10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is :

Solution

Physics sir · 3 steps

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

1. Size of one VSD. 11 Vernier divisions cover the same length as 10 main scale divisions. So 1 VSD = 10/11 MSD.

   LC = 1 MSD - 1 VSD = (1 - 10/11)MSD = 1/11 MSD

2. Put in units. Each main scale division is 5 units, so LC = 1/11 × 5 = 5/11 units.
3. So. The least count is 5/11 units, which is option (a).

The idea to keep. If n Vernier divisions match n-1 main scale divisions, the least count is 1/n of one main scale division. Always convert MSD into real units at the end.

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