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Physics · chapter PYQs

Units and Measurements, on the exam screen.

9 questions from NEET 2026, Re-NEET 2026, NEET 2020, 9 minutes, marked +4 / −1. See them with solutions.

Units and Measurements · PYQs

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  1. Question 1 · NEET 2026 · Q1 · +4 / −1

    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:

    Answer and explanation

    Answer: (4)

    In the new system the speed of light itself is 1 unit, so distance = speed ×\times time = 1×t1 \times t. Converting the time, 6 min 40 s = 360 + 40 = 400 s, giving a distance of 400 new units, so (4) is correct. Options 3×1083\times10^{8} and 3×10103\times10^{10} wrongly re-insert the SI value of c even though c is now unity, and 500 comes from mis-converting the time.

  2. Question 2 · NEET 2026 · Q14 · +4 / −1

    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−3X \times 10^3\ \text{kg m}^{-3} where the value of XX is:

    Answer and explanation

    Answer: (4)

    Density = mass/volume = 5.580(9.0)3\dfrac{5.580}{(9.0)^3} which numerically gives about 7.654×103 kg m−37.654 \times 10^3\ \text{kg m}^{-3}. The side is measured to only two significant figures (9.0 cm), so the final answer must be rounded to two significant figures, giving 7.7×1037.7 \times 10^3. Hence X = 7.7 (option (4)); the other values retain more significant figures than the data justify.

  3. Question 3 · NEET 2026 · Q25 · +4 / −1

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

    Answer and explanation

    Answer: (3)

    Least count = 1 main scale division (MSD) - 1 VSD. Since 20 VSD = 16 MSD, one VSD equals 1620\dfrac{16}{20} MSD, so L.C. = (1−1620)\left(1 - \dfrac{16}{20}\right) MSD = 420\dfrac{4}{20} MSD = 0.2×10.2 \times 1 mm = 0.2 mm = 0.02 cm (option (3)). Forgetting to convert millimetres to centimetres gives the distractor 0.2 cm.

  4. Question 4 · Re-NEET 2026 · Q26 · +4 / −1

    One main scale division of a Vernier calliper is equal to 1 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 4th Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be 1 cm, the actual length of the wire is :

    Answer and explanation

    Answer: (4)

    NTA dropped this question because no printed option matched the correct answer; option (4) is corrected here to 1.06 cm. Working: least count =1 MSDN=1 mm10=0.1 mm= \frac{1\,\text{MSD}}{N} = \frac{1\,\text{mm}}{10} = 0.1\,\text{mm}. The Vernier zero lies to the left of the main-scale zero, so the zero error is negative. With the 4th Vernier division coinciding, zero error =−(N−n)×LC=−(10−4)(0.1)=−0.6 mm= -(N-n)\times LC = -(10-4)(0.1) = -0.6\,\text{mm}. Therefore true length =observed reading−zero error=10.0−(−0.6)=10.6 mm=1.06 cm= \text{observed reading} - \text{zero error} = 10.0 - (-0.6) = 10.6\,\text{mm} = 1.06\,\text{cm}. NCERT Class 11, Units and Measurements (least count and instrumental error).

  5. Question 5 · Re-NEET 2026 · Q37 · +4 / −1

    Consider that σs\sigma_s, kBk_B, bb represents Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of σskB−1b\sigma_s k_B^{-1} b is

    Answer and explanation

    Answer: (1)

    Take each constant from its defining equation. Stefan's law E=σsT4E = \sigma_s T^4 makes σs\sigma_s a power per unit area per K4K^4, so [σs]=[MT−3K−4][\sigma_s] = [MT^{-3}K^{-4}]. Boltzmann's constant is an energy per kelvin, [kB]=[ML2T−2K−1][k_B] = [ML^2T^{-2}K^{-1}], and Wien's constant is a wavelength times a temperature, [b]=[LK][b] = [LK]. Multiplying, [MT−3K−4][M−1L−2T2K][LK]=[L−1T−1K−2][MT^{-3}K^{-4}][M^{-1}L^{-2}T^{2}K][LK] = [L^{-1}T^{-1}K^{-2}]. NCERT Class 11, Units and Measurements; the constants themselves come from Thermal Properties of Matter.

  6. Question 6 · NEET 2020 · Q159 · +4 / −1

    Dimensions of stress are

    Answer and explanation

    Answer: (3)

    Stress is defined as force per unit area, with dimensions [FA]=[MLT−2L2]=[ML−1T−2]\left[ \frac{F}{A} \right] = \left[ \frac{M L T^{-2}}{L^2} \right] = [M L^{-1} T^{-2}]. Chapter: Units and Measurements.

  7. Question 7 · NEET 2020 · Q160 · +4 / −1

    The energy required to break one bond in DNA is 10−20^{-20} J. This value in eV is nearly

    Answer and explanation

    Answer: (2)

    To convert joules to electron-volts, use the conversion factor 1 eV=1.6×10−19 J1\ eV = 1.6 \times 10^{-19}\ J. For 10−20 J10^{-20}\ J, the energy in eV is 10−201.6×10−19≈0.06 eV\frac{10^{-20}}{1.6 \times 10^{-19}} \approx 0.06\ eV. Chapter: Units and Measurements.

  8. Question 8 · NEET 2020 · Q171 · +4 / −1

    Taking into account of the significant figures, what is the value of 9.99 m - 0.0099 m?

    Answer and explanation

    Answer: (1)

    Subtracting the values, 9.99\ m - 0.0099\ m = 9.9801\ m. Considering significant figures, the result should be rounded to three significant figures, giving 9.98\ m. Chapter: Units and Measurements.

  9. Question 9 · NEET 2020 · Q176 · +4 / −1

    A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is :

    Answer and explanation

    Answer: (2)

    The least count of a screw gauge is given by LC=pitchnumber of divisionsLC = \frac{\text{pitch}}{\text{number of divisions}}. Given LC=0.01 mmLC = 0.01\ \text{mm} and 50 divisions, pitch = LC × number of divisions = 0.01×50=0.5 mm0.01 \times 50 = 0.5\ \text{mm}. Chapter: Units and Measurements.