JEE Main 2024 · 30 January · evening · Single correct · +4 / −1

If 50 Vernier divisions are equal to 49 main scale divisions of a traveling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of traveling microscope is

  1. (a) 0.01 mm
  2. (b) 0.01 cm
  3. (c) 0.1 mm
  4. (d) 0.1 cm
Answer and solution

Answer (a) 0.01 mm

Solution

Physics sir · 3 steps

Find how much shorter one Vernier division is than one main scale division. That gap is the Vernier constant.

  1. Size of one VSD. 50 Vernier divisions cover 49 main scale divisions, so 1 VSD = 49/50 MSD.

    VC = 1 MSD - 1 VSD = (1 - 49/50)MSD = 1/50 MSD

  2. Put in numbers. One main scale division is 0.5 mm, so VC = 0.5/50 mm = 0.01 mm.
  3. So. The Vernier constant of the travelling microscope is 0.01 mm, which is option (a).

The idea to keep. Vernier constant is 1 MSD - 1 VSD. When n Vernier divisions match (n-1) main scale divisions, it is just MSD/n.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 60 times in JEE Main since 2020. This question is from 2024.