Q.A calorie is a unit of heat (energy in transit) and it equals about $4.2\ \text{J}$ where $1\,\text{J} = 1\ \text{kg m}^2\,\text{s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha$ kg, the unit of length equals $\beta$ m, the unit of time is $\gamma$ s. Show that a calorie has a magnitude $4.2\ \alpha^{-1}\,\beta^{-2}\,\gamma^{2}$ in terms of the new units.

Showing the 12 most recent of 17 on this concept.

1. GUJCET 2026Set x1 markMCQ
   Q.Out of the following physical quantities which quantity has the same unit as that of Planck's constant? (A) moment of force (B) power (C) angular momentum (D) moment of inertia

   ›Reveal solutionSolution

   $[h]=\text{J·s}$, identical to the unit of angular momentum.

   Planck's constant appears in $E=h\nu$, so $[h]=\dfrac{[E]}{[\nu]}=\dfrac{\text{J}}{\text{s}^{-1}}=\text{J·s}=\text{kg·m}^2\text{s}^{-1}$.

   Angular momentum $L=mvr$ has units $\text{kg}\cdot(\text{m/s})\cdot\text{m}=\text{kg·m}^2\text{s}^{-1}=\text{J·s}$.

   The other choices differ: moment of force (torque) is $\text{N·m}=\text{J}$, power is $\text{J/s}=\text{W}$, and moment of inertia is $\text{kg·m}^2$.

   ✓Final answer

   Option (C) angular momentum

   ANSWER: (C)

2. GUJCET 2025Set 031 markMCQ
   Q.The dimensional formula of current sensitivity of moving coil galvanometer is (A) $[L^{2}]$ (B) $[M^{1}L^{2}T^{-2}A^{-1}]$ (C) $[A^{-1}]$ (D) $[M^{1}L^{2}T^{-2}]$

   ›Reveal solutionSolution

   [!TLDR]

   Current sensitivity $=\theta/I$; since $\theta$ is dimensionless, its dimension is $[A^{-1}]$.

   Concept

   The current sensitivity of a moving-coil galvanometer is the deflection produced per unit current, $S_I=\dfrac{\theta}{I}=\dfrac{NBA}{k}$. An angular deflection is dimensionless.

   Solution

   Write the dimensions:

   $[S_I] = \frac{[\theta]}{[I]} = \frac{[\text{dimensionless}]}{[A]} = [A^{-1}]$

   [!ANSWER]

   (C) $[A^{-1}]$

3. GUJCET 2024Set 131 markMCQ
   Q.Js is the unit of ________ physical quantity. (A) Angular momentum (B) Work function (C) Moment of Inertia (D) Rydberg constant

   ›Reveal solutionSolution

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

   Concept. $\text{J·s} = \text{kg m}^2\text{s}^{-1}$, which is the dimension of angular momentum (and of Planck's constant). Work function has unit J, moment of inertia $\text{kg m}^2$, Rydberg constant $\text{m}^{-1}$ — none of these is J·s.

   ✓Final answer

   Option (A) Angular momentum

   ANSWER: (A)

4. GUJCET 2023Set 091 markMCQ
   Q.Unit of mobility in terms of fundamental units is ______. (A) $kg^{-1}s^{-2}A$ (B) $kgs^2A$ (C) $kg^{-1}s^2A$ (D) $kg^{-1}s^2A^{-1}$

   ›Reveal solutionSolution

   [!TLDR]

   Reducing $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ to fundamental units gives $\text{kg}^{-1}\text{s}^2\text{A}$.

   Concept

   Mobility $\mu=\dfrac{v_d}{E}$ has units of $\dfrac{\text{m/s}}{\text{V/m}}=\text{m}^2\text{V}^{-1}\text{s}^{-1}$. Convert the volt to base units to express $\mu$ in fundamental units.

   Solution

   $\text{V}=\frac{\text{J}}{\text{C}}=\frac{\text{kg}\,\text{m}^2\text{s}^{-2}}{\text{A}\,\text{s}}=\text{kg}\,\text{m}^2\text{A}^{-1}\text{s}^{-3}.$

   Then

   $[\mu]=\frac{\text{m}^2}{\text{V}\cdot\text{s}}=\frac{\text{m}^2}{(\text{kg}\,\text{m}^2\text{A}^{-1}\text{s}^{-3})\,\text{s}}=\frac{\text{m}^2}{\text{kg}\,\text{m}^2\text{A}^{-1}\text{s}^{-2}}=\text{kg}^{-1}\text{A}\,\text{s}^2.$

   So the unit is $\text{kg}^{-1}\text{s}^2\text{A}$.

   [!ANSWER]

   (C) $kg^{-1}s^2A$

5. GUJCET 2023Set 091 markMCQ
   Q.The dimensional formula of self inductance is ______. (A) $M^1L^1T^{-2}A^{-2}$ (B) $M^1L^2T^{-2}A^{-2}$ (C) $M^{-1}L^{-1}T^2A^2$ (D) $M^1L^{-1}T^{-1}A^{-2}$

   ›Reveal solutionSolution

   [!TLDR] $[L]=M^1L^2T^{-2}A^{-2}$ → (B).

   Concept

   Self-inductance is defined by $\varepsilon=-L\dfrac{dI}{dt}$, so $L=\dfrac{\varepsilon\,dt}{dI}$. (NCERT Electromagnetic Induction.)

   Solution

   EMF $[\varepsilon]=\dfrac{\text{energy}}{\text{charge}}=\dfrac{ML^2T^{-2}}{AT}=ML^2T^{-3}A^{-1}$.

   Then $[L]=\dfrac{[\varepsilon][t]}{[I]}=\dfrac{ML^2T^{-3}A^{-1}\cdot T}{A}=ML^2T^{-2}A^{-2}.$

   [!ANSWER] (B)

6. GUJCET 2023Set 091 markMCQ
   Q.The earth takes 24 h to rotate once about its axis. How much time does the Sun takes to shift by 1 minute viewed from the earth. (A) 4 minutes (B) 40 s (C) 4 s (D) 40 minutes

   ›Reveal solutionSolution

   [!TLDR]

   The Sun moves 1° in 4 minutes, so 1 arcminute of apparent shift takes 4 seconds.

   Concept

   Earth's rotation makes the Sun appear to sweep $360^\circ$ in 24 hours; a small angular shift corresponds to a proportional time.

   Solution

   Angular rate of the Sun's apparent motion:

   $\frac{360^\circ}{24\ \text{h}} = 15^\circ/\text{h} = \frac{1^\circ}{4\ \text{min}}$

   So the Sun shifts $1^\circ$ in 4 minutes. A shift of 1 minute of arc is $\frac{1}{60}$ of a degree:

   $t = \frac{1}{60}\times 4\ \text{min} = \frac{4}{60}\ \text{min} = 4\ \text{s}$

   [!ANSWER]

   (C) 4 s

7. GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
   Q.The dimensional formula of electric flux is ___.

   (a) M^1 L^-3 T^-3 A^-1

   (b) M^1 L^3 T^-3 A^-1

   (c) M^-1 L^3 T^-3 A^-1

   (d) M^1 L^3 T^3 A^-1

   ›Reveal solutionSolution

   Flux = E x area, i.e. (V/m) x m^2 = volt-metre; expressing volt in base units gives M L^3 T^-3 A^-1.

   Electric flux Phi = E x A, with units (V/m)(m^2) = V.m.

   Volt = J/C = (kg m^2 s^-2)/(A s) = kg m^2 s^-3 A^-1, i.e. M L^2 T^-3 A^-1.

   So V.m = M L^2 T^-3 A^-1 x L = M L^3 T^-3 A^-1.

   ✓Final answer

   (b) M^1 L^3 T^-3 A^-1.

8. GUJCET 2021Set 151 markMCQ
   Q.Which of the following option gives the Dimensional Formula of Electrical Potential? (A) $[M^{-1} L^{2} T^{-3} A^{1}]$ (B) $[M^{0} L^{3} T^{3} A^{-1}]$ (C) $[M^{-1} L^{-2} T^{-4} A^{2}]$ (D) $[M^{1} L^{2} T^{-3} A^{-1}]$

   ›Reveal solutionSolution

   Electric potential is energy per unit charge.

   Concept:

   $[V]=\frac{[\text{energy}]}{[\text{charge}]}=\frac{ML^2T^{-2}}{AT}=M^{1}L^{2}T^{-3}A^{-1}.$

   ✓Final answer

   (D) $[M^{1} L^{2} T^{-3} A^{-1}]$

   ANSWER: (D)

9. GUJCET 2020Set 071 markMCQ
   Q.The earth rotates on its axis takes 24 hours to complete one revolution. How much time it takes at sun from earth to have shift of $1^\circ$? (A) 4 sec. (B) 4 hrs. (C) 4 min. (D) 24 hrs.

   ›Reveal solutionSolution

   $24$ hours per $360^\circ$ gives $4$ minutes per degree.

   Concept: Earth turns $360^\circ$ in $24$ h, so the angular rate is $360^\circ/24\,\text{h}=15^\circ/\text{h}$.

   $t=\frac{24\ \text{h}}{360}=\frac{24\times60\ \text{min}}{360}=4\ \text{min}$

   ✓Final answer

   (C) 4 min.

   ANSWER: (C)

10. GUJCET 2020Set 071 markMCQ
    Q.Which is not the unit of Inductance? (A) H (B) $V\cdot s\cdot A^{-1}$ (C) $Wb\,A^{-1}$ (D) $Wb\cdot s\cdot A^{-1}$

    ›Reveal solutionSolution

    $1\,\text{H}=\text{Wb/A}=\text{V·s/A}$; $\text{Wb·s·A}^{-1}$ has an extra second and is not inductance.

    Concept — units of inductance. From $\varepsilon=-L\,dI/dt$ and $\Phi=LI$: $L=\Phi/I$, so henry $=\text{Wb/A}=\text{V·s/A}$. Options A, B, C are all equal to the henry; option D, $\text{Wb·s·A}^{-1}=\text{H·s}$, has dimensions of henry-second, not inductance.

    ✓Final answer

    (D) $Wb\cdot s\cdot A^{-1}$

    ANSWER: (D)

11. GUJCET 2019Set 131 markMCQ
    Q.The dimensional formula of effective torsional constant of spring is.......... (A) $M^0L^0T^0$ (B) $M^1L^2T^{-2}$ (C) $M^1L^2T^{-2}A^{-2}$ (D) $M^1L^2T^{-3}$

    ›Reveal solutionSolution

    Torsional constant has the dimensions of torque, $M^1L^2T^{-2}$.

    Concept: For a torsion spring, restoring torque $\tau = C\theta$, so $C = \tau/\theta$. Angle $\theta$ is dimensionless, hence $C$ has the dimensions of torque (energy).

    Steps:

    • Torque dimension: $[\tau] = M L^2 T^{-2}$.
    • $\theta$ dimensionless.
    • $[C] = M^1 L^2 T^{-2}$.

    ✓Final answer

    Option (B) — $M^1L^2T^{-2}$

    ANSWER: (B)

12. GUJCET 2019Set 131 markMCQ
    Q.The dimensional formula of JWL is ................. Take Q as the dimension of charge. (A) $M^1L^2T^1Q^{-2}$ (B) $M^1L^2T^{-1}Q^{-2}$ (C) $M^1L^{-2}T^{-1}Q^{-2}$ (D) $M^{-1}L^2T^{-1}Q^{-2}$

    ›Reveal solutionSolution

    The dimensional formula asked for is that of resistance (equivalently inductive reactance $\omega L$): $M^1L^2T^{-1}Q^{-2}$.

    Concept: Resistance $R=\dfrac{V}{I}$. Expressing potential and current in terms of charge $Q$: $V=\dfrac{\text{energy}}{Q}=\dfrac{ML^2T^{-2}}{Q}$ and $I=\dfrac{Q}{T}$.

    Steps:

    • $[R]=\dfrac{[V]}{[I]}=\dfrac{ML^2T^{-2}Q^{-1}}{Q\,T^{-1}}$.
    • $=ML^2T^{-2}\cdot T\cdot Q^{-2}=M^1L^2T^{-1}Q^{-2}$.
    • This is the only option carrying the resistance/reactance signature.

    ✓Final answer

    Option (B) — $M^1L^2T^{-1}Q^{-2}$

    ANSWER: (B)