Consider a Vernier calipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale .In the Vernier calipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:

Vernier Calipers                                                            Screw Gauge
 $1MSD=\frac{1}{8}cm$                                                         Pitch of the screw  $P=\frac{2}{1}$
4MSD = 5 VSD                                                               P = 2HSD
$\Rightarrow 1VSD=\frac{4}{5}MSD$                                          Least count $=\frac{P}{100}=\frac{1}{50}HSD$
$\Rightarrow 1VSD=\frac{4}{5}\times \frac{1}{8}cm=\frac{1}{10}cm=1mm$ 
$Least count = 1MSD – 1VSD$
$\Rightarrow LC=\left(\frac{1}{8}-\frac{1}{10}\right)cm=\frac{1}{40}cm=\frac{1}{4}mm$ 
(B)  $$\begin{gathered} Pitch of the screw gauge is twice the least count of the Vernier calipers \\ P=2HSD=\left(\frac{1}{4}mm\right)2 \\ \Rightarrow 1HSD=\frac{1}{4}mm \end{gathered}$$
Least count of screw Gauge  $=\left(\frac{1}{50}\right)\left(\frac{1}{4}mm\right)=0.005mm$
(C) L.C of linear scale of screw gauge = 1HSD
Given that 1HSD =  $2\left[\frac{1}{4}mm\right]=\frac{1}{2}mm$
L.C of screw gauge  $=\frac{1}{50}HSD=\frac{1}{50}\left[\frac{1}{2}mm\right]=0.01mm$