In a vernier callipers, $(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1\, MSD$ represents $0.1\, mm$, the vernier constant (in $cm$ ) is:
Answer & explanation
Correct answer: option 2
The correct answer is option (2) : $\frac{1}{100( N+1)}$
$V.C=MSD-VSD$ .........(1)
Given : $(N+1)VSD=N\, MSD$
$VSD=\left(\frac{N}{N+1}\right)MSD $ .......... (2)
From (1) and (2)
$V.C=(MSD) -\frac{N}{N+1}(MSD)$
$= MSD \left(1-\frac{N}{N+1}\right) =\frac{MSD}{N+1}$
$=\frac{0.01}{N+1}=\frac{1}{100(N+1)}$