Units and Dimensions JEE Main previous year questions with solutions

5 solved JEE Main questions on Units and Dimensions, free to read — no sign-in needed. The full chapter has 148 questions; sign in to attempt the remaining 143 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Dimensions
   Consider the equation $H = \dfrac{x^p \epsilon^q E^r}{t^s}$, where $H=$ magnetic field; $E=$ electric field, $\epsilon=$ permittivity, $x=$ distance, $t=$ time. The values of $p, q, r$ and $s$ respectively are:
   1. A.$1, 1, 1, 1$
   2. B.$-1, 1, 2, 1$
   3. C.$1, -1, -2, 1$
   4. D.$-1, -2, -2, 1$
   Show answer & solution

   Answer: (A)

   The dimensional formulas for the given physical quantities are: Magnetic field intensity $[H] = [M^0 L^{-1} T^0 A^1]$ Distance $[x] = [M^0 L^1 T^0 A^0]$ Permittivity $[\epsilon] = [M^{-1} L^{-3} T^4 A^2]$ Electric field $[E] = [M^1 L^1 T^{-3} A^{-1}]$ Time $[t] = [M^0 L^0 T^1 A^0]$ Given the equation $H = \dfrac{x^p \epsilon^q E^r}{t^s}$, we can write it in terms of dimensions as: $[M^0 L^{-1} T^0 A^1] = [L]^p [M^{-1} L^{-3} T^4 A^2]^q [M L T^{-3} A^{-1}]^r [T]^{-s}$ Combining the powers of $M, L, T$, and $A$ on the right side: $[M^0 L^{-1} T^0 A^1] = [M^{-q+r} L^{p-3q+r} T^{4q-3r-s} A^{2q-r}]$ Equating the powers of corresponding fundamental quantities from both sides: For $M$: $0 = -q + r \implies q = r$ For $A$: $1 = 2q - r$ Substituting $r = q$ into the equation for $A$: $1 = 2q - q \implies q = 1$ Since $q = r$, we get $r = 1$. For $L$: $-1 = p - 3q + r$ Substituting $q = 1$ and $r = 1$: $-1 = p - 3(1) + 1 \implies -1 = p - 2 \implies p = 1$ For $T$: $0 = 4q - 3r - s$ Substituting $q = 1$ and $r = 1$: $0 = 4(1) - 3(1) - s \implies 0 = 1 - s \implies s = 1$ Thus, the values are $p = 1$, $q = 1$, $r = 1$, and $s = 1$. Answer: $1, 1, 1, 1$
2. Q2JEE Main 2025 (02 Apr, Shift 2)Units
   Match List-I with List-II. $\begin{array}{ll} \text{List-I} & \text{List-II} \\ \text{(A) Heat capacity of body} & \text{(I) } \mathrm{J} \mathrm{kg}^{-1} \\ \text{(B) Specific heat capacity of body} & \text{(II) } \mathrm{JK}^{-1} \\ \text{(C) Latent heat} & \text{(III) } \mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1} \\ \text{(D) Thermal conductivity} & \text{(IV) } \mathrm{Jm}^{-1} \mathrm{~K}^{-1} \mathrm{~s}^{-1} \end{array}$ Choose the correct answer from the options given below :
   1. A.(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
   2. B.(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
   3. C.(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
   4. D.(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
   Show answer & solution

   Answer: (D)

   $\begin{aligned} & \mathrm{C}^{\prime}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{T}}=\mathrm{JK}^{-1} \\ & \mathrm{~S}=\frac{\Delta \mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}=\mathrm{Jkg}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{~L}=\frac{\Delta \mathrm{Q}}{\mathrm{m}}=\mathrm{Jkg}^{-1} \\ & \Delta \mathrm{Q}=\frac{\mathrm{KA} \Delta \mathrm{T}}{\mathrm{L}} \Rightarrow \mathrm{K}=\frac{\Delta \mathrm{Q}(\mathrm{L})}{\mathrm{A} \Delta \mathrm{T}}=\mathrm{Jm}^{-1} \mathrm{~K}^{-1} \mathrm{~s}^{-1}\end{aligned}$
3. Q3JEE Main 2026 (06 Apr, Shift 2)Dimensions
   Match List - I with List - II.List - IList - IIA. Boltzmann constantI. $[M^{-1}L^3T^{-2}]$B. Stefan's constantII. $[ML^2T^{-1}]$C. Planck's constantIII. $[ML^2T^{-2}K^{-1}]$D. Gravitational constantIV. $[ML^0T^{-3}K^{-4}]$Choose the correct answer from the options given below :
   1. A.A-I, B-II, C-III, D-IV
   2. B.A-IV, B-III, C-II, D-I
   3. C.A-III, B-IV, C-II, D-I
   4. D.A-II, B-I, C-IV, D-III
   Show answer & solution

   Answer: (C)

   Boltzmann constant $k$: $E = k T \Rightarrow [k] = \dfrac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$ Stefan's constant $\sigma$: $\dfrac{E}{At} = \sigma T^4 \Rightarrow [\sigma] = \dfrac{[ML^2T^{-2}]}{[L^2][T][K^4]} = [ML^0T^{-3}K^{-4}]$ Planck's constant $h$: $E = h\nu \Rightarrow [h] = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ Gravitational constant $G$: $F = \dfrac{Gm_1m_2}{r^2} \Rightarrow [G] = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$ Thus, A-III, B-IV, C-II, D-I. Answer: A-III, B-IV, C-II, D-I
4. Q4JEE Main 2026 (06 Apr, Shift 1)Dimensions
   The potential energy of a particle changes with distance $x$ from a fixed origin as $V = \dfrac{A\sqrt{x}}{x + B}$, where $A$ and $B$ are constant with appropriate dimensions. The dimensions of $AB$ are _______.
   1. A.$[M^1 L^{5/2} T^{-2}]$
   2. B.$[M^{3/2} L^{5/2} T^{-2}]$
   3. C.$[M^1 L^2 T^{-2}]$
   4. D.$[M^1 L^{7/2} T^{-2}]$
   Show answer & solution

   Answer: (D)

   By the principle of dimensional homogeneity, quantities added or subtracted must have the same dimensions. In the denominator, $B$ is added to $x$ (distance). $[B] = [x] = [L]$ The dimensions of potential energy $V$ are $[M^1 L^2 T^{-2}]$. From the given equation $V = \dfrac{A\sqrt{x}}{x + B}$, we can write the dimensional formula as: $[V] = \dfrac{[A][x]^{1/2}}{[x + B]}$ $[M^1 L^2 T^{-2}] = \dfrac{[A][L]^{1/2}}{[L]}$ $[M^1 L^2 T^{-2}] = [A][L]^{-1/2}$ $[A] = [M^1 L^2 T^{-2}][L]^{1/2} = [M^1 L^{5/2} T^{-2}]$ Now, the dimensions of $AB$ are: $[AB] = [A][B] = [M^1 L^{5/2} T^{-2}][L] = [M^1 L^{7/2} T^{-2}]$ Answer: $[M^1 L^{7/2} T^{-2}]$
5. Q5JEE Main 2026 (05 Apr, Shift 2)Dimensions
   Match List-I with List-II. List-IList-IIA. Meter (L)I. $\sqrt{\dfrac{hc}{G}}$B. Second (S)II. $\sqrt{\dfrac{Gh}{c^5}}$C. Kilogram (M)III. $\sqrt{\dfrac{K^2L^2c^3}{Gh}}$D. Kelvin (K)IV. $\sqrt{\dfrac{Gh}{c^3}}$ where $h$ (Planck's constant), $G$ (gravitational constant) and $c$ (speed of light in vacuum) as fundamental units. Choose the correct answer from the options given below :
   1. A.A-II, B-IV, C-I, D-III
   2. B.A-IV, B-II, C-I, D-III
   3. C.A-IV, B-I, C-II, D-III
   4. D.A-III, B-I, C-II, D-IV
   Show answer & solution

   Answer: (B)

   The dimensional formulas for the given fundamental constants are: $[h] = M L^2 T^{-1}$ $[c] = L T^{-1}$ $[G] = M^{-1} L^3 T^{-2}$ Evaluating the dimensions of the expressions in List-II: For I: $\sqrt{\dfrac{hc}{G}} = \sqrt{\dfrac{(M L^2 T^{-1})(L T^{-1})}{M^{-1} L^3 T^{-2}}} = \sqrt{M^2} = M$ (Kilogram). Thus, C $\rightarrow$ I. For II: $\sqrt{\dfrac{Gh}{c^5}} = \sqrt{\dfrac{(M^{-1} L^3 T^{-2})(M L^2 T^{-1})}{(L T^{-1})^5}} = \sqrt{\dfrac{L^5 T^{-3}}{L^5 T^{-5}}} = \sqrt{T^2} = T$ (Second). Thus, B $\rightarrow$ II. For IV: $\sqrt{\dfrac{Gh}{c^3}} = \sqrt{\dfrac{(M^{-1} L^3 T^{-2})(M L^2 T^{-1})}{(L T^{-1})^3}} = \sqrt{\dfrac{L^5 T^{-3}}{L^3 T^{-3}}} = \sqrt{L^2} = L$ (Meter). Thus, A $\rightarrow$ IV. For III: Substituting $L^2 = \dfrac{Gh}{c^3}$ from IV into the expression gives $\sqrt{\dfrac{K^2 L^2 c^3}{Gh}} = \sqrt{K^2 \left(\dfrac{Gh}{c^3}\right) \dfrac{c^3}{Gh}} = \sqrt{K^2} = K$ (Kelvin). Thus, D $\rightarrow$ III. The correct matching is A-IV, B-II, C-I, D-III. Answer: A-IV, B-II, C-I, D-III