In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4ᵗʰ division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is 0.04 mm then how many main scale divisions are there in 1 cm ?

Solution

Physics sir · 4 steps

Zero error gives you the least count, and the 50-to-49 rule turns that into the main scale division.

1. Least count from error. With the jaws closed, the 4th vernier division lines up. So the zero error is 4 least counts, and its size is 0.04 mm. That gives 4 × LC = 0.04 mm, so LC = 0.01 mm.
2. Least count from scales. 50 VSD = 49 MSD, so 1 VSD = 49/50 MSD. The least count is 1 MSD minus 1 VSD.

   LC = 1 MSD - 49/50 MSD = (1 MSD)/50

3. Find one MSD. 1 MSD = 50 × 0.01 mm = 0.5 mm.
4. So. 1 cm = 10 mm holds 10/0.5 = 20 main scale divisions, which is option (d).

The idea to keep. Zero error equals the number of coinciding divisions times the least count. Then LC = 1 MSD / N, where N vernier divisions match N - 1 main divisions. These two facts link everything.

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