Moment of Inertia Geometry

8 MCQs3 revision cards9-step worked example

Source: NCERT System of Particles and Rotational MotionPYQ coverage: NEET 2021Official key: NTA-verifiedLast updated: 25 Sep 2026

Moment of Inertia Geometry, explained for NEET

The coefficient swap between solid and hollow bodies is a high-frequency trap in NEET rotational mechanics. If you've ever written 2/3 MR² for a solid sphere or 2/5 MR² for a hollow sphere, this lesson is your repair session.

The core table you must own:

Geometry	Axis	I
Solid sphere	Diameter	(2/5)MR²
Hollow sphere (thin shell)	Diameter	(2/3)MR²
Disc / solid cylinder	Central axis (⊥ to face)	(1/2)MR²
Ring / hollow cylinder (thin)	Central axis (⊥ to plane)	MR²
Uniform rod	Centre, ⊥ to length	(1/12)ML²
Uniform rod	One end, ⊥ to length	(1/3)ML²

The solid sphere, disc, ring and rod-about-centre values are in NCERT Class 11 Physics Chapter 6, Table 6.1, page 116. The hollow sphere and the rod about one end are not in the current table: the hollow sphere is a standard result the NEET (UG) 2026 syllabus covers, and the rod-end value follows from the parallel axes theorem (Example 7.11 in NCERT Class 11 Physics (pre-2023 edition), Chapter 7, page 167). Every formula assumes uniform mass distribution and rotation about the stated symmetry axis.

Why the coefficients differ — the physical logic: Moment of inertia measures how mass is distributed relative to the rotation axis. A solid sphere packs mass throughout its volume, including near the centre — so its MOI coefficient (2/5) is smaller than the hollow sphere's (2/3), which has all mass at the outer surface, far from the axis. Same logic: a disc (1/2) has less I than a ring (1) of equal M and R, because the disc has mass spread from centre to rim.

Shifting axes: When NEET asks for I about a non-standard axis, apply the parallel axes theorem: I = I_cm + Md². For planar bodies, the perpendicular axes theorem gives I_z = I_x + I_y. Both theorems are in the NEET (UG) 2026 syllabus; NCERT removed them from the current Class 11 book in 2023, so they are cited to NCERT Class 11 Physics (pre-2023 edition), Chapter 7, pages 165–167. Neither theorem changes the base coefficients — it adds Md² or combines two in-plane values.

Watch out: The most common distractor in NEET geometry-ratio problems swaps the solid and hollow coefficients. Before marking an answer, ask: "Is mass closer to the axis or farther?" Closer → smaller coefficient.

Can you answer these Moment of Inertia Geometry 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

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

Show answer and why every option is right or wrong

Answer: D. The standard NCERT result for a solid sphere about a diameter is (2/5)MR² (NCERT Class 11 Physics Chapter 6, Table 6.1, page 116).

Why A is wrong: A gives (2/3)MR², which is the MOI of a hollow sphere (thin shell), not a solid sphere — a common coefficient swap (trap: solid vs hollow swap).

Why B is wrong: B gives MR², which is the MOI of a ring about its central axis perpendicular to the plane.

Why C is wrong: C gives (1/2)MR², which is the MOI of a disc or solid cylinder about its central axis, not a sphere.

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MCQ 2Easy RecallPractice

The moment of inertia of a uniform thin-walled hollow sphere of mass M and radius R about a diameter is:

Show answer and why every option is right or wrong

Answer: A. For a thin hollow sphere (spherical shell) about a diameter, I = (2/3)MR² (not in NCERT's moment-of-inertia table in either the current or the pre-2023 edition; a standard result the NEET (UG) 2026 syllabus covers under moments of inertia of simple geometrical objects).

Why B is wrong: B gives (1/2)MR², which is the disc/solid cylinder value about its symmetry axis.

Why C is wrong: C gives (2/5)MR², which is the solid sphere coefficient — the classic swap between solid and hollow (trap: coefficient swap).

Why D is wrong: D gives (1/3)MR², which is the MOI of a uniform rod about one end.

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MCQ 3Easy RecallPractice

A uniform disc of mass M and radius R rotates about an axis through its centre and perpendicular to its plane. Its moment of inertia is:

Show answer and why every option is right or wrong

Answer: C. The standard result for a uniform disc about the perpendicular central axis is (1/2)MR² (NCERT Class 11 Physics Chapter 6, Table 6.1, page 116).

Why A is wrong: A gives MR², which is the ring's MOI, not the disc's — the ring has all mass at radius R, the disc distributes mass from centre to rim (trap: disc vs ring swap).

Why B is wrong: B gives (1/4)MR², which is the MOI of a disc about a diameter (an in-plane axis), not about the perpendicular axis.

Why D is wrong: D gives (2/5)MR², which is the solid sphere's value, not relevant for a disc.

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MCQ 4Direct ApplicationPractice

A solid sphere and a hollow sphere have the same mass M and the same radius R. The ratio of their moments of inertia about their respective diameters, I_solid : I_hollow, is:

Show answer and why every option is right or wrong

Answer: B. I_solid = (2/5)MR², I_hollow = (2/3)MR². Ratio = (2/5)/(2/3) = 3/5, so I_solid : I_hollow = 3 : 5 (the solid sphere's value is in NCERT Class 11 Physics Chapter 6, Table 6.1, page 116; the hollow sphere's is not in NCERT's moment-of-inertia table in either the current or the pre-2023 edition; a standard result the NEET (UG) 2026 syllabus covers under moments of inertia of simple geometrical objects).

Why A is wrong: A reverses the ratio — this would mean the solid sphere has a larger MOI, but since more of its mass is near the centre, the solid sphere has the smaller MOI (trap: coefficient swap giving inverted ratio).

Why C is wrong: C gives 2 : 3 — this is the hollow sphere's own coefficient, 2/3, written as a ratio, not the ratio of the two moments of inertia.

Why D is wrong: D gives 2 : 5 — this is the solid sphere's own coefficient, 2/5, written as a ratio; it compares the solid sphere with MR², not with the hollow sphere.

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MCQ 5Direct ApplicationPractice

A uniform ring of mass M and radius R lies in the xy-plane with its centre at the origin. The moment of inertia about a diameter (say, the x-axis) is:

Show answer and why every option is right or wrong

Answer: D. By the perpendicular axes theorem, I_z = I_x + I_y. For a ring, I_z = MR² (about the perpendicular central axis). By symmetry I_x = I_y, so I_x = MR²/2. NCERT Class 11 Physics Chapter 6, Table 6.1, page 116 lists this result; for the theorem itself, NCERT Class 11 Physics (pre-2023 edition), Chapter 7, page 165 (NCERT removed this theorem from the current book in 2023; the NEET (UG) 2026 syllabus still lists it).

Why A is wrong: A gives MR², which is the ring's MOI about the perpendicular axis through its centre, not about a diameter — the perpendicular axes theorem halves this for each in-plane axis.

Why B is wrong: B gives 2MR², which has no standard geometric basis — likely a multiplication error.

Why C is wrong: C gives (1/4)MR², which would apply to a disc about its diameter ((1/2)MR² ÷ 2 = (1/4)MR²), not a ring.

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MCQ 6Direct ApplicationPractice

A uniform disc of mass M and radius R has its MOI about the central perpendicular axis equal to (1/2)MR². Using the perpendicular axes theorem, the MOI of the disc about a diameter is:

Show answer and why every option is right or wrong

Answer: A. I_z = I_x + I_y. For the disc, I_z = (1/2)MR². By symmetry I_x = I_y, so I_x = (1/2)MR² ÷ 2 = (1/4)MR². NCERT Class 11 Physics Chapter 6, Table 6.1, page 116 lists this result (disc about a diameter, MR²/4); for the theorem itself, NCERT Class 11 Physics (pre-2023 edition), Chapter 7, page 166 (NCERT removed this theorem from the current book in 2023; the NEET (UG) 2026 syllabus still lists it); this exact case is Example 7.10 there.

Why B is wrong: B gives MR², which is the ring's perpendicular-axis value, not applicable here.

Why C is wrong: C gives (1/2)MR², which is the disc's MOI about the perpendicular axis, not about a diameter.

Why D is wrong: D gives (3/4)MR², which has no basis — possibly a miscombination of the parallel and perpendicular axis results.

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MCQ 7CalculationPractice

A uniform solid sphere of mass M and radius R rotates about an axis tangent to the sphere. The moment of inertia about this tangent axis is:

Show answer and why every option is right or wrong

Answer: C. The tangent axis is parallel to a diameter and a distance d = R from the centre. By the parallel axes theorem: I = I_cm + Md² = (2/5)MR² + MR² = (7/5)MR². The solid sphere's (2/5)MR² is in NCERT Class 11 Physics Chapter 6, Table 6.1, page 116; for the theorem itself, NCERT Class 11 Physics (pre-2023 edition), Chapter 7, page 167 (NCERT removed this theorem from the current book in 2023; the NEET (UG) 2026 syllabus still lists it).

Why A is wrong: A gives (2/5)MR², which is only the MOI about the diameter — this forgets the Md² term from the parallel axes theorem entirely.

Why B is wrong: B gives (5/3)MR² — this would result from using the hollow sphere value (2/3)MR² + MR² = (5/3)MR² instead of the solid sphere value (trap: solid vs hollow coefficient swap).

Why D is wrong: D gives (3/5)MR² — this is not derivable from any standard formula combination; likely a numerical error in the fraction addition.

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MCQ 8CalculationPractice

A uniform thin ring of mass M and radius R lies in the xy-plane. Its moment of inertia about an axis tangent to the ring and lying in its plane is:

Show answer and why every option is right or wrong

Answer: B. First, find I about a diameter: by the perpendicular axes theorem, I_diameter = MR²/2 (since I_z = MR² and I_x = I_y). Then apply the parallel axes theorem for the tangent axis (in-plane, distance R from centre): I = MR²/2 + MR² = (3/2)MR². The ring's MR²/2 about a diameter is in NCERT Class 11 Physics Chapter 6, Table 6.1, page 116; for the theorems, see NCERT Class 11 Physics (pre-2023 edition), Chapter 7, pages 165–167 (the ring about a tangent is Example 7.12 on page 167); NCERT removed them from the current book in 2023, but the NEET (UG) 2026 syllabus still lists them.

Why A is wrong: A gives MR², which is the ring's MOI about the perpendicular central axis — this ignores both the perpendicular-to-diameter step and the parallel axis shift.

Why C is wrong: C gives 2MR², which would result from incorrectly using I_diameter = MR² (the perpendicular-axis value) and adding MR² — forgetting that the in-plane diameter value is MR²/2, not MR².

Why D is wrong: D gives (1/2)MR², which is the diameter MOI alone — this forgets the parallel axes theorem's Md² = MR² term for shifting to the tangent.

Report an error · Every fix is public: corrections log

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Moment of Inertia Geometry: quick recall before you leave

How do you solve a Moment of Inertia Geometry question? A worked example

Pattern: Comparing MOI of two standard geometries (NEET 2022/2023 pattern — ratio of moments of inertia for solid sphere vs hollow sphere of same mass and radius).

1
Given
A solid sphere and a thin hollow sphere each have mass M = 2.0 kg and radius R = 0.10 m. Both rotate about a diameter.
2
Required
Find (a) I for each sphere, (b) the ratio I_solid : I_hollow.
3
Concept
The moment of inertia depends on how mass is distributed relative to the axis. Standard tabulated values apply for uniform bodies about their symmetry axes (NCERT Class 11 Physics Chapter 6, Table 6.1, page 116).
4
Formula
• Solid sphere: I_solid = (2/5)MR²• Hollow sphere: I_hollow = (2/3)MR²
5
Substitution
• I_solid = (2/5) × 2.0 × (0.10)² = (2/5) × 2.0 × 0.010• I_hollow = (2/3) × 2.0 × (0.10)² = (2/3) × 2.0 × 0.010
6
Calculation
• I_solid = (2/5) × 0.020 = 0.0080 kg·m²• I_hollow = (2/3) × 0.020 = 0.01333… kg·m²
Note: The fractions 2/5 and 2/3 are exact geometric constants derived from integration; they do not affect significant-figure counting. The data values (M = 2.0 kg, R = 0.10 m) each have 2 significant figures, so the final answer is reported to 2 significant figures.
7
Final answer
• I_solid = 8.0 × 10⁻³ kg·m²• I_hollow = 1.3 × 10⁻² kg·m²• Ratio I_solid : I_hollow = (2/5)/(2/3) = 3/5 = 3 : 5
8
Common trap
Swapping the coefficients — writing 2/3 for the solid and 2/5 for the hollow — inverts the ratio to 5 : 3. The physical anchor: solid sphere has mass distributed throughout, including near the centre, so its coefficient is smaller. Hollow sphere has all mass at the maximum distance R, so its coefficient is larger.
9
Similar NEET-style question
A solid cylinder and a thin hollow cylinder have equal mass M and equal radius R. Find the ratio of their moments of inertia about their central axes.

*(Answer: I_solid_cyl / I_hollow_cyl = (1/2)MR² / MR² = 1 : 2)*

---

What to remember before solving Moment of Inertia Geometry questions

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

Which Moment of Inertia Geometry formulas do you need for NEET?

1 formula — click to collapse

Moment of inertia for common rigid bodies

I_{sphere} = \frac{2}{5} M R^{2}; I_{hollow} = \frac{2}{3} M R^{2}; I_{rod,c} = \frac{1}{12} M L^{2}; I_{ring} = M R^{2}; I_{disc} = \frac{1}{2} M R^{2}

Standard moments of inertia about the symmetry axis. For other axes use parallel/perpendicular axes theorems.

Symbol	Quantity	SI Unit
M	mass	kg
R	radius	m
L	length	m
I	moment of inertia	kg*m^2

Valid when

Uniform mass distribution
Rotation about symmetry axis (unless noted)

Where do students lose marks on Moment of Inertia Geometry?

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

2 items — click to collapse

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.

Mistake

Confuses MOI coefficients (e.g. solid sphere 2/5 vs hollow sphere 2/3).

Root cause: concept gap

Correction

Memorise standard moments of inertia. Solid sphere has more mass near axis (smaller MOI = 2MR²/5); hollow sphere has all mass at radius R (larger MOI = 2MR²/3).

More in System of Particles and Rotational Motion: 5 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.

Moment of Inertia Geometry questions from past NEET papers

1 question from NEET 2021. Answers verified against NTA official keys. — click to collapse

NEET 2021

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times 'MR²'. Then the value of 'K' is

11/8

23/4

37/8

41/4

NTA Answer: Option 2(final)

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All 11 past-paper questions from System of Particles and Rotational Motion →

How does NEET ask about Moment of Inertia Geometry?

1 recurring pattern from past papers — click to collapse

Compare moments of inertia (or radii of gyration) of two standard rigid bodies (e.g. solid sphere vs hollow sphere; disc vs ring) about their natural axes. Apply tabulated I formulas; take ratio.

Direct ApplicationEasy

Common distractors

swap solid hollow coefficients

Confusing 2/5 (solid sphere) with 2/3 (hollow sphere)

Sources

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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Moment of Inertia Parallel Perpendicular Axes Theorems

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Moment of Inertia Geometry

Moment of Inertia Geometry, explained for NEET

Can you answer these Moment of Inertia Geometry MCQs?

Free NEET study resources

Moment of Inertia Geometry: quick recall before you leave

Standard MOI formulas

Parallel axes theorem

Perpendicular axes theorem (planar bodies only)

How do you solve a Moment of Inertia Geometry question? A worked example

What to remember before solving Moment of Inertia Geometry questions

Common moments of inertia

Which Moment of Inertia Geometry formulas do you need for NEET?

Moment of inertia for common rigid bodies

Valid when

Where do students lose marks on Moment of Inertia Geometry?

MOI coefficient swap (solid vs hollow)

When it triggers

How to avoid

Confuses MOI coefficients (e.g. solid sphere 2/5 vs hollow sphere 2/3).

Correction

Moment of Inertia Geometry questions from past NEET papers

How does NEET ask about Moment of Inertia Geometry?

Compare moments of inertia (or radii of gyration) of two standard rigid bodies (e.g. solid sphere vs hollow sphere; disc vs ring) about their natural axes. Apply tabulated I formulas; take ratio.

Common distractors

Sources

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