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. COMEDK 2026Set 2026-A1 markMCQ
   Q.The dimensional formula for specific resistance is: (A) $\left[M L^3 T^3 A^2\right]$ (B) $\left[M L^3 T^{-3} A^{-2}\right]$ (C) $\left[M L^{-3} T^{-2} A^{-2}\right]$ (D) $\left[M L^3 T^{-3} A^2\right]$

   ›Reveal solutionSolution

   Specific resistance (resistivity) is derived from resistance using $R = \rho \frac{L}{A}$, so its dimensional formula is $[M L^3 T^{-3} A^{-2}]$, which corresponds to option (B).

   Concept & Intuition

   Specific resistance, or resistivity ($\rho$), is a material property that quantifies how strongly it opposes current flow. The key is to start from something familiar: resistance $R$, which obeys Ohm’s law $V = IR$. We know the dimensions of voltage ($V$), current ($I$), and length ($L$), and the relation $R = \rho \frac{L}{A}$ ties them together. By finding the dimensions of $R$ first, then isolating $\rho$, we get the answer cleanly.

   Step-by-step derivation

   1. Recall the formula linking resistance and resistivity For a uniform conductor of length $L$ and cross-sectional area $A$,

   $R = \rho \frac{L}{A}.$

   So resistivity is

   $\rho = R \cdot \frac{A}{L}.$

   If we find the dimensions of $R$, $A$, and $L$, we can combine them.

   2. Find the dimensions of resistance $R$ from Ohm’s law Ohm’s law: $V = I R$, so $R = V / I$.
      • Current $I$ has dimension $[A]$ (amperes).
      • Voltage $V$ is work per unit charge: $V = W / Q$. Work $W$ has dimensions of energy: $[M L^2 T^{-2}]$. Charge $Q = I \cdot t$, so $[Q] = [A T]$. Hence $[V] = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$.
      • Therefore,

   $[R] = \frac{[V]}{[I]} = \frac{[M L^2 T^{-3} A^{-1}]}{[A]} = [M L^2 T^{-3} A^{-2}].$

   3. Combine with area and length From $\rho = R \cdot \frac{A}{L}$:
      • Area $A$ has dimension $[L^2]$.
      • Length $L$ has dimension $[L]$. So

   $[\rho] = [M L^2 T^{-3} A^{-2}] \cdot \frac{[L^2]}{[L]} = [M L^2 T^{-3} A^{-2}] \cdot [L] = [M L^3 T^{-3} A^{-2}].$

   4. Match with the options The result $[M L^3 T^{-3} A^{-2}]$ is exactly option (B).

   Watch out

   A common mistake is to forget that voltage involves charge, not current directly, leading to an incorrect exponent on $T$ or $A$. Always break $V$ into work per charge.

   Tip

   You can also remember that resistance has dimensions $[M L^2 T^{-3} A^{-2}]$ (like $V/I$), and since $\rho = R \times \text{(length)}$, the $L$ exponent increases by 1, giving $L^3$.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. COMEDK 2026Set 2026-A1 markMCQ
   Q.In the equation $X=G^{-1 / 2} h^{1 / 2} c^{5 / 2}$, where G- universal gravitation constant, $h$ - Planck's constant and c - velocity of light, the dimensions of X are that of (A) Stress (B) Energy (C) Upthrust (D) Momentum

   ›Reveal solutionSolution

   Substituting the dimensions of $G$, $h$ and $c$ into $X=G^{-1/2}h^{1/2}c^{5/2}$ and adding exponents gives $[X]=M L^{2}T^{-2}$, the dimension of energy — option (B).

   Concept

   Every physical constant carries a dimensional formula in terms of mass $M$, length $L$ and time $T$. Raising each to its power in the expression and summing the exponents base by base gives the dimensions of $X$:

   $[G]=M^{-1}L^{3}T^{-2},\qquad [h]=M L^{2}T^{-1},\qquad [c]=L T^{-1}.$

   Solution

   1. Apply each exponent:

   $[G^{-1/2}]=\big(M^{-1}L^{3}T^{-2}\big)^{-1/2}=M^{1/2}L^{-3/2}T^{1},$

   $[h^{1/2}]=\big(M L^{2}T^{-1}\big)^{1/2}=M^{1/2}L^{1}T^{-1/2},$

   $[c^{5/2}]=\big(L T^{-1}\big)^{5/2}=L^{5/2}T^{-5/2}.$

   2. Multiply and collect exponents:

   • Mass: $\tfrac12+\tfrac12=1.$
   • Length: $-\tfrac32+1+\tfrac52=2.$
   • Time: $1-\tfrac12-\tfrac52=-2.$

   $[X]=M^{1}L^{2}T^{-2}.$

   3. Identify: $M L^{2}T^{-2}$ is the dimension of energy (e.g. $E=\tfrac12 mv^{2}\Rightarrow M(LT^{-1})^{2}=M L^{2}T^{-2}$). Stress would be $M L^{-1}T^{-2}$, momentum $M L T^{-1}$ — neither matches.

   So $X$ has the dimensions of energy, option (B).

   Watch out

   A frequent slip is mis-adding the length exponents: $-\tfrac32+1+\tfrac52=2$, not $0$. Getting $L^{0}$ wrongly leads to the "momentum" trap.

   ✓Final answer

   The correct option is (B): $[X]=M L^{2}T^{-2}$ (energy).

3. KCET 2025Set D-41 markMCQ
   Q.Match the following types of nuclei with examples shown Column-I \hspace{1cm} Column-II \ A. Isotopes \hspace{1cm} i. $\ce{Li^7}$, $\ce{Be^7}$ \ B. Isobars \hspace{1cm} ii. $\ce{O^{18}}$, $\ce{F^{19}}$ \ C. Isotopes \hspace{1cm} iii. $\ce{H^1}$, $\ce{H^2}$ (A) A-ii, B-iii, C-i (B) A-i, B-iii, C-ii (C) A-iii, B-ii, C-i (D) A-iii, B-i, C-ii

   ›Reveal solutionSolution

   Classify each nuclide pair by what it holds constant — proton number (isotopes), mass number (isobars), or neutron number (isotones) — and match.

   Step 1 — The three definitions.

   For a nuclide $^{A}_{Z}X$, with $Z$ = proton number, $A$ = mass number and $N = A - Z$ = neutron number:

   Family	Same	Different
   Isotopes	$Z$ (same element)	$A$ (and hence $N$)
   Isobars	$A$	$Z$ and $N$
   Isotones	$N$	$Z$ and $A$

   (Column-I lists "Isotopes" twice; the third entry, C, must be Isotones — otherwise two rows would be identical and only one Column-II pair could satisfy them. The three Column-II pairs are exactly one of each family, which confirms the reading.)

   Step 2 — Analyse each Column-II pair.

   (i) $\mathrm{Li^{7}}$ and $\mathrm{Be^{7}}$

   • $\mathrm{^{7}_{3}Li}$: $Z = 3$, $A = 7$, $N = 4$
   • $\mathrm{^{7}_{4}Be}$: $Z = 4$, $A = 7$, $N = 3$

   Same mass number ($A = 7$), different $Z$ $\Rightarrow$ ISOBARS.

   (ii) $\mathrm{O^{18}}$ and $\mathrm{F^{19}}$

   • $\mathrm{^{18}_{8}O}$: $Z = 8$, $A = 18$, $N = 18 - 8 = \mathbf{10}$
   • $\mathrm{^{19}_{9}F}$: $Z = 9$, $A = 19$, $N = 19 - 9 = \mathbf{10}$

   Different $Z$, different $A$, but the same neutron number ($N = 10$) $\Rightarrow$ ISOTONES.

   (iii) $\mathrm{H^{1}}$ and $\mathrm{H^{2}}$

   • $\mathrm{^{1}_{1}H}$ (protium): $Z = 1$, $A = 1$, $N = 0$
   • $\mathrm{^{2}_{1}H}$ (deuterium): $Z = 1$, $A = 2$, $N = 1$

   Same element / same $Z$, different $A$ $\Rightarrow$ ISOTOPES.

   Step 3 — Assemble the match.

   $\text{A (Isotopes)} \to \text{iii}, \qquad \text{B (Isobars)} \to \text{i}, \qquad \text{C (Isotones)} \to \text{ii}$

   That is A-iii, B-i, C-ii, which is option (D).

   Option (C) (A-iii, B-ii, C-i) gets isotopes right but then wrongly calls the $\mathrm{O^{18}/F^{19}}$ pair isobars — their mass numbers 18 and 19 clearly differ, so it fails immediately.

   ✓Final answer

   The correct option is (D) — A-iii, B-i, C-ii.

   ANSWER: D

4. COMEDK 2025Set 2025-A1 markMCQ
   Q.What is the dimensional formula for electric flux? (A) $\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]$ (B) $\left[\mathrm{M}^2 \mathrm{~L}^3 \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$ (C) $\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$ (D) $\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$

   ›Reveal solutionSolution

   Electric flux is defined as $\Phi_E = \mathbf{E} \cdot \mathbf{A}$, so its dimension is the product of electric field and area. The electric field has dimensions $[M^1 L^1 T^{-3} A^{-1}]$, and area has $[L^2]$, giving electric flux dimensions $[M^1 L^3 T^{-3} A^{-1}]$, which corresponds to option (D).

   The key is to recall that electric flux measures the "flow" of electric field through a surface. Since it's the dot product of electric field $\mathbf{E}$ and area vector $\mathbf{A}$, its dimension is simply the product of their dimensions.

   1. Start with the definition

      Electric flux $\Phi_E = \mathbf{E} \cdot \mathbf{A}$. So $[\Phi_E] = [E] \cdot [A]$.

   2. Find the dimension of electric field $E$

      From Coulomb’s law: $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$, so $E = \frac{F}{q}$.

      Force $F$ has dimensions $[M L T^{-2}]$, and charge $q$ has dimensions $[A T]$ (since current $I = q/t$).

      Thus $[E] = \frac{[M L T^{-2}]}{[A T]} = [M^1 L^1 T^{-3} A^{-1}]$.

   3. Find the dimension of area $A$

      Area is length squared: $[A] = [L^2]$.

   4. Multiply them

      $[\Phi_E] = [M^1 L^1 T^{-3} A^{-1}] \times [L^2] = [M^1 L^3 T^{-3} A^{-1}]$.

   Tip

   A quick check: electric flux can also be written as $\Phi_E = \int \mathbf{E} \cdot d\mathbf{A}$, and in SI units its unit is $\text{V}\cdot\text{m}$. Since volt $= \frac{\text{J}}{\text{C}} = \frac{M L^2 T^{-2}}{A T} = M L^2 T^{-3} A^{-1}$, multiplying by meter gives $M L^3 T^{-3} A^{-1}$ — same result.

   Watch out

   A common mistake is to confuse electric flux with electric field itself. Electric field has $[M L T^{-3} A^{-1}]$, but flux includes an extra factor of area, so it gains $L^2$, not $L^1$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. COMEDK 2025Set 2025-A1 markMCQ
   Q.In the expression $\mathrm{P}=\mathrm{El}^2 \mathrm{~m}^{-5} \mathrm{G}^{-2}$ where $\mathrm{E}, 1, \mathrm{~m}$ and G represent Energy, Angular Momentum, Mass and Gravitational Constant, the dimensions of P are (A) $\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$ (B) $\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$ (C) $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$ (D) $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-2}\right]$

   ›Reveal solutionSolution

   The key idea is to substitute the fundamental dimensions of energy, angular momentum, mass, and the gravitational constant into the given expression and simplify. The result shows that P is dimensionless, so the correct option is (C).

   We are given:

   $\mathrm{P} = \mathrm{E} \, l^{2} \, \mathrm{m}^{-5} \, \mathrm{G}^{-2}$

   where $\mathrm{E}$ = energy, $l$ = angular momentum, $\mathrm{m}$ = mass, and $\mathrm{G}$ = gravitational constant. We need the dimensions of P.

   ---

   Why this approach works

   Every physical quantity can be expressed in terms of the fundamental dimensions: mass $[M]$, length $[L]$, and time $[T]$. By writing each given quantity in these dimensions, we can combine them algebraically to find the dimensions of P. If all dimensions cancel, P is dimensionless.

   ---

   Step-by-step reasoning

   1. Dimensions of energy (E) Energy = work = force × distance. Force = mass × acceleration, so:

   $[E] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$

   2. Dimensions of angular momentum (l) Angular momentum = moment of inertia × angular velocity, or more directly:

   $l = \text{mass} \times \text{velocity} \times \text{radius}$

   Velocity has dimensions $[L T^{-1}]$, so:

   $[l] = [M] \times [L T^{-1}] \times [L] = [M L^2 T^{-1}]$

   3. Dimensions of mass (m) Simply:

   $[m] = [M]$

   4. Dimensions of gravitational constant (G) From Newton’s law: $F = G \frac{m_1 m_2}{r^2}$. Rearranging:

   $G = \frac{F r^2}{m_1 m_2}$

   Force has dimensions $[M L T^{-2}]$, so:

   $[G] = \frac{[M L T^{-2}] [L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$

   5. Combine into the expression for P

   $[P] = [E] \times [l]^2 \times [m]^{-5} \times [G]^{-2}$

   Substitute each:

   $[P] = [M^1 L^2 T^{-2}] \times [M^2 L^4 T^{-2}] \times [M^{-5}] \times [M^{2} L^{-6} T^{4}]$

   (Note: $[l]^2 = [M^2 L^4 T^{-2}]$ and $[G]^{-2} = [M^{2} L^{-6} T^{4}]$)

   6. Simplify exponents
      • Mass: $1 + 2 - 5 + 2 = 0$
      • Length: $2 + 4 - 6 = 0$
      • Time: $-2 - 2 + 4 = 0$ So:

   $[P] = [M^0 L^0 T^0]$

   Thus P is dimensionless.

   Tip

   A quick check: if you ever suspect a combination might be dimensionless, look for known dimensionless constants. Here, the structure resembles a ratio that cancels all base units — a common trick in dimensional analysis problems.

   Watch out

   A common mistake is forgetting that angular momentum has dimensions $[M L^2 T^{-1}]$, not $[M L^2 T^{-2}]$ (which is energy). Mixing these up would give a wrong exponent for time.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. COMEDK 2025Set 2025-E1 markMCQ
   Q.The unit of universal gravitational constant is : (A) $\mathrm{N} \mathrm{m}^2 \mathrm{~kg}^2$ (B) $\mathrm{N} \mathrm{m}^{-2} \mathrm{~kg}^{-2}$ (C) $\mathrm{N} \mathrm{m}^{-2} \mathrm{~kg}^2$ (D) $\mathrm{N} \mathrm{m}^2 \mathrm{~kg}^{-2}$

   ›Reveal solutionSolution

   The universal gravitational constant $G$ appears in Newton’s law $F = G \frac{m_1 m_2}{r^2}$. Solving for $G$ gives units of $\mathrm{N\,m^2\,kg^{-2}}$, so the correct choice is (D).

   The key is to remember that units must balance in any physical equation. Newton’s law of gravitation tells us the force between two masses depends on their product and the inverse square of the distance. The constant $G$ is the proportionality factor that makes the numbers work — so its units are whatever is needed to turn the right‑hand side into newtons.

   1. Start with the defining equation Newton’s law:

   $F = G\,\frac{m_1 m_2}{r^2}$

   Here $F$ is force (in newtons, N), $m_1$ and $m_2$ are masses (in kg), and $r$ is distance (in m).

   2. Rearrange to isolate $G$

   $G = \frac{F\,r^2}{m_1 m_2}$

   3. Substitute the base SI units

      • Force $F$ has units of $\mathrm{N}$ (newton).
      • Distance $r$ has units of $\mathrm{m}$.
      • Masses have units of $\mathrm{kg}$.

      So:

   $\text{Units of } G = \frac{(\mathrm{N})\,(\mathrm{m}^2)}{(\mathrm{kg})(\mathrm{kg})} = \mathrm{N\,m^2\,kg^{-2}}$

   4. Match with the options
      • (A) $\mathrm{N\,m^2\,kg^2}$ → mass squared in numerator, wrong.
      • (B) $\mathrm{N\,m^{-2}\,kg^{-2}}$ → distance in denominator, wrong.
      • (C) $\mathrm{N\,m^{-2}\,kg^2}$ → both distance and mass inverted, wrong.
      • (D) $\mathrm{N\,m^2\,kg^{-2}}$ → exactly what we derived.

   Tip

   A quick sanity check: if $G$ had $\mathrm{kg^2}$ in the numerator, doubling both masses would quadruple the force without changing $G$ — but the law says force quadruples anyway, so $G$ must stay constant. That’s only possible if the mass units cancel properly, which requires $\mathrm{kg^{-2}}$.

   Watch out

   A common mistake is to forget that $r^2$ is in the denominator of the original law, so when solving for $G$ it moves to the numerator. That’s why the answer has $\mathrm{m^2}$, not $\mathrm{m^{-2}}$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. COMEDK 2025Set 2025-E1 markMCQ
   Q.Two physical quantities having the same dimensional formula $\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]$ are (A) Thrust and Strain (B) Moment of force and Thrust (C) Stress and Pressure (D) Work and thrust

   ›Reveal solutionSolution

   The dimensional formula $\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]$ corresponds to pressure or stress — force per unit area. Among the options, only stress and pressure share this formula, so the correct choice is (C).

   The key idea is that dimensional analysis lets us match physical quantities by their fundamental units. Here, the given formula $\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]$ means: one power of mass, one inverse power of length, and two inverse powers of time. That’s exactly the combination you get when you take force (mass × acceleration, $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$) and divide it by area ($\left[\mathrm{L}^2\right]$). So any quantity that is “force per unit area” will have this formula. The classic examples are pressure and stress. Let’s check each option.

   1. Option (A): Thrust and Strain

      • Thrust is a force (e.g., from a rocket engine), so its dimension is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$.
      • Strain is a ratio of lengths (change in length / original length), so it is dimensionless: $\left[\mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^0\right]$.
      • They do not match the given formula. So (A) is wrong.

   2. Option (B): Moment of force and Thrust

      • Moment of force (torque) is force × distance, so its dimension is $\left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}\right]$.
      • Thrust, as above, is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$.
      • These are different from each other and from the target formula. So (B) is wrong.

   3. Option (C): Stress and Pressure

      • Stress = force / area, and pressure = force / area. Both have dimension $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right] / \left[\mathrm{L}^2\right] = \left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right]$.
      • This matches exactly. So (C) is correct.

   4. Option (D): Work and Thrust

      • Work = force × distance, dimension $\left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}\right]$.
      • Thrust is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$.
      • Neither matches the given formula. So (D) is wrong.

   Watch out

   A common mistake is to confuse thrust (a force) with pressure (force per area). Thrust has one less power of length in the denominator, so its dimension is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$, not $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right]$.

   Tip

   If you ever forget the dimension of pressure, just recall the formula $P = F/A$. Since force is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$ and area is $\left[\mathrm{L}^2\right]$, dividing gives $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right]$. Stress is defined the same way, so they share the same dimensions.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. COMEDK 2025Set 2025-M1 markMCQ
   Q.If $R$ and $L$ denote resistance and inductance of a material, then the dimension of $L R$ will be: (A) $M^2 L^4 T^{-5} A^{-4}$ (B) $M L T A^{-1}$ (C) $M^0 L^0 T^0 A^0$ (D) $M^{-1} L^4 T A^{-3}$

   ›Reveal solutionSolution

   Computing $[L]$ and $[R]$ from their defining relations and multiplying gives $M^2L^4T^{-5}A^{-4}$, matching option (A).

   Concept & Intuition

   Inductance appears in the energy stored by an inductor, $U=\tfrac12 LI^2$; resistance appears in Ohm's law, $V=IR$. Using energy ($ML^2T^{-2}$), current ($A$), and voltage as energy per charge, both dimensions follow directly.

   Step-by-step derivation

   1. Dimension of $L$.

      From $U=\tfrac12 LI^2$: $[L] = \dfrac{[U]}{[I]^2} = \dfrac{ML^2T^{-2}}{A^2} = ML^2T^{-2}A^{-2}$.

   2. Dimension of $R$.

      Voltage is energy per charge: $[V] = \dfrac{ML^2T^{-2}}{AT} = ML^2T^{-3}A^{-1}$. From $V=IR$: $[R] = \dfrac{[V]}{[I]} = ML^2T^{-3}A^{-2}$.

   3. Multiply.

   $[LR] = (ML^2T^{-2}A^{-2})(ML^2T^{-3}A^{-2}) = M^2L^4T^{-5}A^{-4}$

   Tip

   Sanity check: the inductive time constant $\tau = L/R$ has dimension of time, so $[L]=[R][T]$, hence $[LR]=[R]^2[T] = (ML^2T^{-3}A^{-2})^2 \cdot T = M^2L^4T^{-6}A^{-4}\cdot T = M^2L^4T^{-5}A^{-4}$ — same result.

   Watch out

   Don't confuse $LR$ with $L/R$: the ratio $L/R$ is dimensionless-in-time (it's a time constant), but the product $LR$ carries an extra factor of $[R]^2$.

   ✓Final answer

   The correct option is (A).

9. COMEDK 2024Set 2024-A1 markMCQ
   Q.If units of mass, length and gravitational constant are chosen to fundamental units, the dimensions of time would be (A) $\mathrm{M}^{1 / 2} \mathrm{~L}^{1 / 2} \mathrm{G}^{-1 / 2}$ (B) $\mathrm{M}^1 \mathrm{~L}^{1 / 2} \mathrm{G}^2$ (C) $\mathrm{M}^{-1 / 2} \mathrm{~L}^{3 / 2} \mathrm{G}^{-1 / 2}$ (D) $\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{G}^{-1}$

   ›Reveal solutionSolution

   We treat mass (M), length (L), and the gravitational constant (G) as fundamental dimensions, then solve for the dimension of time (T) by expressing G in terms of M, L, T and inverting the relation. The result is that time has dimensions $\mathrm{M}^{-1/2} \mathrm{L}^{3/2} \mathrm{G}^{-1/2}$, which corresponds to option (C).

   The key idea is that we are redefining the set of fundamental units. Normally, mass (M), length (L), and time (T) are fundamental, and the gravitational constant G has derived dimensions. Here, the problem says: choose M, L, and G as the fundamental units. That means we must express the dimension of time in terms of M, L, and G. So we need to find exponents $a, b, c$ such that:

   $[T] = M^a L^b G^c$

   We know the usual dimensions of G from Newton’s law of gravitation: $F = G \frac{m_1 m_2}{r^2}$. Force has dimensions $MLT^{-2}$, so:

   $[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{(MLT^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}$

   Now we treat this as an equation relating dimensions. We want to solve for $[T]$ in terms of $M, L, G$.

   1. Write the dimensional equation. We have $[T] = M^a L^b G^c$. Substitute the known dimensions of G:

   $[T] = M^a L^b \left( M^{-1} L^3 T^{-2} \right)^c$

   Simplify:

   $[T] = M^{a - c} \, L^{b + 3c} \, T^{-2c}$

   2. Match exponents on both sides.

      The left side is $T^1$, which in terms of M, L, T is $M^0 L^0 T^1$. So we equate exponents:

      • For M: $a - c = 0$
      • For L: $b + 3c = 0$
      • For T: $-2c = 1$

   3. Solve the system.

      From the T-equation: $-2c = 1 \Rightarrow c = -\frac{1}{2}$.

      Then from M: $a - (-\frac{1}{2}) = 0 \Rightarrow a = -\frac{1}{2}$.

      From L: $b + 3(-\frac{1}{2}) = 0 \Rightarrow b - \frac{3}{2} = 0 \Rightarrow b = \frac{3}{2}$.

   4. Write the result.

      So $[T] = M^{-1/2} L^{3/2} G^{-1/2}$. That is exactly option (C).

   Tip

   A quick check: If G is fundamental, then time should have a negative exponent on G because G contains $T^{-2}$ — to get $T^1$ we need $G^{-1/2}$. This immediately eliminates options (A) and (B) which have positive exponents on G.

   Watch out

   A common mistake is to forget that we are solving for T in terms of G, not the other way around. Students sometimes write $[G] = M^{-1}L^3T^{-2}$ and then try to match it to the given forms directly, instead of inverting the relation.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

10. COMEDK 2024Set 2024-E1 markMCQ
    Q.The dimension $[\mathrm{ML}^{-1} \mathrm{~T}^{-2}]$ is the physical quantity of (A) Pressure $\times$ Area (B) $\mathrm{\frac{Force}{Pressure}}$ (C) Power $\times$ Time (D) Energy density

    ›Reveal solutionSolution

    The dimension $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$ matches energy density (energy per unit volume). The correct option is (D).

    We are given the dimensional formula $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$ and asked which physical quantity it represents. The key is to recall the dimensions of common physical quantities and see which one matches exactly.

    Concept and intuition:

    Dimensions are like the "DNA" of a physical quantity — they tell us how it depends on mass (M), length (L), and time (T). If we know the dimensions of a quantity, we can identify it by comparing with known formulas. Here, $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$ looks like pressure (force per area) but let's check each option carefully.

    ---

    1. Option (A): Pressure × Area Pressure has dimensions $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$ (force per area). Multiplying by area ($[L^2]$) gives:

    $[\mathrm{ML}^{-1} \mathrm{T}^{-2}] \times [L^2] = [\mathrm{M} L^{1} \mathrm{T}^{-2}]$

    That is the dimension of force, not $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$. So (A) is incorrect.

    2. Option (B): Force / Pressure Force has dimensions $[\mathrm{MLT}^{-2}]$. Dividing by pressure $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$ gives:

    $\frac{[\mathrm{MLT}^{-2}]}{[\mathrm{ML}^{-1} \mathrm{T}^{-2}]} = [L^2]$

    That is area, not the given dimension. So (B) is incorrect.

    3. Option (C): Power × Time Power has dimensions $[\mathrm{ML}^2 \mathrm{T}^{-3}]$ (energy per time). Multiplying by time $[T]$ gives:

    $[\mathrm{ML}^2 \mathrm{T}^{-3}] \times [T] = [\mathrm{ML}^2 \mathrm{T}^{-2}]$

    That is energy (or work), not $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$. So (C) is incorrect.

    4. Option (D): Energy density Energy has dimensions $[\mathrm{ML}^2 \mathrm{T}^{-2}]$. Energy density is energy per unit volume, so divide by volume $[L^3]$:

    $\frac{[\mathrm{ML}^2 \mathrm{T}^{-2}]}{[L^3]} = [\mathrm{ML}^{-1} \mathrm{T}^{-2}]$

    This matches exactly. Energy density is also the same as pressure (in fact, pressure and energy density share the same dimensions). So (D) is correct.

    Tip

    A quick shortcut: Pressure and energy density always have the same dimensions. If you see $[\mathrm{ML}^{-1} \mathrm{T}^{-2}]$, it could be pressure, stress, or energy density. Here only energy density appears among the options.

    Watch out

    A common mistake is to confuse "energy density" with "energy" itself. Energy has $[L^2]$ in the numerator, but energy density has $[L^{-1}]$ because volume divides it.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. COMEDK 2024Set 2024-E1 markMCQ
    Q.Joule second is the unit of (A) Energy (B) Power (C) Angular momentum (D) Linear momentum

    ›Reveal solutionSolution

    Joule-second $=kg\,m^2\,s^{-1}$, the dimensions of angular momentum (and of Planck's constant).

    Energy has units $J=kg\,m^2\,s^{-2}$, so

    $J\cdot s=kg\,m^2\,s^{-1}$

    Angular momentum $L=I\omega$ has units $(kg\,m^2)(s^{-1})=kg\,m^2\,s^{-1}=J\cdot s$.

    (Energy is $J$, power $J\,s^{-1}$, linear momentum $kg\,m\,s^{-1}$ — none equal $J\cdot s$.)

    ✓Final answer

    The correct option is (C) — Angular momentum

12. COMEDK 2024Set 2024-M1 markMCQ
    Q.Find the value of '$n$' in the given equation $P=\rho^n v^2$ where '$P$' is the pressure, '$\rho$' density and '$v$' velocity. (A) $\mathrm{n}=\frac{1}{2}$ (B) $\mathrm{n}=1$ (C) $\mathrm{n}=3$ (D) $\mathrm{n}=2$

    ›Reveal solutionSolution

    The problem is solved by dimensional analysis: pressure has dimensions $[M L^{-1} T^{-2}]$, density $[M L^{-3}]$, and velocity $[L T^{-1}]$; equating exponents gives $n=1$, so the correct option is (B).

    Concept & Intuition

    When an equation relates physical quantities, the dimensions on both sides must match — this is the principle of dimensional homogeneity. Here we’re told $P = \rho^n v^2$, but we don’t yet know $n$. By writing each quantity in terms of mass ($M$), length ($L$), and time ($T$), we can solve for the exponent $n$ that makes the dimensions balance. This is a classic trick: instead of memorizing formulas, let the units guide you.

    Step-by-step reasoning

    1. Write the dimensions of each quantity

       • Pressure $P$ is force per area. Force = mass × acceleration, so $[P] = \frac{[M L T^{-2}]}{[L^2]} = M L^{-1} T^{-2}$.
       • Density $\rho$ is mass per volume: $[\rho] = M L^{-3}$.
       • Velocity $v$ is length per time: $[v] = L T^{-1}$.

    2. Set up the dimensional equation

       The given equation is $P = \rho^n v^2$. Taking dimensions:

    $[P] = [\rho]^n \cdot [v]^2$

    Substitute the dimensional forms:

    $M^1 L^{-1} T^{-2} = (M^1 L^{-3})^n \cdot (L^1 T^{-1})^2$

    3. Simplify the right-hand side

    $M^1 L^{-1} T^{-2} = M^{n} L^{-3n} \cdot L^{2} T^{-2}$

    Combine the length terms:

    $M^1 L^{-1} T^{-2} = M^{n} L^{-3n + 2} T^{-2}$

    4. Equate exponents for each base

       • For mass $M$: $1 = n$ → $n = 1$.
       • For length $L$: $-1 = -3n + 2$. Substitute $n=1$: $-1 = -3(1) + 2 = -1$, which checks.
       • For time $T$: $-2 = -2$, automatically satisfied.

    5. Conclusion

       The only consistent value is $n = 1$.

    Tip

    You could also notice that $P = \rho v^2$ is reminiscent of the dynamic pressure formula in fluid mechanics ($P = \frac{1}{2}\rho v^2$), which differs only by a dimensionless constant. That’s a quick sanity check: the exponent on density must be 1.

    Watch out

    A common mistake is to treat $v^2$ as having dimensions of $L^2 T^{-2}$ but forget that $v$ itself is $L T^{-1}$, so $v^2$ is $L^2 T^{-2}$. Also, don’t confuse pressure with energy density — they share dimensions, but here the structure is different.

    ✓Final answer

    The correct option is (B).

    ANSWER: B