Chapter Exam

Chapter-6 System of Particles and Rotational Motion — Online MCQ Test

Q: The centre of mass of a system of particles is defined as the point where the entire mass of the system can be considered to be concentrated. Which of the following correctly represents the position of centre of mass?

Answer: A. r_cm = (Σm_i r_i)/(Σm_i) — The position of centre of mass is the weighted average of positions of all particles, given by r_cm = (Σm_i r_i)/(Σm_i), where m_i is the mass of each particle and r_i is its position vector.

Q: Torque is defined as the rotational effect produced by a force. The SI unit of torque is:

Answer: B. Newton-metre (N·m) — Torque is the cross product of force and perpendicular distance (τ = r × F), so its unit is Newton-metre (N·m), not Joule, even though both have the same dimensions.

Q: Angular momentum (L) of a rigid body is related to angular velocity (ω) by the equation:

Answer: A. L = Iω — Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity. This is the rotational analogue of linear momentum p = mv.

Q: A rigid body has both translational and rotational motion. Its total kinetic energy is:

Answer: C. KE_total = (1/2)Mv_cm² + (1/2)Iω² — Total kinetic energy of a rigid body is the sum of translational kinetic energy (1/2)Mv_cm² and rotational kinetic energy (1/2)Iω² about the centre of mass.

Q: Two particles of equal mass are placed at distances r₁ and r₂ from a fixed axis. Their moments of inertia about the axis are I₁ and I₂ respectively. If r₂ = 2r₁, then the ratio I₁:I₂ is:

Answer: A. 1:4 — Moment of inertia I = mr². For equal masses, I₁ = mr₁² and I₂ = m(2r₁)² = 4mr₁². Therefore, I₁:I₂ = 1:4.

Q: A uniform disc of mass M and radius R rolls down an incline without slipping. The fraction of kinetic energy that is rotational is:

Answer: B. 1/3 — For a disc, I = (1/2)MR². Total KE = (1/2)Mv² + (1/2)Iω² = (1/2)Mv² + (1/4)Mv² = (3/4)Mv². Rotational KE = (1/4)Mv², so the fraction is 1/3.

Q: In a collision between two particles, the centre of mass experiences:

Answer: A. No acceleration if external force is zero — The centre of mass acceleration depends only on external forces (a_cm = F_external/M_total), not on internal collision forces. If F_external = 0, then a_cm = 0.

Q: A flywheel of moment of inertia I rotates with angular velocity ω. A brake applies a constant torque τ_b opposing the rotation. The time taken to stop the flywheel is:

Answer: A. t = Iω/τ_b — Using τ = Iα, we get α = -τ_b/I. From ω = ω₀ + αt, setting ω = 0: 0 = ω - (τ_b/I)t, giving t = Iω/τ_b.

PHYSICS · CLASS 11 INTER I YEAR · Andhra State Board

Practice Chapter-6 System of Particles and Rotational Motion with a free chapter-wise online MCQ test for Andhra State Board CLASS 11 INTER I YEAR PHYSICS. This chapter covers: Centre of mass - Momentum conservation - Centre of mass motion - Torque - Angular momentum - Conservation of angular momentum - Equilibrium of rigid bodies - Moment of inertia - Ra.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-6 System of Particles and Rotational Motion — Important Questions & Answers (FAQ)

Frequently asked questions from Andhra State Board CLASS 11 INTER I YEAR PHYSICS — Chapter-6 System of Particles and Rotational Motion, with answers and explanations. These are sample questions; the exam has its own separate question set.

The centre of mass of a system of particles is defined as the point where the entire mass of the system can be considered to be concentrated. Which of the following correctly represents the position of centre of mass?

A. r_cm = (Σm_i r_i)/(Σm_i) ✓
B. r_cm = (Σr_i)/(Σm_i)
C. r_cm = (Σm_i)/(Σm_i r_i)
D. r_cm = (Σm_i r_i²)/(Σm_i)

Answer: A. r_cm = (Σm_i r_i)/(Σm_i)
The position of centre of mass is the weighted average of positions of all particles, given by r_cm = (Σm_i r_i)/(Σm_i), where m_i is the mass of each particle and r_i is its position vector.

Which one of the following is NOT a consequence of conservation of momentum?

A. In the absence of external force, the momentum of a system remains constant
B. The centre of mass of an isolated system moves with constant velocity
C. The velocity of each particle in a system always remains constant ✓
D. Newton's third law can be derived from momentum conservation

Answer: C. The velocity of each particle in a system always remains constant
While momentum of a system is conserved in the absence of external force, individual particles can change their velocities as long as the total momentum remains constant. Option C incorrectly claims all particles maintain constant velocity.

Torque is defined as the rotational effect produced by a force. The SI unit of torque is:

A. Newton (N)
B. Newton-metre (N·m) ✓
C. Kilogram-metre per second (kg·m/s)
D. Joule (J)

Answer: B. Newton-metre (N·m)
Torque is the cross product of force and perpendicular distance (τ = r × F), so its unit is Newton-metre (N·m), not Joule, even though both have the same dimensions.

Angular momentum (L) of a rigid body is related to angular velocity (ω) by the equation:

A. L = Iω ✓
B. L = Iα
C. L = Iτ
D. L = I/ω

Answer: A. L = Iω
Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity. This is the rotational analogue of linear momentum p = mv.

A rigid body has both translational and rotational motion. Its total kinetic energy is:

A. KE_total = (1/2)Mv_cm²
B. KE_total = (1/2)Iω²
C. KE_total = (1/2)Mv_cm² + (1/2)Iω² ✓
D. KE_total = Mv_cm² + Iω²

Answer: C. KE_total = (1/2)Mv_cm² + (1/2)Iω²
Total kinetic energy of a rigid body is the sum of translational kinetic energy (1/2)Mv_cm² and rotational kinetic energy (1/2)Iω² about the centre of mass.

Two particles of equal mass are placed at distances r₁ and r₂ from a fixed axis. Their moments of inertia about the axis are I₁ and I₂ respectively. If r₂ = 2r₁, then the ratio I₁:I₂ is:

A. 1:4 ✓
B. 1:2
C. 2:1
D. 4:1

Answer: A. 1:4
Moment of inertia I = mr². For equal masses, I₁ = mr₁² and I₂ = m(2r₁)² = 4mr₁². Therefore, I₁:I₂ = 1:4.

For a rigid body in pure rolling motion on a horizontal surface, which statement is correct?

A. The instantaneous axis of rotation is at the geometrical centre
B. The instantaneous axis of rotation is at the point of contact with the surface ✓
C. There is no instantaneous axis of rotation
D. The instantaneous axis of rotation changes with time

Answer: B. The instantaneous axis of rotation is at the point of contact with the surface
In pure rolling, the point of contact with the ground has zero instantaneous velocity, making it the instantaneous axis of rotation about which the body appears to be purely rotating.

A uniform disc of mass M and radius R rolls down an incline without slipping. The fraction of kinetic energy that is rotational is:

A. 1/2
B. 1/3 ✓
C. 1/4
D. 2/3

Answer: B. 1/3
For a disc, I = (1/2)MR². Total KE = (1/2)Mv² + (1/2)Iω² = (1/2)Mv² + (1/4)Mv² = (3/4)Mv². Rotational KE = (1/4)Mv², so the fraction is 1/3.

In a collision between two particles, the centre of mass experiences:

A. No acceleration if external force is zero ✓
B. Infinite acceleration due to collision forces
C. An acceleration due to internal collision forces
D. A deceleration proportional to the impact

Answer: A. No acceleration if external force is zero
The centre of mass acceleration depends only on external forces (a_cm = F_external/M_total), not on internal collision forces. If F_external = 0, then a_cm = 0.

A flywheel of moment of inertia I rotates with angular velocity ω. A brake applies a constant torque τ_b opposing the rotation. The time taken to stop the flywheel is: