In a vernier calliper, when both jaws touch each other, the zero of the vernier scale shifts towards the left and its $4^{\text{th}}$ division coincides exactly with a certain division on the main scale. If 50 vernier scale divisions equal 49 main scale divisions and the zero error in the instrument is $0.04\ \text{mm}$, then how many main scale divisions are there in $1\ \text{cm}$?
Answer: (C) $20$
The zero error corresponds to 4 vernier divisions: $4\times\text{LC} = 0.04\ \text{mm} \Rightarrow \text{LC} = 0.01\ \text{mm}$.
$50\ \text{VSD} = 49\ \text{MSD} \Rightarrow \text{LC} = \dfrac{1\ \text{MSD}}{50}$, so $1\ \text{MSD} = 0.5\ \text{mm}$.
Number of main scale divisions in $1\ \text{cm}$ $= \dfrac{10}{0.5} = 20$.
Solution by Sreeraj P, M.Sc Physics