The coefficient swap between solid and hollow bodies is the trap that costs marks on moment of inertia (MOI) questions. A solid sphere uses 2/5, a hollow sphere uses 2/3, a disc uses 1/2, and a ring uses 1. The difference comes from how mass is distributed relative to the rotation axis, and mixing these coefficients is a common source of negative marking.
What moment of inertia is. MOI quantifies a rigid body's resistance to angular acceleration about a given axis — the rotational analogue of mass (NCERT Class 11 Physics, Chapter 6, page 114). For a system of point particles: I = Σ mᵢrᵢ², where rᵢ is the perpendicular distance of mass mᵢ from the axis.
Standard geometries. For uniform rigid bodies rotating about their symmetry axis:
The physical logic: a hollow body has all its mass at the maximum distance from the axis, so its MOI is always larger than the corresponding solid body of the same M and R.
Shifting axes — two theorems. When the rotation axis is not the symmetry axis:
Watch out. NEET questions frequently ask you to compare MOI ratios of two geometries (solid sphere vs hollow sphere, disc vs ring) or to apply the parallel axes theorem to shift a known symmetry-axis MOI to an edge or tangent. The coefficient swap trap — using 2/5 where 2/3 is needed, or vice versa — is a high-frequency error. Before substituting, confirm the geometry word-for-word from the stem.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The SI unit of moment of inertia is:
Answer: D. Moment of inertia I = Σmr² has dimensions [M][L²], giving SI unit kg·m² (NCERT Class 11 Physics, Chapter 6, page 115).
Why A is wrong: A is wrong because kg·m has dimensions [M][L] — mass times a distance, the first moment of mass used in locating a centre of mass — not the mass times distance squared that defines MOI.
Why B is wrong: B is wrong because N·m is the unit of torque ([M][L²][T⁻²]), not MOI.
Why C is wrong: C is wrong because kg·m²/s has dimensions [M][L²][T⁻¹], which is angular momentum, not MOI.
Moment of inertia of a rigid body depends on:
Answer: C. MOI depends on the mass of the body and how that mass is distributed relative to the axis of rotation (NCERT Class 11 Physics, Chapter 6, page 115). It does not depend on any kinematic or dynamic quantity.
Why A is wrong: A is wrong because angular velocity is a kinematic quantity; MOI is a geometric property of mass distribution and does not change with ω.
Why B is wrong: B is wrong because angular acceleration is determined by torque and MOI (τ = Iα), not the other way around.
Why D is wrong: D is wrong because torque causes angular acceleration but does not alter the body's mass distribution, so it does not change MOI.
The moment of inertia of a uniform solid sphere about an axis through its centre is (2/5)MR². What is the MOI of a uniform hollow sphere of the same mass and radius about a diameter?
Answer: D. A hollow sphere (thin spherical shell) has all its mass at radius R, giving I = (2/3)MR² about a diameter (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). This is larger than the solid sphere's (2/5)MR² because mass is farther from the axis.
Why A is wrong: A is wrong because (2/5)MR² is the coefficient for a solid sphere, not a hollow one. This is the coefficient-swap trap — a hollow sphere has more mass at the rim, so its MOI must be larger (trap: MOI geometry coefficient swap).
Why B is wrong: B is wrong because (1/2)MR² is the MOI of a uniform disc about its symmetry axis, not a hollow sphere (trap: MOI geometry coefficient swap).
Why C is wrong: C is wrong because MR² is the MOI of a ring (hoop) where all mass is at one radius in a plane, not distributed over a spherical shell.
A uniform disc has moment of inertia I_cm = (1/2)MR² about an axis through its centre and perpendicular to its plane. Using the parallel axes theorem, the MOI about a tangent axis parallel to this central axis is:
Answer: A. Parallel axes theorem: I = I_cm + Md². Here d = R (centre to tangent), so I = (1/2)MR² + MR² = (3/2)MR². The disc's (1/2)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 B is wrong: B is wrong because this equals Md² = MR² alone without including the original I_cm = (1/2)MR². The parallel axes theorem requires the sum of both terms.
Why C is wrong: C is wrong because this is just I_cm without adding the Md² parallel-axis correction. The tangent axis is at distance R from the centre, so Md² = MR² must be added.
Why D is wrong: D is wrong because 2MR² would require d² = (3/2)R², which does not correspond to any standard axis shift for a disc. The tangent is at distance R, not R√(3/2).
A uniform ring has mass M and radius R. Its moment of inertia about a diameter (in-plane axis through centre) is:
Answer: B. By the perpendicular axes theorem for this planar body: I_z = I_x + I_y. The ring's MOI about the axis perpendicular to its plane through the centre is I_z = MR². By symmetry I_x = I_y, so MR² = 2I_x, giving I_x = (1/2)MR². NCERT Class 11 Physics Chapter 6, Table 6.1, page 116 lists this result (ring about a diameter, MR²/2); 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 is wrong because MR² is the MOI about the perpendicular axis (through centre, normal to plane), not a diameter. The perpendicular axes theorem splits this value between two in-plane axes.
Why C is wrong: C is wrong because (1/4)MR² would imply I_z = 2 × (1/4)MR² = (1/2)MR², but I_z for a ring is MR², not (1/2)MR².
Why D is wrong: D is wrong because 2MR² exceeds the perpendicular-axis MOI (MR²). No in-plane axis through the centre of a ring can have MOI greater than the perpendicular axis.
A uniform circular disc of mass M and radius R has I = (1/2)MR² about the axis perpendicular to its plane through the centre. What is its moment of inertia about a tangent line lying in the plane of the disc (i.e., touching the rim, in the same plane as the disc)?
Answer: B. The perpendicular axes theorem gives the diameter value: I_diameter = I_perp/2 = (1/2)MR² ÷ 2 = (1/4)MR², matching the disc-about-diameter value directly tabulated in NCERT Class 11 Physics, Chapter 6, Table 6.1, page 116. The tangent line lies in the plane of the disc, parallel to a diameter and displaced by d = R, so the parallel axes theorem (NCERT Class 11 Physics (pre-2023 edition), Chapter 7, page 167) gives I_tangent = (1/4)MR² + MR² = (5/4)MR².
Why A is wrong: A stops after applying the perpendicular axes theorem (giving the diameter value, (1/4)MR²) but forgets to then shift this to the tangent line using the parallel axes theorem.
Why C is wrong: C applies the parallel axes theorem to the perpendicular-to-plane axis value instead: (1/2)MR² + MR² = (3/2)MR². That combination gives the MOI about a tangent perpendicular to the disc's plane, not a tangent lying in the plane.
Why D is wrong: D correctly starts from the diameter value (1/4)MR² but then adds M(R/√2)² instead of MR², using the wrong distance between the diameter and the tangent line.
A thin uniform rod of mass M and length L has MOI = (1/12)ML² about an axis through its centre perpendicular to its length. The MOI about a parallel axis through one end is:
Answer: B. Parallel axes theorem: I_end = I_cm + M(L/2)² = (1/12)ML² + (1/4)ML² = (1/12 + 3/12)ML² = (4/12)ML² = (1/3)ML². The rod's (1/12)ML² about its centre 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); this exact case is Example 7.11 there.
Why A is wrong: A is wrong because (1/6)ML² would require Md² = (1/6 − 1/12)ML² = (1/12)ML², implying d = L/(2√3), which is not L/2.
Why C is wrong: C is wrong because (1/4)ML² equals Md² = M(L/2)² alone without adding the original I_cm = (1/12)ML². The parallel axes theorem requires summing both terms.
Why D is wrong: D is wrong because (1/2)ML² would require Md² = (1/2 − 1/12)ML² = (5/12)ML², implying d = L√(5/12), which does not correspond to the rod's end.
Two solid spheres A and B have the same mass M. Sphere A has radius R and sphere B has radius 2R. The ratio I_A : I_B about their respective diameters is:
Answer: C. I_A = (2/5)MR². I_B = (2/5)M(2R)² = (2/5)M × 4R² = (8/5)MR². Ratio = (2/5)MR² : (8/5)MR² = 2 : 8 = 1 : 4. MOI scales as R² for fixed mass (NCERT Class 11 Physics Chapter 6, Table 6.1, page 116).
Why A is wrong: A is wrong because 1 : 2 assumes MOI scales linearly with R. Since I ∝ R² (for constant mass), doubling R quadruples I, not doubles it.
Why B is wrong: B is wrong because 2 : 5 conflates the coefficient 2/5 with the ratio. The coefficient cancels in the ratio; only the R² factor matters.
Why D is wrong: D is wrong because 4 : 1 inverts the correct ratio. Sphere B has the larger radius and therefore the larger MOI, so I_A : I_B must be less than 1, not greater.
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Pattern: MOI geometry ratio (based on pattern NEET pattern: moi geometry ratio, observed in NEET 2022 and 2023)
Given
• Solid sphere: mass M, radius R• Disc: mass M, radius R• Both rotate about their symmetry axes
Required
Ratio I_sphere : I_disc
Concept
Each standard uniform rigid body has a tabulated MOI about its symmetry axis. The coefficients differ because mass is distributed differently relative to the axis.
Formula
• Solid sphere: I_sphere = (2/5)MR²• Disc: I_disc = (1/2)MR²
Substitution
Ratio = I_sphere / I_disc = [(2/5)MR²] / [(1/2)MR²]
Calculation
MR² cancels:
Ratio = (2/5) ÷ (1/2) = (2/5) × (2/1) = 4/5
Note: the coefficients 2, 5, and 1, 2 are exact mathematical fractions from integration over uniform mass distributions. They do not limit significant figures.
Final answer
I_sphere : I_disc = 4 : 5
The solid sphere has a slightly smaller MOI than the disc of the same mass and radius, because the sphere distributes mass in three dimensions (some mass lies closer to the axis), while the disc distributes mass in a plane.
Common trap
Swapping the coefficients (using 2/3 for the solid sphere instead of 2/5) would give 4/3 instead of 4/5 — a ratio greater than 1, which should immediately signal an error since the solid sphere must have less rotational inertia than a flat disc of the same M and R.
Similar NEET-style question
A ring and a solid cylinder have equal mass and equal radius. What is the ratio of their MOI about the symmetry axis? (Answer: I_ring : I_cylinder = MR² : (1/2)MR² = 2 : 1.)
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Moment of inertia of a rigid body about an axis: I = Σ m_i r_i² (sum over all particles). For continuous distribution: I = ∫ r² dm. Plays the role of mass in rotational equations: τ = I α.
-- NCERT Class 11 Physics, Ch. 6, p. 114Standard 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 |
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