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Moment of inertia, radius of gyration and standard geometrical bodies: NEET MCQs with answers

89 questions on this topic of Rotational Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The moment of inertia of a uniform circular disc of mass 7 kg and radius 0.5 m about an axis through its centre, perpendicular to its plane, is:

  1. 0.875 kg m²
  2. 0.438 kg m²
  3. 1.31 kg m²
  4. 1.75 kg m²
Show answer and explanation

Answer: (A) 0.875 kg m²

For a disc about its central perpendicular axis, I = MR²/2 = 7 × 0.5²/2 = 0.875 kg m².

Quick tip: Disc (perpendicular axis): MR²/2. Ring: MR². Disc about a diameter: MR²/4.

NCERT Class 11, Chapter 6

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2. A thin spherical shell of mass 8 kg and radius 20 cm rotates about a diameter. Its moment of inertia is:

  1. 0.128 kg m²
  2. 0.16 kg m²
  3. 0.213 kg m²
  4. 0.32 kg m²
Show answer and explanation

Answer: (C) 0.213 kg m²

For a thin spherical shell about a diameter, I = ⅔MR² = 0.667 × 8 × 0.2² = 0.213 kg m².

hard

3. Three identical uniform thin rods, each of mass m and length L, are joined to form an equilateral triangular frame. The moment of inertia of this frame about an axis passing through one of its vertices and perpendicular to the plane of the triangle is:

  1. 5/6 m L²
  2. 5/4 m L²
  3. 2/3 m L²
  4. 3/2 m L²
Show answer and explanation

Answer: (D) 3/2 m L²

For the two rods meeting at the vertex, the axis passes through one of their ends; each contributes (1/3)mL². For the opposite third rod, its centre of mass is at a distance d = L sin(60°) = (√3/2)L from the vertex. Using the parallel axis theorem, its moment of inertia is mL²/12 + m((√3/2)L)² = mL²/12 + 3mL²/4 = 10mL²/12 = (5/6)mL². Summing all three: I = 2 × (1/3)mL² + (5/6)mL² = (2/3 + 5/6)mL² = (9/6)mL² = (3/2)mL².

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4. A solid sphere of mass 5 kg and radius 20 cm rotates about a diameter. Its moment of inertia is:

  1. 0.08 kg m²
  2. 0.133 kg m²
  3. 0.2 kg m²
  4. 0.1 kg m²
Show answer and explanation

Answer: (A) 0.08 kg m²

For a solid sphere about a diameter, I = ⅖MR² = 0.4 × 5 × 0.2² = 0.08 kg m².

easy

5. The moment of inertia of a uniform circular disc of mass 4 kg and radius 0.1 m about an axis through its centre, perpendicular to its plane, is:

  1. 0.04 kg m²
  2. 0.2 kg m²
  3. 0.01 kg m²
  4. 0.02 kg m²
Show answer and explanation

Answer: (D) 0.02 kg m²

For a disc about its central perpendicular axis, I = MR²/2 = 4 × 0.1²/2 = 0.02 kg m².

Quick tip: Disc (perpendicular axis): MR²/2. Ring: MR². Disc about a diameter: MR²/4.

NCERT Class 11, Chapter 6

hard

6. A circular disc of mass M and radius R has radius of gyration k₁ about a tangent perpendicular to the plane of the disc, and radius of gyration k₂ about a tangent in the plane of the disc. The ratio k₁ / k₂ is equal to:

  1. √6 / √5
  2. √3 / √2
  3. 2 / √3
  4. √5 / 2
Show answer and explanation

Answer: (A) √6 / √5

About a perpendicular tangent: I₁ = I_cm + MR² = (1/2)MR² + MR² = (3/2)MR², so k₁² = (3/2)R². About a tangent in the plane: I₂ = I_dia + MR² = (1/4)MR² + MR² = (5/4)MR², so k₂² = (5/4)R². Thus, k₁/k₂ = √[(3/2)/(5/4)] = √(6/5) = √6 / √5.

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7. Consider the following statements regarding the moment of inertia: Statement A: The moment of inertia of a body about an axis is always minimum about an axis passing through its centre of mass parallel to the given axis. Statement B: The theorem of perpendicular axes is strictly applicable only to planar lamina bodies. Choose the correct option:

  1. Both Statement A and Statement B are true.
  2. Statement A is true but Statement B is false.
  3. Statement A is false but Statement B is true.
  4. Both Statement A and Statement B are false.
Show answer and explanation

Answer: (A) Both Statement A and Statement B are true.

According to the parallel axis theorem, I = I_cm + Md², which is minimized when d = 0 (axis passing through CM). Perpendicular axis theorem (I_z = I_x + I_y) holds strictly only for two-dimensional planar bodies (laminas). Hence both statements are true.

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8. A uniform disc of mass 5 kg and radius 10 cm rotates about its central axis perpendicular to its plane. Its moment of inertia is:

  1. 0.0333 kg m²
  2. 0.05 kg m²
  3. 0.025 kg m²
  4. 0.02 kg m²
Show answer and explanation

Answer: (C) 0.025 kg m²

For a uniform disc about its central axis perpendicular to its plane, I = ½MR² = 0.5 × 5 × 0.1² = 0.025 kg m².

easy

9. A uniform disc has mass 5 kg and radius 0.3 m. Its moment of inertia about the axis through its centre and perpendicular to its face is:

  1. 0.112 kg m²
  2. 0.225 kg m²
  3. 0.75 kg m²
  4. 0.45 kg m²
Show answer and explanation

Answer: (B) 0.225 kg m²

For a disc about its central perpendicular axis, I = MR²/2 = 5 × 0.3²/2 = 0.225 kg m².

Quick tip: Disc (perpendicular axis): MR²/2. Ring: MR². Disc about a diameter: MR²/4.

NCERT Class 11, Chapter 6

medium

10. A thin spherical shell of mass 5 kg and radius 30 cm rotates about a diameter. Its moment of inertia is:

  1. 0.18 kg m²
  2. 0.225 kg m²
  3. 0.3 kg m²
  4. 0.45 kg m²
Show answer and explanation

Answer: (C) 0.3 kg m²

For a thin spherical shell about a diameter, I = ⅔MR² = 0.667 × 5 × 0.3² = 0.3 kg m².

medium

11. A thin uniform circular disc has mass M and radius R. An axis passes through a point on the circumference of the disc and is perpendicular to its plane. The radius of gyration of the disc about this axis is:

  1. (√3/2) R
  2. √(5/4) R
  3. √(2/3) R
  4. √(3/2) R
Show answer and explanation

Answer: (D) √(3/2) R

The moment of inertia of the disc about an axis through its centre of mass perpendicular to its plane is I_cm = (1/2)MR². By the parallel-axis theorem, the moment of inertia about a perpendicular axis through the rim is I = (1/2)MR² + MR² = (3/2)MR². Since I = Mk², the radius of gyration is k = √(3/2) R.

medium

12. A uniform disc of mass 3 kg and radius 50 cm rotates about its central axis perpendicular to its plane. Its moment of inertia is:

  1. 0.3 kg m²
  2. 0.375 kg m²
  3. 0.75 kg m²
  4. 0.5 kg m²
Show answer and explanation

Answer: (B) 0.375 kg m²

For a uniform disc about its central axis perpendicular to its plane, I = ½MR² = 0.5 × 3 × 0.5² = 0.375 kg m².

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