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:

$\frac{1}{100( N+1)}$

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)

$V.C=(MSD) -\frac{N}{N+1}(MSD)$

$= MSD \left(1-\frac{N}{N+1}\right) =\frac{MSD}{N+1}$

$=\frac{0.01}{N+1}=\frac{1}{100(N+1)}$