Q.A particle is moving with constant velocity parallel to x-axis. Its angular momentum relative to origin point (A) is zero
(B) remains constant
(C) goes on increasing
(D) goes on decreasing

Showing the 12 most recent of 20 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.Write S. I. unit of Angular Momentum.

   ›Reveal solutionSolution

   $L=I\omega$ has SI unit $\text{kg}\,\text{m}^2\,\text{s}^{-1}$, same as J·s.

   Angular momentum is $L=\vec r \times \vec p = I\omega$, where moment of inertia $I$ has SI unit $\text{kg}\,\text{m}^2$ and angular velocity $\omega$ has SI unit $\text{s}^{-1}$ (rad/s, but radian is dimensionless). Multiplying:

   $[L] = \text{kg}\,\text{m}^2 \times \text{s}^{-1} = \text{kg}\,\text{m}^2\,\text{s}^{-1}$ …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 markMCQ
   Q.A particle is moving with a constant velocity along a straight line parallel to positive x-axis. The magnitude of its angular momentum with respect to origin is:

   (a) decreasing with x

   (b) zero

   (c) remaining constant

   (d) increasing with x

   ›Reveal solutionSolution

   Angular momentum about the origin depends on the perpendicular distance from the origin to the particle's line of motion; for straight-line motion parallel to the x-axis, this perpendicular distance (the particle's fixed y-coordinate) never changes, so L is constant.

   The angular momentum of a particle about the origin is

   L = r x p = m(r x v)

   For a particle moving along a straight line parallel to the positive x-axis at a fixed height y = y0, with constant velocity v = vx-hat, the position vector is r = xx-hat + y0*y-hat.

   L = m(r x v) = m[(xx-hat + y0y-hat) x (vx-hat)] = m[xv*(x-hat x x-hat) + y0v(y-hat x x-hat)] = -mvy0*z-hat

   …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: Moment of linear momentum is called ____.

   ›Reveal solutionSolution

   The moment of linear momentum about a point is called angular momentum.

   Just as the moment of a force about a point is called torque (τ = r × F), the moment of linear momentum p about a point is defined as angular momentum:

   L = r × p …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The correct relation between radius of gyration (k) and moment of inertia (I) is

   (a) (A) k = √(m/I)

   (b) (B) k = √(I/m)

   (c) (C) k = Im²

   (d) (D) k = ml²

   ›Reveal solutionSolution

   [!TLDR]

   (B) k = √(I/m)

   Why

   By definition I = mk², so k = √(I/m …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2024Set ANNUAL1 markMCQ
   Q.A particle is moving with constant velocity parallel to x-axis. Its angular momentum relative to origin point (A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing

   ›Reveal solutionSolution

   A particle in straight-line uniform motion has constant angular momentum about any fixed point (unless it passes through that point along the line).

   Angular momentum about the origin is $\vec L=\vec r\times\vec p$, with magnitude $L=p\cdot d$, where $d$ is the perpendicular distance from the origin to the line along which the particle moves. Since the particle moves with constant velocity, $\vec p=m\vec v$ is constant in both m …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2024Set ANNUAL1 markMCQ
   Q.A body of mass M is moving with uniform angular velocity ω about its axis of rotation. I is its moment of inertia about this axis. Its angular momentum will be (A) ½Iω^2 (B) MIω^2 (C) I^2ω (D) Iω

   ›Reveal solutionSolution

   Angular momentum of a rotating body is $L=I\omega$.

   Just as linear momentum is $p=mv$, angular momentum about the rotation axis is defined as $L=I\omega$, where $I$ is the moment of inertia about that axis and $\omega$ is the a …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2024Set ANNUAL1 markMCQ
   Q.The product of moment of inertia and angular velocity is called

   (a) Torque

   (b) Impulse

   (c) Linear momentum

   (d) Angular momentum

   ›Reveal solutionSolution

   Angular momentum L is defined as the product of moment of inertia I and angular velocity ω: L = Iω.

   This is the rotational analogue of linear momentum p = mv, with mass replaced by moment of inertia and linear velocity replaced by angular velocity. Torque is the rotational analogue of force, impulse is force …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2024Set sz1 markMCQ
   Q.Angular momentum is a: (A) Polar vector (B) Axial vector (C) Scalar (D) None of these

   ›Reveal solutionSolution

   Angular momentum is an axial (pseudo) vector because it is defined as the cross product of two polar vectors, position and linear momentum.

   Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$.

   Both $\vec{r}$ (position) and $\vec{p}$ (linear momentum) are polar (true) vectors — their direction is along an actual physical displacement or motion. The cross product of two polar vectors, however, produces an axial vector (also called a pseudovector): its direction is assigned by convention (the right-hand rule) perpendicular to the plane cont …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2023Set ANNUAL1 markMCQ
   Q.A rigid body rotates with an angular momentum L. If its kinetic energy is halved, the angular momentum becomes :

   (a) 2L

   (b) L

   (c) L/√2

   (d) L/2

   ›Reveal solutionSolution

   Since KE = L^2/(2I), halving KE while keeping I fixed means L is scaled by 1/sqrt(2).

   The rotational kinetic energy of a rigid body is

   KE = (1/2) I omega^2

   Angular momentum is L = I omega, so omega = L/I. Substituting:

   KE = (1/2) I (L/I)^2 = L^2 / (2I)

   So L = sqrt(2 x I x KE), i.e. L is proportional to sqrt(KE) for a fixed I (no external torque changes I here).

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2023Set ANNUAL1 markMCQ
    Q.SI unit of angular momentum is:

    (a) Joule x Second

    (b) Newton x Meter

    (c) kg x m^2

    (d) Newton x m / Sec

    ›Reveal solutionSolution

    Angular momentum is measured in Joule-second (equivalently kg m^2/s), NOT in kg x m^2 alone or in Newton x metre.

    Angular momentum L = I*omega, where moment of inertia I has units kg m^2 and angular velocity omega has units rad/s (dimensionless rad, so effectively 1/s). Hence:

    [L] = kg m^2 x (1/s) = kg m^2/s

    Now check the given options:

    1. Joule x Second = (kg m^2/s^2) x s = kg m^2/s -- matches
    2. Newton x Metre = (kg m/s^2) x m = kg m^2/s^2 -- this is the unit of TORQUE, not angular momentum …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2023Set ANNUAL1 markMCQ
    Q.Match the column: Angular momentum L — match with the correct expression.

    (a) sqrt(2gR)

    (b) sqrt(T/m)

    (c) GMm/r^2

    (d) I*omega

    (e) 2pisqrt(l/g)

    (f) sqrt(gR)

    (g) m*R^2

    ›Reveal solutionSolution

    Angular momentum is L = Iomega, the rotational counterpart of linear momentum p = mv, matching option (d).

    Just as linear momentum p = mv combines mass (inertia for straight-line motion) with linear velocity, angular momentum L combines moment of inertia I (inertia for rotational motion) with angular velocity omega: L = Iomega

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2022Set TERM11 markMCQ
    Q.Moment of linear momentum is

    (1) Couple

    (2) Torque

    (3) Impulse

    (4) Angular momentum

    ›Reveal solutionSolution

    'Moment of linear momentum' is the literal definition of angular momentum, L = r x p (position vector crossed with linear momentum), just as torque is defined as the moment of force.

    In rotational mechanics, taking the 'moment' of a vector quantity about a point means crossing the position vector r (from that point to where the quantity acts) with the quantity itself.

    • Moment of FORCE = r x F = Torque
    • Moment of LINEAR MOMENTUM = r x p = Angular Momentum, denoted L …

    🔒 Start your 14-day free trial to see the full solution →