Units & Measurements Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Units & Measurements with step-by-step KaTeX solutions covering dimensional analysis, error analysis, and physical constants.
Fully solved JEE Main 2026 previous year questions from the Units & Measurements chapter, with concise step-by-step reasoning for every problem.
Solutions are AI-generated and pending review.
Solution
For a cylinder, $\rho = \dfrac{m}{V} = \dfrac{m}{\frac{\pi}{4} d^2 l} = \dfrac{4m}{\pi d^2 l}$.
The maximum fractional error adds contributions from each measured quantity:
$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\,\frac{\Delta d}{d} + \frac{\Delta l}{l}$$Substituting the values:
$$\frac{\Delta \rho}{\rho} = \frac{0.02}{97.42} + 2\cdot\frac{0.02}{20.20} + \frac{0.05}{8.35}$$$$= 0.000205 + 0.001980 + 0.005988 = 0.008173$$$$\frac{\Delta \rho}{\rho} \times 100 = 0.817\% \approx 0.82\%$$Answer: B ($0.82\%$)
Solution
Since $x$ and $B$ are added, $B$ must have the dimension of length:
$$[B] = [L]$$$V$ is a potential energy, so $[V] = [M L^2 T^{-2}]$. From the expression:
$$[V] = \frac{[A]\,[L]^{1/2}}{[L]} = [A]\,[L]^{-1/2}$$$$[A] = [V]\,[L]^{1/2} = [M L^2 T^{-2}][L^{1/2}] = [M L^{5/2} T^{-2}]$$Therefore:
$$[AB] = [M L^{5/2} T^{-2}]\,[L] = [M L^{7/2} T^{-2}]$$Answer: D ($[M^1 L^{7/2} T^{-2}]$)
Solution
The quantity $\dfrac{1}{2}\epsilon_0 E^2$ is the energy density of the electric field, i.e. energy per unit volume:
$$\left[\frac{1}{2}\epsilon_0 E^2\right] = \frac{[\text{Energy}]}{[\text{Volume}]} = \frac{[M L^2 T^{-2}]}{[L^3]} = [M^1 L^{-1} T^{-2}]$$So $a = 1$, $b = -1$, $c = -2$. Then:
$$2a - b + c = 2(1) - (-1) + (-2) = 2 + 1 - 2 = 1$$Answer: B ($1$)
Solution
The dimensions are $[G] = [M^{-1} L^3 T^{-2}]$ and $[h] = [M L^2 T^{-1}]$.
Write $G = h^a\, L^b\, M^c\, T^d$ and match powers using $[h] = [M L^2 T^{-1}]$:
- Mass: $-1 = a + c$
- Length: $3 = 2a + b$
- Time: $-2 = -a + d$
Taking $a = 1$: $c = -2$, $b = 3 - 2 = 1$, and $d = -2 + 1 = -1$.
$$[G] = [h^1\, T^{-1}\, L^1\, M^{-2}] = [hT^{-1}LM^{-2}]$$Answer: B ($[hT^{-1}LM^{-2}]$)
Solution
The volume of a sphere in terms of diameter $d$:
$$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{\pi}{6} d^3 \quad\Rightarrow\quad V \propto d^3$$The fractional error propagates with the power:
$$\frac{\Delta V}{V} = 3\,\frac{\Delta d}{d}$$$$\frac{\Delta V}{V}\times 100 = 3 \times 2\% = 6\%$$Answer: C ($6$)
Solution
Determine each dimensional formula:
A. Boltzmann constant — from $E = k_B T$: $[k_B] = \dfrac{[M L^2 T^{-2}]}{[K]} = [M L^2 T^{-2} K^{-1}]$ → III
B. Stefan’s constant — from $\dfrac{P}{A} = \sigma T^4$: $[\sigma] = \dfrac{[M T^{-3}]}{[K^4]} = [M L^0 T^{-3} K^{-4}]$ → IV
C. Planck’s constant — from $E = h\nu$: $[h] = \dfrac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$ → II
D. Gravitational constant — from $F = \dfrac{G m_1 m_2}{r^2}$: $[G] = \dfrac{[M L T^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$ → I
So the mapping is A-III, B-IV, C-II, D-I.
Answer: C (A-III, B-IV, C-II, D-I)
Solution
Base dimensions of the components:
$$[L] = [M L^2 T^{-2} A^{-2}], \quad [C] = [M^{-1} L^{-2} T^{4} A^{2}], \quad [R] = [M L^2 T^{-3} A^{-2}]$$Test option A, $\dfrac{R}{\sqrt{LC}}$. First $LC$:
$$[LC] = [M L^2 T^{-2} A^{-2}][M^{-1} L^{-2} T^{4} A^{2}] = [T^{2}]$$$$[\sqrt{LC}] = [T]$$Then:
$$\left[\frac{R}{\sqrt{LC}}\right] = \frac{[M L^2 T^{-3} A^{-2}]}{[T]} = [M L^2 T^{-4} A^{-2}]$$This matches the target formula exactly.
(Note: $\sqrt{LC}$ is the time constant of an $LC$ oscillator, and $R/\sqrt{LC}$ therefore carries resistance-per-time dimensions.)
Answer: A ($\frac{R}{\sqrt{LC}}$)
Solution
Use $[h] = [M L^2 T^{-1}]$, $[c] = [L T^{-1}]$, $[G] = [M^{-1} L^3 T^{-2}]$.
Planck length (A, Meter): $\sqrt{\dfrac{Gh}{c^3}}$
$$\left[\frac{Gh}{c^3}\right] = \frac{[M^{-1}L^3T^{-2}][ML^2T^{-1}]}{[L^3 T^{-3}]} = [L^2] \Rightarrow \sqrt{\cdot} = [L]$$→ IV
Planck time (B, Second): $\sqrt{\dfrac{Gh}{c^5}}$
$$\left[\frac{Gh}{c^5}\right] = \frac{[L^5 T^{-3}]}{[L^5 T^{-5}]} = [T^2] \Rightarrow \sqrt{\cdot} = [T]$$→ II
Planck mass (C, Kilogram): $\sqrt{\dfrac{hc}{G}}$
$$\left[\frac{hc}{G}\right] = \frac{[ML^2T^{-1}][LT^{-1}]}{[M^{-1}L^3T^{-2}]} = [M^2] \Rightarrow \sqrt{\cdot} = [M]$$→ I
Kelvin (D): the only expression containing $K$ is III → III
Mapping: A-IV, B-II, C-I, D-III.
Answer: B (A-IV, B-II, C-I, D-III)
Solution
The unit of length is defined as $\alpha = c$ (numerically equal to the speed of light).
Total time taken by light:
$$t = 6\ \text{min}\ 40\ \text{s} = 6\times 60 + 40 = 400\ \text{s}$$Distance travelled by light:
$$d = c \times t = c \times 400\ \text{s}$$Expressing this in units of $\alpha = c$:
$$d = \frac{c \times 400}{c}\,\alpha = 400\,\alpha$$Answer: B ($400\,\alpha$)
Solution
Use base dimensions with current $A$:
$$[H] = [L^{-1} A]\ (\text{A/m}), \quad [x] = [L], \quad [\epsilon] = [M^{-1}L^{-3}T^{4}A^{2}], \quad [E] = [MLT^{-3}A^{-1}], \quad [t] = [T]$$Test option A, $(p,q,r,s) = (1,1,1,1)$, i.e. $\dfrac{x\,\epsilon\,E}{t}$:
$$[x][\epsilon][E] = [L]\,[M^{-1}L^{-3}T^{4}A^{2}]\,[MLT^{-3}A^{-1}]$$Combine: $M$: $-1+1=0$; $L$: $1-3+1=-1$; $T$: $4-3=1$; $A$: $2-1=1$, giving $[L^{-1}T^{1}A^{1}]$.
Divide by $t$:
$$\frac{[L^{-1}T^{1}A^{1}]}{[T]} = [L^{-1}A]$$This equals $[H]$, the dimension of the magnetic field strength ($\text{A/m}$).
Answer: A ($1, 1, 1, 1$)