Select the correct statement about the formula (expression) of moment of inertia (M.I.) in terms of mass M of the object and some of its distance parameter/s, such as R, L, etc.

(A) Different objects must have different expressions for their M.I.
(B) When rotating about their ce

Question

Select the correct statement about the formula (expression) of moment of inertia (M.I.) in terms of mass M of the object and some of its distance parameter/s, such as R, L, etc.

(A) Different objects must have different expressions for their M.I.
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
(C) Expression for the M.I. for a parallelepiped rotating about the transverse axis passing through its centre includes its depth.
(D) Expression for M.I. of a rod and that of a plane sheet is the same about a transverse axis.

Accepted Answer

Concept understanding — Moment of Inertia

Moment of inertia ( $I$ ) is the rotational analogue of mass: just as mass measures a body's inherent resistance to a change in its state of linear motion (inertia), moment of inertia measures a body's resistance to a change in its state of rotational motion, for rotation about a specific axis. Its SI unit is kg m $^2$ , and its dimension is $[ML^2]$ .

For a single point mass $m_i$ at perpendicular distance $r_i$ from a fixed axis, the moment of inertia about that axis is $I=m_ir_i^2$ . For a rigid body made up of many such point masses, the moment of inertia about the axis is the sum over all of them:

I=\sum_i m_ir_i^2.

Key contrast with mass. Ordinary mass is (to excellent approximation, ignoring relativistic effects) a fixed, invariant property of a body. Moment of inertia is emphatically not invariant — it depends not only on how much mass a body has, but crucially on exactly how that mass is distributed relative to the chosen axis. The same rigid body has a different moment of inertia for every different axis of rotation one might choose, even axes lying entirely outside the physical body itself.

Computing $I$ for a continuous body. Treating an infinitesimally small mass element $dm$ , at perpendicular distance $r$ from the axis, as a point mass gives $dI=(dm)r^2$ ; integrating over the whole body gives the moment of inertia of any continuous, uniformly-distributed bulk object:

I=\int dI=\int r^2\,dm.

This integral, carried out with the mass element $dm$ expressed via the object's linear density $\lambda=M/\ell$ (for a rod), surface density $\sigma=M/(\pi R^2)$ (for a disc), or similar, yields the standard results for common shapes:

Uniform rod (mass $M$ , length $\ell$ ), axis through the center, perpendicular to the rod: $I=\dfrac{1}{12}M\ell^2$ ; axis through one end: $I=\dfrac{1}{3}M\ell^2$ .
Uniform ring (mass $M$ , radius $R$ ), axis through the center, perpendicular to the plane: $I=MR^2$ (every mass element is at the same distance $R$ from the axis, so the integral is trivial). …

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Check each option against the standard moment-of-inertia expressions (Table 3). (A) is false: different objects can share the SAME formula, as (B) itself demonstrates -- the underlying shape does not uniquely determine the formula, only how mass is distributed relative to the axis does. (B) is TRUE: Table 3 lists 'Uniform disc or solid cylinder, Central: $I=\frac{1}{2}MR^2$ ' and separately 'Uniform hollow right circular cone, Central: $I=\frac{1}{2}MR^2$ ' -- identical expressions, even though a flat disc and a hollow cone are visually very different shapes; this happens because, for both shapes, an infinitesimal mass element's contribution works out to be proportional to $r\,dr$ in exactly the same way when integrated from the axis out to the rim, so the coefficient comes out equal. (C) is false: Table 3's parallelepiped entry, $I=\frac{1}{12}M(L^2+b^2)$ about the CENTRAL TRANSVERSE axis, involves only the length L and breadth b -- the object's DEPTH (its extent along the rotation axis itself) does not enter the formula at all for this particular axis. (D) is false as a general statement: while a thin uniform ROD and a thin RECTANGULAR PLATE do share the formula $I=\frac{1}{12}ML^2$ about a transverse central axis (both being effectively 'long and …

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

CBSE 2025Set ANNUAL1 markMCQ
$Q. The moment of inertia of a uniform rod about an axis which is perpendicular to the rod and touches any one end of the rod is ____.$
(a) I = MR^2
(b) I = (1/12) Ml^2
(c) I = (1/2) MR^2
(d) I = (1/3) Ml^2
›Reveal solutionSolution

The moment of inertia of a uniform rod of mass M and length l about a perpendicular axis through one end is (1/3)Ml^2, obtained from the centre-of-mass value (1/12)Ml^2 via the parallel axis theorem.

Step 1: Moment of inertia about the centre.

For a uniform rod of mass M and length l, rotating about a perpendicular axis through its geometric centre (centre of mass), the standard result (derived by integrating dI = x^2 dm over the length) is:

I_cm = (1/12) M l^2

Step 2: Apply the parallel axis theorem. …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The moment of inertia (MI) of a disc of radius R and mass M about its central axis is ______.$
(a) MR²/4
(b) MR²/2
(c) MR²
(d) 3MR²/2
›Reveal solutionSolution

Moment of inertia of a uniform disc about its central (symmetry) axis.

Treat the disc as a set of concentric thin rings of radius $x$ , thickness $dx$ , mass $dm=\dfrac{2M}{R^2}x\,dx$ (surface density $\times$ ring area). Each ring's MI about the central axis is $dm\,x^2$ . Integrating from $0$ to $R$ : …

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CBSE 2024Set ANN1 mark
$Q. The moment of inertia of a disc about an axis passing through the centre and perpendicular to its plane.$
(a) MR²
(b) ½ MR²
(c) ⅔ MR²
(d) ⅖ MR²
›Reveal solutionSolution

A disc's moment of inertia about the central perpendicular axis is ½MR², derived by integrating over concentric rings.

Consider a uniform disc of mass M and radius R, lying in a plane, with the axis passing through its centre O and perpendicular to the plane (the axis a student would picture as sticking straight out of the disc's face).

Divide the disc into thin concentric rings of radius x and thickness dx. The mass of this ring is

dm = (M / πR²) × (2πx dx) = (2M/R²) x dx

…

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CBSE 2022Set ANNUAL1 mark
$Q. Calculate the moment of inertia of a uniform disc of mass 10 kg and radius 60 cm about an axis perpendicular to its length and passing through its centre.$
›Reveal solutionSolution

MI of a uniform disc about the central axis perpendicular to its plane is (1/2)MR².

For a uniform disc about an axis through its centre, perpendicular to its plane: …

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CBSE 2021Set ANN1 mark
$Q. What is the analogue of mass in rotational motion ?$
›Reveal solutionSolution

Moment of inertia (I) is the rotational analogue of mass — it measures a rigid body's resistance to a change in its angular velocity, just as mass measures resistance to a change in linear velocity.

In translational (straight-line) motion, Newton's second law is F = ma: mass m is the measure of inertia — how strongly a body resists being accelerated by a force. In rotational motion, the corresponding law is τ = Iα, where torque τ plays the role of force, angular acceleration α plays the role of linear acceleration, and moment of in …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. A closed cylindrical container is partially filled with water. As the container rotates in a horizontal plane about a perpendicular bisector, its moment of inertia :$
(a) remains constant
(b) depends on the direction of rotation
(c) increases
(d) decreases
›Reveal solutionSolution

As the rotating cylindrical container's water redistributes outward under the rotation, the moment of inertia of the water-container system increases.

The moment of inertia of a mass distribution about an axis is I = sum(m_i r_i^2), where r_i is each mass element's perpendicular distance from the axis. It depends on HOW mass is distributed relative to the axis, not just the total mass.

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. A thin ring has mass 0.25 kg and radius 0.5 m. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is ____.$
(a) $0.0625\ \text{kg m}^2$
(b) $0.625\ \text{kg m}^2$
(c) $6.25\ \text{kg m}^2$
(d) $62.5\ \text{kg m}^2$
›Reveal solutionSolution

For a thin ring about its central axis (perpendicular to its plane), all the mass lies at the same distance $R$ from the axis, so $I = MR^2$ directly.

For a thin circular ring, every mass element lies at radius $R$ from the axis passing through the centre, perpendicular to the plane of the ring. Hence the moment of inertia is

$I = \sum m_i r_i^2 = MR^2$ …

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CBSE 2018Set ANN1 mark
$Q. State the theorem of parallel axes on moment of inertia.$
›Reveal solutionSolution

Parallel axis theorem: I = I_cm + Md².

The theorem of parallel axes states that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis passing through its centre of mass, plus the product of the mass of the body and the square of the perpendicular distance between the two axes.

…

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CBSE 2016Set ANNUAL1 markMCQ
$Q. A body of moment of inertia 5 kg m² rotating with an angular velocity 6 rad/s has the same kinetic energy as a mass of 20 kg moving with a velocity of ______. (A) 5 m/s (B) 4 m/s (C) 3 m/s (D) 2 m/s$
›Reveal solutionSolution

Equate rotational KE ( $\tfrac12 I\omega^2$ ) to translational KE ( $\tfrac12 mv^2$ ) and solve for $v$ .

Rotational KE of the rotating body:

$KE_{rot}=\frac12 I\omega^2=\frac12\times 5\times 6^2=\frac12\times 5\times 36=90\text{ J}$

This must equal the translational KE of the 20 kg mass: …

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