Units & Measurement — NEET UG practice

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Sample questions with solutions

Q1 · 2026

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

  • A.

    0.2 cm

  • B.

    0.01 cm

  • C.

    0.02 cm

  • D.

    0.1 cm

Answer: C
  1. Recall the least count formula for a vernier caliper.

LC=1 MSD−1 VSDLC = 1\,\text{MSD} - 1\,\text{VSD}

  1. Express 1 VSD in terms of MSD, since the problem states that 20 VSD equal 16 MSD.

1 VSD=1620 MSD1\,\text{VSD} = \frac{16}{20}\,\text{MSD}

  1. Substitute this into the least count formula.

LC=1 MSD−1620 MSD=420 MSD=0.2 MSDLC = 1\,\text{MSD} - \frac{16}{20}\,\text{MSD} = \frac{4}{20}\,\text{MSD} = 0.2\,\text{MSD}

  1. Convert to centimetres, since 1 MSD=1 mm1\,\text{MSD} = 1\,\text{mm}.

LC=0.2 mm=0.02 cmLC = 0.2\,\text{mm} = 0.02\,\text{cm}

Hence, the least count is 0.02 cm, matching option C.

Q2 · 2026

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 4th4^{\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 cm, the actual length of the wire is:

  • A.

    1.04 cm

  • B.

    0.60 cm

  • C.

    0.96 cm

  • D.

    1.06 cm

Answer: D
  1. Identify the least count (LC) of the vernier caliper.

Given 1 MSD=1 mm1\,\text{MSD} = 1\,\text{mm} and 10 divisions on the vernier scale, so

LC=1 MSD10=0.01 cmLC = \frac{1\,\text{MSD}}{10} = 0.01\,\text{cm}

  1. Determine the type of zero error.

Since the vernier's zero lies to the left of the main scale's zero when the jaws are closed, this is a negative zero error.

  1. Find the magnitude of the zero error using the coinciding division.

Given the 4th vernier division coincides (out of 10), therefore

Zero error=−(10−4)×LC=−(6)(0.01)=−0.06 cm\text{Zero error} = -(10-4)\times LC = -(6)(0.01) = -0.06\,\text{cm}

  1. Convert the zero error into a zero correction (opposite sign of the zero error).

Zero correction=+0.06 cm\text{Zero correction} = +0.06\,\text{cm}

  1. Add the zero correction to the observed reading to get the actual length.

Given the observed reading R=1 cmR = 1\,\text{cm}, therefore

R′=R+Zero correction=1+0.06=1.06 cmR' = R + \text{Zero correction} = 1 + 0.06 = 1.06\,\text{cm}

Q3 · 2026

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:

  • A.

    3×1083 \times 10^8

  • B.

    500

  • C.

    3×10103 \times 10^{10}

  • D.

    400

Answer: D
  1. Convert the given time into seconds, since in the new unit system the speed of light is taken as c=1c=1.

Given t=6 min 40 st = 6\,\text{min}\,40\,\text{s}, so

t=6×60+40=400 st = 6\times 60 + 40 = 400\,\text{s}

  1. Use the relation between distance, speed, and time, since distance d=c td = c\,t.

Given c=1c=1, therefore

d=1×400=400d = 1\times 400 = 400

Hence, the distance between the Sun and the Earth in the new unit is 400, matching option D.

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