Q.A ring, a disc and a sphere, all having the same radius and mass, roll down an inclined plane from the same height. Which of these will reach the bottom first? Explain why. OR Define Moment of Inertia and radius of gyration. If the moment of inertia of a thin rod of mass 'M' and length 'L' about an axis through its centre and perpendicular to its lenght is ML^2/12, calculate its radius of gyration.

Showing the 12 most recent of 21 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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) √3:√5 (D) √5:√3

   ›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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9. 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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10. 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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11. 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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12. 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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