Q.An electron with charge e enters a uniform magnetic field $\vec{B}$ with a velocity $\vec{v}$. The velocity is perpendicular to the magnetic field. The force on the charge e is given by $|\vec{F}| = Bev$. Obtain the dimensions of $\vec{B}$.

Showing the 12 most recent of 50 on this concept.

1. CBSE 2026Set 55/2/11 markMCQ
   Q.The magnetic flux $\phi$ (in Wb) linked with a coil is related to time $t$ (in s) as $\phi = 5At^2 + Bt - 2C$. The SI units of $A$ and $B$ are respectively (A) $\text{Wb s}^2,\ \text{Wb s}$ (B) $\text{Wb s}^{-1},\ \text{Wb}$ (C) $\text{Wb s}^{-2},\ \text{Wb s}^{-1}$ (D) $\text{Wb s}^{-1},\ \text{Wb s}^{-2}$

   ›Reveal solutionSolution

   The key idea is dimensional consistency: each term in $\phi = 5At^2 + Bt - 2C$ must have the same unit as $\phi$ (Wb). This forces $A$ to have units $\text{Wb s}^{-2}$ and $B$ to have units $\text{Wb s}^{-1}$, so the correct option is (C).

   The problem gives you a relation between magnetic flux $\phi$ (in webers) and time $t$ (in seconds):

   $\phi = 5At^2 + Bt - 2C.$

   You’re asked for the SI units of $A$ and $B$. The constants $5$ and $2$ are pure numbers — they have no units. So the only way this equation makes physical sense is if every term on the right-hand side has the same unit as $\phi$, which is the weber (Wb). This is the principle of dimensional homogeneity, and it’s the entire foundation of the solution.

   Let’s apply it term by term.

   1. First term: $5At^2$ Since $5$ is dimensionless, the unit of $5At^2$ is the unit of $A$ multiplied by the unit of $t^2$. Time $t$ is in seconds, so $t^2$ has unit $\text{s}^2$. For this term to equal a flux in Wb, we need:

   $\text{unit of }A \times \text{s}^2 = \text{Wb}.$

   Therefore:

   $\text{unit of }A = \frac{\text{Wb}}{\text{s}^2} = \text{Wb s}^{-2}.$

   2. Second term: $Bt$ Here $B$ is multiplied by $t$ (unit s). So:

   $\text{unit of }B \times \text{s} = \text{Wb}.$

   Hence:

   $\text{unit of }B = \frac{\text{Wb}}{\text{s}} = \text{Wb s}^{-1}.$

   3. Third term: $-2C$ The constant $2$ is dimensionless, so $C$ must itself have units of Wb. This isn’t asked in the question, but it confirms the pattern: $C$ is just a constant flux offset.

   Watch out

   A common mistake is to think that because $t^2$ appears, $A$ must have units like $\text{Wb s}^2$ — but that’s backwards. The term $At^2$ must equal a flux, so $A$ must divide by $t^2$, not multiply it. Always check: if $t^2$ is in the numerator, its unit goes to the denominator of the constant’s unit.

   So the units are:

   $A \to \text{Wb s}^{-2}$, $B \to \text{Wb s}^{-1}$.

   Tip

   You can also think of it as: the coefficient of $t^2$ must have units of flux divided by time-squared, and the coefficient of $t$ must have units of flux divided by time. This pattern holds for any polynomial in time.

   ✓Final answer

   The correct option is (C).

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following physical quantity has no dimensions?

   (1) Angular Velocity

   (2) Angular Acceleration

   (3) Angular Displacement

   (4) Stress

   ›Reveal solutionSolution

   Angular displacement is dimensionless because it is defined as the ratio of two lengths (arc / radius).

   Check each option's dimensional formula:

   • Angular velocity ω = dθ/dt → dimension [T^-1] (angle is dimensionless, so only 1/time survives).
   • Angular acceleration α = dω/dt → dimension [T^-2].
   • Angular displacement θ = arc length / radius = a ratio of two lengths → the length dimensions cancel, leaving a pure number (dimensionless), measured in radians.
   • Stress = force / area → dimension [M L^-1 T^-2].

   Only angular displacement has no dimensions at all.

   ✓Final answer

   (3) Angular Displacement — being a ratio of two lengths (arc/radius), it is dimensionless (measured in radians, which is not a true unit).

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The dimensional formula of work done is the same as the dimensional formula of

   (a) Momentum

   (b) Power

   (c) Energy

   (d) Torque

   ›Reveal solutionSolution

   Work done and energy share the exact same dimensional formula [ML^2T^-2], because work is defined as a mode of energy transfer.

   Work done W = Force x displacement = [MLT^-2] x [L] = [ML^2T^-2].

   Check each option:

   • Momentum p = mv = [M][LT^-1] = [MLT^-1] -- different.

   • Power P = Work/time = [ML^2T^-2]/[T] = [ML^2T^-3] -- different.

   • Energy (kinetic or potential) = (1/2)mv^2 or mgh = [ML^2T^-2] -- matches exactly.

   • Torque = Force x perpendicular distance = [MLT^-2][L] = [ML^2T^-2] -- also matches numerically, but it is not the standard listed match here since torque and work are dimensionally identical yet physically distinct (torque is a vector-like quantity, work is scalar). The question asks which quantity SHARES the dimensional formula of work in the conventional sense taught in this chapter, which is Energy.

   ✓Final answer

   (c) Energy -- both work and energy have dimensional formula [ML^2T^-2].

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.What is the dimensional formula of angular momentum?

   (a) [ML^2T^-1]

   (b) [MLT^-2]

   (c) [MLT^-1]

   (d) [M^-1L^3T^2]

   ›Reveal solutionSolution

   Angular momentum L = mvr, so its dimensions are mass x velocity x radius = [ML^2T^-1].

   Angular momentum is defined as L = r x p = mvr (magnitude, for a point mass moving with linear velocity v at perpendicular distance r from the axis).

   Dimensions:

   • Mass, m = [M]
   • Velocity, v = [LT^-1]
   • Radius, r = [L]

   So [L_ang] = [M] x [LT^-1] x [L] = [ML^2T^-1].

   ✓Final answer

   (a) [ML^2T^-1].

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following physical quantities has the same dimensions as impulse?

   (a) Force

   (b) Momentum

   (c) Work

   (d) Power

   ›Reveal solutionSolution

   Impulse J = F x delta t has dimensions [MLT^-1], identical to momentum p = mv.

   Impulse J = Force x time = [MLT^-2] x [T] = [MLT^-1].

   Compare:

   • Force = [MLT^-2] -- different.
   • Momentum = mv = [M][LT^-1] = [MLT^-1] -- matches exactly.
   • Work = [ML^2T^-2] -- different.
   • Power = [ML^2T^-3] -- different.

   This dimensional match is not a coincidence -- Newton's second law in impulse form states impulse = change in momentum (J = delta p), so they must share dimensions.

   ✓Final answer

   (b) Momentum.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.The pair of physical quantities not having same dimension is:

   (a) Torque and Energy

   (b) Surface Tension and Impulse

   (c) Angular momentum and Planck's constant

   (d) None of the above

   ›Reveal solutionSolution

   Surface tension ($MT^{-2}$) and impulse ($MLT^{-1}$) have different dimensions; the other two pairs match.

   Work out each pair's dimensional formula:

   • Torque $= r \times F$, dimension $ML^2T^{-2}$. Energy also has dimension $ML^2T^{-2}$. Same.
   • Surface tension $=$ force/length, dimension $MLT^{-2}/L = MT^{-2}$. Impulse $=$ force $\times$ time, dimension $MLT^{-2}\cdot T = MLT^{-1}$. Different — an $L$ appears in impulse but not in surface tension.
   • Angular momentum $= mvr$, dimension $M \cdot LT^{-1} \cdot L = ML^2T^{-1}$. Planck's constant $h$ (from $E=h\nu$) has dimension (energy)/(frequency) $= ML^2T^{-2}/T^{-1} = ML^2T^{-1}$. Same.

   So the pair that does NOT share the same dimension is surface tension and impulse.

   ✓Final answer

   (b) Surface Tension and Impulse.

7. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: Dimensional formula of ____ is same as the dimensional formula of angular momentum.

   ›Reveal solutionSolution

   Planck's constant has the same dimensional formula, [M L² T⁻¹], as angular momentum.

   Angular momentum L = mvr has dimensions [M][LT⁻¹][L] = [M L² T⁻¹].

   Planck's constant is defined through E = hν, so h = E/ν. Energy E has dimensions [M L² T⁻²] and frequency ν has dimensions [T⁻¹], so:

   h = [M L² T⁻²] / [T⁻¹] = [M L² T⁻¹]

   This matches angular momentum exactly. Both quantities are, in fact, examples of the physical concept of 'action', which is why they share the same dimensional formula.

   ✓Final answer

   Planck's constant (h) has the same dimensional formula as angular momentum: [M L² T⁻¹].

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following has dimensional formula [ML^2T^-2] ?

   (a) Acceleration

   (b) Force

   (c) Work

   (d) Linear momentum

   ›Reveal solutionSolution

   [ML^2T^-2] is the dimension of energy/work, so the answer is (C) Work.

   We find the dimensions of each option:

   • Acceleration = velocity/time = [LT^-2]
   • Force = mass x acceleration = [MLT^-2]
   • Work = force x displacement = [MLT^-2] x [L] = [ML^2T^-2]
   • Linear momentum = mass x velocity = [MLT^-1]

   Only work matches [ML^2T^-2]. (Kinetic energy ½mv^2 gives the same dimensions, confirming that any form of energy has [ML^2T^-2].)

   ✓Final answer

   (C) Work — its dimensional formula is [ML^2T^-2].

9. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following couples of quantities have same dimension? (A) Work and energy (B) Force and momentum (C) Force and power (D) Momentum and energy

   ›Reveal solutionSolution

   Work and energy share the same dimension $[ML^2T^{-2}]$, both measured in joules.

   Check each pair:

   • Work $W = Fd$, dimension $[MLT^{-2}][L] = [ML^2T^{-2}]$. Kinetic energy $= \tfrac{1}{2}mv^2$, dimension $[M][LT^{-1}]^2 = [ML^2T^{-2}]$. Same dimension.
   • Force $[MLT^{-2}]$ vs momentum $[MLT^{-1}]$ — different.
   • Force $[MLT^{-2}]$ vs power $[ML^2T^{-3}]$ — different.
   • Momentum $[MLT^{-1}]$ vs energy $[ML^2T^{-2}]$ — different.

   Only work and energy match.

   ✓Final answer

   (A) Work and energy.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The velocity of a particle

    (v) at an instant t is given by v = at + bt^2. The dimension of b is

    (a) [L]

    (b) [LT^-1]

    (c) [LT^-2]

    (d) [LT^-3]

    ›Reveal solutionSolution

    By the principle of dimensional homogeneity, every term added to v must have the dimension of velocity; this forces [b] = [LT^-3].

    Given v = at + bt^2, where v is velocity, [v] = [LT^-1].

    By dimensional homogeneity, each term on the right side must independently have the dimension of velocity (you cannot add quantities of different dimensions).

    For the term bt^2:

    [b][t]^2 = [LT^-1]

    [b][T^2] = [LT^-1]

    [b] = [LT^-1] / [T^2] = [LT^-1-2] = [LT^-3]

    ✓Final answer

    (d) [LT^-3].

11. CBSE 2025Set ANN1 markMCQ
    Q.Find out the fundamental quantity from among the physical quantities given below:

    (a) velocity

    (b) temperature

    (c) force

    (d) density

    ›Reveal solutionSolution

    Temperature is the fundamental quantity here; velocity, force and density are all derived from base quantities.

    The seven SI base (fundamental) quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity. A derived quantity is built from these using multiplication or division.

    Checking each option:

    • velocity = displacement / time = length / time -> derived.

    • force = mass x acceleration = mass x length / time squared -> derived.

    • density = mass / volume = mass / length cubed -> derived.

    • temperature -> one of the seven SI base quantities, so it is fundamental.

    ✓Final answer

    (b) Temperature is the fundamental quantity.

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The physical quantity having the dimensions [M¹T⁻²] is

    (a) (A) Density

    (b) (B) Tension

    (c) (C) Surface tension

    (d) (D) Viscosity

    ›Reveal solutionSolution

    [!TLDR]

    (C) Surface tension

    Why

    Surface tension = force/length = (MLT⁻²)/L = MT⁻², exactly matching the given dimensions.

    [!ANSWER]

    (C) Surface tension