Q 11-01-155JEE MainJEE Main 2022 (26 Jun, Shift 1)Medium
In a vernier calliper, each cm on the main scale is divided into $20$ equal parts. If the tenth vernier scale division coincides with the ninth main scale division, then the value of the vernier constant will be ______ $\times10^{-2}\ \text{mm}$.
Numerical value type. Enter your answer.
Answer: 5
$1\ \text{MSD} = \dfrac{10\ \text{mm}}{20} = 0.5\ \text{mm}$. $10\ \text{VSD} = 9\ \text{MSD}\Rightarrow1\ \text{VSD} = 0.9\ \text{MSD}$.
$$\text{VC} = 1\ \text{MSD} - 1\ \text{VSD} = 0.1\times0.5 = 0.05\ \text{mm} = 5\times10^{-2}\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics