Q.Explain this statement clearly: "To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison". In view of this, reframe the following statements wherever necessary:

Showing the 12 most recent of 24 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Match the physical quantities with their Dimensional formulae:

   	Physical Quantity		Dimensional Formula
   a)	Coefficient of Viscosity (η)

   (i) | $M^{-1}L^3T^4A^2$ | | b) | Young's Modulus (Y) |

   (ii) | $ML^{-1}T^{-1}$ | | c) | Permittivity (ε) |

   (iii) | $ML^{-1}T^{-2}$ | | d) | Universal Gravitational constant (G) |

   (iv) | $M^{-1}L^3T^{-2}$ | (A) a-(ii), b-(iii), c-(iv), d-(i) (B) a-(ii), b-(iii), c-(i), d-(iv) (C) a-(iii), b-(ii), c-(i), d-(iv) (D) a-(iv), b-(ii), c-(i), d-(iii)

   ›Reveal solutionSolution

   Matching each quantity to its known SI dimensional formula: viscosity-(ii), Young's modulus-(iii), permittivity-(i), gravitational constant-(iv).

   Concept and Intuition

   Each physical quantity has a fixed dimensional formula derivable from its defining equation. Recognizing a couple of these outright (especially $G$, which has a very distinctive $L^3T^{-2}$ signature, and viscosity/Young's modulus, which are both mechanical but differ by one power of $T$) lets the rest fall into place by elimination.

   Step-by-Step Solution

   1. Young's modulus $Y$: stress/strain = (Force/Area)/(dimensionless) = $\dfrac{MLT^{-2}}{L^2}=ML^{-1}T^{-2}$ — matches (iii).
   2. Coefficient of viscosity $\eta$: from Newton's law of viscosity, $F=\eta A\dfrac{dv}{dx}$, so $\eta=\dfrac{F}{A}\cdot\dfrac{L}{v}=\dfrac{MLT^{-2}}{L^2}\cdot\dfrac{L}{LT^{-1}}=ML^{-1}T^{-1}$ — matches (ii).
   3. Universal gravitational constant $G$: from $F=\dfrac{Gm_1m_2}{r^2}$, $G=\dfrac{Fr^2}{m_1m_2}=\dfrac{MLT^{-2}\cdot L^2}{M^2}=M^{-1}L^3T^{-2}$ — this matches (iv) exactly. …

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2. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Match the physical quantities in List I with the corresponding SI units in List II List I | List II A. Torque | I. N m s$^{-1}$ B. Stress | II. N m kg$^{-1}$ C. Latent heat | III. N m D. Power | IV. N m$^{-2}$ (A) A – III, B – II, C – I, D – IV (B) A – III, B – IV, C – II, D – I (C) A – IV, B – I, C – III, D – II (D) A – II, B – III, C – I, D - IV

   ›Reveal solutionSolution

   This tests whether you can derive the SI unit of each quantity from its defining formula, then read it off in the $N,m,kg,s$ combination given in List II. Answer: (B).

   Concept and Intuition

   Every mechanical quantity's unit can be built from its defining equation. Torque is a force times a lever arm, so its unit is simply force × length. Stress is force per unit area, the inverse geometry of torque. Latent heat is energy delivered per unit mass (no time involved — it's not a rate). Power is energy delivered per unit time. Keeping the defining relation in mind (not memorising units) lets you rebuild any of these from $N$, $m$, $kg$, $s$.

   Step-by-Step Solution

   1. Torque $\tau = F \times r$: unit $= N \cdot m$ → matches III.
   2. Stress $= F/A$: unit $= N/m^2 = N\,m^{-2}$ → matches IV.
   3. Latent heat $L = Q/m$ (heat per unit mass): unit $= J/kg = (N\,m)/kg = N\,m\,kg^{-1}$ → matches II. …

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3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Of the following, the pair of physical quantities not having the same dimensional formula is (A) work and torque (B) angular momentum and Planck's constant (C) stress and linear momentum (D) surface tension and force constant

   ›Reveal solutionSolution

   This tests recall/derivation of dimensional formulas for several physical-quantity pairs to find the one mismatch. Answer: (C).

   Concept and Intuition

   Many pairs of physical quantities share a dimensional formula even though they measure conceptually different things — that's exactly why "dimensional formula" problems are useful for unit-consistency checks but can't distinguish physically different quantities. Here we must actually compute each pair's dimensions.

   Step-by-Step Solution

   1. Work and torque: Work $= F\cdot d$, torque $=F\cdot d$ (force times a perpendicular distance) — both give $[ML^2T^{-2}]$. Same.
   2. Angular momentum and Planck's constant: Angular momentum $L = mvr$ has dimensions $[M][LT^{-1}][L]=[ML^2T^{-1}]$. Planck's constant from $E=h\nu$: $h = E/\nu$, dimensions $[ML^2T^{-2}]/[T^{-1}]=[ML^2T^{-1}]$. Same.
   3. Stress and linear momentum: Stress $=$ force/area $= [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$. Linear momentum $= mv = [M][LT^{-1}]=[MLT^{-1}]$. These are not the same (different powers of $L$ and $T$). …

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4. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.The electron gain enthalpy ($\Delta_{eg}H$) of chlorine is $-3.7\,\mathrm{eV\,mol^{-1}}$. How much of energy (in k cal mol$^{-1}$) is released when 7.1 g of chlorine atoms are completely converted into $\mathrm{Cl^-}$ ions in gaseous state ? (1 eV = 23 k cal) (A) 1.072 (B) 10.72 (C) 17.02 (D) 1.702

   ›Reveal solutionSolution

   Converting the given electron gain enthalpy to kcal/mol and scaling by the moles of chlorine atoms present gives 17.02 kcal released.

   Concept and Intuition

   Electron gain enthalpy is the energy change when a gaseous atom gains an electron. Since it's negative (energy released) for chlorine, converting to a consistent energy unit and multiplying by the number of moles gives the total heat released.

   Step-by-Step Solution

   1. Moles of Cl atoms: $n=\dfrac{7.1\ \text{g}}{35.5\ \text{g/mol}}=0.2$ mol.
   2. Convert $\Delta_{eg}H$ to kcal/mol: $3.7\ \text{eV}\times23\ \text{kcal/eV}=85.1$ kcal/mol (magnitude; the process releases this energy). …

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5. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.If σ denotes Stefan constant and S denotes heat capacity, then the dimensional formula of $\dfrac{S}{\sigma}$ is (A) $[M^0L^2T^{-1}K^3]$ (B) $[M^0L^2TK^3]$ (C) $[ML^2T^{-1}K^{-4}]$ (D) $[M^0L^2T^{-1}K^{-3}]$

   ›Reveal solutionSolution

   Using $\sigma = [MT^{-3}K^{-4}]$ from Stefan's law and $S = [ML^2T^{-2}K^{-1}]$ for heat capacity, dividing gives $S/\sigma = [M^0L^2T K^3]$. Answer: (B).

   Concept and Intuition

   Dimensional analysis lets us find the units of a derived physical constant purely from the physical law that defines it, without needing to remember the units by rote. The Stefan-Boltzmann law $E = \sigma T^4$ (power radiated per unit area by a black body, proportional to the fourth power of absolute temperature) directly gives us $\sigma$'s dimensions once we know the dimensions of power and area. Heat capacity, defined as $S = Q/\Delta T$ (heat energy needed per unit rise in temperature), similarly follows directly from the dimensions of energy and temperature.

   Step-by-Step Solution

   1. Write Stefan's law: $E = \sigma T^4$, where $E$ is the power emitted per unit surface area, so $E$ has dimensions of $\dfrac{\text{Power}}{\text{Area}} = \dfrac{[ML^2T^{-3}]}{[L^2]} = [MT^{-3}]$.
   2. So $\sigma = \dfrac{E}{T^4} = \dfrac{[MT^{-3}]}{[K^4]} = [M L^0 T^{-3} K^{-4}]$.
   3. Write the definition of heat capacity: $S = \dfrac{Q}{\Delta T}$, where $Q$ (heat energy) has dimensions of energy, $[ML^2T^{-2}]$, and $\Delta T$ has dimensions $[K]$.
   4. So $S = \dfrac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$. …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If force $= \dfrac{\alpha}{\text{density} + \beta^3}$, then the dimensional formulae of $\alpha$ and $\beta$ are respectively (A) $[M L^2 T^{-2}], [M L^{-1/3} T^0]$ (B) $[M^2 L^4 T^{-2}], [M^{1/3} L^{-1} T^0]$ (C) $[M^2 L^{-2} T^{-2}], [M^{1/3} L^{-1} T^0]$ (D) $[M^2 L^{-2} T^{-2}], [M L^{-3} T^0]$

   ›Reveal solutionSolution

   Tests the principle of dimensional homogeneity — only like quantities can be added. The answer is (C).

   Concept and Intuition

   In any physically valid equation, terms being added or subtracted must have identical dimensions — you cannot add mass to length. Here $\beta^3$ is added to density inside the denominator, so $\beta^3$ must itself carry the dimensions of density.

   Step-by-Step Solution

   1. Density has dimensions $[ML^{-3}]$.
   2. Since $\beta^3$ is added to density, $[\beta^3] = [ML^{-3}]$, so $[\beta] = [M^{1/3}L^{-1}T^0]$.
   3. Because $\beta^3$ matches density's dimensions, the whole denominator $(\text{density}+\beta^3)$ also has dimensions $[ML^{-3}]$. …

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7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.Match the following

   a) Thermal conductivity i) $MLT^{-3}K^{-1}$

   b) Boltzman constant ii) $M^0L^2T^{-2}K^{-1}$

   c) Latent heat iii) $ML^2T^{-2}K^{-1}$

   d) Specific heat iv) $M^0L^2T^{-2}$ (A) a-i, b-iii, c-iv, d-ii (B) a-i, b-ii, c-iv, d-iii (C) a-iii, b-ii, c-i, d-iv (D) a-ii, b-i, c-iii, d-iv

   ›Reveal solutionSolution

   Matching each thermal quantity's SI unit to its dimensional formula gives a-i, b-iii, c-iv, d-ii.

   Concept and Intuition

   Each thermal quantity's dimensional formula follows directly from its defining equation and SI unit; recognizing whether mass and temperature appear (and with what power) quickly distinguishes the four formulas given.

   Step-by-Step Solution

   1. Thermal conductivity $k$: defined via $Q=\dfrac{kA\Delta T\,t}{d}$, with SI unit $\text{W}\,\text{m}^{-1}\text{K}^{-1} = \text{kg}\,\text{m}\,\text{s}^{-3}\text{K}^{-1}$ → dimension $MLT^{-3}K^{-1}$, matching (i).
   2. Boltzmann constant $k_B$: appears in $E=k_BT$ (energy = $k_B\times$ temperature), so its unit is $J/K=\text{kg}\,\text{m}^2\text{s}^{-2}\text{K}^{-1}$ → dimension $ML^2T^{-2}K^{-1}$, matching (iii).
   3. Latent heat $L$: defined via $Q=mL$, so unit is $J/\text{kg}=\text{m}^2\text{s}^{-2}$ → dimension $M^0L^2T^{-2}$ (mass cancels out), matching (iv). …

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8. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.E, m, L, G represent energy, mass, angular momentum and gravitational constant respectively. The dimensions of $\dfrac{EL^2}{m^5G^2}$ will be that of (A) Angle (B) Length (C) Mass (D) Time

   ›Reveal solutionSolution

   A dimensional-analysis question: substitute the known dimensional formulas for energy, angular momentum, mass and $G$, and simplify.

   Concept and Intuition

   Quantities like plane angle, solid angle, strain and refractive index are dimensionless — recognizing that an expression's dimensions cancel completely to $[M^0L^0T^0]$ is the signal that it represents an angle (or another dimensionless ratio), never a physical quantity like length or time.

   Step-by-Step Solution

   1. $E$ (energy) $=[ML^2T^{-2}]$; $L$ (angular momentum) $=[ML^2T^{-1}]$; $m=[M]$; $G$ (from $F=Gm_1m_2/r^2$) $=[M^{-1}L^3T^{-2}]$.
   2. Numerator: $EL^2=[ML^2T^{-2}]\cdot[ML^2T^{-1}]^2=[ML^2T^{-2}]\cdot[M^2L^4T^{-2}]=[M^3L^6T^{-4}]$.
   3. Denominator: $m^5G^2=[M^5]\cdot[M^{-1}L^3T^{-2}]^2=[M^5]\cdot[M^{-2}L^6T^{-4}]=[M^3L^6T^{-4}]$. …

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9. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.Which of the following is not a unit of permeability (A) Henry meter$^{-1}$ (B) Weber ampere$^{-1}$ meter$^{-1}$ (C) Ohm second meter$^{-1}$ (D) Volt second meter$^{-1}$

   ›Reveal solutionSolution

   Permeability is measured in Henry/meter; check each option by converting it to Henry/meter using $1\,\text{H}=1\,\Omega\cdot\text{s}=1\,\text{Wb/A}=1\,\text{V·s/A}$ — the option missing the per-ampere factor is not valid.

   Concept and Intuition

   Permeability $\mu$ (as in $\mu_0$) is defined via $B=\mu H$ or via inductance formulas, and its SI unit is the Henry per metre (H/m). Since the Henry itself has several equivalent unit expressions ($\text{Wb/A}$, $\Omega\cdot\text{s}$, $\text{T·m/A}$), several of the listed options are just disguised forms of H/m. The trick is to convert every option back to base SI units and see which one is actually dimensionally inequivalent.

   Step-by-Step Solution

   1. Standard unit: $\mu_0$ is in Henry per metre, $\text{H/m}$.
   2. (A) Henry meter$^{-1}$ = H/m — this IS the standard unit.
   3. (B) Weber ampere$^{-1}$ meter$^{-1}$ = $\text{Wb}/(\text{A·m})$. Since $1\,\text{H}=1\,\text{Wb/A}$, this is $\text{H/m}$ — valid.
   4. (C) Ohm second meter$^{-1}$ = $\Omega\cdot\text{s}/\text{m}$. Since $\Omega=\text{V/A}$, this is $(\text{V·s/A})/\text{m}=\text{H/m}$ (because $\text{H}=\text{V·s/A}$ from $V=L\,dI/dt$) — valid. …

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10. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
    Q.If Young's modules of elasticity is $Y = \dfrac{2mgl^{x}}{5bt^{3}e}$, where 'g' is the acceleration due to gravity, 'm' is the mass, 'l' is the length, 'b' is the breadth, 't' is the thickness and 'e' is the elongation, then the value of x is (A) 1 (B) 2 (C) 3 (D) 4

    ›Reveal solutionSolution

    A dimensional-analysis question: matching the powers of length on both sides fixes $x=3$. Answer: (C) 3.

    Concept and Intuition

    Young's modulus is stress divided by strain, i.e. $\dfrac{\text{Force}/\text{Area}}{\text{dimensionless strain}}$, giving dimensions $[\mathrm{ML^{-1}T^{-2}}]$ (same as pressure). Any correct physical formula for $Y$ must reduce to exactly this dimension regardless of what the individual symbols mean — this lets us solve for an unknown exponent purely from dimensional consistency, without needing to know the physical derivation of the formula (which in this case is the standard bending-of-a-beam Young's modulus experiment, where $l$ is the length between supports, $b$ the breadth, $t$ the thickness, and $e$ the elongation/depression).

    Step-by-Step Solution

    1. Write the dimension of each quantity: $[m]=\mathrm{M}$, $[g]=\mathrm{LT^{-2}}$, $[l]=\mathrm{L}$, $[b]=\mathrm{L}$, $[t]=\mathrm{L}$, $[e]=\mathrm{L}$ (the numeric factors $2,5$ are dimensionless).
    2. Numerator dimension: $[m][g][l^{x}] = \mathrm{M}\cdot \mathrm{LT^{-2}}\cdot \mathrm{L}^{x} = \mathrm{M\,L^{1+x}\,T^{-2}}$. …

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11. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
    Q.The energy consumed by a 1000 W electric bulb when it is used for 1 hour is (A) $3.6 \times 10^{5} \, \mathrm{W}$ (B) $3.6 \times 10^{6} \, \mathrm{J}$ (C) $3.6 \times 10^{6} \, \mathrm{W}$ (D) $3.6 \times 10^{5} \, \mathrm{J}$

    ›Reveal solutionSolution

    Energy consumed equals power times time: $1000\ \mathrm{W} \times 1\ \mathrm{hour} = 3.6\times10^6$ J.

    Concept and Intuition

    Electrical energy consumed by an appliance is the product of its power rating and the duration of use: $E = Pt$. Power (watts) is energy per unit time, so multiplying by time (in seconds) recovers energy in joules. This is also the basis for the commercial unit 'kWh' (1 kWh = 1000 W $\times$ 3600 s = $3.6\times10^6$ J).

    Step-by-Step Solution

    1. Given: $P = 1000$ W, $t = 1$ hour $= 3600$ s.
    2. $E = Pt = 1000 \times 3600 = 3.6 \times 10^{6}$ J.
    3. Options (A) and (C) are expressed in watts (a unit of power, not energy) — dimensionally wrong for 'energy consumed,' so they can be eliminated on units alone. …

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12. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.Among the following, the unit of permeability is NOT represented by (A) henry/metre (B) weber/ampere (C) ohm-second/metre (D) volt-second/metre$^2$

    ›Reveal solutionSolution

    Permeability's SI unit is $H/m$, equally expressible as $Wb/(A\cdot m)$,

    $T\cdot m/A$, or $\Omega\cdot s/m$. "Volt-second per square metre" is dimensionally

    just $Wb/m^2=$ tesla — the unit of magnetic flux density, not permeability — so it

    is the one option that does NOT represent permeability.

    Concept and Intuition

    Permeability $\mu$ appears in $B = \mu H$, relating flux density $B$ (tesla) to the

    magnetising field $H$ (ampere/metre). So dimensionally,

    $[\mu] = \frac{[B]}{[H]} = \frac{\text{tesla}}{A/m} = \text{tesla}\cdot m \cdot A^{-1}.$

    Every correct unit of permeability must carry this exact combination: one length

    in the numerator, one ampere in the denominator (along with whatever combination of

    kg, m, s reproduces tesla). Recognising which listed unit is "one ampere short" (or

    has an extra/misplaced length power) is the key skill being tested.

    Step-by-Step Solution

    1. Recall the standard equivalent forms of permeability's unit, all equal to $kg\cdot m \cdot s^{-2}\cdot A^{-2}$:
       • Henry/metre ($H/m$)
       • Weber/(Ampere·metre) ($Wb\,A^{-1}m^{-1}$)
       • Ohm·second/metre ($\Omega\,s\,m^{-1}$)
       • Tesla·metre/Ampere ($T\,m\,A^{-1}$)
    2. Check (A) Henry/metre: this is the textbook SI unit itself — correct.
    3. Check (C) Ohm-second/metre: $\Omega = V/A$, so $\Omega\cdot s/m = (V\cdot s)/(A\cdot m) = Wb/(A\cdot m) = H/m$ — correct (uses $1\ Wb = 1\ V\cdot s$).
    4. Check (D) Volt-second/metre²: $1\ V\cdot s = 1\ Wb$, so this unit is $Wb/m^2$, which is exactly the definition of the tesla — the unit of magnetic flux density $B$, not of permeability $\mu$. It is missing the $/\text{ampere}$ …

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