JEE Main 2026 · 24 January · evening · Single correct · +4 / −1
In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division =0.05 mm, then the least count of the vernier callipers is ____ mm.
- (a) 0.002
- (b) 0.02
- (c) 0.05
- (d) 0.005
Answer and solution
Answer (a) 0.002
Solution
Physics sir · 3 steps
Least count is 1 MSD minus 1 VSD, so first work out how big one vernier division is.
- Size of one VSD. 50 vernier divisions cover 48 main scale divisions. So 1 VSD = 48/50 MSD = 0.96 MSD.
LC = 1 MSD - 1 VSD = (1 - 0.96) MSD = 0.04 MSD
- Put in numbers. One main scale division is 0.05 mm, so LC = 0.04 × 0.05 mm = 0.002 mm.
- So. The least count of the vernier callipers is 0.002 mm, which is option (a).
The idea to keep. When n vernier divisions equal m main scale divisions, the least count is (1 - m/n) times one MSD. It is not always 1 MSD divided by n.
Answer from the NTA final answer key. Solution by PrepWiser.