Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
The dimensional formula of $\frac{1}{2}\varepsilon_0 E^2$ ($\varepsilon_0=$ permittivity of vacuum and E= electric field) is $M^aL^bT^c$. The value of $2a-b+c=$ __________.
(1) 0
(2) 1
(3) $-1$
(4) 2
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Answer: Option (2)
Key idea: Dimensional Analysis of Energy Density
Find dimensions of electric field: $[E] = \frac{[F]}{[q]} = \frac{[MLT^{-2}]}{[AT]} = [MLT^{-3}A^{-1}]$
Find dimensions of $\varepsilon_0$ from Coulomb's law: $[\varepsilon_0] = \frac{[q^2]}{[F][L^2]} = [M^{-1}L^{-3}T^4A^2]$
In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50$^{\text{th}}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is __________ mm.
(1) 5.045
(2) 5.055
(3)
(4)
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Answer: Option (1)
Key idea: Screw Gauge: Zero Error and Zero Correction
Calculate least count: $LC = \frac{\text{Pitch}}{\text{No. of divisions}} = \frac{0.1}{100} = 0.001$ mm
Find zero error: 5th division coincides when studs touch, so zero error $= +5 \times 0.001 = +0.005$ mm
In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm. If the circular scale has 100 divisions, the least count of screw gauge is ______ mm.
(1) $1 \times 10^{-2}$
(2) $1 \times 10^{-3}$
(3) $5 \times 10^{-2}$
(4) $5 \times 10^{-3}$
Show answer and solution
Answer: Option (4)
Key idea: Least Count of Screw Gauge
Given: 5 complete rotations move the screw linearly by 2.5 mm, and circular scale has 100 divisions
Calculate the pitch (linear distance per rotation): Pitch = $\frac{2.5 \text{ mm}}{5} = 0.5$ mm
Apply the least count formula: LC = $\frac{\text{Pitch}}{\text{Number of divisions}}$
Substitute values: LC = $\frac{0.5}{100} = 0.005$ mm
Express in scientific notation: LC = $5 \times 10^{-3}$ mm
Match the LIST-I with LIST-II
| List-I | List-II |
|--------|--------|
| A. Planck's constant | I. $M L^2 T^{-2}$ |
| B. Stopping potential | II. $T^{-1}$ |
| C. Work function | III. $M L^2 T^{-1}$ |
| D. Threshold frequency | IV. $M L^2 T^{-3} A^{-1}$ |
Choose the correct answer from the options given below:
(1) A-III, B-IV, C-I, D-II
(2) A-I, B-II, C-III, D-IV
(3) A-IV, B-III, C-I, D-II
(4) A-I, B-IV, C-III, D-II
Show answer and solution
Answer: Option (1)
Key idea: Dimensional Analysis of Physical Quantities in Photoelectric Effect
**Planck's constant:** From $E = h\nu$, we get $h = \frac{E}{\nu} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ → Match with III
**Stopping potential:** Voltage $= \frac{\text{Energy}}{\text{Charge}} = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$ → Match with IV
**Work function:** It is energy required to eject electron $= [ML^2T^{-2}]$ → Match with I
**Threshold frequency:** It is a frequency $= [T^{-1}]$ → Match with II
In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{th}$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has
(1) 0.07 cm negative zero error
(2) 0.7 cm negative zero error
(3) 0.07 cm positive zero error
(4) 0.7 cm positive zero error
Show answer and solution
Answer: Option (3)
Key idea: Zero Error in Vernier Calipers
Calculate Least Count: $LC = 1 ext{ MSD} - 1 ext{ VSD}$. Since $10 ext{ VSD} = 9 ext{ MSD}$, we get $1 ext{ VSD} = 0.9 ext{ mm}$
$L$, $C$ and $R$ represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $ML^2T^{-4}A^{-2}$ corresponds to ______.
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Dimensional Analysis of Electrical Quantities
Match List - I with List - II.
List - I
A. Meter (L)
B. Second (S)
C. Kilogram (M)
D. Kelvin (K)
List - II
I. $\sqrt{\frac{\hbar c}{G}}$
II. $\sqrt{\frac{G\hbar}{c^5}}$
III. $\sqrt{\frac{K^2 L^2 c^3}{G\hbar}}$
IV. $\sqrt{\frac{G\hbar}{c^3}}$
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:
The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$, length $l$ and diameter $d$. The measured values of $m$, $l$ and $d$ are $97.42 \pm 0.02$ g, $8.35 \pm 0.05$ mm and $20.20 \pm 0.02$ mm, respectively. Calculated percentage fractional error in $\rho$ is _______.
(1) 0.63%
(2) 0.82%
(3) 0.72%
(4) 0.25%
Show answer and solution
Answer: Option (2)
Key idea: Error propagation in derived quantities (density of cylinder)
Write the formula for density of cylinder: $\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}$
Apply error propagation formula: $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$ (note: power of $d$ is 2)
The potential energy of a particle changes with distance $x$ from a fixed origin as $V = \frac{A\sqrt{x}}{x+B}$, where $A$ and $B$ are constant with appropriate dimensions. The dimensions of $AB$ are _______.
(1) $[M^1 L^{5/2} T^{-2}]$
(2) $[M^{3/2} L^{5/2} T^{-2}]$
(3) $[M^1 L^2 T^{-2}]$
(4) $[M^1 L^{7/2} T^{-2}]$
Show answer and solution
Answer: Option (4)
Key idea: Dimensional Analysis
Step 1: Identify that potential energy $V$ has dimensions $[M^1 L^2 T^{-2}]$ and distance $x$ has dimensions $[L]$
Step 2: Since $(x+B)$ must be dimensionally consistent, $[B] = [L]$
Match List - I with List - II.
List - I
A. Boltzmann constant
B. Stefan's constant
C. Planck's constant
D. Gravitational constant
List - II
I. $[M^{-1}L^3T^{-2}]$
II. $[ML^2T^{-1}]$
III. $[ML^2T^{-2}K^{-1}]$
IV. $[ML^0T^{-3}K^{-4}]$
Choose the correct answer from the options given below:
(1) A-I, B-II, C-III, D-IV
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-IV, C-II, D-I
(4) A-II, B-I, C-IV, D-III
Show answer and solution
Answer: Option (3)
Key idea: Dimensional Analysis of Physical Constants
**Boltzmann constant:** From $PV = Nk_BT$, we get $k_B = \frac{PV}{NT} = \frac{[ML^{-1}T^{-2}][L^3]}{[K]} = [ML^2T^{-2}K^{-1}]$ → **III**
**Stefan's constant:** From $P = \sigma AT^4$, we get $\sigma = \frac{P}{AT^4} = \frac{[ML^2T^{-3}]}{[L^2][K^4]} = [ML^0T^{-3}K^{-4}]$ → **IV**
**Planck's constant:** From $E = h\nu$, we get $h = \frac{E}{\nu} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ → **II**
**Gravitational constant:** From $F = \frac{Gm_1m_2}{r^2}$, we get $G = \frac{Fr^2}{m_1m_2} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$ → **I**
A new unit ($\alpha$) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of $\alpha$ units if light takes 6 min. 40 s to cover this distance?
(1) $200\alpha$
(2) $400\alpha$
(3) $300\alpha$
(4) $500\alpha$
Show answer and solution
Answer: Option (2)
Key idea: Dimensional Analysis and Units; Distance = Speed × Time
Given: New unit $\alpha$ is equal to speed of light, so $\alpha = c$
Convert time: $t = 6$ min $40$ s $= 6 \times 60 + 40 = 400$ s
Use distance formula: $d = c \times t = c \times 400$ s
Since $\alpha = c$, we can substitute: $d = 400\alpha$
Answer: Option (2) $400\alpha$
FAQs
Frequently asked questions
How many questions come from Units, Dimensions and Errors in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 12 questions on it, about 1.3 per shift.
Where can I practise Units, Dimensions and Errors PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
