / (3M/4) = (−R/8) × (4/3) = −R/6. The centre of gravity lies at a distance R/6 from the centre of the original disc, on the side opposite to the cut-out hole (the negative sign). ✓ matches the NCERT key.

6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

ANSWER The stick’s own weight acts at its centre, the 50.0 cm mark. Total coin mass = 2 × 5 g = 10 g, placed at 12.0 cm. The new balance point (pivot) is at 45.0 cm. Take torques about the new pivot (45.0 cm). The coins lie to the left (clockwise side) and the stick’s weight (at 50.0 cm) lies to the right. Coins’ torque arm = 45.0 − 12.0 = 33.0 cm; stick’s torque arm = 50.0 − 45.0 = 5.0 cm. 10 g × 33.0 cm = m × 5.0 cm ⇒ m = (10 × 33.0) / 5.0 = 330 / 5.0 = 66.0 g. ✓ matches the NCERT key.

6.17 The oxygen molecule has a mass of 5.30 × 10−26 kg and a moment of inertia of 1.94 × 10−46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

ANSWER Given m = 5.30 × 10−26 kg, I = 1.94 × 10−46 kg m2, v = 500 m/s, and KErot = (2/3)KEtrans. ½Iω2 = (2/3) × ½mv2 ⇒ Iω2 = (2/3)mv2 ⇒ ω2 = (2mv2) / (3I). ω2 = (2 × 5.30 × 10−26 × (500)2) / (3 × 1.94 × 10−46) = (2 × 5.30 × 10−26 × 2.5 × 105) / (5.82 × 10−46) = (2.65 × 10−20) / (5.82 × 10−46) = 4.553 × 1025. ω = √(4.553 × 1025) ≈ 6.75 × 1012 rad s−1. ✓ matches the NCERT key.

Extra Practice Questions

Short Answer Type Questions

Q1. Why does the centre of mass of a system move as if all external forces act at that single point?

ANSWERInternal forces between the particles occur in equal and opposite pairs (Newton’s third law) and cancel in the total. Only external forces survive in the sum, giving MA = Fext. So the CM responds to the net external force exactly as a single particle of mass M would.

Q2. State two differences between the scalar product and the vector product of two vectors.

ANSWER(i) The scalar (dot) product gives a scalar (ab cosθ) while the vector (cross) product gives a vector (ab sinθ, perpendicular to the plane). (ii) The dot product is commutative (a.b = b.a), but the cross product is anti-commutative (a×b = −b×a).

Q3. Why does a diver curl up her body during a somersault and stretch out before entering the water?

ANSWERDuring the dive there is no external torque about her CM, so angular momentum L = Iω is conserved. Curling up reduces her moment of inertia I, so ω increases and she spins faster; stretching out increases I, reducing ω for a clean, slow entry.

Q4. Define moment of inertia and state the two factors on which it depends.

ANSWERMoment of inertia I = Σmiri2 is the rotational analogue of mass — a body’s resistance to angular acceleration. It depends on (i) the mass of the body and how that mass is distributed, and (ii) the position and orientation of the axis of rotation.

Q5. The angular momentum of a particle is zero. Give two distinct situations in which this can happen.

ANSWERl = rp sinθ is zero when (i) the particle passes through the origin (r = 0) or its line of momentum passes through the origin (so θ = 0° or 180°), and (ii) when the particle is at rest or has no momentum (p = 0).

Long Answer Type Questions

Q1. Derive the relation between torque and angular acceleration (τ = Iα) for a rigid body rotating about a fixed axis.

ANSWERConsider a rigid body rotating about a fixed axis with angular acceleration α. A particle of mass mi at perpendicular distance ri from the axis has tangential acceleration ai = riα, so the tangential force on it is Fi = miriα. The torque this produces about the axis is τi = Firi = miri2α. Summing over all particles, the total external torque is τ = Σmiri2α = (Σmiri2)α. Since Σmiri2 = I, the moment of inertia about that axis, we obtain τ = Iα. This is the rotational form of Newton’s second law: torque plays the role of force, moment of inertia the role of mass, and angular acceleration the role of linear acceleration.

Q2. State and explain the law of conservation of angular momentum with two real-life examples.

ANSWERThe law states that if the net external torque on a system is zero, its total angular momentum L = Iω remains constant: dL/dt = τext = 0. This follows directly from τext = dL/dt. Example 1: an ice-skater spinning with arms outstretched pulls them in, reducing I; to keep Iω constant, ω rises and she spins much faster. Example 2: a planet in an elliptical orbit moves faster near the Sun (small r) and slower when far (large r), because gravity exerts no torque about the Sun, conserving the planet’s angular momentum — this is the physics behind Kepler’s law of equal areas.

Q3. Distinguish between the translational and rotational motion of a rigid body, listing the analogous physical quantities.

ANSWERIn pure translation every particle of the body has the same velocity at any instant and the body is described by the motion of its centre of mass. In rotation about a fixed axis every particle moves in a circle and shares the same angular velocity. Each translational quantity has a rotational analogue: displacement s ↔ angular displacement θ; velocity v ↔ angular velocity ω; acceleration a ↔ angular acceleration α; mass m ↔ moment of inertia I; force F ↔ torque τ; linear momentum p = mv ↔ angular momentum L = Iω; and Newton’s law F = ma ↔ τ = Iα. The kinetic energy is ½mv2 for translation and ½Iω2 for rotation. General motion of a rigid body is the sum of the translation of its CM and rotation about the CM.

MCQs & Assertion–Reason

1. The centre of mass of a uniform ring lies:

(a) on the rim    (b) at the geometric centre, in empty space    (c) outside the plane of the ring    (d) it has no centre of mass

2. The SI unit of torque is:

(a) N s    (b) N m−1    (c) N m    (d) kg m s−1

3. The moment of inertia of a solid sphere of mass M and radius R about a diameter is:

(a) MR2    (b) ½MR2    (c) (2/5)MR2    (d) (2/3)MR2

4. When the net external torque on a system is zero, the conserved quantity is:

(a) linear momentum only    (b) angular momentum    (c) kinetic energy    (d) moment of inertia

5. For a particle moving only in the x-y plane, its angular momentum about the origin has:

(a) only an x-component    (b) only a y-component    (c) only a z-component    (d) all three components

6. The relation between linear velocity and angular velocity for a particle of a rotating rigid body is:

(a) v = ω/r    (b) v = ωr    (c) v = r/ω    (d) v = ω2r

7. Power delivered to a body rotating at angular speed ω under a torque τ is:

(a) τ/ω    (b) τω    (c) τω2    (d) ω/τ

8. Equal torques act on a hollow cylinder and a solid sphere of equal mass and radius. After a given time, the greater angular speed is gained by the:

(a) hollow cylinder    (b) solid sphere    (c) both equal    (d) cannot be decided

9. The vector product a × b is:

(a) commutative    (b) equal to b × a    (c) anti-commutative (= −b × a)    (d) a scalar

10. A solid cylinder (I = ½MR2) of mass 20 kg, radius 0.25 m spins at 100 rad s−1. Its angular momentum about the axis is:

(a) 31.25 J s    (b) 62.5 J s    (c) 125 J s    (d) 3125 J s

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(c), 6-(b), 7-(b), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The centre of mass of a body may lie outside its material.

Reason: For hollow bodies such as a ring, the geometric centre falls in the empty region.

A-R 2. Assertion: An ice-skater spins faster when she pulls her arms inward.

Reason: Pulling the arms in reduces her moment of inertia and, with no external torque, angular momentum is conserved.

A-R 3. Assertion: The moment of inertia of a body is a fixed property of the body alone.

Reason: Moment of inertia depends only on the total mass of the body.

A-R 4. Assertion: If the total external force on a system is zero, the velocity of its centre of mass stays constant.

Reason: Internal forces cancel in pairs and cannot change the total momentum of the system.

A-R 5. Assertion: The angular momentum of a particle moving in a straight line through the origin is zero.

Reason: When the line of motion passes through the origin, the perpendicular distance r⊥ is zero.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Confusing the moment of inertia of a solid cylinder/disc (½MR2) with a hollow cylinder (MR2) or a solid sphere ((2/5)MR2).
  • Treating angular momentum as a scalar — both torque and angular momentum are vectors, and a×b ≠ b×a.
  • Forgetting to convert units — radius in cm to m, rev/min versus rad/s — before substituting in formulas.
  • Assuming kinetic energy is conserved when angular momentum is conserved; in problem 6.12 the KE actually increases because the child does internal work.
  • Using r instead of the perpendicular distance r⊥ when computing torque or angular momentum.
  • Placing the centre of gravity in problem 6.15 on the same side as the hole — the negative-mass method shifts it to the opposite side.

Exam Tips

How to score full marks in this chapter

Memorise the standard moments of inertia and the linear-to-rotational analogy table — most numericals reduce to picking the right I and applying τ = Iα, L = Iω or KE = ½Iω2. For equilibrium problems (6.8, 6.9, 6.16) always write both the force balance and a torque balance about a chosen point; choosing the pivot at an unknown force eliminates it. Show every substitution with units and round the final answer sensibly. For conceptual marks, quote the reason (zero external torque → angular momentum conserved) rather than just the result.

Frequently Asked Questions

How many exercises are there in Class 11 Physics Chapter 6?

The NCERT textbook chapter System of Particles and Rotational Motion has 17 numbered exercises (6.1 to 6.17), a mix of conceptual proofs and numerical problems on centre of mass, torque, angular momentum and moment of inertia. All 17 are solved step by step on this page.

Is the centre of mass always inside the body?

No. For solid uniform bodies it lies at the geometric centre inside the body, but for hollow bodies such as a ring, hollow sphere or hollow cylinder the geometric centre falls in empty space, so the centre of mass lies outside the material of the body.

Why does a child’s rotation speed up when arms are folded (Exercise 6.12)?

The turntable is frictionless, so there is no external torque and angular momentum L = Iω is conserved. Folding the arms reduces the moment of inertia I to 2/5 of its value, so ω rises to 5/2 times, i.e. from 40 rev/min to 100 rev/min.

Are these Class 11 Physics Chapter 6 solutions free?

Yes. All solutions are free, follow the official NCERT textbook for session 2026–27, and every numerical answer is cross-checked against the NCERT answer key.

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