Q 11-01-121JEE MainJEE Main 2024 (30 Jan, Shift 2)Easy
If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of the main scale is $0.5\ \text{mm}$, the Vernier constant of the travelling microscope is:
Answer: (D) $0.01\ \text{mm}$
$1\ \text{VSD} = \dfrac{49}{50}\ \text{MSD}$
$$\text{VC} = 1\ \text{MSD} - 1\ \text{VSD} = \frac{1}{50}\times0.5\ \text{mm} = 0.01\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics