Units and Measurements NEET PYQs (2005–2026) with Solutions

10 solved Units and Measurements PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyUnits & SI System

   The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is:

   1. AOption A: 3 × 10⁸
   2. BOption B: 500
   3. COption C: 3 × 10¹⁰
   4. DOption D: 400

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   Correct answer: (D) 400

   Explanation

   Speed of light in the new unit system is taken as 1. Time taken = 6 min 40 s = 6×60 + 40 = 400 s Distance = speed × time Since v = 1, d = 1 × 400 = 400 Hence the correct answer is option (D).

2. Question 2 (NEET 2026)

   NEET 2026MediumErrors in Measurement

   Each side of a metallic cube of mass 5.580 kg is measured to be 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as X × 10³ kg m⁻³ where the value of X is:

   1. AOption A: 7.654
   2. BOption B: 7.6
   3. COption C: 7.65
   4. DOption D: 7.7

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   Correct answer: (D) 7.7

   Explanation

   Density: ρ = Mass / Volume Given: Mass = 5.580 kg Side = 9.0 cm = 9.0 × 10⁻² m Volume: V = (9.0 × 10⁻²)³ V = 729 × 10⁻⁶ m³ Density: ρ = 5.580 / (729 × 10⁻⁶) ρ = 7.654 × 10³ kg m⁻³ Applying significant figures: Mass has 4 significant figures, Side measurement (9.0 cm) has 2 significant figures. Therefore final answer must have 2 significant figures. Thus: ρ ≈ 7.7 × 10³ kg m⁻³ Hence X = 7.7 and option (D) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumErrors in Measurement

   In a vernier caliper, 20 VSD coincide with 16 MSD (each division length 1 mm). The least count of the vernier calipers is:

   1. AOption A: 0.2 cm
   2. BOption B: 0.01 cm
   3. COption C: 0.02 cm
   4. DOption D: 0.1 cm

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   Correct answer: (C) 0.02 cm

   Explanation

   20 VSD = 16 MSD ⇒ 1 VSD = 16/20 MSD. Least count = 1 MSD − 1 VSD = 1 − 16/20 = 4/20 mm = 0.2 mm = 0.02 cm. Hence option (C) is correct.

4. Question 4 (NEET 2026)

   NEET 2026MediumZero error of Vernier callipers

   One main scale division of a Vernier calliper is equal to $1\ \mathrm{mm}$ and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{\text{th}}$ Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be $1\ \mathrm{cm}$, the actual length of the wire is :

   1. AOption A: $0.60\ \mathrm{cm}$
   2. BOption B: $0.96\ \mathrm{cm}$
   3. COption C: $1.00\ \mathrm{cm}$
   4. DOption D: $1.04\ \mathrm{cm}$

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   Correct answer: (A) $0.60\ \mathrm{cm}$, (B) $0.96\ \mathrm{cm}$, (C) $1.00\ \mathrm{cm}$, (D) $1.04\ \mathrm{cm}$

   Explanation

   NTA dropped this question (no option is correct), so every candidate got the marks. Least count $= \frac{1\ \mathrm{MSD}}{10} = 0.1\ \mathrm{mm}$. With the jaws closed, the Vernier zero lies to the left of the main-scale zero, so the zero error is negative. When the $4^{\text{th}}$ Vernier division coincides, the Vernier zero is $(10 - 4) \times 0.1 = 0.6\ \mathrm{mm}$ to the left of the main zero: zero error $= -0.6\ \mathrm{mm} = -0.06\ \mathrm{cm}$. Actual length $=$ reading $-$ zero error $= 1\ \mathrm{cm} - (-0.06\ \mathrm{cm}) = 1.06\ \mathrm{cm}$. 1.06 cm is not among the options (taking the error wrongly as $-0.4\ \mathrm{mm}$ gives 1.04 cm, option (D)), which is why no option is correct.

5. Question 5 (NEET 2026)

   NEET 2026MediumDimensions of physical constants

   Consider that $\sigma_s$, $k_B$, $b$ represents Stefan-Boltzmann constant, Boltzmann constant and Wien’s displacement law constant, respectively. The dimension of $\sigma_s k_B^{-1} b$ is

   1. AOption A: $[\mathrm{L^{-1}T^{-1}K^{-2}}]$
   2. BOption B: $[\mathrm{L^{-1}K^{-2}}]$
   3. COption C: $[\mathrm{L^{-1}T^{-1}K^{-3}}]$
   4. DOption D: $[\mathrm{L^{-1}T^{-1}K^{-4}}]$

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   Correct answer: (A) $[\mathrm{L^{-1}T^{-1}K^{-2}}]$

   Explanation

   $\sigma_s$: from $\frac{P}{A} = \sigma_s T^4$, its unit is $\mathrm{W\,m^{-2}K^{-4}}$, so $[\sigma_s] = [\mathrm{MT^{-3}K^{-4}}]$. $k_B$: from $E = k_BT$, $[k_B] = [\mathrm{ML^2T^{-2}K^{-1}}]$. $b$: from $\lambda_m T = b$, $[b] = [\mathrm{LK}]$. $$[\sigma_s k_B^{-1} b] = \frac{[\mathrm{MT^{-3}K^{-4}}][\mathrm{LK}]}{[\mathrm{ML^2T^{-2}K^{-1}}]} = [\mathrm{L^{-1}T^{-1}K^{-2}}]$$ Option (A).

6. Question 6 (NEET 2025)

   NEET 2025MediumErrors in Measurement

   Consider the diameter of a spherical object being measured with the help of Vernier callipers. Suppose this 10 Vernier Scale Divisions (V.S.D.) are equal to 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at x = 0.1 cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is M = 5 cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is

   1. AOption A: 5.18 cm
   2. BOption B: 5.08 cm
   3. COption C: 4.98 cm
   4. DOption D: 5.00 cm

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   Correct answer: (C) 4.98 cm

   Explanation

   Given 10 VSD = 9 MSD and 1 MSD = 0.1 cm. Therefore, least count = 0.1 - 0.09 = 0.01 cm. Observed reading = 5 + (8 × 0.01) = 5.08 cm. Since zero of vernier is at 0.1 cm when closed, positive zero error = 0.1 cm. Corrected reading = 5.08 - 0.10 = 4.98 cm. Hence option (C) is correct.

7. Question 7 (NEET 2025)

   NEET 2025EasyErrors in Measurement

   A physical quantity $P$ is related to four observations $a$, $b$, $c$ and $d$ as follows: $$P = a^3 b^2 c \sqrt{d}$$ The percentage errors of measurement in a, b, c and d are 1%, 3%, 2%, and 4% respectively. The percentage error in the quantity P is

   1. AOption A: 10%
   2. BOption B: 2%
   3. COption C: 13%
   4. DOption D: 15%

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   Correct answer: (C) 13%

   Explanation

   For multiplication and powers, percentage errors add according to powers: $$\frac{\Delta P}{P} = 3\left(\frac{\Delta a}{a}\right) + 2\left(\frac{\Delta b}{b}\right) + \left(\frac{\Delta c}{c}\right) + \frac{1}{2}\left(\frac{\Delta d}{d}\right)$$ Substituting the given percentage errors: $$= 3(1\%) + 2(3\%) + 2\% + \frac{1}{2}(4\%)$$ $$= 3\% + 6\% + 2\% + 2\%$$ $$= 13\%$$ Therefore, the correct answer is option (C).

8. Question 8 (NEET 2024)

   NEET 2024MediumDimensions & Dimensional Formulae

   The quantities which have the same dimensions as those of solid angle are:

   1. AOption A: strain and arc
   2. BOption B: angular speed and stress
   3. COption C: strain and angle
   4. DOption D: stress and angle

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   Correct answer: (C) strain and angle

   Explanation

   Solid angle is dimensionless. Strain is the ratio of two lengths, so it is dimensionless. Plane angle is also dimensionless. Thus, strain and angle have the same dimensions as solid angle. Hence, option (C) is correct.

9. Question 9 (NEET 2024)

   NEET 2024HardDimensional Analysis & Applications

   A force defined by F = αt² + βt acts on a particle at a given time t. The factor which is dimensionless, if α and β are constants, is:

   1. AOption A: αβt
   2. BOption B: αβ/t
   3. COption C: βt/α
   4. DOption D: αt/β

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   Correct answer: (D) αt/β

   Explanation

   Given: $$F = \alpha t^2 + \beta t$$ Both terms must have dimensions of force. Therefore: $$[\alpha] = \frac{[F]}{[t^2]} = MLT^{-4}$$ $$[\beta] = \frac{[F]}{[t]} = MLT^{-3}$$ Now, $$\frac{\alpha t}{\beta}$$ has dimensions: $$\frac{(MLT^{-4})(T)}{MLT^{-3}} = 1$$ Thus it is dimensionless. Hence, option (D) is correct.

10. Question 10 (NEET 2022)

    NEET 2022EasySignificant Figures

    The area of a rectangular field (in m²) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is:

    1. AOption A: 14 × 10²
    2. BOption B: 138 × 10¹
    3. COption C: 1382
    4. DOption D: 1382.5

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    Correct answer: (A) 14 × 10²

    Explanation

    Area = length × breadth = 55.3 × 25 = 1382.5 m² The least number of significant figures is 2 (from 25). Thus rounded value: = 1.4 × 10³ m² = 14 × 10² m² Hence option (A) is correct.