System of Particles and Rotational Motion — Class 11 MCQs with Answers

System of Particles and Rotational Motion — important questions & MCQs with answers (Class 11 Physics)

13 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 53 are timed and scored when you sign in.

1. Q1Easy

   Centre of mass of a uniform rod is at:

   A.One end

   B.The geometric centre✓ Correct

   C.1/3 from one end

   D.Outside the rod

   Answer: B. The geometric centre

   Explanation: Uniform rod ⇒ mass symmetric about its midpoint, so centre of mass sits at the geometric centre — e.g. a 2 m rod balances at 1 m from either end. COM is not at an end, and it never lies outside a rod.

2. Q2Easy

   Torque τ =

   A.F·r²

   B.F/r

   C.F + r

   D.r × F✓ Correct

   Answer: D. r × F

   Explanation: Torque τ = r×F, the vector cross product of position and force (magnitude rF sinθ, direction by right-hand rule). Adding F+r or dividing F/r ignores that torque has units N·m, not N or 1/m.

3. Q3Easy

   Torque τ = ?

   A.r × F✓ Correct

   B.F × r

   C.|r||F|

   D.r + F

   Answer: A. r × F

   Explanation: Torque is defined as τ = r×F, position vector crossed into force — order matters for direction (r×F = −F×r), and torque is never the plain magnitude product |r||F| or a vector sum r+F.

4. Q4Easy

   Angular momentum L = ?

   A.r + p

   B.p · r

   C.|r||p|

   D.r × p✓ Correct

   Answer: D. r × p

   Explanation: Angular momentum is L = r×p, the cross product of position and linear momentum p = mv — not a dot product p·r (which gives a scalar) or the plain magnitude product |r||p|.

5. Q5Easy

   Moment of inertia I = ?

   A.Σ m_i

   B.Σ m_i r_i

   C.Σ m_i r_i²✓ Correct

   D.Σ r_i²

   Answer: C. Σ m_i r_i²

   Explanation: Moment of inertia is I = Σmᵢrᵢ², weighting each mass element by the square of its distance from the axis — squaring rᵢ (not just summing rᵢ or mᵢ) is why I depends so strongly on how mass is spread out.

6. Q6Medium

   Velocity of COM = ?

   A.Heaviest particle's v

   B.Σ v_i

   C.Σ m_i v_i / Σ m_i✓ Correct

   D.Average v

   Answer: C. Σ m_i v_i / Σ m_i

   Explanation: Velocity of the centre of mass is the mass-weighted average of individual velocities, v_cm = Σmᵢvᵢ/Σmᵢ — not the plain sum Σvᵢ or the heaviest particle's velocity alone, both of which ignore the other masses.

7. Q7Medium

   Moment of inertia of a thin ring about its central axis (mass M, radius R):

   A.⅓MR²

   B.½MR²

   C.MR²✓ Correct

   D.⅔MR²

   Answer: C. MR²

   Explanation: Every mass element of a thin ring sits at the same distance R from the central axis, so I = ΣmᵢR² = MR² exactly — no ⅓ or ½ factor, unlike a disc (½MR²) where mass spreads inward.

8. Q8Medium

   F = 10 N at distance 2 m, perpendicular to displacement. Torque:

   A.10 N·m

   B.20 N·m✓ Correct

   C.2 N·m

   D.100 N·m

   Answer: B. 20 N·m

   Explanation: Torque τ = rF sinθ; with the force perpendicular to r, sinθ = sin90° = 1, so τ = 2 m × 10 N = 20 N·m — using sinθ instead of assuming θ=0° (which would wrongly give zero) is the key step.

9. Q9Medium

   For particle m moving in straight line, angular momentum about any point on its line of motion is:

   A.Zero✓ Correct

   B.mvr

   C.Maximum

   D.Variable

   Answer: A. Zero

   Explanation: For a particle moving in a straight line, its velocity v is parallel to the position vector r drawn to any point on that line, so sinθ = 0 and L = mvr sinθ = 0 — not mvr, which needs r and v non-parallel.

10. Q10Medium

    Parallel axis theorem: I = ?

    A.Md²

    B.I_cm

    C.I_cm + Md²✓ Correct

    D.I_cm − Md²

    Answer: C. I_cm + Md²

    Explanation: Parallel axis theorem: I = I_cm + Md², adding the centre-of-mass moment of inertia to Md², where d is the distance to the new parallel axis — dropping I_cm and keeping only Md² undercounts I.

11. Q11Hard

    For half-disc (semi-circular plate), COM is at distance from straight edge:

    A.4R/(3π)✓ Correct

    B.R/2

    C.R/3

    D.2R/3

    Answer: A. 4R/(3π)

    Explanation: For a uniform semicircular plate of radius R, integrating y over the area gives the centroid at distance 4R/(3π) from the straight (diameter) edge — roughly 0.42R, closer to the edge than R/2 or 2R/3.

12. Q12Hard

    Moment of inertia of a solid sphere about diameter:

    A.MR²

    B.(2/3)MR²

    C.(2/5)MR²✓ Correct

    D.½MR²

    Answer: C. (2/5)MR²

    Explanation: For a solid sphere spinning about a diameter, integrating mr² over the volume gives I = (2/5)MR² — smaller than a hollow sphere's (2/3)MR² because solid mass is concentrated closer to the axis.

13. Q13Hard

    Disc of mass M, radius R has tangential force F. Angular acceleration:

    A.F/M

    B.F/(MR)

    C.FR/M

    D.2F/(MR)✓ Correct

    Answer: D. 2F/(MR)

    Explanation: τ = FR = Iα for a disc, with I = MR²/2, so FR = (MR²/2)α, giving α = 2F/(MR) — using the disc's ½MR² (not a point mass's MR²) is what produces the factor of 2 in the numerator.

System of Particles and Rotational Motion — FAQs

More Class 11 Physics chapters

• 01Physical World
• 02Units and Measurements
• 03Motion in a Straight Line
• 04Motion in a Plane
• 05Laws of Motion
• 06Work, Energy and Power

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