Question

In an instrument, 25 divisions of the vernier scale, coincide with 24 divisions of the main scale. One cm on main scale, is divided into 20 equal parts. The least count of the vernier scale is _____ cm.

Answer 1

Given that,

$25VSD=24MSD$

or, $1VSD=\frac{24}{25}MSD$

or, $1VSD=0.96MSD$

Now, Least count of Vernier calipers, is $LC=1MSD-1VSD

or, $LC=0.04MSD$

$1cm$ is divided into 20 parts. Hence, 1 main scale division is $\frac{1}{5}cm$, or $2mm$

Using this, $LC=\frac{0.04}{2}$

or, $LC=0.02mm$ is our required least count.