JEE Main 2025 · 4 April · evening · Single correct · +4 / −1

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm ) :

  1. (a) 0.002
  2. (b) 0.0025
  3. (c) 0.0005
  4. (d) 0.001
Answer and solution

Answer (a) 0.002

Solution

Physics sir · 3 steps

Find one main scale division first, then the least count is 1 MSD minus 1 VSD.

  1. One main division. 300 divisions cover 15 cm, so 1 MSD = 15/300 = 0.05 cm.
  2. One vernier division. 25 vernier divisions equal 24 main divisions, so 1 VSD = 24/25 × 0.05 = 0.048 cm.

    LC = 1 MSD - 1 VSD = (1 MSD)/25 = 0.05/25 cm

  3. So. LC = 0.05 - 0.048 = 0.002 cm, which is option (a).

The idea to keep. If n vernier divisions match (n-1) main divisions, the least count is one main scale division divided by n. Always work out the size of one MSD first.

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 2025.