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 155 questions; sign in to attempt the remaining 150 in the exam simulator.

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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 2023 (13 Apr, Shift 1)Energy
   A body of mass $(5\pm 0.5)kg$ is moving with a velocity of $(20\pm 0.4)m{s}^{-1}.$ Its kinetic energy will be
   1. A.$(1000\pm 0.14)J$
   2. B.$(500\pm 0.14)J$
   3. C.$(500\pm 140)J$
   4. D.$(1000\pm 140)J$
   Show answer & solution

   Answer: (D)

   The kinetic energy is given by, $K=\dfrac{1}{2}m{v}^{2}$ Differentiating the above equation, $dK=\dfrac{1}{2}({v}^{2}dm+2mvdv)$ Dividing by $K$, $\Rightarrow \dfrac{dK}{K}=\dfrac{dm}{m}+\dfrac{2dv}{v}$ $\Rightarrow \dfrac{dK}{1000}=\dfrac{0.5}{5}+\dfrac{2\times 0.4}{20}$ $\Rightarrow dK=100+40=140$ $\Rightarrow KE=(1000\pm 140)J$
4. Q4JEE Main 2022 (27 Jun, Shift 2)Errors of Measurement
   The distance of the Sun from earth is $1.5\times {10}^{11}m$ and its angular diameter is $2000''$ when observed from the earth. The diameter of the Sun will be
   1. A.$2.45\times {10}^{10}m$
   2. B.$1.45\times {10}^{10}m$
   3. C.$1.45\times {10}^{9}m$
   4. D.$0.14\times {10}^{9}m$
   Show answer & solution

   Answer: (C)

   Diameter of Sun is $D=r\theta$ $\begin{matrix}=1.5\times {10}^{11}\times (2000)\times \left(\dfrac{1}{3600}\right)\times \left(\dfrac{\pi }{180}\right) \\ =1.45\times {10}^{9}m\end{matrix}$
5. Q5JEE Main 2020 (05 Sep, Shift 1)Basics of Mechanical Waves
   Assume that the displacement (s) of air is proportional to the pressure difference $\left(\Delta p\right)$ created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air$\left(\rho \right)$ and the frequency (f). If $\Delta p~10Pa,n~300m/s,p~1kg/{m}^{3}$ $f~1000Hz$, then s will be of the order of (take the multiplicative constant to be 1 )
   1. A.$\dfrac{3}{100}mm$
   2. B.$10mm$
   3. C.$\dfrac{1}{10}mm$
   4. D.$1mm$
   Show answer & solution

   Answer: (A)

   $S=\dfrac{{P}_{0}}{BK}$ $=\dfrac{{P}_{0}}{P{V}^{2}\dfrac{\omega }{V}}$ $=\dfrac{{P}_{0}}{P{V}_{\omega }}$ $=\dfrac{10}{1\times 1000\times 300}m$ $≃\dfrac{3}{100}mm$