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 |
These are NCERT-tabulated values (Class 11 Physics Chapter 6, page 10). 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 (NCERT Chapter 6, page 12). For planar bodies, the perpendicular axes theorem gives I_z = I_x + I_y (same page). 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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The moment of inertia of a uniform solid sphere of mass M and radius R about its diameter is:
The moment of inertia of a uniform thin-walled hollow sphere of mass M and radius R about a diameter is:
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:
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:
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:
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:
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:
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:
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).
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.
Required
Find (a) I for each sphere, (b) the ratio I_solid : I_hollow.
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, page 10).
Formula
- Solid sphere: I_solid = (2/5)MR² - Hollow sphere: I_hollow = (2/3)MR²
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
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.
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**
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.
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)* ---
Thin rod (length L, axis through centre, perpendicular): I = (1/12) M L². Solid sphere: I = (2/5) M R². Hollow sphere: I = (2/3) M R². Solid cylinder/disc: I = (1/2) M R² (axis through axis of symmetry). Ring: I = M R².
-- NCERT Class 11 Physics, Ch. 6, p. 10For a particle: L = r x p. For a rigid body about its rotation axis: L = I omega. Vector quantity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| L | angular momentum | kg*m^2/s |
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
The position of the centre of mass equals the mass-weighted average of particle positions. For continuous bodies use integral form.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R_cm | CoM position | m |
| m_i | mass of i-th particle | kg |
| r_i | position of i-th particle | m |
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 |
Moment of inertia about any axis = moment about parallel axis through CM + Md^2.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | MOI about given axis | kg*m^2 |
| I_cm | MOI about parallel CM axis | kg*m^2 |
| M | total mass | kg |
| d | perpendicular distance | m |
For planar lamina: MOI about axis perpendicular to plane = sum of MOI about two perpendicular in-plane axes through same point.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_z | MOI perp to plane | kg*m^2 |
| I_x, I_y | MOI in plane | kg*m^2 |
Rotational analogues of linear kinematic equations under constant angular acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| omega | angular velocity | rad/s |
| alpha | angular acceleration | rad/s^2 |
| theta | angular displacement | rad |
| t | time | s |
Energy of rotation about an axis. Adds to translational KE for rolling bodies.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
Cross product of position vector and force vector. Magnitude r F sin(theta).
| Symbol | Quantity | SI Unit |
|---|---|---|
| tau | torque | N*m |
| r | position from pivot | m |
| F | force | N |
| theta | angle between r and F | rad |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student answers L/2 for two-particle CoM regardless of mass ratio.
Question gives two masses on rigid rod and asks for CoM distance.
R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.
Category: Similar Terms
Student conserves rotational KE when angular momentum is conserved (or vice versa). When I changes, L = Iω is conserved but KE = ½Iω² is NOT (it depends on I and ω together).
Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).
L conservation requires zero external torque. KE conservation requires no work done — different criteria. When I changes via internal forces, L conserved, ω increases, KE increases.
Category: Similar Terms
Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).
Question gives a specific geometry and asks for I or radius of gyration.
Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.
Category: Unit Conversion
Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.
Question gives ω in rpm and asks for kinematic quantities in SI units.
Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.
Root cause: concept gap
L = Iω is conserved when external torque is zero. KE = ½Iω² is NOT conserved when I changes (since ω changes too). When skater pulls arms in, L conserved, ω increases, KE increases (work done by muscles).
Root cause: concept gap
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).
Root cause: unit error
Convert: ω(rad/s) = (2π/60) × ω(rpm). For example, 1200 rpm = 1200 × 2π/60 = 125.66 rad/s.
The angular acceleration of a body, moving along the circumference of a circle, is
forgets conversion rpm to rad s
Treats rpm as rad/s without 2*pi/60 conversion
uses equal distribution
Default to L/2 regardless of mass ratio
uses energy conservation instead
Confusing L conservation with KE conservation
swap solid hollow coefficients
Confusing 2/5 (solid sphere) with 2/3 (hollow sphere)
Test yourself on this topic with real past-paper questions:
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