A uniform cube of edge $a$ and mass $m$ has its centre at the point O. Three mutually perpendicular axes $x$, $y$, $z$ pass through O parallel to the cube's edges (the $z$-axis being perpendicular to the top and bottom faces). Two further axes are drawn parallel to the $z$-axis: $z'$ passes vertically through one edge (a vertical edge) of the cube, and $z''$ passes vertically through the diagonally opposite vertical edge. State whether each of the following statements is true or false; more than one may be correct. (a) $I_z = I_x + I_y$ (b) The moment of inertia about $z'$ is $I'_z = I_z + \dfrac{ma^2}{2}$ (c) The moment of inertia about $z''$ is $I''_z = I_z + \dfrac{ma^2}{2}$ (d) $I_x = I_y$

> [!TLDR] > The perpendicular-axis theorem $I_z=I_x+I_y$ applies only to planar laminae, not to a three-dimensional cube — so (A) is false (indeed, by symmetry $I_x=I_y=I_z$, so $I_x+I_y=2I_z\neq I_z$). The cube's symmetry makes $I_x=I_y$ (D true). Both $z'$ and $z''$ are parallel to $z$ through vertical edges, each a perpendicular distance $a/\sqrt2$ from the central axis, so the parallel-axis theorem gives $I_z+ma^2/2$ for each (B and C true). ### Concepts - **Perpendicular-axis theorem** ($I_z=I_x+I_y$) is valid **only for a flat (2-D) lamina** in the $x$-$y$ plane. A cube is a solid body, so it does not apply. - **Parallel-axis theorem**: $I = I_{cm} + md^2$, where $d$ is the perpendicular distance between the parallel axes. - **Symmetry**: a cube is identical under interchange of $x$, $y$, $z$, so $I_x=I_y=I_z$. ### Statement by statement - **(A) False.** The cube is 3-D; the perpendicular-axis theorem does not hold. Moreover $I_x=I_y=I_z$, so $I_x+I_y=2I_z$, which is not $I_z$. …

A uniform cube of edge a and mass m has its centre at the point O. Three mutually perpendicular axes x, y, z p

NCERT Exemplar · Q13

a

(a)

I_z = I_x + I_y

(b) The moment of inertia about

z'

I'_z = I_z + \dfrac{ma^2}{2}

z''

I''_z = I_z + \dfrac{ma^2}{2}

(d)

I_x = I_y

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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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The perpendicular-axis theorem $I_z=I_x+I_y$ applies only to planar laminae, not to a three-dimensional cube — so (A) is false (indeed, by symmetry $I_x=I_y=I_z$ , so $I_x+I_y=2I_z\neq I_z$ ). The cube's symmetry makes $I_x=I_y$ (D true). Both $z'$ and $z''$ are parallel to $z$ through vertical edges, each a perpendicular distance $a/\sqrt2$ from the central axis, so the parallel-axis theorem gives $I_z+ma^2/2$ for each (B and C true).

Concepts

Perpendicular-axis theorem ( $I_z=I_x+I_y$ ) is valid only for a flat (2-D) lamina in the $x$ - $y$ plane. A cube is a solid body, so it does not apply.
Parallel-axis theorem: $I = I_{cm} + md^2$ , where $d$ is the perpendicular distance between the parallel axes.
Symmetry: a cube is identical under interchange of $x$ , $y$ , $z$ , so $I_x=I_y=I_z$ .

Statement by statement

(A) False. The cube is 3-D; the perpendicular-axis theorem does not hold. Moreover $I_x=I_y=I_z$ , so $I_x+I_y=2I_z$ , which is not $I_z$ . …

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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 19 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$Q. A thin circular disc and a uniform thin circular ring have their masses and radii in the ratio 2 : 1 and 1 : 2 respectively. The ratio of their moments of inertia about their respective diameters is (A) 4 : 1 (B) 1 : 1 (C) 1 : 4 (D) 2 : 1 (E) 1 : 2$
›Reveal solutionSolution

Using $I_{disc}=\tfrac14 MR^2$ and $I_{ring}=\tfrac12 MR^2$ about a diameter, with mass ratio $2:1$ and radius ratio $1:2$ , the ratio is $1:4$ .

About a diameter, a disc has $I_d=\tfrac14 M_d R_d^2$ and a ring $I_r=\tfrac12 M_r R_r^2$ . With $\dfrac{M_d}{M_r}=2$ and $\dfrac{R_d}{R_r}=\dfrac12$ , …
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KEAM 2026Set eng-2026-04194 marksMCQ
$cm$
›Reveal solutionSolution

Equate sphere-about-diameter $\tfrac25 MR^2$ to cylinder-about-axis $\tfrac12 Mr^2$ and solve for $r$ .

Moment of inertia of a solid sphere about a diameter: $I_s=\dfrac25 MR^2$ .

Moment of inertia of a solid cylinder about its axis: $I_c=\dfrac12 Mr^2$ .

Equal masses and equal moments of inertia: …
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KEAM 2026Set eng-2026-04204 marksMCQ
$Q. A flywheel in a steam engine ensures a smooth ride for the passengers on the vehicle because it has (A) low mass (B) high speed (C) low acceleration (D) large moment of inertia (E) high friction$
›Reveal solutionSolution

A flywheel stores rotational kinetic energy owing to its large moment of inertia, damping speed fluctuations.

A flywheel is a heavy wheel with mass concentrated at the rim, giving it a large moment of inertia $I$ . Because the angular acceleration $\alpha=\dfrac{\tau}{I}$ is small for a given torque fluctuation, the flywheel resists sudden changes in angular velocity. It absorbs energy when the eng …
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KEAM 2026Set eng-2026-04214 marksMCQ
$x$
›Reveal solutionSolution

Moment of inertia about the $z$ -axis is $\sum m_i r_i^2=2\cdot 4+4\cdot 16=72\ kg\,m^2$ .

For point masses at perpendicular distances $r$ from the axis, $I=\sum m_i r_i^2$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$6\ \text{kg}$
›Reveal solutionSolution

Treating both cylinders as the same material (tight fit, no air gap), the mass of the hollow cylinder works out to 2 kg.

Reasoning

Let the solid cylinder (mass $m_s=6$ kg) have radius $r$ ; the hollow cylinder has outer radius $R=0.6$ m and inner radius $r$ .

Axial moments of inertia:
$I_{solid}=\tfrac12 m_s r^2=3r^2,\qquad I_{hollow}=\tfrac12 m_h(R^2+r^2).$
Equal density $\rho$ (no gap): with $k=\rho\pi\ell$ , $\;m_s=kr^2=6\Rightarrow k=6/r^2$ , and $m_h=k(R^2-r^2)=\dfrac{6(R^2-r^2)}{r^2}$ .

Total: …
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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. If a thin uniform circular ring and a thin uniform circular disc have the same mass and radius, then the ratio of their moments of inertia about their central axes normal to their planes is (A) 3 : 2 (B) 2 : 3 (C) 1 : 4 (D) 1 : 2 (E) 2 : 1$
›Reveal solutionSolution
Ring $I=MR^2$ , disc $I=\tfrac12 MR^2$ ; ratio = 2:1.

About the central axis perpendicular to the plane:

Ring: $I_{ring} = MR^2$ .

Disc: $I_{disc} = \tfrac{1}{2}MR^2$ .

For the same mass and radius, …
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KEAM 2025Set eng-2025-04254 marksMCQ
$Q. If the moment of inertia of a solid sphere of mass M and radius R about its diameter is I, then that of another sphere of mass 2M and radius 2R about its diameter is (A) 2I (B) 4I (C) 8I (D) 16I (E) I$
›Reveal solutionSolution

Moment of inertia of a solid sphere is $\tfrac{2}{5}MR^2$ ; doubling mass and radius multiplies it by $2\times 4 = 8$ , giving $8I$ .

Moment of inertia of a solid sphere about a diameter:
$I = \frac{2}{5}MR^2.$
For the second sphere (mass $2M$ , radius $2R$ ): …
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KEAM 2025Set eng-2025-04284 marksMCQ
$\frac{R}{8}$
›Reveal solutionSolution

$I_{\text{diameter}}=\tfrac14 MR^2$ , so $k=\sqrt{I/M}=R/2$ .

The moment of inertia of a uniform circular disc about a diameter is
$I=\frac{1}{4}MR^2.$
The radius of gyration $k$ is defined by $I=Mk^2$ , so …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$\sqrt{2}:\sqrt{3}$
›Reveal solutionSolution

Radius of gyration $k=\sqrt{I/M}$ . For a ring $I=MR^2\Rightarrow k=R$ ; for a disc $I=\tfrac12 MR^2\Rightarrow k=R/\sqrt2$ . The ratio is $\sqrt2:1$ .

The radius of gyration is defined by $I=Mk^2$ , so $k=\sqrt{I/M}$ .

For a circular ring about its central axis, $I_{ring}=MR^2$ , so
$k_{ring}=\sqrt{\frac{MR^2}{M}}=R.$
For a circular disc about its central axis, $I_{disc}=\tfrac12 MR^2$ , so …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$I$
›Reveal solutionSolution

Central axis: $I=\tfrac12 MR^2$ . About a diameter: $I_d=\tfrac14 MR^2 = \dfrac{I}{2}$ (perpendicular-axis theorem, $I_z=I_x+I_y=2I_d$ ). …

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KEAM 2024Set eng-2024-06064 marksMCQ
$Q. A flywheel ensures a smooth ride on the vehicle because of its (A) larger speed (B) zero moment of inertia (C) large moment of inertia (D) lesser mass with smaller radius (E) small moment of inertia$
›Reveal solutionSolution

A flywheel's large moment of inertia opposes fluctuations in angular velocity, giving a smooth ride.

Moment of inertia is the rotational analogue of mass and measures resistance to changes in angular velocity. A flywheel is built with a large moment of inertia (mass concentrated at large radius). It therefore stores rotational kinetic energy and resi …
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KEAM 2024Set eng-2024-06084 marksMCQ
$K$
›Reveal solutionSolution

Hollow cylinder about its symmetry axis has $I=MR^2$ , hence radius of gyration $K=R$ .

The radius of gyration $K$ is defined by $I = MK^2$ . …
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