From a disc of radius $R$ and mass $M$, a circular hole of diameter $R$, whose rim passes through the center, is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the original center? (a) $15MR^2/32$ (b) $13MR^2/32$ (c) $11MR^2/32$ (d) $9MR^2/32$

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the center, is cut

I. Multiple Choice Questions · Q12

R

(a)

15MR^2/32

(b)

13MR^2/32

(c)

11MR^2/32

(d)

9MR^2/32

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Concept understanding — Rotational Inertia Comparison

Rotational Inertia Comparison: From Intuition to Precision

Imagine pushing a shopping cart that's nearly empty, then pushing the same cart loaded with bricks. The loaded cart is harder to get moving — it resists changes to its motion more. That resistance is inertia, and it depends only on how much mass is there.

Now imagine spinning a bicycle wheel. If you hold the axle and try to tilt the spinning wheel, it fights you. But here's the twist: a lightweight wheel that's large in diameter can be harder to spin or stop than a heavy wheel that's small in diameter, even if the heavy wheel has more mass. Why? Because rotational inertia depends not just on how much mass, but on where that mass is placed relative to the axis of rotation.

Note

Rotational inertia (also called moment of inertia) is the rotational equivalent of mass. It measures how difficult it is to change an object's rotational motion — to start it spinning, stop it, or change its spin speed.

The Core Idea: Mass × Distance²

The precise statement is this:

I = \sum m_i r_i^2

For a collection of point masses, rotational inertia $I$ is the sum of each mass $m_i$ multiplied by the square of its perpendicular distance $r_i$ from the axis of rotation.

That square is crucial. Doubling the distance from the axis quadruples the rotational inertia. A mass far from the axis contributes much more to rotational inertia than the same mass close to the axis.

Why Comparison Matters

When you compare two objects, you're asking: Which is harder to spin? The answer depends on both mass and shape.

Example 1: A ring vs. a disk of the same mass and radius

A ring has all its mass at the outer edge ( $r = R$ for all mass). Its rotational inertia is $I_{\text{ring}} = MR^2$ .
A solid disk has mass spread evenly from center to edge. Its rotational inertia is $I_{\text{disk}} = \frac{1}{2}MR^2$ .

The ring has twice the rotational inertia of the disk. Same mass, same radius — but the ring is harder to spin because its mass is concentrated farther from the axis.

Example 2: A long rod vs. a short rod of the same mass

A rod spun about its center: $I = \frac{1}{12}ML^2$ .
A rod spun about one end: $I = \frac{1}{3}ML^2$ .

The same rod, same mass — but spinning it about the end is four times harder than spinning it about the center. The mass is, on average, farther from the axis.

Watch out

A common mistake is to think that rotational inertia depends only on mass. It does not. Two objects with the same mass can have wildly different rotational inertias depending on how their mass is distributed.

The Intuition Behind the Square

Why distance squared? Think of a spinning object. A mass far from the axis has to travel a longer path in the same time — it has a higher linear speed for the same angular speed. To change that speed (to accelerate or decelerate the rotation), you need to apply a force over that longer distance. The square comes from the geometry: the work required scales with distance, and the lever-arm effect also scales with distance. The two factors multiply.

A Quick Comparison Table

Object	Axis location	Rotational inertia $I$	Relative difficulty to spin
Point mass $m$ at distance $R$	Through point	$mR^2$	Baseline

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Step 1. Work out the hole's size and position.

The full disc has radius $R$ and mass $M$ , uniform surface density $\sigma = \dfrac{M}{\pi R^2}$ . A circular hole of diameter $R$ (so hole radius $r=R/2$ ) is cut such that the hole's rim passes through the disc's original center — meaning the hole's own center sits at distance $d = r = R/2$ from the disc's center (the hole reaches exactly up to the center from the far side).

Step 2. Find the mass of the removed (hole) piece.

m_{hole} = \sigma \cdot \pi r^2 = \dfrac{M}{\pi R^2}\cdot \pi \left(\dfrac{R}{2}\right)^2 = \dfrac{M}{\pi R^2}\cdot \dfrac{\pi R^2}{4} = \dfrac{M}{4}.

Step 3. Find the hole's moment of inertia about its own center, then shift it to the disc's center.

About its own center, a disc of mass $m_{hole}$ and radius $r=R/2$ has

I_{hole,own} = \dfrac12 m_{hole} r^2 = \dfrac12 \cdot \dfrac{M}{4}\cdot\left(\dfrac{R}{2}\right)^2 = \dfrac12\cdot\dfrac{M}{4}\cdot\dfrac{R^2}{4} = \dfrac{MR^2}{32}.

Using the parallel axis theorem to shift this to the disc's center, a distance $d=R/2$ away:

I_{hole,disc-center} = I_{hole,own} + m_{hole}d^2 = \dfrac{MR^2}{32} + \dfrac{M}{4}\left(\dfrac{R}{2}\right)^2 = \dfrac{MR^2}{32} + \dfrac{M}{4}\cdot\dfrac{R^2}{4} = \dfrac{MR^2}{32} + \dfrac{MR^2}{16}.

Converting to a common denominator of 32: $\dfrac{MR^2}{32} + \dfrac{2MR^2}{32} = \dfrac{3MR^2}{32}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Moment of inertia of a uniform circular ring about a diameter of a ring is ............. .$
›Reveal solutionSolution

Using the perpendicular axis theorem on a ring's axis ( $I=MR^2$ ) and its two diameters gives $I_{\text{diameter}}=\tfrac12MR^2$ .

For a uniform circular ring of mass $M$ and radius $R$ , the moment of inertia about the axis perpendicular to its plane, through its centre, is $I_z=MR^2$ (all the mass lies at distance $R$ from that axis).

By the perpendicular axis theorem, for any two mutually perpendicular diameters $x$ and $y$ lying in the plane of the ring:

$I_z = I_x + I_y$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. (C) The M.I. of a Solid sphere of mass 5 kg and radius 1 m about its diameter is :$
(a) 1/2 Kg m^2
(b) 2 Kg m^2
(c) 1/5 Kg m^2
(d) 5 Kg m^2.
›Reveal solutionSolution

Using I = (2/5) M R² for a solid sphere about its diameter, with M = 5 kg and R = 1 m, gives I = 2 kg m².

The standard formula for the moment of inertia of a uniform solid sphere about an axis through its diameter (centre) is:

I = (2/5) M R²

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. (D) The moment of Inertia of a body, depends upon :$
(a) Distribution of Mass from axis of Rotation.
(b) Angular velocity of the Body.
(c) Angular acceleration of the Body.
(d) None of the above.
›Reveal solutionSolution
Since I = Σ m r², moment of inertia depends purely on how mass is distributed relative to the chosen axis, not on how fast the body happens to be rotating.

Moment of Inertia is defined as I = Σ mi ri², where ri is the perpendicular distance of each mass element from the axis of rotation.

This makes I depend entirely on:

the total mass of the body, and

how that mass is distributed with respect to the particular axis chosen (mass farther from the axis contributes more, since r² appears in the formula). …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The moment of inertia of a circular disc of mass M and radius R about its diameter is$
(a) MR^2
(b) MR^2/4
(c) 3MR^2/4
(d) MR^2/2
›Reveal solutionSolution

Moment of inertia of a disc about a diameter is MR^2/4. Answer (B).

About the central axis perpendicular to the disc, I_z = ½ MR^2.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the mass of a body is M and moment of inertia about an axis is I, then the radius of gyration about that axis is$
(a) MI^2
(b) I^2/M
(c) √(I/M)
(d) √(M/I)
›Reveal solutionSolution

Radius of gyration k = sqrt(I/M). Answer (C).

The radius of gyration k is the distance from the axis at which the whole mass could be concentrated to give the same moment of inert …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the moment of inertia of a cylindrical shell of mass M and radius R about its own axis?$
(a) MR^2
(b) (1/2) MR^2
(c) 2MR^2
(d) (2/3) MR^2
›Reveal solutionSolution

A cylindrical (hollow) shell has its entire mass concentrated at a fixed distance R from the axis, so its moment of inertia is simply MR^2, exactly like a ring or hoop.

Moment of inertia I = Integral of r^2 dm, where r is the perpendicular distance of each mass element from the axis.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Define radius of gyration.$
›Reveal solutionSolution

Radius of gyration is a way of representing how the mass of an extended body is distributed relative to an axis, using a single equivalent 'point-mass' distance.

For a rigid body of total mass $M$ and moment of inertia $I$ about a given axis, the radius of gyration $k$ is defined as the distance from the axis at which, if the entire mass $M$ of the body were concentrated as a single point mass, the moment of inertia about that axis would be the same as the actual moment of inertia of the body.

That is, $I = Mk^2$ , so $k=\sqrt{I/M}$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Mass and moment of inertia about diameter of a solid sphere and a thin spherical shell is same. The ratio between their radii is (A) 3:5 (B) 5:3 (C) \sqrt3:\sqrt5 (D) \sqrt5:\sqrt3$
›Reveal solutionSolution

Equal $I$ and equal $M$ for a solid sphere and shell gives radii ratio $\sqrt5:\sqrt3$ .

Moment of inertia about a diameter: solid sphere $I_1=\tfrac25MR_1^2$ ; thin spherical shell $I_2=\tfrac23MR_2^2$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. What is the unit of the radius of gyration?$
(a) kg-m^2
(b) kg
(c) m
(d) kg-m
›Reveal solutionSolution

Radius of gyration k is defined by I = M k^2, so k = sqrt(I/M), which works out to a length — its SI unit is the metre.

Moment of inertia I has units kg·m^2. Setting I = Mk^2 and solving for k:

k = sqrt(I/M), units = sqrt(kg·m^2 / kg) = sqrt(m^2) = m

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. What is the moment of inertia of a solid cylinder of mass M and radius R about its own axis?$
(a) MR^2
(b) (1/2) MR^2
(c) (3/2) MR^2
(d) (5/2) MR^2
›Reveal solutionSolution

The moment of inertia of a solid cylinder (or disc) of mass M and radius R about its own central axis is (1/2) MR^2 — a standard result derived by integrating r^2 dm over circular mass shells.

Model the solid cylinder as a stack of thin circular discs. Integrating I = ∫ r^2 dm over the cross-section (using dm = (2M/R^2) r dr for a uniform disc of radius R) gives:

I = (1/2) M R^2

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Analogue of mass in rotational motion is$
(1) moment of inertia
(2) torque
(3) radius of gyration
(4) angular momentum
›Reveal solutionSolution

Moment of inertia (I) plays the same role in rotational dynamics that mass (m) plays in linear dynamics - both quantify inertia, i.e. resistance to a change in motion.

In linear motion, Newton's second law is F = ma - mass m measures how much a body resists linear acceleration for a given force.

In rotational motion, the analogous law is tau = I alpha - moment of inertia I measures how much a body resists angular acceleration for a given torque. It depends not just on the mass of the body but on how that mass is distributed relative to the axis of rotation (I = sum of m_i r_i^2).

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. M is mass and R is radius of a circular ring. The moment of Inertia about an axis passing through the centre and perpendicular to the plane is:$
(a) MR²
(b) (1/2) MR²
(c) (2/5) MR²
(d) (2/3) MR²
›Reveal solutionSolution

Every mass element of a ring is at distance $R$ from the central perpendicular axis, so $I = MR^2$ exactly, with no fractional factor.

Moment of inertia about an axis is $I = \sum m_i r_i^2$ , where $r_i$ is each mass element's perpendicular distance from the axis. For a thin circular ring of mass $M$ and radius $R$ , every particle of the ring lies exactly on the circumference, at distance $R$ from the centre, and the axis considered passes through the centre perpendicular to the plane of the ring. Hence every mass element has the same $r_i = R$ :

$I = \sum m_i R^2 = R^2 \sum m_i = MR^2$ …

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Q.From a disc of radius RRR and mass MMM, a circular hole of diameter RRR, whose rim passes through the center, is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the original center?

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Q.From a disc of radius $R$ and mass $M$ , a circular hole of diameter $R$ , whose rim passes through the center, is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the original center?

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