Radius of Gyration

8 MCQs9-step worked example
Source: NCERT System of Particles and Rotational MotionPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 7 Oct 2026

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For a body of mass M and moment of inertia I about an axis, which relation defines the radius of gyration k?
  1. A.I = Mk²
  2. B.I = Mk
  3. C.I = M²k
  4. D.I = M/k²
Tap to see the answer

Answer: A. A is correct: NCERT writes I = Mk², where k has the dimension of length and is called the radius of gyration (NCERT Class 11 Physics, Chapter 6, page 115).

B is wrong: B is wrong because a moment of inertia has dimensions of mass times length squared, so k must enter squared.

C is wrong: C is wrong because the mass appears once in I = Mk², not squared, and k is squared.

D is wrong: D is wrong because k appears in the numerator, not the denominator; a larger k means a larger I.

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Radius of Gyration, explained for NEET

The trap is the coefficient swap: the radius of gyration k depends on the shape and on the axis, and the same body gives a different k about a different axis. Solid versus hollow, and a disc about its normal versus about a diameter, are the usual swaps.

For a body of mass M and moment of inertia I about an axis, NCERT writes I = Mk², where k has the dimension of length (NCERT Class 11 Physics, Chapter 6, page 115). So k = √(I/M). NCERT calls k a geometric property of the body and the axis of rotation.

The definition: the radius of gyration of a body about an axis is the distance from the axis of a mass point whose mass equals the mass of the whole body and whose moment of inertia equals that of the body about the axis (NCERT Class 11 Physics, Chapter 6, page 115).

The values come from Table 6.1 (NCERT Class 11 Physics, Chapter 6, page 116). Ring about the axis perpendicular to its plane at the centre: I = MR², so k = R. Disc about the axis perpendicular to it at the centre: I = MR²/2, so k = R/√2. Disc about a diameter: I = MR²/4, so k = R/2 (the page-115 text gives the same R/2). Thin rod of length L about the perpendicular axis at its midpoint: I = ML²/12, so k = L/√12. Solid sphere about a diameter: I = 2MR²/5, so k = R√(2/5). Hollow and solid cylinder about their axes: MR² and MR²/2.

NEET has asked solid versus thin hollow sphere comparisons. NCERT's table does not give the thin hollow sphere: I = 2MR²/3 is learned from outside the table, and it is the number that gets swapped with 2/5.

Watch-out: k is a length. Take the square root last, and check which axis the table row refers to before reading its coefficient.

How do you solve a Radius of Gyration question? A worked example

  1. 1

    Given

    A thin ring and a circular disc each have mass 3.0 kg and radius 0.40 m. Each rotates about the axis perpendicular to its plane through its centre.

  2. 2

    Required

    The radius of gyration of each body and the ratio k(ring) : k(disc).

  3. 3

    Concept

    The radius of gyration follows from I = Mk², so k = √(I/M). The table values are I = MR² for the ring and I = MR²/2 for the disc about this axis (NCERT Class 11 Physics, Chapter 6, pages 115 and 116).

  4. 4

    Formula

    k = √(I/M), with I(ring) = MR² and I(disc) = MR²/2.

  5. 5

    Substitution

    I(ring) = 3.0 × (0.40)²; I(disc) = (1/2) × 3.0 × (0.40)². Then k = √(I / 3.0) for each.

  6. 6

    Calculation

    I(ring) = 3.0 × 0.16 = 0.48 kg m², so k(ring) = √(0.48 / 3.0) = √0.16 = 0.40 m. I(disc) = 0.24 kg m², so k(disc) = √(0.24 / 3.0) = √0.080 = 0.28 m. The factor 1/2 in the disc coefficient and the 2 in the square root are exact and do not affect the significant figures.

  7. 7

    Final answer

    k(ring) = 0.40 m and k(disc) = 0.28 m, so k(ring) : k(disc) = 0.40 / 0.28 = 1.4, which is √2 : 1.

  8. 8

    Common trap

    Writing the ratio of the moments of inertia (2 : 1) as the ratio of the radii of gyration. The radius of gyration is the square root, so the ratio is √2 : 1. A second slip is using the disc's coefficient 1/2 for the ring.

  9. 9

    Similar NEET-style question

    A solid sphere and a solid cylinder have the same mass M and the same radius R. The sphere rotates about a diameter and the cylinder about its axis. What is the ratio k(sphere) : k(cylinder)? (Answer: k = √(I/M), so k(sphere) = R√(2/5) and k(cylinder) = R√(1/2); the ratio is √((2/5) / (1/2)) = √(4/5) = 0.89.)

    ---

Can you answer these Radius of Gyration MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

For a body of mass M and moment of inertia I about an axis, which relation defines the radius of gyration k?

Show answer and why every option is right or wrong

Answer: A. A is correct: NCERT writes I = Mk², where k has the dimension of length and is called the radius of gyration (NCERT Class 11 Physics, Chapter 6, page 115).

Why B is wrong: B is wrong because a moment of inertia has dimensions of mass times length squared, so k must enter squared.

Why C is wrong: C is wrong because the mass appears once in I = Mk², not squared, and k is squared.

Why D is wrong: D is wrong because k appears in the numerator, not the denominator; a larger k means a larger I.

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MCQ 2Easy RecallPractice

Which statement correctly defines the radius of gyration of a body about an axis?

Show answer and why every option is right or wrong

Answer: C. C is correct: this is the NCERT definition of the radius of gyration (NCERT Class 11 Physics, Chapter 6, page 115).

Why A is wrong: A is wrong because the centre of mass distance is a different quantity; k depends on how the mass is distributed about the axis.

Why B is wrong: B is wrong because a plain average distance ignores the weighting that makes I = Mk².

Why D is wrong: D is wrong because an enclosing circle is a size of the body, whereas k is defined through the moment of inertia.

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MCQ 3Easy RecallPractice

What is the dimension of the radius of gyration?

Show answer and why every option is right or wrong

Answer: B. B is correct: in I = Mk², k has the dimension of length (NCERT Class 11 Physics, Chapter 6, page 115).

Why A is wrong: A is wrong because mass times length squared is the dimension of I, not of k.

Why C is wrong: C is wrong because length squared is the dimension of k², the ratio I / M.

Why D is wrong: D is wrong because mass is the other factor in I = Mk²; k itself is a length.

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MCQ 4Direct ApplicationPractice

A thin ring of mass M and radius R rotates about the axis perpendicular to its plane through its centre. What is its radius of gyration?

Show answer and why every option is right or wrong

Answer: D. D is correct: Table 6.1 gives I = MR² for the ring about this axis, so k = √(I/M) = R (NCERT Class 11 Physics, Chapter 6, page 116).

Why A is wrong: A is wrong because R/2 is the radius of gyration of a disc about a diameter, not of a ring about its normal axis.

Why B is wrong: B is wrong because R/√2 comes from the disc coefficient 1/2; a ring has all its mass at R, so its coefficient is 1.

Why C is wrong: C is wrong because no row of Table 6.1 gives I = MR²/3.

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MCQ 5Direct ApplicationPractice

A circular disc of mass M and radius R rotates about one of its diameters. What is its radius of gyration?

Show answer and why every option is right or wrong

Answer: A. A is correct: for a disc about a diameter, I = MR²/4, so k = R/2 (NCERT Class 11 Physics, Chapter 6, page 115, and Table 6.1, page 116).

Why B is wrong: B is wrong because R/√2 uses I = MR²/2, which is the row for the axis perpendicular to the disc, not for a diameter.

Why C is wrong: C is wrong because R/4 forgets the square root: it is the coefficient 1/4 used directly as k.

Why D is wrong: D is wrong because k = R is the ring value, with all the mass at distance R.

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MCQ 6Direct ApplicationPractice

A thin uniform rod of length 1.20 m rotates about an axis perpendicular to the rod through its midpoint. What is its radius of gyration?

Show answer and why every option is right or wrong

Answer: C. C is correct: I = ML²/12 about this axis, so k = L/√12 = 1.20 / 3.464 = 0.346 m (NCERT Class 11 Physics, Chapter 6, page 115, and Table 6.1, page 116).

Why A is wrong: A is wrong because 0.100 m is L/12, the coefficient of I used directly as k with no square root.

Why B is wrong: B is wrong because 0.600 m is L/2, the distance from the midpoint to an end, not the radius of gyration.

Why D is wrong: D is wrong because 0.693 m is L/√3, which would need I = ML²/3 (an axis through one end), not the midpoint axis.

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MCQ 7CalculationPractice

A solid cylinder and a hollow cylinder have the same mass and the same radius R. Each rotates about its own axis. What is the ratio of the radius of gyration of the solid cylinder to that of the hollow cylinder?

Show answer and why every option is right or wrong

Answer: B. B is correct: Table 6.1 gives MR²/2 for the solid cylinder and MR² for the hollow one, so the radii of gyration are R/√2 and R, and their ratio is 1 : √2 (NCERT Class 11 Physics, Chapter 6, page 116).

Why A is wrong: A is wrong because 1 : 2 is the ratio of the moments of inertia; the radius of gyration is the square root of I / M.

Why C is wrong: C is wrong because it inverts the ratio: the solid cylinder has its mass nearer the axis, so its k is the smaller one.

Why D is wrong: D is wrong because 2 : 1 inverts the moments of inertia and skips the square root.

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MCQ 8CalculationPractice

A solid sphere of mass 5.0 kg and radius 0.20 m rotates about a diameter. What is its radius of gyration?

Show answer and why every option is right or wrong

Answer: D. D is correct: I = 2MR²/5 = (2/5)(5.0)(0.20)² = 0.080 kg m², so k = √(I/M) = √(0.080 / 5.0) = 0.13 m (NCERT Class 11 Physics, Chapter 6, page 116).

Why A is wrong: A is wrong because k = R applies to a ring or hollow cylinder, not to a solid sphere.

Why B is wrong: B is wrong because 0.16 m uses the hollow-sphere coefficient 2/3 in place of the solid-sphere 2/5, the coefficient swap.

Why C is wrong: C is wrong because 0.080 is the value of I in kg m², not of k; the unit and the square root are missing.

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What to remember before solving Radius of Gyration questions

3 NCERT lines
Formula

Common moments of inertia

Table 6.1 (moments of inertia of some regular shaped bodies about specific axes, printed page 116): thin circular ring, radius R, about the axis perpendicular to its plane at the centre: MR²; about a diameter: MR²/2. Thin rod, length L, about the axis perpendicular to the rod at its midpoint: ML²/12. Circular disc, radius R, about the axis perpendicular to the disc at the centre: MR²/2; about a diameter: MR²/4. Hollow cylinder about its axis: MR². Solid cylinder about its axis: MR²/2. Solid sphere about a diameter: 2MR²/5. The table gives no hollow (thin-shell) sphere; the text notes derivations are beyond the book's scope.

-- NCERT Class 11 Physics, Ch. 6, p. 116
Key Fact

I = Mk^2: k (radius of gyration) has the dimension of length; rod L/sqrt(12), disc R/2

Notice from the Table 6.1 that in all cases, we can write I = Mk2, where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint, i.e. 2 2 12, k L = = 12 k L . Similarly, k = R/2 for the circular disc about its diameter. The length k is a geometric property of the body and axis of rotation. It is called the radius of gyration.

-- NCERT Class 11 Physics, Ch. 6, p. 115

Where do students lose marks on Radius of Gyration?

1 trap or mistake

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

MOI coefficient swap (solid vs hollow)

Category: Similar Terms

Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).

When it triggers

Question gives a specific geometry and asks for I or radius of gyration.

How to avoid

Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.

Radius of Gyration: NEET previous year questions (PYQs) with answers

1 question from NEET 2022, answers verified against NTA official keys
NEET 2022

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

11 : √2
22 : 1
3√2 : 1
44 : 1
NTA Answer: Option 3(final)

Why the other options are wrong

  • Option 1: Inverted. About the normal axis I = MR²/2, so k = R/√2; about a diameter I = MR²/4, so k = R/2. The ratio is (R/√2) : (R/2) = √2 : 1.
  • Option 2: 2 : 1 is the ratio of the moments of inertia (MR²/2 to MR²/4). The radius of gyration goes as the square root, giving √2 : 1.
  • Option 4: 4 : 1 is neither ratio: the moments of inertia are in ratio 2 : 1 and the radii of gyration in ratio √2 : 1.

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All 12 past-paper questions from System of Particles and Rotational Motion →

How does NEET ask about Radius of Gyration?

1 recurring pattern from past papers

More in System of Particles and Rotational Motion: 6 exam traps and mistakes · 8 formulas · 3 question patterns from its other lessons.

Sources

NCERT refs: Class 11 Physics Chapter 6, p.115 | Class 11 Physics Chapter 6, p.116

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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Every past-paper question from this chapter:

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