Parallel Perpendicular Axes Theorems

8 MCQs3 revision cards9-step worked example

Source: NCERT System of Particles and Rotational MotionPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

You know the standard moments of inertia — disc ½MR², ring MR², rod about centre (1/12)ML². But NEET rarely asks for those values directly. The high-frequency trap in this topic is applying the theorems to shift axes and getting the geometry wrong.

The trap: students reach for the perpendicular axes theorem on a solid sphere, forgetting it applies only to planar laminae. Or they apply the parallel axes theorem but use the wrong distance — plugging in the radius when the shift is to a tangent, or confusing the distance to the edge with the distance to the centre of mass.

Parallel axes theorem (NCERT Class 11 Physics Chapter 7, page 12): the moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass plus Md², where d is the perpendicular distance between the two axes. Both axes must be parallel. I_cm must be the value about the CM axis specifically — not about some other convenient axis.

Perpendicular axes theorem (NCERT Class 11 Physics Chapter 7, page 12): for a planar body (lamina) only, I_z = I_x + I_y, where z is perpendicular to the plane and x, y are two mutually perpendicular axes in the plane, all three intersecting at the same point. This theorem does not apply to three-dimensional bodies like spheres or cylinders.

Bridge to NEET: questions test whether you can combine the standard MOI formula with one or both theorems. A disc about a tangent in its plane requires both theorems in sequence: first perpendicular axes to get an in-plane diameter MOI, then parallel axes to shift to the tangent.

Watch-out: always identify whether the body is planar before invoking the perpendicular axes theorem. If the problem says "sphere" or "cylinder" (solid 3D body), the perpendicular axes theorem does not apply.

Practice 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 perpendicular axes theorem (I_z = I_x + I_y) is valid for which type of body?

MCQ 2Easy RecallPractice

In the parallel axes theorem I = I_cm + Md², the quantity d represents:

MCQ 3Direct ApplicationPractice

A uniform disc of mass M and radius R has moment of inertia ½MR² about its central axis (perpendicular to its plane). Using the perpendicular axes theorem, what is its moment of inertia about a diameter?

MCQ 4Direct ApplicationPractice

A uniform thin ring of mass M and radius R has I = MR² about an axis through its centre, perpendicular to its plane. What is the moment of inertia about a tangent perpendicular to its plane?

MCQ 5Direct ApplicationPractice

A uniform disc of mass M and radius R rotates about a tangent in its plane. Its moment of inertia about this axis is:

MCQ 6Easy RecallPractice

The parallel axes theorem requires that one of the two parallel axes must pass through:

MCQ 7CalculationPractice

A uniform solid sphere of mass M and radius R has I_cm = (2/5)MR² about a diameter. What is its moment of inertia about a tangent to the sphere?

MCQ 8CalculationPractice

A uniform rectangular lamina has moment of inertia I₁ about an axis along its length (through its centre, in its plane) and I₂ about an axis along its breadth (through its centre, in its plane). If I₁ = 4.0 × 10⁻² kg·m² and I₂ = 9.0 × 10⁻² kg·m², what is its moment of inertia about an axis through its centre, perpendicular to its plane?

Quick recall before you leave

Worked Example

Pattern: Combination of perpendicular and parallel axes theorems applied to a disc (highest-relevance in-scope application — the dossier's only topic-specific pattern, `NEET pattern: moi geometry ratio`, involves MOI geometry comparisons; this worked example extends the core theorem application to a multi-step axis shift).

1
Given
A uniform disc has mass M = 2.0 kg and radius R = 0.20 m. Its moment of inertia about the central perpendicular axis is I_z = ½MR².
2
Required
Find the moment of inertia about a tangent to the disc lying in its plane.
3
Concept
This requires two theorem applications in sequence: 1. Perpendicular axes theorem to find I about a diameter (in-plane axis through centre). 2. Parallel axes theorem to shift from the diameter to a parallel tangent (distance d = R).
4
Formula
Perpendicular axes theorem: I_z = I_x + I_y, where I_x = I_y = I_diameter for the disc (by symmetry). Parallel axes theorem: I_tangent = I_diameter + MR².
5
Substitution
I_z = ½MR² = ½ × 2.0 × (0.20)² = 0.040 kg·m² By symmetry: I_diameter = I_z / 2 = 0.040 / 2 = 0.020 kg·m² I_tangent = I_diameter + MR² = 0.020 + 2.0 × (0.20)² = 0.020 + 0.080
6
Calculation
I_tangent = 0.020 + 0.080 = 0.100 kg·m² In terms of MR²: I_tangent = ¼MR² + MR² = ⁵⁄₄MR². Here, the factor ⁵⁄₄ is an exact rational number arising from the geometry and does not affect significant-figure counting. The given values (2.0 kg, 0.20 m) each have 2 significant figures.
7
Final answer
I_tangent = 0.10 kg·m² (2 significant figures, consistent with the given data). The exact constants (½, ¼, and the integer 2 in the perpendicular axes theorem splitting) are mathematical identities and do not limit the significant-figure count.
8
Common trap
Students often skip the perpendicular axes theorem step and use I_diameter = ½MR² (the central perpendicular axis value) instead of ¼MR². This gives I_tangent = ½MR² + MR² = ³⁄₂MR² = 0.12 kg·m² — wrong by 20%.
9
Similar NEET-style question
A uniform circular disc of mass 5.0 kg and radius 0.10 m is rotated about an axis tangent to the disc and perpendicular to its plane. Find the moment of inertia about this axis. (Answer: I = I_cm + Md² = ½MR² + MR² = ³⁄₂MR² = ³⁄₂ × 5.0 × (0.10)² = 0.075 kg·m². Note the perpendicular axes theorem is not needed here because the tangent is already perpendicular to the plane — only the parallel axes theorem applies.) ---

Before solving, remember these

Law

Parallel axes theorem

I = I_cm + M d², where I is moment of inertia about an axis at perpendicular distance d from a parallel axis through the centre of mass, and I_cm is the moment about the parallel axis through CM.

-- NCERT Class 11 Physics, Ch. 6, p. 12

Law

Perpendicular axes theorem

For a planar lamina (2D): I_z = I_x + I_y, where I_z is the moment of inertia about an axis perpendicular to the plane and I_x, I_y are about two perpendicular axes in the plane (all axes intersecting at one point).

-- NCERT Class 11 Physics, Ch. 6, p. 12

Formulas

8 formulas — click to collapse

Angular momentum

L = r \times p; L = I ω

For 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

Valid when

Reference point/axis chosen
I about same axis as omega

Centre of mass of n-particle system

R_{c m} = \frac{\sum m _{i} r _{i}}{\sum m _{i}}

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

Valid when

System of point particles or rigid body
Inertial reference frame

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)

Parallel axes theorem

I = I_{c m} + M d^{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

Valid when

Both axes parallel
I_cm known about CM axis

Perpendicular axes theorem (planar)

I_{z} = I_{x} + I_{y}

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

Valid when

Body is planar (2D lamina)
All three axes intersect at one point

Rotational kinematic equations (constant alpha)

ω = ω_{0} + α t; θ = ω_{0} t + \frac{1}{2} α t^{2}; ω^{2} = ω_{0}^{2} + 2 α θ

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

Valid when

Constant alpha
Single rotation axis

Rotational kinetic energy

K E_{rot} = \frac{1}{2} I ω^{2}

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

Valid when

Rotation about fixed axis
I and omega about same axis

Torque (moment of force)

τ = r \times F

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

Valid when

Rigid body or extended object
r measured from chosen pivot/axis

Exam Traps & Common Mistakes

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

7 items — click to collapse

Trap

Centre of mass: equal-distance default

Category: Overthinking

Student answers L/2 for two-particle CoM regardless of mass ratio.

When it triggers

Question gives two masses on rigid rod and asks for CoM distance.

How to avoid

R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.

Trap

Angular momentum vs rotational KE conservation

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).

When it triggers

Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).

How to avoid

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.

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.

Trap

RPM vs rad/s confusion

Category: Unit Conversion

Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.

When it triggers

Question gives ω in rpm and asks for kinematic quantities in SI units.

How to avoid

Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.

Mistake

Conserves rotational KE when only L is conserved (or vice versa) when MOI changes.

Root cause: concept gap

Correction

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).

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).

Mistake

Plugs ω in rpm directly into kinematic formulas requiring rad/s.

Root cause: unit error

Correction

Convert: ω(rad/s) = (2π/60) × ω(rpm). For example, 1200 rpm = 1200 × 2π/60 = 125.66 rad/s.

Past Year Questions

10 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

1108 days

2100 days

3105 days

4115 days

NTA Answer: Option 1(final)

NEET 2024Revised key

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm2. The length of the 400 g rod is nearly:

18.5 cm

217.5 cm

320.7 cm

472.0 cm

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is ν in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively)?

1Point P moves slower than point Q

2Point P moves faster than point Q

3Both the points P and Q move with equal speed

4Point P has zero speed

NTA Answer: Option 2(revised_final)

NEET 2023

The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is

15 : 3

22 : 5

35 : 2

43 : 5

NTA Answer: Option 5(final)

NEET 2023

The angular acceleration of a body, moving along the circumference of a circle, is

1Along the radius towards the centre

2Along the tangent to its position

3Along the axis of rotation

4Along the radius, away from centre

NTA Answer: Option 3(final)

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

1- 3 - 1 : 2

22 : 1

32 : 1

44 : 1

NTA Answer: Option 3(final)

NEET 2022

Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is 10

15 m

2m 3

3- 4 - 2 0 3 m

410 m

NTA Answer: Option 3(final)

NEET 2022

The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s2 is

1104

22

34

412

NTA Answer: Option 3(final)

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' 'MR2'. times Then the value of 'K' is 1

18 3

24 7

38 1

NTA Answer: Option 2(final)

NEET 2021

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in m/s2) equilibrium. (g = 10 20 cm 40 cm 160 cm 0 m 2 kg 1

1kg 12 1

2kg 2 1

3kg 3 1

4kg 6

NTA Answer: Option 1(final)

How NEET usually asks this

4 recurring patterns from past papers — click to collapse

Flywheel undergoing uniform angular acceleration; given initial/final omega and time, find alpha or theta.

Direct ApplicationEasy

Common distractors

forgets conversion rpm to rad s

Treats rpm as rad/s without 2*pi/60 conversion

Two-particle system on a rigid massless rod; find distance of CM from one of the masses. R_cm from m1 = m2*L/(m1+m2).

Direct ApplicationEasy

Common distractors

uses equal distribution

Default to L/2 regardless of mass ratio

Body's angular speed changes when its moment of inertia changes (e.g. star collapses, skater pulls in arms). Apply L = I*omega = constant when no external torque.

Multi StepMedium

Common distractors

uses energy conservation instead

Confusing L conservation with KE conservation

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

NCERT refs: Class 11 Physics Chapter 7, p.12

Test yourself on this topic with real past-paper questions:

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Parallel Perpendicular Axes Theorems

Lesson

Practice MCQs

Quick recall before you leave

I = I_cm + Md²

I_z = I_x + I_y

I_disc(diameter) = ¼MR²

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources