Q 11-01-071JEE MainJEE Main 2026 (24 Jan, Shift 2)Easy
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.
Answer: (A) 0.002
$1\ \text{VSD} = \dfrac{48}{50}\ \text{MSD}$.
$$\text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = \left(1 - \frac{48}{50}\right)\times0.05 = \frac{2}{50}\times0.05 = 0.002\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics