One of the combinations from the fundamental physical constants is $\dfrac{hc}{G}$. The unit of this expression is (a) kg$^2$ (b) m$^3$ (c) s$^{-1}$ (d) m

One of the combinations from the fundamental physical constants is (hc)/(G). The unit of this expression is (a

I. Multiple Choice Questions · Q1

\dfrac{hc}{G}

(a) kg

^2

(b) m

^3

^{-1}

(d) m

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Concept understanding — Dimensional Formula, Dimensional Constants and Homogeneity Principle

Every derived quantity reduces to some combination of powers of the seven fundamental (base) quantities -- length $[L]$ , mass $[M]$ , time $[T]$ , electric current $[A]$ , temperature $[K]$ , amount of substance $[\text{mol}]$ , luminous intensity $[\text{cd}]$ -- and this combination, written in square brackets, is called the quantity's dimensional formula. Writing that formula as an equation (e.g. $[v]=[M^0LT^{-1}]$ ) is the dimensional equation.

Worked derivation (velocity): $\text{velocity}=\dfrac{\text{displacement}}{\text{time}}=\dfrac{[L]}{[T]}=[M^0LT^{-1}]$ -- dimension $0$ in mass, $1$ in length, $-1$ in time.

Dimensional formulas of common quantities follow directly from their defining relation: force $=$ mass $\times$ acceleration $=[MLT^{-2}]$ ; work $=$ force $\times$ distance $=[ML^2T^{-2}]$ ; Planck's constant $h=$ energy/frequency $=[ML^2T^{-2}]/[T^{-1}]=[ML^2T^{-1}]$ ; the gravitational constant $G$ (from Newton's law $F=Gm_1m_2/r^2$ ) $=[\text{force}\times\text{distance}^2]/\text{mass}^2=[M^{-1}L^3T^{-2}]$ ; torque, being force $\times$ distance just like work, shares work's dimensional formula $[ML^2T^{-2}]$ -- so torque and energy, despite measuring physically different things, have the same dimensions.

Classifying by dimension. Every physical quantity falls into exactly one of four boxes:

Dimensional variables -- have dimensions, and take variable values (length, velocity, acceleration).
Dimensionless variables -- no dimensions, but variable values (specific gravity, strain, refractive index).
Dimensional constants -- have dimensions, but a fixed value ( $G$ , $h$ ).
Dimensionless constants -- neither dimensioned nor variable ( $\pi$ , $e$ , pure counting numbers).

Principle of homogeneity of dimensions. Every additive term in a physically valid equation must carry the same dimensions -- you cannot add a length to a time. E.g. in $v^2=u^2+2as$ , the terms $v^2$ , $u^2$ and $2as$ all reduce to $[L^2T^{-2}]$ . This principle is exactly what makes the checking use of dimensional analysis possible: if two sides (or two additive terms) of a proposed equation come out with different dimensions, the equation is definitely wrong -- though matching dimensions alone does not guarantee the equation is numerically correct (see 'Dimensional Analysis').

Step 1. Dimensional formulas: $[h]=[ML^2T^{-1}]$ , $[c]=[LT^{-1}]$ , $[G]=[M^{-1}L^3T^{-2}]$ .

Step 2. $[hc]=[ML^2T^{-1}][LT^{-1}]=[ML^3T^{-2}]$ .

Step 3. $\left[\dfrac{hc}{G}\right]=\dfrac{[ML^3T^{-2}]}{[M^{-1}L^3T^{-2}]}=[M^{1-(-1)}L^{3-3}T^{-2-(-2)}]=[M^2]$ .

Step 4. A dimension of $[M^2]$ means the expression has units of mass-squared, i.e. kg $^2$ .

✓Final answer

^2

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

CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following has the dimension of (mu0 epsilon0)^(-1/2)?$
(a) Velocity
(b) Length
(c) Force
(d) Time
›Reveal solutionSolution

(mu0 epsilon0)^(-1/2) is exactly the speed of light c, so its dimension is that of velocity, [M^0 L^1 T^-1].

Maxwell's electromagnetic theory gives the speed of an electromagnetic wave in vacuum as:

c = 1 / sqrt(mu0 epsilon0)

where mu0 is the permeability of free space and epsilon0 is the permittivity of free space.

Rearranging, (mu0 epsilon0)^(-1/2) = 1/sqrt(mu0 epsilon0) = c.

Since c is a speed (distance per unit time), its dimensional formula is [L T^-1] (i.e., [M^0 L^1 T^-1 A^0]) -- the dimension of velocity. This can also be verified directly from the individual dimensional formulas of mu0 ([M L T^-2 A^-2]) and epsilon0 ([M^-1 L^-3 T^4 A^2]), whose product's inverse square root works out to [L T^-1].

✓Final answer
$The correct option is (a) Velocity.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following pairs of physical quantities have same dimensions?$
(a) torque and power
(b) force and power
(c) force and torque
(d) torque and energy
›Reveal solutionSolution
Torque and energy share the same dimensional formula [ML^2T^-2], because both are computed as (force) x (distance), even though torque is a vector (cross product) and energy/work is a scalar (dot product).

Work out the dimensional formula of each quantity:

Force F = mass x acceleration -> [M][LT^-2] = [MLT^-2]

Torque tau = force x perpendicular distance -> [MLT^-2][L] = [ML^2T^-2]

Energy/Work W = force x displacement (along force) -> [MLT^-2][L] = [ML^2T^-2]

Power P = work/time -> [ML^2T^-2]/[T] = [ML^2T^-3]

Now compare the four given pairs:

torque [ML^2T^-2] vs power [ML^2T^-3] -> different (T exponent differs)

force [MLT^-2] vs power [ML^2T^-3] -> different

force [MLT^-2] vs torque [ML^2T^-2] -> different (L exponent differs)

torque [ML^2T^-2] vs energy [ML^2T^-2] -> IDENTICAL So torque (measured in N.m) and energy (measured in joules, which is also N.m) are dimensionally identical, even though they represent physically different concepts.

✓Final answer
$The correct option is (d) torque and energy.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. The Dimensional formula for strain :$
(a) ML^-2 T^-1
(b) M^0 L^0 T^0
(c) ML^-1 T^-2
(d) M^0 L T^0
›Reveal solutionSolution

Strain is always a ratio of two quantities of the same kind (for example, change in length to original length, or change in volume to original volume), so all units cancel out and the dimensional formula is M^0 L^0 T^0.

Definition of (longitudinal) strain:

Strain = (change in length, ΔL) / (original length, L)

Dimensionally:

[Strain] = [L] / [L] = [L^0] = dimensionless

The same is true for other kinds of strain used in this course (volumetric strain = ΔV/V, shear strain = a ratio of a displacement to a length) — in every case it is a ratio of two quantities with the same dimension, so the dimensions always cancel.

Hence the dimensional formula of strain is written as M^0 L^0 T^0, i.e., it has no dimensions in mass, length, or time.

✓Final answer
$The correct option is (b) M^0 L^0 T^0.$
CBSE 2019Set ANNUAL1 markMCQ
$Q. Which of the following pairs of physical quantities have the same dimensions ?$
(a) Torque and Power
(b) Force and Torque
(c) Force and Power
(d) Torque and Energy
›Reveal solutionSolution
Torque and Energy have identical dimensions [ML^2T^-2], even though they represent physically different concepts.

Dimensional formulas:

Torque, tau = F x r (perpendicular distance) -> [MLT^-2][L] = [ML^2T^-2]

Energy (work), W = F x d -> [MLT^-2][L] = [ML^2T^-2]

Power, P = W/t -> [ML^2T^-2]/[T] = [ML^2T^-3]

Force, F -> [MLT^-2]

Checking each option:

Torque [ML^2T^-2] and Power [ML^2T^-3] -- different.

Force [MLT^-2] and Torque [ML^2T^-2] -- different.

Force [MLT^-2] and Power [ML^2T^-3] -- different.

Torque [ML^2T^-2] and Energy [ML^2T^-2] -- SAME. Although torque (a vector, measured in N m) and energy (a scalar, measured in joules) are conceptually distinct, their dimensional formulas and even their SI base-unit combination coincide.

✓Final answer
$The correct option is (d) Torque and Energy -- both have dimensions [ML^2T^-2].$

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Q.One of the combinations from the fundamental physical constants is hcG\dfrac{hc}{G}Ghc​. The unit of this expression is

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Q.One of the combinations from the fundamental physical constants is $\dfrac{hc}{G}$ . The unit of this expression is

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