If n main scale divisions coincide with (n + 1) vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into five equal parts, will be :

Solution

Physics sir · 3 steps

First find one MSD in mm, then the least count is that MSD divided by n+1.

1. One main division. One centimetre is cut into 5 equal parts, so 1 MSD = (10 mm)/5 = 2 mm.
2. One vernier division. n main divisions cover the same length as (n+1) vernier divisions. So 1 VSD = n/(n+1) MSD, which is a little shorter than one MSD.

   LC = 1 MSD - 1 VSD = (1 - n/(n+1))MSD = 1/(n+1) MSD

3. So. LC = 1/(n+1) × 2 mm = 2/(n+1) mm, which is option (a).

The idea to keep. Least count is always 1 MSD minus 1 VSD. Work out the size of one MSD in mm first, then express the VSD in terms of it.

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